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Quantum dynamics as a pseudo-density matrix

James Fullwood
Abstract

While in relativity theory space evolves over time into a single entity known as spacetime, quantum theory lacks a standard notion of how to encapsulate the dynamical evolution of a quantum state into a single "state over time". Recently it was emphasized in the work of Fitzsimons, Jones and Vedral that if such a state over time is to encode not only spatial but also temporal correlations which exist within a quantum dynamical process, then it should be represented not by a density matrix, but rather, by a pseudo-density matrix. A pseudo-density matrix is a hermitian matrix of unit trace whose marginals are density matrices, and in this work, we make use a factorization system for quantum channels to associate a pseudo-density matrix with a quantum system which is to evolve according to a finite sequence of quantum channels. We then view such a pseudo-density matrix as the quantum analog of a local patch of spacetime, and we make an in-depth mathematical analysis of such quantum dynamical pseudo-density matrices and the properties they satisfy. We also show how to explicitly extract quantum dynamics from a given pseudo-density matrix, thus solving an open problem posed in the literature.

1 Introduction

In 1907 Hermann Minkowski showed that the work of Maxwell, Lorentz and Einstein could be viewed geometrically as a 4-dimensional theory of spacetime [MKWSKI], after which he boldly predicted that space by itself, and time by itself, were then "doomed to fade away in mere shadows". Minkowski’s prediction was indeed correct, and ever since we have only furthered our understanding of the inextricable connection between space and time. By 1927 quantum theory had been established through the revolutionary work Bohr, Heisenberg, Schrödinger, Born and others, where the fundamental view was that at the sub-atomic realm, reality was appropriately described by "quantum states" evolving in time. While debates have raged through the ages regarding the ontological status of a quantum state, the one thing that most can agree upon is that at a fundamental level a quantum state consists of information, and there is an emerging viewpoint — famously coined by Wheeler as "it from bit" — that such quantum information is the basis of our physical reality. But while our macroscopic view of the world has blossomed from Minkowski’s unification of space and time, for the most part there has not been an analogous unification of information and time in quantum theory, where we seemed to have left the insights of Minkowski behind. In particular, while in relativity theory space evolves in time to form a single mathematical object called spacetime, quantum theory lacks a standard notion of how to encapsulate the global evolution of a quantum state over successive instances of time into a single entity, or rather, as a "state over time". In this letter, we then provide an answer to the "?" in the following analogy:

General Relativity:space + time\displaystyle\text{\text@underline{General Relativity}}:\quad\,\text{space + time} = spacetime\displaystyle=\,\text{ spacetime}
Quantum Theory:state + time\displaystyle\text{\text@underline{Quantum Theory}}:\quad\,\text{state + time} =?\displaystyle=\hskip 5.69054pt\text{?}

Moreover, as spacetime is fundamental to our understanding of gravity, it seems only natural that a mathematically precise formulation of the above analogy will yield insights into the nature of quantum gravity.

To summarize our approach to such an analogy, consider a local patch SS of spacetime, which we may view as a fibration φ:S[t0,t1]\varphi:S\to[t_{0},t_{1}] of spatial slices over an interval of time [t0,t1][t_{0},t_{1}]. In such a case, we may view SS as a cobordism, i.e., as representing the process of the spatial slice S0=φ1(t0)S_{0}=\varphi^{-1}(t_{0}) evolving over time into the spatial slice S1=φ1(t1)S_{1}=\varphi^{-1}(t_{1}). As such, in general relativity, the objects encoding evolution over time are of the same class of entity as the objects which are evolving over time, as SS, S0S_{0} and S1S_{1} are all manifolds. However, a crucial distinction between the process manifold SS and the spatial slices S0S_{0} and S1S_{1}, is that while S0S_{0} and S1S_{1} are of euclidean signature, the signature of the process manifold SS picks up a negative sign as it extends over time.

For our quantum analogy of a local patch of spacetime, suppose there is a quantum system in state ρ𝒜0\rho\in\mathcal{A}_{0} at time t=t0t=t_{0}, which is to evolve according to a quantum channel :𝒜0𝒜1\mathscr{E}:\mathcal{A}_{0}\to\mathcal{A}_{1}, so that (ρ)𝒜1\mathscr{E}(\rho)\in\mathcal{A}_{1} is then viewed as the state of the system at time t=t1t=t_{1}. We also assume that Δt=t1t0\Delta t=t_{1}-t_{0} is on the order of the Planck time, so that the interval [t0,t1][t_{0},t_{1}] may be viewed as a single discrete time-step. Then in accordance with our spacetime analogy, we wish to associate a "state over time"

ψ(ρ,)𝒜0𝒜1\psi(\rho,\mathscr{E})\in\mathcal{A}_{0}\otimes\mathcal{A}_{1}

encoding the dynamical evolution of ρ\rho according to the channel :𝒜0𝒜1\mathscr{E}:\mathcal{A}_{0}\to\mathcal{A}_{1}. Moreover, in analogy with the spatial slices S0S_{0} and S1S_{1} being the source and target of the process manifold SS, the states ρ\rho and (ρ)\mathscr{E}(\rho) should be the reduced density matrices of the state over time ψ(ρ,)\psi(\rho,\mathscr{E}) with respect to tracing out 𝒜1\mathcal{A}_{1} and 𝒜0\mathcal{A}_{0} respectively. In the case that 𝒜0\mathcal{A}_{0} and 𝒜1\mathcal{A}_{1} admit large tensor factorizations

𝒜0=i=1m𝕄i&𝒜1=j=1n𝕄j\mathcal{A}_{0}=\bigotimes_{i=1}^{m}\mathbb{M}_{i}\qquad\&\qquad\mathcal{A}_{1}=\bigotimes_{j=1}^{n}\mathbb{M}_{j}

(where 𝕄k\mathbb{M}_{k} denotes the matrix algebra of k×kk\times k matrices with complex entries), each tensor factor 𝕄i\mathbb{M}_{i} and 𝕄j\mathbb{M}_{j} may be identified with a region of space at times t=t0t=t_{0} and t=t1t=t_{1} respectively, and in such a case, we view the state over time ψ(ρ,)\psi(\rho,\mathscr{E}) as a quantum mechanical unit of spacetime.

While various formulations of dynamical quantum states have appeared in the literature, including the pseudo-density operator (PDO) formalism for systems of qubits first appearing in the work of Fitzsimons, Jones and Vedral [FJV15, LQDV, ZPTGVF18, ZhangV20, Marletto2021], the causal states of Leifer and Spekkens [Le06, Le07, LeSp13], the right bloom construction of Parzygnat and Russo [PaRuBayes, PaRu19], the super-density operators of Cotler et alia [Cotler2018], the compound states of Ohya [Ohya1983], the generalized conditional expectations of Tsang [Ts22], the Wigner function approach of Wooters [Woot87] and the process states of Huang and Guo [GuoZ1], in this work we focus mainly on the state over time ψ(ρ,)\psi(\rho,\mathscr{E}) given by

ψ(ρ,)=12{(ρ𝟙)𝒥[]},\psi(\rho,\mathscr{E})=\frac{1}{2}\left\{(\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}]\right\}, (1.1)

where {,}\{*,*\} denotes the anti-commutator, and

𝒥[]=i,j|ij|(|ji|)\mathscr{J}[\mathscr{E}]=\sum_{i,j}|i\rangle\langle j|\otimes\mathscr{E}(|j\rangle\langle i|)

is the Jamiołkowski matrix111The Jamiołkowski matrix is not to be confused with the Choi matrix 𝒞[]=i,j|ij|(|ij|)\mathscr{C}[\mathscr{E}]=\sum_{i,j}|i\rangle\langle j|\otimes\mathscr{E}(|i\rangle\langle j|). associated with the channel :𝒜0𝒜1\mathscr{E}:\mathcal{A}_{0}\to\mathcal{A}_{1}. The state over time given by (1.1) was shown in [FuPa22, FuPa22a] to satisfy a list of desiderata for states over time put forth in [HHPBS17], and at present it is the only known state over time construction to satisfy such properties. We note however that while the state over time given by (1.1) is hermitian and of unit trace, it is not positive in general, and as such, is often referred to as a pseudo-density matrix. In our spacetime analogy, the negative eigenvalues of a state over time are analogous to the negative sign appearing in the time-component of the spacetime metric, and further justification for such a state over time to have negative eigenvalues appears in [FJV15], where it is argued that negative eigenvalues serve as a witness to temporal correlations encoded in ψ(ρ,)\psi(\rho,\mathscr{E}).

One desideratum for states over time is a property often referred to as "associativity", as it allows one to unambiguously associate a "2-step state over time"

ψ(ρ,,)𝒜0𝒜1𝒜2\psi(\rho,\mathscr{E},\mathscr{F})\in\mathcal{A}_{0}\otimes\mathcal{A}_{1}\otimes\mathcal{A}_{2}

associated with the evolution of a state ρ\rho according to a 2-chain 𝒜0𝒜1𝒜2\mathcal{A}_{0}\overset{\mathscr{E}}{\to}\mathcal{A}_{1}\overset{\mathscr{F}}{\to}\mathcal{A}_{2} of quantum channels. The potential ambiguity stems from the fact that there are two parenthezations (𝒜0𝒜1)𝒜2(\mathcal{A}_{0}\otimes\mathcal{A}_{1})\otimes\mathcal{A}_{2} and 𝒜0(𝒜1𝒜2)\mathcal{A}_{0}\otimes(\mathcal{A}_{1}\otimes\mathcal{A}_{2}) of the matrix algebra 𝒜0𝒜1𝒜2\mathcal{A}_{0}\otimes\mathcal{A}_{1}\otimes\mathcal{A}_{2}, which lead to 2 fundamentally different constructions of a 2-step state over time ψ(ρ,,)\psi(\rho,\mathscr{E},\mathscr{F}). The associativity condition then ensures that these two distinct constructions of 2-step states over time are in fact equal, leading to a well-defined notion of a 2-step state over time.

In this work, we consider the case of a quantum state ρ\rho evolving according to an nn-chain of quantum channels

𝒜01𝒜1𝒜n1n𝒜n,\mathcal{A}_{0}\overset{\mathscr{E}_{1}}{\longrightarrow}\mathcal{A}_{1}\longrightarrow\cdots\longrightarrow\mathcal{A}_{n-1}\overset{\mathscr{E}_{n}}{\longrightarrow}\mathcal{A}_{n}, (1.2)

with which we wish to associate an "nn-step" state over time

ψ(ρ,1,,n)𝒜0𝒜n.\psi(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}.

Similar to how the ambiguity for 2-step states over time arises from the two different parenthezations of the matrix algebra 𝒜0𝒜1𝒜2\mathcal{A}_{0}\otimes\mathcal{A}_{1}\otimes\mathcal{A}_{2}, there is an ambiguity for nn-step states over time corresponding to the nnth Catalan number cn=1n+1(2nn)c_{n}=\frac{1}{n+1}\binom{2n}{n} of ways to parenthesize the matrix algebra 𝒜0𝒜n\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}, leading to cnc_{n} fundamentally distinct constructions of an nn-step state over time. We then prove that the cnc_{n} different constructions of such an nn-step state over time are all in fact equal (see item ii of Theorem 7.4), leading to a well-defined notion of an nn-step state over time. In accordance with our spacetime analogy, we then view an nn-step state over time ψ(ρ,1,,n)\psi(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) as a quantum analog of a local patch of spacetime fibered over an interval of time

[t0,tn]=[t0,t1][tn1,tn],[t_{0},t_{n}]=[t_{0},t_{1}]\cup\cdots\cup[t_{n-1},t_{n}],

where Δt=titi1\Delta t=t_{i}-t_{i-1} is on the order Planck time, so that [ti1,ti][t_{i-1},t_{i}] is viewed as a discrete time-step corresponding to the evolution of the quantum system according to the channel i:𝒜i1𝒜i\mathscr{E}_{i}:\mathcal{A}_{i-1}\to\mathcal{A}_{i}.

As the nn-step extension of the state over time given by (1.1) is the only known construction which always yields a hermitian state over time, we denote the associated state over time by \Yinyang(ρ,1,,n)\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}), and refer to the mapping

(ρ,1,,n)\Yinyang(ρ,1,,n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\longmapsto\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})

as the \Yinyang-function. In Example 8.20 we show that in the case of dynamically evolving systems of qubits, the output of the \Yinyang-function coincides with the coarse-grained pseudo-density operator (PDO) associated with the system, which was introduced in [FJV15] to treat temporal and spatial correlations in quantum theory on equal footing. As such, the \Yinyang-function may be viewed as a generalization to arbitrary finite-dimensional quantum systems of the pseudo-density operator formalism for systems of qubits. We also prove that given a unit-trace hermitian element τ𝒜0𝒜n\tau\in\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n} satisfying some technical conditions, one may explicitly extract a dynamical process (ρ,1,,n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) such that \Yinyang(ρ,1,,n)=τ\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})=\tau (see Theorem 9.5). This solves an open question recently posed in [jia2023], where it is stated "Another interesting and closely relevant open question is, for a given PDO (pseudo-density operator), how to find a quantum process to realize it.".

In what follows, we provide the necessary definitions and notation which will be used throughout in Section 2. In Section 3, we motivate the definition of quantum state over time by first analyzing the classical case of a random variable XX stochastically evolving into a random variable YY. In particular, we show that classical joint distribution (x,y)\mathbb{P}(x,y) associated with such stochastic evolution arises from a unique factorization of the associated classical channel, and moreover, that (x,y)\mathbb{P}(x,y) may be re-written in a way that is valid for any 1-step quantum process (ρ,)(\rho,\mathscr{E}). In Section 4 we then define a factorization system for quantum channels in terms of a quantum bloom map, and then we use such a factorization system to define quantum states over time associated with 1-step processes (ρ,)(\rho,\mathscr{E}). In Section 5 we recall the associativity property for quantum bloom maps which allow states over time for 1-step processes to extend to well-defined states over time for 2-step processes (ρ,,)(\rho,\mathscr{E},\mathscr{F}), and prove some general results regarding 2-step states over time. In Section 6 we then show how the associativity property for quantum bloom maps yields uniquely-defined extensions to quantum bloom maps for nn-chains of quantum channels. In Section 7 we then use the quantum bloom maps for nn-chains to define states over time ψ(ρ,1,,n)\psi(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) associated with nn-step processes (ρ,1,,n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) for arbitrary n>0n>0. In Section 8 we then focus on the nn-step states over time associated with the \Yinyang-function, and we prove some general properties of the \Yinyang-function. We also two distinct formulas for \Yinyang(ρ,1,,n)\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) in terms of ρ\rho and the states 𝒥[1],,𝒥[n]\mathscr{J}[\mathscr{E}_{1}],...,\mathscr{J}[\mathscr{E}_{n}] which are Jamiołkowski-isomorphic to the channels 1,,n\mathscr{E}_{1},...,\mathscr{E}_{n}. In Section 9 we then show that the \Yinyang-function yields a bijection when restricted to a suitably nice set of nn-step processes, and we prove an explicit formula for its inverse in such a case.

Acknowledgements. We thank Arthur J. Parzygnat and Franceso Buscemi for many useful discussions. This work is supported in part by the Blaumann Foundation.

2 Preliminaries

In this section we provide the basic definitions, notation and terminology which will be used throughout.

Definition 2.1.

Let XX be a finite set. A function p:Xp:X\to{{\mathbb{R}}} will be referred to as a quasi-probability distribution if and only if xXp(x)=1\sum_{x\in X}p(x)=1. In such a case, p(x)p(x)\in{{\mathbb{R}}} will be denoted by pxp_{x} for all xXx\in X. If px[0,1]p_{x}\in[0,1] for all xXx\in X, then pp will be referred to as a probability distribution.

Definition 2.2.

Let XX and YY be finite sets. A stochastic map f:X Yf:X\mathrel{\leavevmode\hbox to10.66pt{\vbox to6.76pt{\pgfpicture\makeatletter\hbox{\hskip 5.33047pt\lower-4.60277pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } {{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.77771pt}{0.0pt}\pgfsys@invoke{ }\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{\;$\scriptstyle$\;}} }}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} { {}}{}{{}}{}{{}} {}{} {}{{}{}}{}{}{}{{}}{{}}{{}{}}{{}{}} {{{{}{}{{}} }}{{}} }{{{{}{}{{}} }}{{}}}{{{{}{}{{}} }}{{}}{{}}} {{{{}{}{{}} }}{{}} {}{}{} }{{{{}{}{{}} }}{{}} {{}{}{}{}} {{}{}{}{}} }{{{{}{}{{}} }}{{}} {{}} } {{{{}{}{{}} }}{{}}{{}}} {}{{ {\pgfsys@beginscope \pgfsys@setdash{}{0.0pt}\pgfsys@roundcap\pgfsys@roundjoin{} {}{}{} {}{}{} \pgfsys@moveto{-2.6pt}{2.05pt}\pgfsys@curveto{-2.125pt}{0.81998pt}{-1.06648pt}{0.23917pt}{0.0pt}{0.0pt}\pgfsys@curveto{-1.06648pt}{-0.23917pt}{-2.125pt}{-0.81998pt}{-2.6pt}{-2.05pt}\pgfsys@stroke\pgfsys@endscope}} }{}{}{{}}\pgfsys@moveto{4.73048pt}{-2.35277pt}\pgfsys@lineto{4.13048pt}{-2.35277pt}\pgfsys@curveto{3.3836pt}{-2.35277pt}{3.01016pt}{-3.20634pt}{2.26328pt}{-3.20634pt}\pgfsys@curveto{1.72255pt}{-3.20634pt}{1.2565pt}{-2.7898pt}{0.76952pt}{-2.35277pt}\pgfsys@curveto{0.28255pt}{-1.91576pt}{-0.1835pt}{-1.49919pt}{-0.72424pt}{-1.49919pt}\pgfsys@lineto{-5.13048pt}{-2.35277pt}\pgfsys@lineto{-5.13048pt}{-2.35277pt}\pgfsys@stroke\pgfsys@invoke{ }{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{4.93048pt}{-2.35277pt}\pgfsys@invoke{ }\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}{{}}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{{ {}{}{}{}{}}{}{{}{}}}{}{}\hss}\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}Y consists of the assignment of a probability distribution fx:Y[0,1]f_{x}:Y\to[0,1] for every xXx\in X. In such a case, (fx)y(f_{x})_{y} will be denoted by fyxf_{yx} for all xXx\in X and yYy\in Y, which is interpreted as the probability of yy given xx. 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The set of stochastic maps from XX to YY will be denoted by 𝕊𝕥𝕠𝕔𝕙(X,Y)\mathbb{Stoch}(X,Y).

Remark 2.3.

A stochastic map f:X Yf:X\mathrel{\leavevmode\hbox to10.66pt{\vbox to6.76pt{\pgfpicture\makeatletter\hbox{\hskip 5.33047pt\lower-4.60277pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } {{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.77771pt}{0.0pt}\pgfsys@invoke{ }\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{\;$\scriptstyle$\;}} }}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} { {}}{}{{}}{}{{}} {}{} {}{{}{}}{}{}{}{{}}{{}}{{}{}}{{}{}} {{{{}{}{{}} }}{{}} }{{{{}{}{{}} }}{{}}}{{{{}{}{{}} }}{{}}{{}}} {{{{}{}{{}} }}{{}} {}{}{} }{{{{}{}{{}} }}{{}} {{}{}{}{}} {{}{}{}{}} }{{{{}{}{{}} }}{{}} {{}} } {{{{}{}{{}} }}{{}}{{}}} {}{}{}{}{{}}\pgfsys@moveto{4.73048pt}{-2.35277pt}\pgfsys@lineto{4.13048pt}{-2.35277pt}\pgfsys@curveto{3.3836pt}{-2.35277pt}{3.01016pt}{-3.20634pt}{2.26328pt}{-3.20634pt}\pgfsys@curveto{1.72255pt}{-3.20634pt}{1.2565pt}{-2.7898pt}{0.76952pt}{-2.35277pt}\pgfsys@curveto{0.28255pt}{-1.91576pt}{-0.1835pt}{-1.49919pt}{-0.72424pt}{-1.49919pt}\pgfsys@lineto{-5.13048pt}{-2.35277pt}\pgfsys@lineto{-5.13048pt}{-2.35277pt}\pgfsys@stroke\pgfsys@invoke{ }{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{4.93048pt}{-2.35277pt}\pgfsys@invoke{ }\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}{{}}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{{ {}{}{}{}{}}{}{{}{}}}{}{}\hss}\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}Y is also commonly referred to as a Markov kernel, or a discrete memoryless channel.

Notation and Terminology 2.4.

Given a natural number mm\in{{\mathbb{N}}}, the set of m×mm\times m matrices with complex entries will be denoted by 𝕄m\mathbb{M}_{m}, and will be referred to as a matrix algebra. As the matrix algebra 𝕄1\mathbb{M}_{1} is simply the complex numbers, it will be denoted by {{\mathbb{C}}}. The matrix units in 𝕄m\mathbb{M}_{m} will be denoted by Eij(m)E_{ij}^{(m)} (or simply EijE_{ij} if mm is clear from the context), and for every ρ𝕄m\rho\in\mathbb{M}_{m}, ρ𝕄m\rho^{{\dagger}}\in\mathbb{M}_{m} denotes the conjugate-transpose of ρ\rho. Given a finite set XX, a direct sum xX𝕄mx\bigoplus_{x\in X}\mathbb{M}_{m_{x}} will be referred to as a multi-matrix algebra, whose multiplication and addition are defined component-wise. If 𝒜\mathcal{A} and \mathcal{B} are multi-matrix algebras, then the vector space of all linear maps from 𝒜\mathcal{A} to \mathcal{B} will be denoted by Hom(𝒜,)\text{Hom}(\mathcal{A},\mathcal{B}). The trace of an element A=xXAxxX𝕄mxA=\bigoplus_{x\in X}A_{x}\in\bigoplus_{x\in X}\mathbb{M}_{m_{x}} is the complex number tr(A)\text{tr}(A) given by tr(A)=xXtr(Ax)\text{tr}(A)=\sum_{x\in X}{\rm tr}(A_{x}) (where tr()\text{tr}(*) is the usual trace on matrices), and the dagger of AA is the element AxX𝕄mxA^{{\dagger}}\in\bigoplus_{x\in X}\mathbb{M}_{m_{x}} given by A=xXAxA^{{\dagger}}=\bigoplus_{x\in X}A_{x}^{{\dagger}}. Given Hom(𝒜,)\mathscr{E}\in\text{Hom}(\mathcal{A},\mathcal{B}) with 𝒜\mathcal{A} and \mathcal{B} multi-matrix algebras, we let Hom(,𝒜)\mathscr{E}^{*}\in\text{Hom}(\mathcal{B},\mathcal{A}) denote the Hilbert–Schmidt dual (or adjoint) of \mathscr{E}, which is uniquely determined by the condition

tr((A)B)=tr(A(B))\text{tr}\left(\mathscr{E}(A)^{{\dagger}}B\right)=\text{tr}\left(A^{{\dagger}}\mathscr{E}^{*}(B)\right)

for all A𝒜A\in\mathcal{A} and BB\in\mathcal{B}. The identity map between algebras will be denoted by id\mathrm{id}, while the unit element in an algebra will be denoted by 𝟙\mathds{1} (subscripts such as id𝒜\mathrm{id}_{{{\mathcal{A}}}} and 𝟙𝒜\mathds{1}_{{{\mathcal{A}}}} will be used if deemed necessary).

Remark 2.5.

Every finite-dimensional CC^{*}-algebra is isomorphic to a multi-matrix algebra [Fa01].

Notation 2.6.

Given a finite set XX, the multi-matrix algebra xX\bigoplus_{x\in X}{{\mathbb{C}}} is canonically isomorphic to algebra X{{\mathbb{C}}}^{X} of complex-valued functions on XX. As such, we will write an element ρxXX\rho\in\bigoplus_{x\in X}{{\mathbb{C}}}\cong{{\mathbb{C}}}^{X} as

ρ=xXρxδxX,\rho=\sum_{x\in X}\rho_{x}\delta_{x}\in{{\mathbb{C}}}^{X},

where δxX\delta_{x}\in{{\mathbb{C}}}^{X} is the Dirac-delta of xx, which is the function taking the value 1 at xx and 0 otherwise. In such a case, ρx\rho_{x}\in{{\mathbb{C}}} will be referred to as the xx-component of ρ\rho for all xXx\in X.

Definition 2.7.

Let XX be a finite set and let 𝒜=xX𝕄mx\mathcal{A}=\bigoplus_{x\in X}\mathbb{M}_{m_{x}} be a multi-matrix algebra. An element A=xXAx𝒜A=\bigoplus_{x\in X}A_{x}\in\mathcal{A} is said to be

  • self-adjoint if and only if Ax=AxA_{x}^{{\dagger}}=A_{x} for all xXx\in X.

  • positive if and only if Ax𝕄mxA_{x}\in\mathbb{M}_{m_{x}} is self-adjoint and has non-negative eigenvalues for all xXx\in X.

  • a state if and only if AA is positive and of unit trace. If XX contains only one element, then a state AA will often be referred to as a density matrix. The set of all states on 𝒜\mathcal{A} will be denoted by 𝒮(𝒜)\mathcal{S}(\mathcal{A}).

Remark 2.8.

If XX is a finite set and 𝒜=X\mathcal{A}={{\mathbb{C}}}^{X}, then a state ρ𝒮(X)\rho\in\mathcal{S}({{\mathbb{C}}}^{X}) is of the form ρ=xXρxδx\rho=\sum_{x\in X}\rho_{x}\delta_{x} with ρx0\rho_{x}\geq 0 and xXρx=1\sum_{x\in X}\rho_{x}=1. As such, any state on X{{\mathbb{C}}}^{X} may be identified with a probability distribution on XX.

Definition 2.9.

Let 𝒜=xX𝕄mx\mathcal{A}=\bigoplus_{x\in X}\mathbb{M}_{m_{x}} and =yY𝕄ny\mathcal{B}=\bigoplus_{y\in Y}\mathbb{M}_{n_{y}}. The tensor product of 𝒜\mathcal{A} and \mathcal{B} is the multi-matrix algebra 𝒜\mathcal{A}\otimes\mathcal{B} given by

𝒜=(x,y)X×Y𝕄mx𝕄ny,\mathcal{A}\otimes\mathcal{B}=\bigoplus_{(x,y)\in X\times Y}\mathbb{M}_{m_{x}}\otimes\mathbb{M}_{n_{y}},

where 𝕄mx𝕄ny\mathbb{M}_{m_{x}}\otimes\mathbb{M}_{n_{y}} is the usual tensor product of matrix algebras. Given elements xXAx𝒜\bigoplus_{x\in X}A_{x}\in\mathcal{A} and yYBy\bigoplus_{y\in Y}B_{y}\in\mathcal{B}, the element (xXAx)(yYBy)𝒜(\bigoplus_{x\in X}A_{x})\otimes(\bigoplus_{y\in Y}B_{y})\in\mathcal{A}\otimes\mathcal{B} is the multi-matrix given by

(xXAx)(yYBy)=(x,y)X×YAxBy,\left(\bigoplus_{x\in X}A_{x}\right)\otimes\left(\bigoplus_{y\in Y}B_{y}\right)=\bigoplus_{(x,y)\in X\times Y}A_{x}\otimes B_{y},

where AxByA_{x}\otimes B_{y} is the usual tensor product of matrices. Given maps Hom(𝒜,𝒜)\mathscr{E}\in\text{Hom}(\mathcal{A},\mathcal{A}^{\prime}) and Hom(,)\mathscr{F}\in\text{Hom}(\mathcal{B},\mathcal{B}^{\prime}), then Hom(𝒜,𝒜)\mathscr{E}\otimes\mathscr{F}\in\text{Hom}(\mathcal{A}\otimes\mathcal{B},\mathcal{A}^{\prime}\otimes\mathcal{B}^{\prime}) is the map corresponding to the linear extension of the assignment ()(AB)=(A)(B)(\mathscr{E}\otimes\mathscr{F})(A\otimes B)=\mathscr{E}(A)\otimes\mathscr{F}(B).

Definition 2.10.

Given multi-matrix algebras 𝒜=xX𝕄mx\mathcal{A}=\bigoplus_{x\in X}\mathbb{M}_{m_{x}} and =yY𝕄my\mathcal{B}=\bigoplus_{y\in Y}\mathbb{M}_{m_{y}}, an element Hom(𝒜,)\mathscr{E}\in\text{Hom}(\mathcal{A},\mathcal{B}) consists of elements yxHom(𝕄mx,𝕄ny)\mathscr{E}_{yx}\in\text{Hom}(\mathbb{M}_{m_{x}},\mathbb{M}_{n_{y}}) such that

(xXρx)=yY(xXyx(ρx)).\mathscr{E}\left(\bigoplus_{x\in X}\rho_{x}\right)=\bigoplus_{y\in Y}\left(\sum_{x\in X}\mathscr{E}_{yx}(\rho_{x})\right).

In such a case, the yx\mathscr{E}_{yx} will be referred to as the component functions of \mathscr{E}.

Definition 2.11.

Let 𝒜\mathcal{A} and \mathcal{B} be multi-matrix algebras. A map Hom(𝒜,)\mathscr{E}\in\text{Hom}(\mathcal{A},\mathcal{B}) is said to be

  • {\dagger}-preserving if and only if (A)=(A)\mathscr{E}(A)^{{\dagger}}=\mathscr{E}(A^{{\dagger}}) for all A𝒜A\in\mathcal{A}.

  • trace-preserving if and only if tr((A))=tr(A)\text{tr}(\mathscr{E}(A))=\text{tr}(A) for all A𝒜A\in\mathcal{A}.

  • positive if and only if (A)\mathscr{E}(A) is positive whenever A𝒜A\in\mathcal{A} is positive.

  • completely positive if and only if id𝒞:𝒜𝒞𝒞\mathscr{E}\otimes\mathrm{id}_{\mathcal{C}}:\mathcal{A}\otimes\mathcal{C}\to\mathcal{B}\otimes\mathcal{C} is positive for every multi-matrix algebra 𝒞\mathcal{C}.

Notation 2.12.

Let (𝒜,)(\mathcal{A},\mathcal{B}) be pair of multi-matrix algebras. The subset of Hom(𝒜,)\text{Hom}(\mathcal{A},\mathcal{B}) consisting of completely positive, trace-preserving maps will be denoted by 𝕋(𝒜,)\mathbb{CPTP}(\mathcal{A},\mathcal{B}), while the subset of Hom(𝒜,)\text{Hom}(\mathcal{A},\mathcal{B}) consisting of trace-preserving maps will be denoted by 𝕋(𝒜,)\mathbb{TP}(\mathcal{A},\mathcal{B}). An element of 𝕋(𝒜,)\mathbb{CPTP}(\mathcal{A},\mathcal{B}) with 𝒜\mathcal{A} and \mathcal{B} matrix algebras will often be referred to as a quantum channel.

Definition 2.13.

Let XX and YY be finite sets. An element 𝕋(X,Y)\mathscr{E}\in\mathbb{CPTP}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}) will be referred to as a classical channel. In such a case, we let yx[0,1]\mathscr{E}_{yx}\in[0,1] denote the elements such that

(δx)=yYyxδy.\mathscr{E}(\delta_{x})=\sum_{y\in Y}\mathscr{E}_{yx}\delta_{y}.

In such a case, the elements yx\mathscr{E}_{yx} will be referred to as the conditional probabilities associated with \mathscr{E}.

Definition 2.14.

Given a pair (𝒜,)(\mathcal{A},\mathcal{B}) of multi-matrix algebras, an element (ρ,)𝒮(𝒜)×𝕋(𝒜,)(\rho,\mathscr{E})\in\mathcal{S}(\mathcal{A})\times\mathbb{CPTP}(\mathcal{A},\mathcal{B}) will be referred to as a process, and the set of processes 𝒮(𝒜)×𝕋(𝒜,)\mathcal{S}(\mathcal{A})\times\mathbb{CPTP}(\mathcal{A},\mathcal{B}) will be denoted by 𝒫(𝒜,)\mathscr{P}(\mathcal{A},\mathcal{B}). The subset of 𝒫(𝒜,)\mathscr{P}(\mathcal{A},\mathcal{B}) consisting of processes (ρ,)(\rho,\mathscr{E}) with ρ\rho invertible will be denoted by 𝒫+(𝒜,)\mathscr{P}_{+}(\mathcal{A},\mathcal{B}). When 𝒜=X\mathcal{A}={{\mathbb{C}}}^{X} and =Y\mathcal{B}={{\mathbb{C}}}^{Y} for finite sets XX and YY, then (ρ,)𝒫(X,Y)(\rho,\mathscr{E})\in\mathscr{P}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}) will be referred to as a classical process.

Definition 2.15.

Let (𝒜,)(\mathcal{A},\mathcal{B}) be a pair of multi-matrix algebras, and let Hom(𝒜,)\mathscr{E}\in\text{Hom}(\mathcal{A},\mathcal{B}). The channel state of \mathscr{E} is the element 𝒥[]𝒜\mathscr{J}[\mathscr{E}]\in\mathcal{A}\otimes\mathcal{B} given by

𝒥[]=(id𝒜)(μ(𝟙)),\mathscr{J}[\mathscr{E}]=(\text{id}_{\mathcal{A}}\otimes\mathscr{E})(\mu^{*}(\mathds{1})),

where μ:𝒜𝒜𝒜\mu^{*}:\mathcal{A}\to\mathcal{A}\otimes\mathcal{A} is the Hilbert-Schmidt dual of the multiplication map μ:𝒜𝒜𝒜\mu:\mathcal{A}\otimes\mathcal{A}\to\mathcal{A}.

Remark 2.16.

Given a pair (𝒜,)(\mathcal{A},\mathcal{B}) of multi-matrix algebras, the map Hom(𝒜,)𝒜\text{Hom}(\mathcal{A},\mathcal{B})\longrightarrow\mathcal{A}\otimes\mathcal{B} given by 𝒥[]\mathscr{E}\longmapsto\mathscr{J}[\mathscr{E}] is a linear isomorphism, which we refer to as the Jamiołkowski isomorphism [Jam72]. In this work we will also make heavy use of the the inverse 𝒥1:𝒜Hom(𝒜,)\mathscr{J}^{-1}:\mathcal{A}\otimes\mathcal{B}\longrightarrow\text{Hom}(\mathcal{A},\mathcal{B}) of the Jamiołowski isomorphism, which is given by

𝒥1(τ)(ρ)=tr𝒜((ρ𝟙)τ).\mathscr{J}^{-1}(\tau)(\rho)=\text{tr}_{\mathcal{A}}((\rho\otimes\mathds{1})\tau).
Definition 2.17.

If 𝒜\mathcal{A}, \mathcal{B} and 𝒞\mathcal{C} are matrix algebras, there exists an associator isomorphism 𝒜(𝒞)(𝒜)𝒞\mathcal{A}\otimes(\mathcal{B}\otimes\mathcal{C})\longrightarrow(\mathcal{A}\otimes\mathcal{B})\otimes\mathcal{C} which then allows one to define tensor products of of a finite number of matrix algebras iteratively. We can then extend such a constriction to multi-matrix algebras as follows. Let {𝒜x}xX\{\mathcal{A}_{x}\}_{x\in X} be a collection of multi-matrix algebras indexed by a finite set XX, where 𝒜x=λxΛx𝒜λx\mathcal{A}_{x}=\bigoplus_{\lambda_{x}\in\Lambda_{x}}\mathcal{A}_{\lambda_{x}}. The tensor product of the algebras 𝒜x\mathcal{A}_{x} is the multi-matrix algebra xX𝒜x\bigotimes_{x\in X}\mathcal{A}_{x} indexed by the set Γ=xXΛx\Gamma=\prod_{x\in X}\Lambda_{x} given by

xX𝒜x=(λx)xXΓ(xXAλx).\bigotimes_{x\in X}\mathcal{A}_{x}=\bigoplus_{(\lambda_{x})_{x\in X}\in\Gamma}\left(\bigotimes_{x\in X}A_{\lambda_{x}}\right).

If the index set XX is of the form X={0,1,,n}X=\{0,1,...,n\}, then we will write xX𝒜x\bigotimes_{x\in X}\mathcal{A}_{x} as 𝒜0𝒜n\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}.

Definition 2.18.

Let 𝒜=𝒜0𝒜n\mathcal{A}=\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n} be a tensor product of multi-matrix algebras. The iith partial trace for i{0,,n}i\in\{0,...,n\} is the map tri:𝒜𝒜i\text{tr}_{i}:\mathcal{A}\longrightarrow\mathcal{A}_{i} given by the linear extension of the assignment

tri(A0An)=tr(A0Ai^An)Ai𝒜i,\text{tr}_{i}(A_{0}\otimes\cdots\otimes A_{n})=\text{tr}\left(A_{0}\otimes\cdots\otimes\widehat{A_{i}}\otimes\cdots\otimes A_{n}\right)A_{i}\in\mathcal{A}_{i},

where Ai^\widehat{A_{i}} denotes the empty matrix for all i{0,,n}i\in\{0,...,n\}.

Notation 2.19.

If 𝒞=𝒜\mathcal{C}=\mathcal{A}\otimes\mathcal{B}, then the partial trace maps tr1:𝒜𝒜\text{tr}_{1}:\mathcal{A}\otimes\mathcal{B}\longrightarrow\mathcal{A} and tr2:𝒜\text{tr}_{2}:\mathcal{A}\otimes\mathcal{B}\longrightarrow\mathcal{B} will be denoted by tr\text{tr}_{\mathcal{B}} and tr𝒜\text{tr}_{\mathcal{A}} respectively. If 𝒜=X\mathcal{A}={{\mathbb{C}}}^{X} and =Y\mathcal{B}={{\mathbb{C}}}^{Y}, then the partial trace maps trX:XYY\text{tr}_{{{\mathbb{C}}}^{X}}:{{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y}\longrightarrow{{\mathbb{C}}}^{Y} and trY:XYX\text{tr}_{{{\mathbb{C}}}^{Y}}:{{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y}\longrightarrow{{\mathbb{C}}}^{X} will be denoted by trX\text{tr}_{X} and trY\text{tr}_{Y}. If 𝒜=𝒜0𝒜n\mathcal{A}=\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}, then we will frequently make use of the nnth partial trace trn:𝒜𝒜n\text{tr}_{n}:\mathcal{A}\longrightarrow\mathcal{A}_{n}, and as such, we will often denote trn\text{tr}_{n} simply by tr.

Definition 2.20.

Let XX be a finite set and let 𝒜=xX𝕄mx\mathcal{A}=\bigoplus_{x\in X}\mathbb{M}_{m_{x}} be a multi-matrix algebra. A self-adjoint element τ𝒜\tau\in\mathcal{A} is said to be a pseudo-density operator with respect to the factorization 𝒜=𝒜0𝒜n\mathcal{A}=\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n} if and only if for all i{0,,n}i\in\{0,...,n\} we have

tri(τ)𝒮(𝒜i).\text{tr}_{i}(\tau)\in\mathcal{S}(\mathcal{A}_{i}).

If 𝒜\mathcal{A} is a matrix algebra (i.e., when XX consists of a single element), then a pseudo-density operator τ𝒜\tau\in\mathcal{A} with respect to the factorization 𝒜=𝒜0𝒜n\mathcal{A}=\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n} will be referred to as a pseudo-density matrix. The set of all pseudo-density operators with respect to the factorization 𝒜=𝒜0𝒜n\mathcal{A}=\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n} will be denoted by 𝒯(𝒜0𝒜n)\mathscr{T}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}).

Remark 2.21.

A pseudo-density operator with respect to the trivial factorization 𝒜=𝒜0\mathcal{A}=\mathcal{A}_{0} is simply a state, so that 𝒯(𝒜0)=𝒮(𝒜0)\mathscr{T}(\mathcal{A}_{0})=\mathcal{S}(\mathcal{A}_{0}).

Remark 2.22.

Since all of the marginals of a pseudo-density operator are of unit trace, it follows that a pseudo-density operator is necessarily of unit trace as well.

3 Classical probability as CPTP dynamics

In this section we recall how classical probability may be recast in terms of CPTP maps between multi-matrix algebras. In particular, we will show how the joint distribution associated with a classical channel :XY\mathscr{E}:{{\mathbb{C}}}^{X}\longrightarrow{{\mathbb{C}}}^{Y} arises from a unique factorization of the channel of the form =trX!\mathscr{E}={\rm tr}_{X}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}, where !:XX×YXY\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}:{{\mathbb{C}}}^{X}\longrightarrow{{\mathbb{C}}}^{X\times Y}\cong{{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y} is a map referred to as the bloom of \mathscr{E}. If ρX\rho\in{{\mathbb{C}}}^{X} is a state, then the element !(ρ)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho) is the joint distribution associated with the classical process (ρ,)(\rho,\mathscr{E}), which in a more dynamical language we refer to as the state over time associated with (ρ,)(\rho,\mathscr{E}). We will then re-write the classical state over time !(ρ)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho) in a way that is valid for any quantum process (ρ,)𝒫(𝒜,)(\rho,\mathscr{E})\in\mathscr{P}(\mathcal{A},\mathcal{B}) with 𝒜\mathcal{A} and \mathcal{B} arbitrary multi-matrix algebras, thus paving the way for a quantum generalization of the classical state over time !(ρ)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho).

The following proposition shows how classical stochastic maps may be recast in terms of CPTP dynamics.

Proposition 3.1.

Let (X,Y)(X,Y) be a pair of finite sets, and let :𝕊𝕥𝕠𝕔𝕙(X,Y)𝕋(X,Y)\mathbb{Q}:\mathbb{Stoch}(X,Y)\longrightarrow\mathbb{CPTP}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}) be the map given by

(f)(δx)=yYfyxδy.\mathbb{Q}(f)(\delta_{x})=\sum_{y\in Y}f_{yx}\delta_{y}.

Then \mathbb{Q} is a continuous bijection.

Proof.

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The next proposition shows that factorizations of classical channels are essentially unique, which we will use to characterize joint distributions associated with stochastic dynamics.

Proposition 3.2.

Let (X,Y)(X,Y) be a pair of finite sets, let 𝕋(X,Y)\mathscr{E}\in\mathbb{CPTP}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}), and let !𝕋(X,XY)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}\in\mathbb{CPTP}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y}) be such that =trX!\mathscr{E}=\emph{tr}_{X}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}} and idX=trY!\mathrm{id}_{{{\mathbb{C}}}^{X}}=\emph{tr}_{Y}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}. Then !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}} is the linear map given by

!(xXρxδx)=(x,y)X×Yρxyx(δxδy),\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}\left(\sum_{x\in X}\rho_{x}\delta_{x}\right)=\sum_{(x,y)\in X\times Y}\rho_{x}\mathscr{E}_{yx}(\delta_{x}\otimes\delta_{y}), (3.3)

where yx\mathscr{E}_{yx} are the conditional probabilities associated with the map \mathscr{E}.

Proof.

We show that given x0Xx_{0}\in X, we have

!(δx0)=(x,y)X×Yδxx0yx(δxδy).\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\delta_{x_{0}})=\sum_{(x,y)\in X\times Y}\delta_{xx_{0}}\mathscr{E}_{yx}(\delta_{x}\otimes\delta_{y}).

The result then follows as !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}} is assumed to be linear and Dirac deltas form a basis of X{{\mathbb{C}}}^{X}. So let x0Xx_{0}\in X be arbitrary, and let ~(x,y)x0\widetilde{\mathscr{E}}_{(x,y)x_{0}} be the conditional probabilities associated with the map !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}, so that

!(δx0)=(x,y)X×Y~(x,y)x0(δxδy).\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\delta_{x_{0}})=\sum_{(x,y)\in X\times Y}\widetilde{\mathscr{E}}_{(x,y)x_{0}}(\delta_{x}\otimes\delta_{y}).

From the conditions =trX!\mathscr{E}=\text{tr}_{X}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}} and idX=trY!\mathrm{id}_{{{\mathbb{C}}}^{X}}=\text{tr}_{Y}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}} we then have

yYyx0δy=(δx0)=trX(!(δx0))=yY(xX~(x,y)x0)δy\sum_{y\in Y}\mathscr{E}_{yx_{0}}\delta_{y}=\mathscr{E}(\delta_{x_{0}})=\text{tr}_{X}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\delta_{x_{0}}))=\sum_{y\in Y}\left(\sum_{x\in X}\widetilde{\mathscr{E}}_{(x,y)x_{0}}\right)\delta_{y}

and

xXδxx0δx=δx0=trY(!(δx0))=xX(yY~(x,y)x0)δx,\sum_{x\in X}\delta_{xx_{0}}\delta_{x}=\delta_{x_{0}}=\text{tr}_{Y}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\delta_{x_{0}})\right)=\sum_{x\in X}\left(\sum_{y\in Y}\widetilde{\mathscr{E}}_{(x,y)x_{0}}\right)\delta_{x},

where δxx0\delta_{xx_{0}} is the Kronecker delta. For all yYy\in Y we then have

xX~(x,y)x0=yx0,\sum_{x\in X}\widetilde{\mathscr{E}}_{(x,y)x_{0}}=\mathscr{E}_{yx_{0}},

and for all xXx\in X we have

yY~(x,y)x0=δxx0.\sum_{y\in Y}\widetilde{\mathscr{E}}_{(x,y)x_{0}}=\delta_{xx_{0}}.

It then follows that for all (x,y)X×Y(x,y)\in X\times Y we have ~(x,y)x0=δxx0yx\widetilde{\mathscr{E}}_{(x,y)x_{0}}=\delta_{xx_{0}}\mathscr{E}_{yx}, thus

!(δx0)=(x,y)X×Y~(x,y)x0(δxδy)=(x,y)X×Yδxx0yx(δxδy),\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\delta_{x_{0}})=\sum_{(x,y)\in X\times Y}\widetilde{\mathscr{E}}_{(x,y)x_{0}}(\delta_{x}\otimes\delta_{y})=\sum_{(x,y)\in X\times Y}\delta_{xx_{0}}\mathscr{E}_{yx}(\delta_{x}\otimes\delta_{y}),

as desired. ∎

Definition 3.4.

Let (X,Y)(X,Y) be a pair of finite sets and let (ρ,)𝒫(X,Y)(\rho,\mathscr{E})\in\mathscr{P}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}) be a classical process.

  • The map !𝕋(X,XY)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}\in\mathbb{CPTP}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y}) defined by (3.3) will be referred to as the bloom of \mathscr{E}.

  • The element !(ρ)𝒮(XY)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho)\in\mathcal{S}({{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y}) will be referred to as the state over time associated with the classical process (ρ,)(\rho,\mathscr{E}).

  • The map on classical processes given by (ρ,)!(ρ)(\rho,\mathscr{E})\longmapsto\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho) will be referred to as the classical state over time function.

Refer to caption Refer to caption
Figure 1: The bloom map "opens up" the classical channel \mathscr{E}.
Remark 3.5.

Given a pair of finite sets (X,Y)(X,Y) and a classical process (ρ,)𝒫(X,Y)(\rho,\mathscr{E})\in\mathscr{P}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}), if we identify yx\mathscr{E}_{yx} with conditional probabilities (y|x)\mathbb{P}(y|x), ρ𝒮(X)\rho\in\mathcal{S}({{\mathbb{C}}}^{X}) with a probability distribution (x)\mathbb{P}(x) on XX, and !(ρ)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho) with a joint probability distribution (x,y)\mathbb{P}(x,y) on X×YX\times Y, then the equation

!(ρ)(x,y)=ρxyx\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho)_{(x,y)}=\rho_{x}\mathscr{E}_{yx}

may be re-written as

(x,y)=(x)(y|x),\mathbb{P}(x,y)=\mathbb{P}(x)\mathbb{P}(y|x), (3.6)

which is the classical equation relating a joint distribution with a conditional distribution and a prior. As such, Proposition 3.2 may be interpreted as saying that the joint distribution (x,y)\mathbb{P}(x,y) associated with a classical process (ρ,)𝒫(X,Y)(\rho,\mathscr{E})\in\mathscr{P}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}) arises as an intermediary step in a canonical factorization =trX!\mathscr{E}=\text{tr}_{X}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}} of the associated channel. This observation will be crucial for generalizing joint distributions to the quantum domain.

Next we show that in the classical domain, joint distributions are in a bijective correspondence with classical processes.

Notation 3.7.

Given a pair (X,Y)(X,Y) of finite sets, let 𝒮+X(XY)𝒮(XY)\mathcal{S}^{X}_{+}({{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y})\subset\mathcal{S}({{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y}) denote the subset given by

𝒮+X(XY)={τ𝒮(XY)|trY(τ) is invertible}.\mathcal{S}^{X}_{+}({{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y})=\left\{\tau\in\mathcal{S}({{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y})\hskip 2.84526pt|\hskip 2.84526pt\text{$\text{tr}_{Y}(\tau)$ is invertible}\right\}.
Proposition 3.8.

Let (X,Y)(X,Y) be a pair of finite sets, and let

𝔖:𝒫+(X,Y)𝒮+X(XY)\mathfrak{S}:\mathscr{P}_{+}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y})\longrightarrow\mathcal{S}^{X}_{+}({{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y})

be the map given by 𝔖(ρ,)=!(ρ)\mathfrak{S}(\rho,\mathscr{E})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho). Then 𝔖\mathfrak{S} is a bijection, whose inverse is determined by the condition

𝔖1(τ)=(ρ,)yx=τ(x,y)ρx(x,y)X×Y.\mathfrak{S}^{-1}(\tau)=(\rho,\mathscr{E})\implies\mathscr{E}_{yx}=\frac{\tau_{(x,y)}}{\rho_{x}}\quad\quad\forall(x,y)\in X\times Y.
Proof.

Let (X,Y)(X,Y) be a pair of finite sets, and let :𝒮+X(XY)𝒫+(X,Y)\mathfrak{C}:\mathcal{S}^{X}_{+}({{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y})\longrightarrow\mathscr{P}_{+}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}) be the map uniquely determined by the condition

(τ)=(ρ,)yx=τ(x,y)ρx(x,y)X×Y.\mathfrak{C}(\tau)=(\rho,\mathscr{E})\implies\mathscr{E}_{yx}=\frac{\tau_{(x,y)}}{\rho_{x}}\quad\quad\forall(x,y)\in X\times Y.

We will show 𝔖=id\mathfrak{C}\circ\mathfrak{S}=\text{id} and 𝔖=id\mathfrak{S}\circ\mathfrak{C}=\text{id}. For the former case, let (ρ,)𝒫+(X,Y)(\rho,\mathscr{E})\in\mathscr{P}_{+}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}), and let (ρ~,~)=(𝔖(ρ,))(\widetilde{\rho},\widetilde{\mathscr{E}})=\mathfrak{C}\left(\mathfrak{S}(\rho,\mathscr{E})\right). By the definition of \mathfrak{C} we have

(τ)=(ρ,)\displaystyle\mathfrak{C}(\tau)=(\rho,\mathscr{E}) \displaystyle\implies yx=τ(x,y)ρx(x,y)X×Y\displaystyle\mathscr{E}_{yx}=\frac{\tau_{(x,y)}}{\rho_{x}}\quad\quad\hskip 75.39963pt\forall(x,y)\in X\times Y
\displaystyle\implies yYτ(x,y)=yYρxyx=ρxxX\displaystyle\sum_{y\in Y}\tau_{(x,y)}=\sum_{y\in Y}\rho_{x}\mathscr{E}_{yx}=\rho_{x}\quad\quad\forall x\in X
\displaystyle\implies trY(τ)=ρ,\displaystyle\text{tr}_{Y}(\tau)=\rho,

thus

ρ~=trY(𝔖(ρ,))=trY(!(ρ))=ρ.\widetilde{\rho}=\text{tr}_{Y}\left(\mathfrak{S}(\rho,\mathscr{E})\right)=\text{tr}_{Y}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho)\right)=\rho.

Moreover, by definition of \mathfrak{C}, for all (x,y)X×Y(x,y)\in X\times Y we have

~yx=!(ρ)(x,y)ρ~x=ρxyxρx=yx~=,\widetilde{\mathscr{E}}_{yx}=\frac{\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho)_{(x,y)}}{\widetilde{\rho}_{x}}=\frac{\rho_{x}\mathscr{E}_{yx}}{\rho_{x}}=\mathscr{E}_{yx}\implies\widetilde{\mathscr{E}}=\mathscr{E},

thus 𝔖=id\mathfrak{C}\circ\mathfrak{S}=\text{id}.

Now let τ𝒮+X(XY)\tau\in\mathcal{S}^{X}_{+}({{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y}), let (ρ,)=(τ)(\rho,\mathscr{E})=\mathfrak{C}(\tau), and let τ~=𝔖((τ))\widetilde{\tau}=\mathfrak{S}(\mathfrak{C}(\tau)). Then for all (x,y)X×Y(x,y)\in X\times Y we have

τ~(x,y)=𝔖(ρ,)(x,y)=ρxyx=ρx(τ(x,y)ρx)=τ(x,y)τ~=τ,\widetilde{\tau}_{(x,y)}=\mathfrak{S}(\rho,\mathscr{E})_{(x,y)}=\rho_{x}\mathscr{E}_{yx}=\rho_{x}\left(\frac{\tau_{(x,y)}}{\rho_{x}}\right)=\tau_{(x,y)}\implies\widetilde{\tau}=\tau,

thus 𝔖=id\mathfrak{S}\circ\mathfrak{C}=\text{id}, as desired. ∎

Remark 3.9.

In light of Proposition 3.8, it follows that a state τ𝒮+X(XY)\tau\in\mathcal{S}^{X}_{+}({{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y}) may either be viewed as a joint distribution associated with two space-like separated random variables XX and YY occurring in parallel, or as a state over time !(ρ)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho) associated with the process (ρ,)=(τ)𝒫+(X,Y)(\rho,\mathscr{E})=\mathfrak{C}(\tau)\in\mathscr{P}_{+}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}). These two equivalent viewpoints are what we refer to as the static-dynamic duality for classical joint states in 𝒮+X(XY)\mathcal{S}^{X}_{+}({{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y}). Another way to think of the static-dynamic duality for classical joint states, is that classical probability does not distinguish between temporal correlations and spatial correlations, revealing a certain symmetry between space and time for classical random variables. For quantum systems however it turns out that such a symmetry between space and time does not exist, as there exists quantum processes for which temporal correlations may not be viewed as spatial correlations and vice-versa [FJV15]. As such, the static-dynamic duality for classical joint states does not generalize to the quantum domain, and we will see that this is manifested in the fact that quantum states over time are not positive in general. This fact will play a crucial role in the general theory of states over time for quantum processes.

In the next proposition, we show how the state over time !(ρ)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho) associated with a classical process (ρ,)𝒫(X,Y)(\rho,\mathscr{E})\in\mathscr{P}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}) may be re-written in a way that is valid for all quantum processes, which we will use in the next section for a quantum generalization of states over time.

Proposition 3.10.

Let (X,Y)(X,Y) be a pair of finite sets, and let 𝕋(X,Y)\mathscr{E}\in\mathbb{CPTP}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}). Then

!(ρ)=(ρ𝟙)𝒥[]\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho)=(\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}] (3.11)

for all ρX\rho\in{{\mathbb{C}}}^{X}. In particular, if ρ𝒮(X)\rho\in\mathcal{S}({{\mathbb{C}}}^{X}), then (ρ𝟙)𝒥[](\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}] is the state over time associated with the classical process (ρ,)(\rho,\mathscr{E}).

Proof.

Let μ:XXX\mu:{{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{X}\longrightarrow{{\mathbb{C}}}^{X} denote the multiplication map. From the definition of the Hilbert-Schmidt dual we have

μ(𝟙)=xXδxδx,\mu^{*}(\mathds{1})=\sum_{x\in X}\delta_{x}\otimes\delta_{x}\hskip 0.7113pt,

thus

𝒥[]\displaystyle\mathscr{J}[\mathscr{E}] =\displaystyle= (idX)(μ(𝟙))=(idX)(xXδxδx)=xXδx(δx)\displaystyle\left(\text{id}_{{{\mathbb{C}}}^{X}}\otimes\mathscr{E}\right)(\mu^{*}(\mathds{1}))=\left(\text{id}_{{{\mathbb{C}}}^{X}}\otimes\mathscr{E}\right)\left(\sum_{x\in X}\delta_{x}\otimes\delta_{x}\right)=\sum_{x\in X}\delta_{x}\otimes\mathscr{E}(\delta_{x})
=\displaystyle= xXδx(yYyxδy)=(x,y)X×Yyx(δxδy).\displaystyle\sum_{x\in X}\delta_{x}\otimes\left(\sum_{y\in Y}\mathscr{E}_{yx}\delta_{y}\right)=\sum_{(x,y)\in X\times Y}\mathscr{E}_{yx}(\delta_{x}\otimes\delta_{y}).

And since for all ρX\rho\in{{\mathbb{C}}}^{X} we have

ρ𝟙=(xXρxδx)(yYδy)=(x,y)X×Yρx(δxδy),\rho\otimes\mathds{1}=\left(\sum_{x\in X}\rho_{x}\delta_{x}\right)\otimes\left(\sum_{y\in Y}\delta_{y}\right)=\sum_{(x,y)\in X\times Y}\rho_{x}(\delta_{x}\otimes\delta_{y}),

it follows that

(ρ𝟙)𝒥[]\displaystyle(\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}] =\displaystyle= ((x,y)X×Yρx(δxδy))((x,y)X×Yyx(δxδy))\displaystyle\left(\sum_{(x,y)\in X\times Y}\rho_{x}(\delta_{x}\otimes\delta_{y})\right)\left(\sum_{(x,y)\in X\times Y}\mathscr{E}_{yx}(\delta_{x}\otimes\delta_{y})\right)
=\displaystyle= (x,y)X×Yρxyx(δxδy)\displaystyle\sum_{(x,y)\in X\times Y}\rho_{x}\mathscr{E}_{yx}(\delta_{x}\otimes\delta_{y})
=\displaystyle= !(ρ),\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho),

as desired. ∎

Remark 3.12.

While up to now the left-hand side of equation (3.11) is only defined for classical processes, the right-hand side is defined for any quantum process (ρ,)(\rho,\mathscr{E}). Moreover, since for classical processes we have [𝒥[],(ρ𝟙)]=0\left[\mathscr{J}[\mathscr{E}],(\rho\otimes\mathds{1})\right]=0, the right-hand side of equation (3.11) coincides with

λ(ρ𝟙)𝒥[]+(1λ)𝒥[](ρ𝟙)\lambda(\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}]+(1-\lambda)\mathscr{J}[\mathscr{E}](\rho\otimes\mathds{1}) (3.13)

for all λ\lambda\in{{\mathbb{C}}}, which is also well-defined for any quantum process (ρ,)(\rho,\mathscr{E}). However when [(ρ𝟙),𝒥[]]0\left[(\rho\otimes\mathds{1}),\mathscr{J}[\mathscr{E}]\right]\neq 0 and λ1\lambda\neq 1 the expression (3.13) is distinct from (ρ𝟙)𝒥[](\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}], thus in the quantum domain, (3.13) yields a parametric family of states over time associated with a general quantum process (ρ,)(\rho,\mathscr{E}) which generalizes the classical state over time !(ρ)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho). Another crucial point, is that while the multi-matrix (3.13) is of unit trace for all λ\lambda\in{{\mathbb{C}}}, in the quantum domain it is rarely positive (even for λ\lambda\in{{\mathbb{R}}}), and is only guaranteed to be self-adjoint for λ=1/2\lambda=1/2. As such, when generalizing states over time to the quantum domain, we will relax the positivity condition and only require states over time to be self-adjoint elements of unit trace. In fact, as first pointed out in [FJV15], it is the negative eigenvalues of a quantum state over time which act as a witness to causal correlations which exist in the associated process, and are a feature of the theory rather than a defect.

4 In quantum bloom

The classical bloom map !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}} from Proposition 3.2 may be viewed as the output of a mapping that associates every pair (X,Y)(X,Y) of finite sets with a function

!():𝕋(X,Y)𝕋(X,XY)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{(*)}:\mathbb{CPTP}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y})\longrightarrow\mathbb{CPTP}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{X}\otimes{{\mathbb{C}}}^{Y}) (4.1)

such that trX!=\text{tr}_{X}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}=\mathscr{E} and trY!=idX\text{tr}_{Y}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}=\mathrm{id}_{{{\mathbb{C}}}^{X}} for all 𝕋(X,Y)\mathscr{E}\in\mathbb{CPTP}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}). Moreover, the bloom map !()\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{(*)} naturally extends to a map on all of Hom(X,Y)\text{Hom}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}), which together with the partial trace provides a factorization system on Hom(X,Y)\text{Hom}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}) which yields the state over time !(ρ)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho) associated with every classical process (ρ,)𝒫(X,Y)(\rho,\mathscr{E})\in\mathscr{P}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}). In this section, we extend the classical bloom map (4.1) to the quantum domain. Though there are various such extensions, the extension referred to as the symmetric bloom yields a state over time which is always self-adjoint, and will be the focus of our work.

Definition 4.2.

A bloom map associates every pair (𝒜,)(\mathcal{A},\mathcal{B}) of multi-matrix algebras with a map !:𝕋(𝒜,)𝕋(𝒜,𝒜)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}:\mathbb{TP}(\mathcal{A},\mathcal{B})\longrightarrow\mathbb{TP}(\mathcal{A},\mathcal{A}\otimes\mathcal{B}) such that tr𝒜!()=\text{tr}_{\mathcal{A}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})=\mathscr{E} and tr!()=id𝒜\text{tr}_{\mathcal{B}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})=\text{id}_{\mathcal{A}} for all 𝕋(𝒜,)\mathscr{E}\in\mathbb{TP}(\mathcal{A},\mathcal{B}).

Definition 4.3.

If (ρ,)𝒫(𝒜,)(\rho,\mathscr{E})\in\mathscr{P}(\mathcal{A},\mathcal{B}) is a quantum process, and !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} is a bloom map, then !()(ρ)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})(\rho) will be referred to as the state over time associated with the process (ρ,)(\rho,\mathscr{E}) and the bloom map !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}, and will be denoted by !(ρ,)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E}). The map on quantum processes given by

(ρ,)!(ρ,)(\rho,\mathscr{E})\longmapsto\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E})

will then be referred to as the state over time function associated with !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}.

Definition 4.4.

Given a bloom map !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}, the factorization =tr𝒜!()\mathscr{E}=\text{tr}_{\mathcal{A}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}) will be referred to as the bloom-shriek factorization of 𝕋(𝒜,)\mathscr{E}\in\mathbb{TP}(\mathcal{A},\mathcal{B}).

Remark 4.5.

An analogue of bloom-shriek factorization for free gs-monoidal categories appears in [fritz2023free] under the name bloom-circuitry factorization.

Example 4.6.

Let (𝒜,)(\mathcal{A},\mathcal{B}) be a pair of multi-matrix algebras, and let

!:𝕋(𝒜,)𝕋(𝒜,𝒜)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}:\mathbb{TP}(\mathcal{A},\mathcal{B})\longrightarrow\mathbb{TP}(\mathcal{A},\mathcal{A}\otimes\mathcal{B})

be the map given by !()(ρ)=(ρ𝟙)𝒥[]\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})(\rho)=(\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}]. When 𝒜\mathcal{A} is a matrix algebra it was shown in [PaRuBayes] that μ𝒜(𝟙)=i,jEijEji\mu_{\mathcal{A}}^{*}(\mathds{1})=\sum_{i,j}E_{ij}\otimes E_{ji}, where EijE_{ij} are the matrix units in 𝒜\mathcal{A}. We then have

!()(ρ)\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})(\rho) =\displaystyle= (ρ𝟙)𝒥[]=(ρ𝟙)(id𝒜)(μ𝒜(𝟙))\displaystyle(\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}]=(\rho\otimes\mathds{1})(\mathrm{id}_{\mathcal{A}}\otimes\mathscr{E})\left(\mu_{\mathcal{A}}^{*}(\mathds{1})\right)
=\displaystyle= (ρ𝟙)(id𝒜)(i,jEijEji)=(ρ𝟙)(i,jEij(Eji))\displaystyle(\rho\otimes\mathds{1})(\mathrm{id}_{\mathcal{A}}\otimes\mathscr{E})\left(\sum_{i,j}E_{ij}\otimes E_{ji}\right)=(\rho\otimes\mathds{1})\left(\sum_{i,j}E_{ij}\otimes\mathscr{E}(E_{ji})\right)
=\displaystyle= i,jρEij(Eji),\displaystyle\sum_{i,j}\rho E_{ij}\otimes\mathscr{E}(E_{ji}),

from which it follows that

tr𝒜(!()(ρ))=i,jtr(ρEij)(Eji)=i,jρji(Eji)=(ρ),\text{tr}_{\mathcal{A}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})(\rho)\right)=\sum_{i,j}\text{tr}(\rho E_{ij})\mathscr{E}(E_{ji})=\sum_{i,j}\rho_{ji}\mathscr{E}(E_{ji})=\mathscr{E}(\rho),

and

tr(!()(ρ))=i,jtr((Eji))ρEij=i,jδjiρEij=iρEii=ρ,\text{tr}_{\mathcal{B}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})(\rho)\right)=\sum_{i,j}\text{tr}(\mathscr{E}(E_{ji}))\rho E_{ij}=\sum_{i,j}\delta_{ji}\rho E_{ij}=\sum_{i}\rho E_{ii}=\rho,

thus !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} defines a bloom map in this case. When 𝒜=xX𝕄mx\mathcal{A}=\bigoplus_{x\in X}\mathbb{M}_{m_{x}} and =yY𝕄ny\mathcal{B}=\bigoplus_{y\in Y}\mathbb{M}_{n_{y}} are both multi-matrix algebras, then

ρ=xXρx&𝒥[]=(x,y)X×Y𝒥[yx],\rho=\bigoplus_{x\in X}\rho_{x}\quad\&\quad\mathscr{J}[\mathscr{E}]=\bigoplus_{(x,y)\in X\times Y}\mathscr{J}[\mathscr{E}_{yx}],

where yx\mathscr{E}_{yx} are the component functions of Hom(𝒜,)\mathscr{E}\in\text{Hom}(\mathcal{A},\mathcal{B}). We then have

!()(ρ)=(ρ𝟙)𝒥[]=(x,y)X×Y(ρx𝟙)𝒥[yx],\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})(\rho)=(\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}]=\bigoplus_{(x,y)\in X\times Y}(\rho_{x}\otimes\mathds{1})\mathscr{J}[\mathscr{E}_{yx}],

thus the above arguments in the case when 𝒜\mathcal{A} is a matrix algebra may be applied component-wise to deduce tr𝒜!()=\text{tr}_{\mathcal{A}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})=\mathscr{E} and tr!()=id𝒜\text{tr}_{\mathcal{B}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})=\text{id}_{\mathcal{A}}, thus !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} defines a bloom map. Moreover, it follows from Proposition 3.10 that this bloom map recovers the classical state over time !(ρ)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\mathscr{E}}(\rho) associated with classical processes (ρ,)𝒫(X,Y)(\rho,\mathscr{E})\in\mathscr{P}({{\mathbb{C}}}^{X},{{\mathbb{C}}}^{Y}).

In light of the previous example, the following definition provides three examples of bloom maps.

Definition 4.7.

Let (𝒜,)(\mathcal{A},\mathcal{B}) be a pair of multi-matrix algebras.

  • The right bloom is the map !R:𝕋(𝒜,)𝕋(𝒜,𝒜)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}:\mathbb{TP}(\mathcal{A},\mathcal{B})\longrightarrow\mathbb{TP}(\mathcal{A},\mathcal{A}\otimes\mathcal{B}) given by !R()\mathscr{E}\longmapsto\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{E}), where

    !R()(ρ)=(ρ𝟙)𝒥[]\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{E})(\rho)=(\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}]

    for all ρ𝒜\rho\in\mathcal{A}.

  • The left bloom is the map !L:𝕋(𝒜,)𝕋(𝒜,𝒜)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}:\mathbb{TP}(\mathcal{A},\mathcal{B})\longrightarrow\mathbb{TP}(\mathcal{A},\mathcal{A}\otimes\mathcal{B}) given by !L()\mathscr{E}\longmapsto\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{E}), where

    !L()(ρ)=𝒥[](ρ𝟙)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{E})(\rho)=\mathscr{J}[\mathscr{E}](\rho\otimes\mathds{1})

    for all ρ𝒜\rho\in\mathcal{A}.

  • The symmetric bloom is the map \Yinyang:𝕋(𝒜,)𝕋(𝒜,𝒜)\text{\Yinyang}:\mathbb{TP}(\mathcal{A},\mathcal{B})\longrightarrow\mathbb{TP}(\mathcal{A},\mathcal{A}\otimes\mathcal{B}) given by \Yinyang()\mathscr{E}\longmapsto\text{\Yinyang}(\mathscr{E}), where

    \Yinyang()(ρ)=12((ρ𝟙)𝒥[]+𝒥[](ρ𝟙))\text{\Yinyang}(\mathscr{E})(\rho)=\frac{1}{2}\left((\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}]+\mathscr{J}[\mathscr{E}](\rho\otimes\mathds{1})\right)

    for all ρ𝒜\rho\in\mathcal{A}.

Remark 4.8.

The symmetric bloom was introduced in [FuPa22], where it was proved that the associated state over time function (ρ,)\Yinyang(ρ,):=\Yinyang()(ρ)(\rho,\mathscr{E})\longmapsto\text{\Yinyang}(\rho,\mathscr{E}):=\text{\Yinyang}(\mathscr{E})(\rho) satisfies a list of axioms set forth in [HHPBS17]. The list of axioms include hermiticity of \Yinyang(ρ,)\text{\Yinyang}(\rho,\mathscr{E}), bilinearity of \Yinyang(,)\text{\Yinyang}(*,*), a classical limit axiom, and an associativity axiom which ensures that \Yinyang extends in a well-defined way to nn-step processes (ρ,1,,n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}). In a later work we will prove that the symmetric bloom is in fact the only state over time function satisfying the aforementioned list of axioms.

The following statement yields an important property of the symmetric bloom which will be useful for our purposes.

Proposition 4.9.

The (𝒜,)(\mathcal{A},\mathcal{B}) be a pair of multi-matrix algebras, and let 𝕋(𝒜,)\mathscr{E}\in\mathbb{TP}(\mathcal{A},\mathcal{B}). Then the following statements are equivalent.

  1. i.

    \mathscr{E} is {\dagger}-preserving.

  2. ii.

    𝒥[]\mathscr{J}[\mathscr{E}] is self-adjoint.

  3. iii.

    \Yinyang()\emph{\Yinyang}(\mathscr{E}) is {\dagger}-preserving.

Proof.

We prove the statement in the case of matrix algebras.

i\implies ii: Let μ:𝒜𝒜𝒜\mu:\mathcal{A}\otimes\mathcal{A}\longrightarrow\mathcal{A} denote the multiplication map. It follows from [PaRuBayes] that

μ(𝟙)=i,jEijEji,\mu^{*}(\mathds{1})=\sum_{i,j}E_{ij}\otimes E_{ji},

thus

𝒥[]=(id𝒜)μ(𝟙)=(id𝒜)(i,jEijEji)=i,jEij(Eji).\mathscr{J}[\mathscr{E}]=\left(\mathrm{id}_{\mathcal{A}}\otimes\mathscr{E}\right)\mu^{*}(\mathds{1})=\left(\mathrm{id}_{\mathcal{A}}\otimes\mathscr{E}\right)\left(\sum_{i,j}E_{ij}\otimes E_{ji}\right)=\sum_{i,j}E_{ij}\otimes\mathscr{E}(E_{ji}).

We then have

𝒥[]\displaystyle\mathscr{J}[\mathscr{E}]^{{\dagger}} =\displaystyle= (i,jEij(Eji))=i,jEij(Eji)\displaystyle\left(\sum_{i,j}E_{ij}\otimes\mathscr{E}(E_{ji})\right)^{{\dagger}}=\sum_{i,j}E_{ij}^{{\dagger}}\otimes\mathscr{E}(E_{ji})^{{\dagger}}
=\displaystyle= i,jEji(Eji)=i,jEji(Eij)\displaystyle\sum_{i,j}E_{ji}\otimes\mathscr{E}(E_{ji}^{{\dagger}})=\sum_{i,j}E_{ji}\otimes\mathscr{E}(E_{ij})
=\displaystyle= 𝒥[],\displaystyle\mathscr{J}[\mathscr{E}],

as desired.

ii\implies iii: Let ρ𝒜\rho\in\mathcal{A}. Then

\Yinyang()(ρ)\displaystyle\text{\Yinyang}(\mathscr{E})(\rho)^{{\dagger}} =\displaystyle= 12((ρ𝟙)𝒥[]+𝒥[](ρ𝟙))\displaystyle\frac{1}{2}\left((\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}]+\mathscr{J}[\mathscr{E}](\rho\otimes\mathds{1})\right)^{{\dagger}}
=\displaystyle= 12(𝒥[](ρ𝟙)+(ρ𝟙)𝒥[])\displaystyle\frac{1}{2}\left(\mathscr{J}[\mathscr{E}]^{{\dagger}}(\rho\otimes\mathds{1})^{{\dagger}}+(\rho\otimes\mathds{1})^{{\dagger}}\mathscr{J}[\mathscr{E}]^{{\dagger}}\right)
=\displaystyle= 12(𝒥[](ρ𝟙)+(ρ𝟙)𝒥[])\displaystyle\frac{1}{2}\left(\mathscr{J}[\mathscr{E}](\rho^{{\dagger}}\otimes\mathds{1})+(\rho^{{\dagger}}\otimes\mathds{1})\mathscr{J}[\mathscr{E}]\right)
=\displaystyle= \Yinyang()(ρ),\displaystyle\text{\Yinyang}(\mathscr{E})(\rho^{{\dagger}}),

thus \Yinyang()\text{\Yinyang}(\mathscr{E}) is {\dagger}-preserving.

iii\implies i: Since

tr𝒜(AB)=(tr(A)B)=tr(A)¯B=tr(A)B=tr𝒜(AB)=tr𝒜((AB)),\text{tr}_{\mathcal{A}}(A\otimes B)^{{\dagger}}=\left(\text{tr}(A)B\right)^{{\dagger}}=\overline{\text{tr}(A)}B^{{\dagger}}=\text{tr}(A^{{\dagger}})B^{{\dagger}}=\text{tr}_{\mathcal{A}}(A^{{\dagger}}\otimes B^{{\dagger}})=\text{tr}_{\mathcal{A}}\left((A\otimes B)^{{\dagger}}\right),

it follows that tr𝒜\text{tr}_{\mathcal{A}} is {\dagger}-preserving. For all ρA\rho\in A we then have

(ρ)=(tr𝒜\Yinyang())(ρ)=tr𝒜(\Yinyang()(ρ))=tr𝒜(\Yinyang()(ρ))=tr𝒜(\Yinyang()(ρ))=(ρ),\mathscr{E}(\rho)^{{\dagger}}=(\text{tr}_{\mathcal{A}}\circ\text{\Yinyang}(\mathscr{E}))(\rho)^{{\dagger}}=\text{tr}_{\mathcal{A}}\left(\text{\Yinyang}(\mathscr{E})(\rho)\right)^{{\dagger}}=\text{tr}_{\mathcal{A}}\left(\text{\Yinyang}(\mathscr{E})(\rho)^{{\dagger}}\right)=\text{tr}_{\mathcal{A}}\left(\text{\Yinyang}(\mathscr{E})(\rho^{{\dagger}})\right)=\mathscr{E}(\rho^{{\dagger}}),

as desired. ∎

Definition 4.10.

A bloom map !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} is said to be

  • classically reducible if and only if [(ρ𝟙),𝒥[]]=0!()(ρ)=(ρ𝟙)𝒥[]\left[(\rho\otimes\mathds{1}),\mathscr{J}[\mathscr{E}]\right]=0\implies\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})(\rho)=(\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}].

  • hermitian if and only if !()\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}) is {\dagger}-preserving whenever \mathscr{E} is {\dagger}-preserving.

Remark 4.11.

By Proposition 4.9 it follows that the symmetric bloom is hermitian, and it follows directly from its definition that the symmetric bloom is classically reducible. These are two key properties one should expect from a quantum bloom map if it is to be viewed as generalization of the classical bloom. While we have reason to believe that the symmetric bloom is the only bloom map which is both hermitian and classically reducible, we have yet to prove such a result.

We now generalize Proposition 3.8 using the symmetric bloom. In particular, given a pseudo-density operator on τ𝒜\tau\in\mathcal{A}\otimes\mathcal{B}, we derive an explicit formula for a process (ρ,)(\rho,\mathscr{E}) such that τ=\Yinyang(ρ,)\tau=\text{\Yinyang}(\rho,\mathscr{E}). For this, we recall that 𝒯(𝒜)\mathscr{T}(\mathcal{A}\otimes\mathcal{B}) denotes the set of pseudo-density operators with respect to the tensor factorization 𝒜0=𝒜\mathcal{A}_{0}=\mathcal{A}\otimes\mathcal{B} (see Definition 2.20).

Lemma 4.12.

Let τ𝒯(𝒜)\tau\in\mathscr{T}(\mathcal{A}\otimes\mathcal{B}) be such that tr(τ)\emph{tr}_{\mathcal{B}}(\tau) is invertible. Then there exists a unique element Xτ𝒜X_{\tau}\in\mathcal{A}\otimes\mathcal{B} such that

(tr(τ)𝟙)Xτ+Xτ(tr(τ)𝟙)=2τ.(\emph{tr}_{\mathcal{B}}(\tau)\otimes\mathds{1})X_{\tau}+X_{\tau}(\emph{tr}_{\mathcal{B}}(\tau)\otimes\mathds{1})=2\tau. (4.13)
Proof.

We prove the statement in the case of matrix algebras, where we will use the fact from linear algebra that if A,B,C𝕄nA,B,C\in\mathbb{M}_{n}, then the Sylvester equation AX+XB=CAX+XB=C has a unique solution X𝕄nX\in\mathbb{M}_{n} if and only if AA and B-B have disjoint spectrums. So let τ𝒯(𝒜)\tau\in\mathscr{T}(\mathcal{A}\otimes\mathcal{B}) be such that tr(τ)\text{tr}_{\mathcal{B}}(\tau) is invertible. Since tr(τ)\text{tr}_{\mathcal{B}}(\tau) is a state which is invertible, it follows that the eigenvalues of tr(τ)\text{tr}_{\mathcal{B}}(\tau) are strictly positive, thus the spectrums of tr(τ)𝟙\text{tr}_{\mathcal{B}}(\tau)\otimes\mathds{1} and (tr(τ)𝟙)-(\text{tr}_{\mathcal{B}}(\tau)\otimes\mathds{1}) are disjoint. It then follows that the Sylvester equation

(tr(τ)𝟙)X+X(tr(τ)𝟙)=2τ(\text{tr}_{\mathcal{B}}(\tau)\otimes\mathds{1})X+X(\text{tr}_{\mathcal{B}}(\tau)\otimes\mathds{1})=2\tau

has a unique solution Xτ𝒜X_{\tau}\in\mathcal{A}\otimes\mathcal{B}, thus proving the statement. ∎

Notation 4.14.

Given a pair (𝒜,)(\mathcal{A},\mathcal{B}) of multi-matrix algebras, we let

𝒯(𝒜)𝒯+𝒜(𝒜)𝒯(𝒜)\mathscr{T}_{*}(\mathcal{A}\otimes\mathcal{B})\subset\mathscr{T}^{\mathcal{A}}_{+}(\mathcal{A}\otimes\mathcal{B})\subset\mathscr{T}(\mathcal{A}\otimes\mathcal{B}) (4.15)

denote the subsets given by

𝒯+𝒜(𝒜)={τ𝒯(𝒜)|tr(τ)is invertible},\mathscr{T}^{\mathcal{A}}_{+}(\mathcal{A}\otimes\mathcal{B})=\left\{\tau\in\mathscr{T}(\mathcal{A}\otimes\mathcal{B})\hskip 2.84526pt|\hskip 2.84526pt\text{tr}_{\mathcal{B}}(\tau)\hskip 2.84526pt\text{is invertible}\right\},

and

𝒯(𝒜)={τ𝒯+𝒜(𝒜)|𝒥1(Xτ)𝕋(𝒜,)},\mathscr{T}_{*}(\mathcal{A}\otimes\mathcal{B})=\left\{\tau\in\mathscr{T}^{\mathcal{A}}_{+}(\mathcal{A}\otimes\mathcal{B})\hskip 2.84526pt|\hskip 2.84526pt\mathscr{J}^{-1}(X_{\tau})\in\mathbb{CPTP}(\mathcal{A},\mathcal{B})\right\},

where Xτ𝒜X_{\tau}\in\mathcal{A}\otimes\mathcal{B} is the unique element satisfying (4.13).

Lemma 4.16.

Let (𝒜,)(\mathcal{A},\mathcal{B}) be a pair of multi-matrix algebras. Then \Yinyang(ρ,)𝒯(𝒜)\text{\Yinyang}(\rho,\mathscr{E})\in\mathscr{T}_{*}(\mathcal{A}\otimes\mathcal{B}) for all (ρ,)𝒫+(𝒜,)(\rho,\mathscr{E})\in\mathscr{P}_{+}(\mathcal{A},\mathcal{B}).

Proof.

Let (ρ,)𝒫+(𝒜,)(\rho,\mathscr{E})\in\mathscr{P}_{+}(\mathcal{A},\mathcal{B}) and let τ=\Yinyang(ρ,)\tau=\text{\Yinyang}(\rho,\mathscr{E}). To show τ𝒯(𝒜)\tau\in\mathscr{T}_{*}(\mathcal{A}\otimes\mathcal{B}) we need to show that τ\tau is a pseudo-density operator such that tr(τ)\text{tr}_{\mathcal{B}}(\tau) is invertible and 𝒥1(Xτ)\mathscr{J}^{-1}(X_{\tau}) is CPTP. To show τ\tau is a pseuso-density operator we first show τ\tau is self-adjoint. Now since \mathscr{E} is CPTP it is {\dagger}-preserving thus \Yinyang()\text{\Yinyang}(\mathscr{E}) is {\dagger}-preserving by Proposition 4.9. We then have

τ=\Yinyang(ρ,)=\Yinyang()(ρ)=\Yinyang()(ρ)=\Yinyang()(ρ)=τ,\tau^{{\dagger}}=\text{\Yinyang}(\rho,\mathscr{E})^{{\dagger}}=\text{\Yinyang}(\mathscr{E})(\rho)^{{\dagger}}=\text{\Yinyang}(\mathscr{E})(\rho^{{\dagger}})=\text{\Yinyang}(\mathscr{E})(\rho)=\tau,

thus τ\tau is self-adjoint. And since the marginals of τ\tau are ρ\rho and (ρ)\mathscr{E}(\rho) with ρ\rho invertible (since (ρ,)𝒫+(𝒜,)(\rho,\mathscr{E})\in\mathscr{P}_{+}(\mathcal{A},\mathcal{B})), it follows that τ𝒯+𝒜(𝒜)\tau\in\mathscr{T}^{\mathcal{A}}_{+}(\mathcal{A}\otimes\mathcal{B}). Moreover, since ρ=tr(τ)\rho=\text{tr}_{\mathcal{B}}(\tau) and

τ=12((ρ𝟙)𝒥[]+𝒥[](ρ𝟙)),\tau=\frac{1}{2}\left((\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}]+\mathscr{J}[\mathscr{E}](\rho\otimes\mathds{1})\right),

it follows that Xτ=𝒥[]X_{\tau}=\mathscr{J}[\mathscr{E}] and 𝒥1(Xτ)=𝕋(𝒜,)\mathscr{J}^{-1}(X_{\tau})=\mathscr{E}\in\mathbb{CPTP}(\mathcal{A},\mathcal{B}), thus τ𝒯(𝒜)\tau\in\mathscr{T}_{*}(\mathcal{A}\otimes\mathcal{B}), as desired. ∎

Theorem 4.17.

Let (𝒜,)(\mathcal{A},\mathcal{B}) be a pair of multi-matrix algebras, and let 𝔖:𝒫+(𝒜,)𝒯(𝒜)\mathfrak{S}:\mathscr{P}_{+}(\mathcal{A},\mathcal{B})\longrightarrow\mathscr{T}_{*}(\mathcal{A}\otimes\mathcal{B}) be the map given by 𝔖(ρ,)=\Yinyang(ρ,)\mathfrak{S}(\rho,\mathscr{E})=\text{\Yinyang}(\rho,\mathscr{E}). Then 𝔖\mathfrak{S} is a bijection, whose inverse is given by

𝔖1(τ)=(tr(τ),𝒥1(Xτ)),\mathfrak{S}^{-1}(\tau)=\left(\emph{tr}_{\mathcal{B}}(\tau),\mathscr{J}^{-1}(X_{\tau})\right), (4.18)

where Xτ𝒜X_{\tau}\in\mathcal{A}\otimes\mathcal{B} is the unique element satisfying (4.13).

Proof.

Let :𝒯(𝒜)𝒫+(𝒜,)\mathfrak{C}:\mathscr{T}_{*}(\mathcal{A}\otimes\mathcal{B})\longrightarrow\mathscr{P}_{+}(\mathcal{A},\mathcal{B}) be the map given by (τ)=(tr(τ),𝒥1(Xτ))\mathfrak{C}(\tau)=(\text{tr}_{\mathcal{B}}(\tau),\mathscr{J}^{-1}(X_{\tau})). We now show 𝔖=id\mathfrak{C}\circ\mathfrak{S}=\mathrm{id} and 𝔖=id\mathfrak{S}\circ\mathfrak{C}=\mathrm{id}. So let (ρ,)𝒫+(𝒜,)(\rho,\mathscr{E})\in\mathscr{P}_{+}(\mathcal{A},\mathcal{B}) and let τ=𝔖(ρ,)\tau=\mathfrak{S}(\rho,\mathscr{E}), so that ρ=tr(τ)\rho=\text{tr}_{\mathcal{B}}(\tau) and

τ=12((ρ𝟙)𝒥[]+𝒥[](ρ𝟙))𝒥[]=Xτ=𝒥1(Xτ).\tau=\frac{1}{2}\left((\rho\otimes\mathds{1})\mathscr{J}[\mathscr{E}]+\mathscr{J}[\mathscr{E}](\rho\otimes\mathds{1})\right)\implies\mathscr{J}[\mathscr{E}]=X_{\tau}\implies\mathscr{E}=\mathscr{J}^{-1}(X_{\tau}).

We then have

(𝔖)(ρ,)=(τ)=(tr(τ),𝒥1(Xτ))=(ρ,),(\mathfrak{C}\circ\mathfrak{S})(\rho,\mathscr{E})=\mathfrak{C}(\tau)=(\text{tr}_{\mathcal{B}}(\tau),\mathscr{J}^{-1}(X_{\tau}))=(\rho,\mathscr{E}),

thus 𝔖=id\mathfrak{C}\circ\mathfrak{S}=\mathrm{id}. Now let τ𝒯(𝒜)\tau\in\mathscr{T}_{*}(\mathcal{A}\otimes\mathcal{B}), so that

12(tr(τ)𝟙)Xτ+Xτ(tr(τ)𝟙)=τ.\frac{1}{2}\left(\text{tr}_{\mathcal{B}}(\tau)\otimes\mathds{1})X_{\tau}+X_{\tau}(\text{tr}_{\mathcal{B}}(\tau)\otimes\mathds{1}\right)=\tau.

We then have

(𝔖)(τ)\displaystyle(\mathfrak{S}\circ\mathfrak{C})(\tau) =\displaystyle= 𝔖((τ))\displaystyle\mathfrak{S}\left(\mathfrak{C}(\tau)\right)
=\displaystyle= 𝔖(tr(τ),𝒥1(Xτ))\displaystyle\mathfrak{S}(\text{tr}_{\mathcal{B}}(\tau),\mathscr{J}^{-1}(X_{\tau}))
=\displaystyle= 12((tr(τ)𝟙)𝒥[𝒥1(Xτ)]+𝒥[𝒥1(Xτ)](tr(τ)𝟙))\displaystyle\frac{1}{2}\left((\text{tr}_{\mathcal{B}}(\tau)\otimes\mathds{1})\mathscr{J}[\mathscr{J}^{-1}(X_{\tau})]+\mathscr{J}[\mathscr{J}^{-1}(X_{\tau})](\text{tr}_{\mathcal{B}}(\tau)\otimes\mathds{1})\right)
=\displaystyle= 12(tr(τ)𝟙)Xτ+Xτ(tr(τ)𝟙)\displaystyle\frac{1}{2}\left(\text{tr}_{\mathcal{B}}(\tau)\otimes\mathds{1})X_{\tau}+X_{\tau}(\text{tr}_{\mathcal{B}}(\tau)\otimes\mathds{1}\right)
=\displaystyle= τ,\displaystyle\tau,

thus 𝔖=id\mathfrak{S}\circ\mathfrak{C}=\mathrm{id}, as desired. ∎

5 Blooming 2-chains

Our next goal is to extend bloom maps to nn-chains, but first we need to consider the case n=2n=2. In this section we will show is that if a bloom map satisfies an associativity condition on 22-chains, then it naturally extends to a unique bloom map on 22-chains. We then show that the right bloom, left bloom and symmetric bloom are all associative, and as such, yield well-defined state over time functions for 2-step processes. In such a case we also prove an explicit formula for the symmetric bloom state over time function, and show that is satisfies a certain compositionality property. This will provide us with the mathematical foundation for using bloom maps to define explicit states over time associated with the dynamics of nn-step quantum processes for all n>0n>0.

Definition 5.1.

Given a triple (𝒜,,𝒞)(\mathcal{A},\mathcal{B},\mathcal{C}) of multi-matrix algebras, a 22-chain consists of a pair (,)𝕋(𝒜,)×𝕋(,𝒞)(\mathscr{E},\mathscr{F})\in\mathbb{TP}(\mathcal{A},\mathcal{B})\times\mathbb{TP}(\mathcal{B},\mathcal{C}). The set of all such 22-chains will be denoted by 𝕋(𝒜,,𝒞)\mathbb{TP}(\mathcal{A},\mathcal{B},\mathcal{C}).

Example 5.2 (Blooming 2-chains).

Let 𝒜𝒞\mathcal{A}\overset{\mathscr{E}}{\longrightarrow}\mathcal{B}\overset{\mathscr{F}}{\longrightarrow}\mathcal{C} be a 2-chain, let !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} be a bloom map, and consider the following diagram.

𝒜\textstyle{\mathcal{A}}\textstyle{\mathcal{B}}𝒞\textstyle{\mathcal{C}}𝒜\textstyle{\mathcal{A}\otimes\mathcal{B}}𝒞\textstyle{\mathcal{B}\otimes\mathcal{C}}\scriptstyle{\mathscr{E}}\scriptstyle{\mathscr{F}}!\scriptstyle{\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}}tr!\scriptstyle{\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}}tr (5.3)

Using only maps in the above diagram, there are two maps one may construct in 𝕋(𝒜,𝒜𝒞)\mathbb{TP}(\mathcal{A},\mathcal{A}\otimes\mathcal{B}\otimes\mathcal{C}), namely, !(!())\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F})\circ\mathscr{E}\right) and !(tr)!()\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}). As the map !(!())\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F})\circ\mathscr{E}\right) has codomain 𝒜(𝒞)\mathcal{A}\otimes(\mathcal{B}\otimes\mathcal{C}) and the map !(tr)!()\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}) has codomain (𝒜)𝒞(\mathcal{A}\otimes\mathcal{B})\otimes\mathcal{C}, there is a bijection between 2-blooms constructed from the above diagram and parenthezations of the multi-matrix algebra 𝒜𝒞\mathcal{A}\otimes\mathcal{B}\otimes\mathcal{C}. It is then a natural to question whether or not the two maps !(!())\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F})\circ\mathscr{E}\right) and !(tr)!()\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}) are equal. If the two maps are in fact equal, then the bloom map !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} naturally extends to a unique bloom map on 2-chains. Such considerations then motivate the following definition.

Definition 5.4.

A bloom map !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} is said to be associative if and only if for every 2-chain

𝒜𝒞\mathcal{A}\overset{\mathscr{E}}{\longrightarrow}\mathcal{B}\overset{\mathscr{F}}{\longrightarrow}\mathcal{C}

we have

!(!())=!(tr)!().\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F})\circ\mathscr{E}\right)=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}). (5.5)

It turns out that the right bloom, left bloom and symmetric bloom are indeed associative, but before giving the proof, we will first prove the following lemma, which will also be useful later on.

Lemma 5.6.

Let (𝒜,,𝒞)(\mathcal{A},\mathcal{B},\mathcal{C}) be a triple of matrix algebras, and let (,)𝕋(𝒜,,𝒞)(\mathscr{E},\mathscr{F})\in\mathbb{TP}(\mathcal{A},\mathcal{B},\mathcal{C}) be a 2-chain. Then the following statements hold.

  1. i.

    𝒥[!R()]=(𝒥[]𝟙)(𝟙𝒥[])\mathscr{J}[\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E}]=\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1}\right)\left(\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right)

  2. ii.

    𝒥[!L()]=(𝟙𝒥[])(𝒥[]𝟙)\mathscr{J}[\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F})\circ\mathscr{E}]=\left(\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right)\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1}\right)

  3. iii.

    𝒥[tr]=𝟙𝒥[]\mathscr{J}[\mathscr{F}\circ{\rm tr}]=\mathds{1}\otimes\mathscr{J}[\mathscr{F}]

  4. iv.

    !L(!R())=!R(tr)!L()\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F}\circ\emph{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{E})

  5. v.

    !R(!L())=!L(tr)!R()\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F})\circ\mathscr{E})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F}\circ\emph{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{E})

Proof.

Item i: Indeed, for all ρ𝒜\rho\in\mathcal{A} we have

𝒥[!R()]\displaystyle\mathscr{J}[\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E}] =\displaystyle= i,jEij𝒜(!R())(Eji𝒜)\displaystyle\sum_{i,j}E_{ij}^{\mathcal{A}}\otimes\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E}\right)(E_{ji}^{\mathcal{A}})
=\displaystyle= i,jEij𝒜!R()((Eji𝒜))\displaystyle\sum_{i,j}E_{ij}^{\mathcal{A}}\otimes\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\left(\mathscr{E}(E_{ji}^{\mathcal{A}})\right)
=\displaystyle= i,jEij𝒜(((Eji𝒜)𝟙)𝒥[])\displaystyle\sum_{i,j}E_{ij}^{\mathcal{A}}\otimes\left((\mathscr{E}(E_{ji}^{\mathcal{A}})\otimes\mathds{1})\mathscr{J}[\mathscr{F}]\right)
=\displaystyle= i,jEij𝒜(((Eji𝒜)𝟙)(k,lEkl(Elk)))\displaystyle\sum_{i,j}E_{ij}^{\mathcal{A}}\otimes\left((\mathscr{E}(E_{ji}^{\mathcal{A}})\otimes\mathds{1})\left(\sum_{k,l}E_{kl}^{\mathcal{B}}\otimes\mathscr{F}(E_{lk}^{\mathcal{B}})\right)\right)
=\displaystyle= i,j,k,lEij𝒜(Eji𝒜)Ekl(Elk)\displaystyle\sum_{i,j,k,l}E_{ij}^{\mathcal{A}}\otimes\mathscr{E}(E_{ji}^{\mathcal{A}})E_{kl}^{\mathcal{B}}\otimes\mathscr{F}(E_{lk}^{\mathcal{B}})
=\displaystyle= (i,jEij𝒜(Eji𝒜)𝟙)(k,l𝟙Ekl(Elk))\displaystyle\left(\sum_{i,j}E^{\mathcal{A}}_{ij}\otimes\mathscr{E}(E^{\mathcal{A}}_{ji})\otimes\mathds{1}\right)\left(\sum_{k,l}\mathds{1}\otimes E^{\mathcal{B}}_{kl}\otimes\mathscr{F}(E^{\mathcal{B}}_{lk})\right)
=\displaystyle= (𝒥[]𝟙)(𝟙𝒥[]),\displaystyle\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1}\right)\left(\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right),

as desired.

Item ii: The proof is similar to that of item i.

Item iii: Indeed,

𝒥[tr]\displaystyle\mathscr{J}[\mathscr{F}\circ{\rm tr}] =\displaystyle= r,s,k,l(Ers𝒜Ekl)((tr)(Esr𝒜Elk))\displaystyle\sum_{r,s,k,l}\left(E_{rs}^{\mathcal{A}}\otimes E_{kl}^{\mathcal{B}}\right)\otimes\left((\mathscr{F}\circ\text{tr})(E_{sr}^{\mathcal{A}}\otimes E_{lk}^{\mathcal{B}})\right)
=\displaystyle= r,s,k,l(Ers𝒜Ekl)(δsr(Elk))\displaystyle\sum_{r,s,k,l}\left(E_{rs}^{\mathcal{A}}\otimes E_{kl}^{\mathcal{B}}\right)\otimes\left(\delta_{sr}\mathscr{F}(E_{lk}^{\mathcal{B}})\right)
=\displaystyle= r,k,l(Err𝒜Ekl)(Elk)\displaystyle\sum_{r,k,l}\left(E_{rr}^{\mathcal{A}}\otimes E_{kl}^{\mathcal{B}}\right)\otimes\mathscr{F}(E_{lk}^{\mathcal{B}})
=\displaystyle= k,l(𝟙Ekl)(Elk)\displaystyle\sum_{k,l}\left(\mathds{1}\otimes E_{kl}^{\mathcal{B}}\right)\otimes\mathscr{F}(E_{lk}^{\mathcal{B}})
=\displaystyle= 𝟙𝒥[],\displaystyle\mathds{1}\otimes\mathscr{J}[\mathscr{F}],

as desired.

Item iv: Indeed, for all ρ𝒜\rho\in\mathcal{A} we have

(!R(tr)!L())(ρ)\displaystyle\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F}\circ{\rm tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{E})\right)(\rho) =\displaystyle= !R(tr)((!L()(ρ))\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F}\circ\text{tr})\left((\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{E})(\rho)\right)
=\displaystyle= (!L()(ρ)𝟙)𝒥[tr]\displaystyle\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{E})(\rho)\otimes\mathds{1}\right)\mathscr{J}[\mathscr{F}\circ{\rm tr}]
=\displaystyle= (𝒥[]𝟙)(ρ𝟙𝟙)(𝟙𝒥[])\displaystyle\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1}\right)\left(\rho\otimes\mathds{1}\otimes\mathds{1}\right)\left(\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right)
=\displaystyle= (𝒥[]𝟙)(𝟙𝒥[])(ρ𝟙𝟙)\displaystyle\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1}\right)\left(\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right)\left(\rho\otimes\mathds{1}\otimes\mathds{1}\right)
=\displaystyle= 𝒥[!R()](ρ𝟙)\displaystyle\mathscr{J}[\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E}](\rho\otimes\mathds{1})
=\displaystyle= !L(!R())(ρ),\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E})(\rho),

where the third and fifth equalities follows from items iii and i. It then follows that !L(!R())=!R(tr)!L()\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F}\circ{\rm tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{E}), as desired.

Item v: The proof is similar to that of item iv. ∎

Proposition 5.7.

The right bloom, left bloom and symmetric bloom are all associative.

Proof.

Right bloom: Indeed, for all ρ𝒜\rho\in\mathcal{A} we have

(!R(tr)!R())(ρ)\displaystyle\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{E})\right)(\rho) =\displaystyle= !R(tr)((!R()(ρ))\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F}\circ\text{tr})\left((\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{E})(\rho)\right)
=\displaystyle= (!R()(ρ)𝟙)𝒥[tr]\displaystyle\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{E})(\rho)\otimes\mathds{1}\right)\mathscr{J}[\mathscr{F}\circ{\rm tr}]
=\displaystyle= (ρ𝟙)(𝒥[]𝟙)(𝟙𝒥[])\displaystyle(\rho\otimes\mathds{1})\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1}\right)\left(\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right)
=\displaystyle= (ρ𝟙)𝒥[!R()]\displaystyle(\rho\otimes\mathds{1})\mathscr{J}[\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E}]
=\displaystyle= !R(!R())(ρ),\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E})(\rho),

where the third and fourth equalities follow from items iii and i of Lemma 5.6. It then follows that !R(!R())=!R(tr)!R()\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E}\right)=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{E}), as desired.

Left bloom: The proof is similar to that for the right bloom.

Symmetric bloom: We first compute

\Yinyang(\Yinyang())\displaystyle\text{\Yinyang}(\text{\Yinyang}(\mathscr{F})\circ\mathscr{E}) =\displaystyle= 12(!R(12(!R()+!L()))+!L(12(!R()+!L())))\displaystyle\frac{1}{2}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}\left(\frac{1}{2}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F})\right)\circ\mathscr{E}\right)+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}\left(\frac{1}{2}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F})\right)\circ\mathscr{E}\right)\right)
=\displaystyle= 14(!R(!R())+!L(!L())+!R(!L())+!L(!R())).\displaystyle\frac{1}{4}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E}\right)+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F})\circ\mathscr{E}\right)+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F})\circ\mathscr{E}\right)+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E}\right)\right).

Now let =\Yinyang(tr)\Yinyang()\aleph=\text{\Yinyang}(\mathscr{F}\circ\text{tr})\circ\text{\Yinyang}(\mathscr{E}). Then

\displaystyle\aleph =\displaystyle= \Yinyang(tr)\Yinyang()\displaystyle\text{\Yinyang}(\mathscr{F}\circ\text{tr})\circ\text{\Yinyang}(\mathscr{E})
=\displaystyle= 12(!R(tr)+!L(tr))12(!R()+!L())\displaystyle\frac{1}{2}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F}\circ\text{tr})+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F}\circ\text{tr})\right)\circ\frac{1}{2}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{E})+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{E})\right)
=\displaystyle= 14(!R(tr)!R()+!L(tr)!L()+!R(tr)!L()+!L(tr)!R())\displaystyle\frac{1}{4}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{E})+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{E})+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{E})+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{E})\right)
=\displaystyle= 14(!R(!R())+!L(!L())+!R(!L())+!L(!R()))\displaystyle\frac{1}{4}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E}\right)+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F})\circ\mathscr{E}\right)+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F})\circ\mathscr{E}\right)+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E}\right)\right)
=\displaystyle= \Yinyang(\Yinyang()),\displaystyle\text{\Yinyang}(\text{\Yinyang}(\mathscr{F})\circ\mathscr{E}),

where the second-to-last equality follows from the associativity of !R\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}, the associativity of !L\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}, and items v and iv of Lemma 5.6. ∎

Associative bloom maps yield well-defined state over time functions on 2-step processes, as we now show.

Definition 5.8.

Given a triple (𝒜,,)(\mathcal{A},\mathcal{B},{{\mathbb{C}}}) of multi-matrix algebras, an element (ρ,,)𝒮(𝒜)×𝕋(𝒜,)×𝕋(,𝒞)(\rho,\mathscr{E},\mathscr{F})\in\mathcal{S}(\mathcal{A})\times\mathbb{CPTP}(\mathcal{A},\mathcal{B})\times\mathbb{CPTP}(\mathcal{B},\mathcal{C}) will be referred to as a 2-step process, and the set of 2-step processes 𝒮(𝒜)×𝕋(𝒜,)×𝕋(,𝒞)\mathcal{S}(\mathcal{A})\times\mathbb{CPTP}(\mathcal{A},\mathcal{B})\times\mathbb{CPTP}(\mathcal{B},\mathcal{C}) will be denoted by 𝒫(𝒜,,𝒞)\mathscr{P}(\mathcal{A},\mathcal{B},\mathcal{C}). The subset of 𝒫(𝒜,,𝒞)\mathscr{P}(\mathcal{A},\mathcal{B},\mathcal{C}) consisting of processes (ρ,,)(\rho,\mathscr{E},\mathscr{F}) with ρ\rho invertible will be denoted by 𝒫+(𝒜,,𝒞)\mathscr{P}_{+}(\mathcal{A},\mathcal{B},\mathcal{C}).

Definition 5.9.

Suppose !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} is associative. If (ρ,,)𝒫(𝒜,,𝒞)(\rho,\mathscr{E},\mathscr{F})\in\mathscr{P}(\mathcal{A},\mathcal{B},\mathcal{C}) is a 2-step process, then

!(!())(ρ)=(!(tr)!())(ρ)𝒜𝒞\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F})\circ\mathscr{E}\right)(\rho)=\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})\right)(\rho)\in\mathcal{A}\otimes\mathcal{B}\otimes\mathcal{C}

will be referred to as the state over time associated with the 2-step process (ρ,,)(\rho,\mathscr{E},\mathscr{F}) and the bloom map !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}, and will be denoted by !(ρ,,)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F}). The map on 2-step quantum processes given by

(ρ,,)!(ρ,,)(\rho,\mathscr{E},\mathscr{F})\longmapsto\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})

will then be referred to as the 2-step state over time function associated with !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}.

The next result shows that the state over time !(ρ,,)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F}) has the expected marginals.

Proposition 5.10.

Suppose !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} is associative, and let (ρ,,)𝒫(𝒜,,𝒞)(\rho,\mathscr{E},\mathscr{F})\in\mathscr{P}(\mathcal{A},\mathcal{B},\mathcal{C}) be a 2-step process. Then the following statements hold.

  1. i.

    tr𝒞(!(ρ,,))=!(ρ,)\emph{tr}_{\mathcal{C}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})\right)=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E})

  2. ii.

    tr𝒜(!(ρ,,))=!((ρ),)\emph{tr}_{\mathcal{A}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})\right)=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}(\rho),\mathscr{F})

  3. iii.

    tr𝒞(!(ρ,,))=ρ\emph{tr}_{\mathcal{B}\otimes\mathcal{C}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})\right)=\rho

  4. iv.

    tr𝒜𝒞(!(ρ,,))=(ρ)\emph{tr}_{\mathcal{A}\otimes\mathcal{C}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})\right)=\mathscr{E}(\rho)

  5. v.

    tr𝒜(!(ρ,,))=((ρ))\emph{tr}_{\mathcal{A}\otimes\mathcal{B}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})\right)=\mathscr{F}(\mathscr{E}(\rho))

Proof.

Associativity is used in the proofs by identifying !(ρ,,)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F}) with either (!(tr)!())(ρ)\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})\right)(\rho) or !(!())(ρ)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F})\circ\mathscr{E})(\rho).

Item i:

tr𝒞(!(ρ,,))\displaystyle\text{tr}_{\mathcal{C}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})\right) =\displaystyle= tr𝒞(!(tr)(!()(ρ)))\displaystyle\text{tr}_{\mathcal{C}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}\circ\text{tr})\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})(\rho)\right)\right)
=\displaystyle= !()(ρ)\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})(\rho)
=\displaystyle= !(ρ,),\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E}),

as desired.

Item ii:

tr𝒜(!(ρ,,))\displaystyle\text{tr}_{\mathcal{A}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})\right) =\displaystyle= tr𝒜(!(!())(ρ))\displaystyle\text{tr}_{\mathcal{A}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F})\circ\mathscr{E})(\rho)\right)
=\displaystyle= (!())(ρ)\displaystyle(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F})\circ\mathscr{E})(\rho)
=\displaystyle= !()((ρ))\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F})(\mathscr{E}(\rho))
=\displaystyle= !((ρ),),\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}(\rho),\mathscr{F}),

as desired.

Item iii: Indeed,

tr𝒞(!(ρ,,))\displaystyle\text{tr}_{\mathcal{B}\otimes\mathcal{C}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})\right) =\displaystyle= tr(tr𝒞(!(ρ,,)))\displaystyle\text{tr}_{\mathcal{B}}\left(\text{tr}_{\mathcal{C}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})\right)\right)
=\displaystyle= tr(!(ρ,))\displaystyle\text{tr}_{\mathcal{B}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E})\right)
=\displaystyle= ρ,\displaystyle\rho,

where the second equality follows from item i.

Item iv: Indeed,

tr𝒜𝒞(!(ρ,,))\displaystyle\text{tr}_{\mathcal{A}\otimes\mathcal{C}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})\right) =\displaystyle= tr𝒜(tr𝒞(!(ρ,,)))\displaystyle\text{tr}_{\mathcal{A}}\left(\text{tr}_{\mathcal{C}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})\right)\right)
=\displaystyle= tr𝒜(!(ρ,))\displaystyle\text{tr}_{\mathcal{A}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E})\right)
=\displaystyle= (ρ),\displaystyle\mathscr{E}(\rho),

where the second equality follows from item i.

Item v: Indeed,

tr𝒜(!(ρ,,))\displaystyle\text{tr}_{\mathcal{A}\otimes\mathcal{B}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})\right) =\displaystyle= tr(tr𝒜(!(ρ,,)))\displaystyle\text{tr}_{\mathcal{B}}\left(\text{tr}_{\mathcal{A}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\rho,\mathscr{E},\mathscr{F})\right)\right)
=\displaystyle= tr(!((ρ),))\displaystyle\text{tr}_{\mathcal{B}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}(\rho),\mathscr{F})\right)
=\displaystyle= ()(ρ),\displaystyle(\mathscr{F}\circ\mathscr{E})(\rho),

where the second equality follows from item ii. ∎

We now derive a formula for the symmetric bloom state over time \Yinyang(ρ,,)\text{\Yinyang}(\rho,\mathscr{E},\mathscr{F}) in terms of ρ\rho, 𝒥[]\mathscr{J}[\mathscr{E}] and 𝒥[]\mathscr{J}[\mathscr{F}].

Definition 5.11.

Let 𝒜\mathcal{A} be a multi-matrix algebra. The normalized Jordan product is the map 𝕁𝕠𝕣:𝒜×𝒜𝒜\mathbb{Jor}:\mathcal{A}\times\mathcal{A}\longrightarrow\mathcal{A} given by

𝕁𝕠𝕣(A,B)=12(AB+BA).\mathbb{Jor}\left(A,B\right)=\frac{1}{2}\left(AB+BA\right).
Lemma 5.12.

Let (,)𝕋(𝒜,,𝒞)(\mathscr{E},\mathscr{F})\in\mathbb{TP}(\mathcal{A},\mathcal{B},\mathcal{C}) be a 2-chain. Then

𝒥[\Yinyang()]=𝕁𝕠𝕣(𝒥[]𝟙,𝟙𝒥[]).\mathscr{J}[\text{\Yinyang}(\mathscr{F})\circ\mathscr{E}]=\mathbb{Jor}\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1},\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right).
Proof.

Indeed,

𝒥[\Yinyang()]\displaystyle\mathscr{J}[\text{\Yinyang}(\mathscr{F})\circ\mathscr{E}] =\displaystyle= 𝒥[12(!R()+!L())]\displaystyle\mathscr{J}\left[\frac{1}{2}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F}))\circ\mathscr{E}\right]
=\displaystyle= 12𝒥[!R()+!L()]\displaystyle\frac{1}{2}\mathscr{J}\left[\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E}+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F})\circ\mathscr{E}\right]
=\displaystyle= 12(𝒥[!R()]+𝒥[!L()])\displaystyle\frac{1}{2}\left(\mathscr{J}\left[\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E}]+\mathscr{J}[\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F})\circ\mathscr{E}\right]\right)
=\displaystyle= 12((𝒥[]𝟙)(𝟙𝒥[])+(𝟙𝒥[])(𝒥[]𝟙))\displaystyle\frac{1}{2}\left(\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1}\right)\left(\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right)+\left(\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right)\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1}\right)\right)
=\displaystyle= 𝕁𝕠𝕣(𝒥[]𝟙,𝟙𝒥[]),\displaystyle\mathbb{Jor}\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1},\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right),

where the second to last equation follows from items i and ii of Lemma 5.6. ∎

Proposition 5.13.

Let (ρ,,)𝒫(𝒜,,𝒞)(\rho,\mathscr{E},\mathscr{F})\in\mathscr{P}(\mathcal{A},\mathcal{B},\mathcal{C}) be a 2-step process. Then

\Yinyang(ρ,,)=𝕁𝕠𝕣(ρ𝟙,𝕁𝕠𝕣(𝒥[]𝟙,𝟙𝒥[])).\text{\Yinyang}(\rho,\mathscr{E},\mathscr{F})=\mathbb{Jor}\left(\rho\otimes\mathds{1},\mathbb{Jor}(\mathscr{J}[\mathscr{E}]\otimes\mathds{1},\mathds{1}\otimes\mathscr{J}[\mathscr{F}])\right). (5.14)
Proof.

Indeed,

𝒥[\Yinyang()])\displaystyle\mathscr{J}[\text{\Yinyang}(\mathscr{F})\circ\mathscr{E}]) =\displaystyle= 𝒥[12(!R()+!L())]\displaystyle\mathscr{J}\left[\frac{1}{2}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F})\right)\circ\mathscr{E}\right]
=\displaystyle= 12(𝒥[!R()]+!L()])\displaystyle\frac{1}{2}\left(\mathscr{J}[\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{R}(\mathscr{F})\circ\mathscr{E}]+\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{L}(\mathscr{F})\circ\mathscr{E}]\right)
=\displaystyle= 12((𝒥[]𝟙)(𝟙𝒥[])+(𝟙𝒥[])(𝒥[]𝟙))\displaystyle\frac{1}{2}\left((\mathscr{J}[\mathscr{E}]\otimes\mathds{1})(\mathds{1}\otimes\mathscr{J}[\mathscr{F}])+(\mathds{1}\otimes\mathscr{J}[\mathscr{F}])(\mathscr{J}[\mathscr{E}]\otimes\mathds{1})\right)
=\displaystyle= 𝕁𝕠𝕣(𝒥[]𝟙,𝟙𝒥[]),\displaystyle\mathbb{Jor}\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1},\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right),

where the third equality follows from items iv and v Lemma 5.6. We then have

\Yinyang(ρ,,)\displaystyle\text{\Yinyang}(\rho,\mathscr{E},\mathscr{F}) =\displaystyle= \Yinyang(\Yinyang())(ρ)\displaystyle\text{\Yinyang}\left(\text{\Yinyang}(\mathscr{F})\circ\mathscr{E}\right)(\rho)
=\displaystyle= 𝕁𝕠𝕣(ρ𝟙,𝒥[\Yinyang()])\displaystyle\mathbb{Jor}(\rho\otimes\mathds{1},\mathscr{J}[\text{\Yinyang}(\mathscr{F})\circ\mathscr{E}])
=\displaystyle= 𝕁𝕠𝕣(ρ𝟙,𝕁𝕠𝕣(𝒥[]𝟙,𝟙𝒥[])),\displaystyle\mathbb{Jor}(\rho\otimes\mathds{1},\mathbb{Jor}\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1},\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right)),

as desired. ∎

To conclude the section, we prove a result which will be useful later on.

Proposition 5.15 (Compositionality).

Let (ρ,,)𝒫(𝒜,,𝒞)(\rho,\mathscr{E},\mathscr{F})\in\mathscr{P}(\mathcal{A},\mathcal{B},\mathcal{C}) be a two-step process. Then

tr(\Yinyang(ρ,,))=\Yinyang(ρ,).\emph{tr}_{\mathcal{B}}\left(\text{\Yinyang}(\rho,\mathscr{E},\mathscr{F})\right)=\text{\Yinyang}(\rho,\mathscr{F}\circ\mathscr{E}). (5.16)
Proof.

It follows by direct computation that

𝕁𝕠𝕣(𝒥[]𝟙,𝟙𝒥[])=i,j,k,lEij𝒜𝕁𝕠𝕣((Eij𝒜),Ekl)(Elk),\mathbb{Jor}\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1},\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right)=\sum_{i,j,k,l}E_{ij}^{\mathcal{A}}\otimes\mathbb{Jor}\left(\mathscr{E}(E_{ij}^{\mathcal{A}}),E_{kl}^{\mathcal{B}}\right)\otimes\mathscr{F}(E_{lk}^{\mathcal{B}}),

thus

tr((ρ𝟙)𝕁𝕠𝕣(𝒥[]𝟙,𝟙𝒥[]))\displaystyle{\rm tr}_{\mathcal{B}}\left((\rho\otimes\mathds{1})\mathbb{Jor}\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1},\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right)\right) =\displaystyle= tr(i,j,k,lρEij𝒜𝕁𝕠𝕣((Eij𝒜),Ekl)(Elk))\displaystyle{\rm tr}_{\mathcal{B}}\left(\sum_{i,j,k,l}\rho E_{ij}^{\mathcal{A}}\otimes\mathbb{Jor}\left(\mathscr{E}(E_{ij}^{\mathcal{A}}),E_{kl}^{\mathcal{B}}\right)\otimes\mathscr{F}(E_{lk}^{\mathcal{B}})\right)
=\displaystyle= i,j,k,lρEij𝒜((Eij𝒜)Ekl(Elk))\displaystyle\sum_{i,j,k,l}\rho E_{ij}^{\mathcal{A}}\otimes\left(\mathscr{E}(E_{ij}^{\mathcal{A}})E_{kl}^{\mathcal{B}}\mathscr{F}(E_{lk}^{\mathcal{B}})\right)
=\displaystyle= i,j,k,lρEij𝒜((Eij𝒜)lk(Elk))\displaystyle\sum_{i,j,k,l}\rho E_{ij}^{\mathcal{A}}\otimes\left(\mathscr{E}(E_{ij}^{\mathcal{A}})_{lk}\mathscr{F}(E_{lk}^{\mathcal{B}})\right)
=\displaystyle= (ρ𝟙)i,jEij𝒜()(Eji𝒜)\displaystyle(\rho\otimes\mathds{1})\sum_{i,j}E_{ij}^{\mathcal{A}}\otimes(\mathscr{F}\circ\mathscr{E})(E_{ji}^{\mathcal{A}})
=\displaystyle= (ρ𝟙)𝒥[].\displaystyle(\rho\otimes\mathds{1})\mathscr{J}[\mathscr{F}\circ\mathscr{E}].

Similarly, we have

tr(𝕁𝕠𝕣(𝒥[]𝟙,𝟙𝒥[])(ρ𝟙))=𝒥[](ρ𝟙),{\rm tr}_{\mathcal{B}}\left(\mathbb{Jor}\left(\mathscr{J}[\mathscr{E}]\otimes\mathds{1},\mathds{1}\otimes\mathscr{J}[\mathscr{F}]\right)(\rho\otimes\mathds{1})\right)=\mathscr{J}[\mathscr{F}\circ\mathscr{E}](\rho\otimes\mathds{1}),

thus

tr(\Yinyang(ρ,,))\displaystyle{\rm tr}_{\mathcal{B}}\left(\text{\Yinyang}(\rho,\mathscr{E},\mathscr{F})\right) =(5.14)\displaystyle\overset{\eqref{2STPFXM91}}{=} tr(𝕁𝕠𝕣(ρ𝟙,𝕁𝕠𝕣(𝒥[]𝟙,𝟙𝒥[])))\displaystyle{\rm tr}_{\mathcal{B}}\left(\mathbb{Jor}\left(\rho\otimes\mathds{1},\mathbb{Jor}(\mathscr{J}[\mathscr{E}]\otimes\mathds{1},\mathds{1}\otimes\mathscr{J}[\mathscr{F}])\right)\right)
=\displaystyle= 𝕁𝕠𝕣(ρ𝟙,𝒥[])\displaystyle\mathbb{Jor}\left(\rho\otimes\mathds{1},\mathscr{J}[\mathscr{F}\circ\mathscr{E}]\right)
=\displaystyle= \Yinyang(ρ,),\displaystyle\text{\Yinyang}(\rho,\mathscr{F}\circ\mathscr{E}),

as desired. ∎

6 Blooming nn-chains

In this section we show how associative bloom maps naturally extend to bloom maps for arbitrary nn-chains.

Definition 6.1.

Given an (n+1)(n+1)-tuple (𝒜0,,𝒜n)(\mathcal{A}_{0},...,\mathcal{A}_{n}) of multi-matrix algebras, an nn-chain consists of a nn-tuple

(1,,n)𝕋(𝒜0,𝒜1)××𝕋(𝒜n1,𝒜n).(\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathbb{TP}(\mathcal{A}_{0},\mathcal{A}_{1})\times\cdots\times\mathbb{TP}(\mathcal{A}_{n-1},\mathcal{A}_{n}).

The set of all such nn-chains will be denoted by 𝕋(𝒜0,,𝒜n)\mathbb{TP}(\mathcal{A}_{0},...,\mathcal{A}_{n}).

Definition 6.2.

An nn-bloom associates every (n+1)(n+1)-tuple (𝒜0,,𝒜n)(\mathcal{A}_{0},...,\mathcal{A}_{n}) of multi-matrix algebras with a map !:𝕋(𝒜0,,𝒜n)𝕋(𝒜0,𝒜0𝒜n)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}:\mathbb{TP}(\mathcal{A}_{0},...,\mathcal{A}_{n})\longrightarrow\mathbb{TP}(\mathcal{A}_{0},\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}) such that

tri!(1,,n)={i1for i{1,,n}id𝒜0for i=0\text{tr}_{i}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1},...,\mathscr{E}_{n})=\begin{cases}\mathscr{E}_{i}\circ\cdots\circ\mathscr{E}_{1}\quad\quad\text{for $i\in\{1,...,n\}$}\\ \mathrm{id}_{\mathcal{A}_{0}}\quad\quad\hskip 34.99677pt\text{for $i=0$}\end{cases} (6.3)

for all (1,,n)𝕋(𝒜0,,𝒜n)(\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathbb{TP}(\mathcal{A}_{0},...,\mathcal{A}_{n}).

Example 6.4 (Blooming 3-chains).

Let !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} be a bloom map, and let

𝒜01𝒜12𝒜23𝒜3\mathcal{A}_{0}\overset{\mathscr{E}_{1}}{\longrightarrow}\mathcal{A}_{1}\overset{\mathscr{E}_{2}}{\longrightarrow}\mathcal{A}_{2}\overset{\mathscr{E}_{3}}{\longrightarrow}\mathcal{A}_{3}

be a 3-chain, which after incorporating bloom-shriek factorizations yields the following diagram:

𝒜0\textstyle{\mathcal{A}_{0}}𝒜1\textstyle{\mathcal{A}_{1}}𝒜2\textstyle{\mathcal{A}_{2}}𝒜3\textstyle{\mathcal{A}_{3}}𝒜0𝒜1\textstyle{\mathcal{A}_{0}\otimes\mathcal{A}_{1}}𝒜1𝒜2\textstyle{\mathcal{A}_{1}\otimes\mathcal{A}_{2}}𝒜2𝒜3\textstyle{\mathcal{A}_{2}\otimes\mathcal{A}_{3}}1\scriptstyle{\mathscr{E}_{1}}2\scriptstyle{\mathscr{E}_{2}}2\scriptstyle{\mathscr{E}_{2}}!\scriptstyle{\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}}tr!\scriptstyle{\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}}tr!\scriptstyle{\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}}tr (6.5)

Similar to the case of 2-chains, there are then five 3-blooms in 𝕋(𝒜0,𝒜0𝒜1𝒜2𝒜3)\mathbb{TP}(\mathcal{A}_{0},\mathcal{A}_{0}\otimes\mathcal{A}_{1}\otimes\mathcal{A}_{2}\otimes\mathcal{A}_{3}) one may construct from the above diagram (6.5), each of which corresponds to a parenthezation of 𝒜0𝒜1𝒜2𝒜3\mathcal{A}_{0}\otimes\mathcal{A}_{1}\otimes\mathcal{A}_{2}\otimes\mathcal{A}_{3}.

For example, the 3-bloom associated with the parenthezation 𝒜0(𝒜1(𝒜2𝒜3))\mathcal{A}_{0}\otimes(\mathcal{A}_{1}\otimes(\mathcal{A}_{2}\otimes\mathcal{A}_{3})) is constructed as follows. We first associate a map with what is inside the outermost parantheses of 𝒜0(𝒜1(𝒜2𝒜3))\mathcal{A}_{0}\otimes(\mathcal{A}_{1}\otimes(\mathcal{A}_{2}\otimes\mathcal{A}_{3})), namely, 𝒜1(𝒜2𝒜3)\mathcal{A}_{1}\otimes(\mathcal{A}_{2}\otimes\mathcal{A}_{3}), which is a parenthezation of the multi-matrix algebra 𝒜1𝒜2𝒜3\mathcal{A}_{1}\otimes\mathcal{A}_{2}\otimes\mathcal{A}_{3}. As we’ve already seen in the case of 2-chains, the 2-bloom associated with 𝒜1(𝒜2𝒜3)\mathcal{A}_{1}\otimes(\mathcal{A}_{2}\otimes\mathcal{A}_{3}) is then

!(!(3)2):𝒜1𝒜1(𝒜2𝒜3),\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{3})\circ\mathscr{E}_{2}\right):\mathcal{A}_{1}\longrightarrow\mathcal{A}_{1}\otimes(\mathcal{A}_{2}\otimes\mathcal{A}_{3}),

and then post-composing with 1\mathscr{E}_{1} yields

!(!(3)2)1:𝒜0𝒜1(𝒜2𝒜3),\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{3})\circ\mathscr{E}_{2}\right)\circ\mathscr{E}_{1}:\mathcal{A}_{0}\longrightarrow\mathcal{A}_{1}\otimes(\mathcal{A}_{2}\otimes\mathcal{A}_{3}),

so that taking a final bloom yields

!(!(!(3)2)1):𝒜0𝒜0(𝒜1(𝒜2𝒜3)),\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{3})\circ\mathscr{E}_{2}\right)\circ\mathscr{E}_{1}\right):\mathcal{A}_{0}\longrightarrow\mathcal{A}_{0}\otimes(\mathcal{A}_{1}\otimes(\mathcal{A}_{2}\otimes\mathcal{A}_{3})),

which is the desired 3-bloom. The 3-blooms associated with the other 4 parenthezations of 𝒜0𝒜1𝒜2𝒜3\mathcal{A}_{0}\otimes\mathcal{A}_{1}\otimes\mathcal{A}_{2}\otimes\mathcal{A}_{3} may then be obtained by considering the pentagon where each vertex corresponds to one of the 5 parenthezations of 𝒜0𝒜1𝒜2𝒜3\mathcal{A}_{0}\otimes\mathcal{A}_{1}\otimes\mathcal{A}_{2}\otimes\mathcal{A}_{3}, and a directed edge is drawn between two vertices if and only if they are related by a single application of the associator transformation 𝒜(𝒞)(𝒜)𝒞\mathcal{A}\otimes(\mathcal{B}\otimes\mathcal{C})\longmapsto(\mathcal{A}\otimes\mathcal{B})\otimes\mathcal{C}:

𝒜0(𝒜1(𝒜2𝒜3))\textstyle{\mathcal{A}_{0}\otimes(\mathcal{A}_{1}\otimes(\mathcal{A}_{2}\otimes\mathcal{A}_{3}))}(𝒜0𝒜1)(𝒜2𝒜3)\textstyle{(\mathcal{A}_{0}\otimes\mathcal{A}_{1})\otimes(\mathcal{A}_{2}\otimes\mathcal{A}_{3})}((𝒜0𝒜1)𝒜2)𝒜3\textstyle{((\mathcal{A}_{0}\otimes\mathcal{A}_{1})\otimes\mathcal{A}_{2})\otimes\mathcal{A}_{3}}(𝒜0(𝒜1𝒜2))𝒜3\textstyle{(\mathcal{A}_{0}\otimes(\mathcal{A}_{1}\otimes\mathcal{A}_{2}))\otimes\mathcal{A}_{3}}𝒜0((𝒜1𝒜2)𝒜3)\textstyle{\mathcal{A}_{0}\otimes((\mathcal{A}_{1}\otimes\mathcal{A}_{2})\otimes\mathcal{A}_{3})}

The associated 3-blooms for the other 4 parenthezations of 𝒜0𝒜1𝒜2𝒜3\mathcal{A}_{0}\otimes\mathcal{A}_{1}\otimes\mathcal{A}_{2}\otimes\mathcal{A}_{3} can then be obtained from !(!(!(3)2)1)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{3})\circ\mathscr{E}_{2}\right)\circ\mathscr{E}_{1}\right) by applying corresponding associator transformations of maps !(!())!(tr)!()\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F})\circ\mathscr{E}\right)\longmapsto\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}) along the pentagon:

!(!(!(3)2)1)\textstyle{\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{3})\circ\mathscr{E}_{2}\right)\circ\mathscr{E}_{1}\right)}!(!(3)2tr)!(1)\textstyle{\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{3})\circ\mathscr{E}_{2}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1})}!(3tr)!(2tr)!(1)\textstyle{\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\mathscr{E}_{3}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\mathscr{E}_{2}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1})}!(3tr)!(!(2)1)\textstyle{\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\mathscr{E}_{3}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2})\circ\mathscr{E}_{1})}!(!(3tr)!(2)1)\textstyle{\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{3}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2})\circ\mathscr{E}_{1}\right)}

This procedure then yields the following 3-blooms associated with each parenthezation of 𝒜0𝒜1𝒜2𝒜3\mathcal{A}_{0}\otimes\mathcal{A}_{1}\otimes\mathcal{A}_{2}\otimes\mathcal{A}_{3}:

  1. i.

    𝒜0(𝒜1(𝒜2𝒜3))¯:!(!(!(3)2)1)\underline{\mathcal{A}_{0}\otimes(\mathcal{A}_{1}\otimes(\mathcal{A}_{2}\otimes\mathcal{A}_{3}))}:\hskip 5.69054pt\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{3})\circ\mathscr{E}_{2}\right)\circ\mathscr{E}_{1}\right)

  2. ii.

    (𝒜0𝒜1)(𝒜2𝒜3)¯:!(!(3)2tr)!(1)\underline{(\mathcal{A}_{0}\otimes\mathcal{A}_{1})\otimes(\mathcal{A}_{2}\otimes\mathcal{A}_{3})}:\hskip 5.69054pt\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{3})\circ\mathscr{E}_{2}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1})

  3. iii.

    ((𝒜0𝒜1)𝒜2)𝒜3¯:!(3tr)!(2tr)!(1)\underline{((\mathcal{A}_{0}\otimes\mathcal{A}_{1})\otimes\mathcal{A}_{2})\otimes\mathcal{A}_{3}}:\hskip 5.69054pt\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\mathscr{E}_{3}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\mathscr{E}_{2}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1})

  4. iv.

    𝒜0((𝒜1𝒜2)𝒜3)¯:!(!(3tr)!(2)1)\underline{\mathcal{A}_{0}\otimes((\mathcal{A}_{1}\otimes\mathcal{A}_{2})\otimes\mathcal{A}_{3})}:\hskip 5.69054pt\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{3}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2})\circ\mathscr{E}_{1}\right)

  5. v.

    (𝒜0(𝒜1𝒜2))𝒜3¯:!(3tr)!(!(2)1)\underline{(\mathcal{A}_{0}\otimes(\mathcal{A}_{1}\otimes\mathcal{A}_{2}))\otimes\mathcal{A}_{3}}:\hskip 5.69054pt\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\mathscr{E}_{3}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2})\circ\mathscr{E}_{1})

If !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} is assumed to be associative, then this implies that these five 3-blooms are all equal:

i == ii: The statement follows from (5.5) with =1\mathscr{E}=\mathscr{E}_{1} and =!(3)2\mathscr{F}=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{3})\circ\mathscr{E}_{2}.

ii == iii: The statement follows from (5.5) with =2tr\mathscr{E}=\mathscr{E}_{2}\circ\text{tr} and =3\mathscr{F}=\mathscr{E}_{3}.

i == iv: The statement follows from (5.5) with =2\mathscr{E}=\mathscr{E}_{2} and =3\mathscr{F}=\mathscr{E}_{3}.

iv == v: The statement follows from (5.5) with =!(2)1\mathscr{E}=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2})\circ\mathscr{E}_{1} and =3tr\mathscr{F}=\mathscr{E}_{3}\circ\text{tr}.

Remark 6.6.

Note that when showing iv==v we have set trtr=tr\text{tr}\circ\text{tr}=\text{tr} as we are eliding the domains and codomains of the partial trace, which we will continue to do throughout.

We now generalize the previous two examples to nn-chains for all n>1n>1. Let !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} be a bloom map, let n>1n>1, and let

𝒜01𝒜1𝒜n1n𝒜n\mathcal{A}_{0}\overset{\mathscr{E}_{1}}{\longrightarrow}\mathcal{A}_{1}\longrightarrow\cdots\longrightarrow\mathcal{A}_{n-1}\overset{\mathscr{E}_{n}}{\longrightarrow}\mathcal{A}_{n} (6.7)

be an nn-chain. With every parenthezation ϑ(𝒜0,,𝒜n)\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{n}) of the multi-matrix algebra 𝒜0𝒜n\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n} we associate a map

!ϑ(1,,n):𝒜0ϑ(𝒜0,,𝒜n)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n}):\mathcal{A}_{0}\longrightarrow\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{n})

constructed from the bloom map !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}, the maps i\mathscr{E}_{i} for i=1,,ni=1,...,n and the partial trace tr. For the case n=2n=2, we know from Example 5.2 that there are two parenthezations of 𝒜0𝒜1𝒜2\mathcal{A}_{0}\otimes\mathcal{A}_{1}\otimes\mathcal{A}_{2}, namely,

ϑ(𝒜0,𝒜1,𝒜2)=(𝒜0𝒜1)𝒜3&ϑ~(𝒜0,𝒜1,𝒜2)=𝒜0(𝒜1𝒜2).\vartheta(\mathcal{A}_{0},\mathcal{A}_{1},\mathcal{A}_{2})=(\mathcal{A}_{0}\otimes\mathcal{A}_{1})\otimes\mathcal{A}_{3}\quad\&\quad\widetilde{\vartheta}(\mathcal{A}_{0},\mathcal{A}_{1},\mathcal{A}_{2})=\mathcal{A}_{0}\otimes(\mathcal{A}_{1}\otimes\mathcal{A}_{2}).

and we let

!ϑ(1,2)=!(2tr)!(1)&!ϑ~(1,2)=!(!(2)1).\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},\mathscr{E}_{2})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1})\quad\&\quad\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\widetilde{\vartheta}}(\mathscr{E}_{1},\mathscr{E}_{2})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2})\circ\mathscr{E}_{1}).

In such a case we have that !ϑ(1,2)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},\mathscr{E}_{2}) may be obtained from !ϑ~(1,2)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\widetilde{\vartheta}}(\mathscr{E}_{1},\mathscr{E}_{2}) by a transformation of the form

!(!())!(tr)!()\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F})\circ\mathscr{E})\longmapsto\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})

with =2\mathscr{F}=\mathscr{E}_{2} and =1\mathscr{E}=\mathscr{E}_{1}. For general n>2n>2 we will build up recursively from this case.

So now let n>2n>2 and let ϑ(𝒜0,,𝒜n)\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{n}) be an arbitrary parenthezation of 𝒜0𝒜n\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}. It then follows that ϑ(𝒜0,,𝒜n)\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{n}) is in one of the three forms:

I:\displaystyle I: ϑ(𝒜0,,𝒜n)\displaystyle\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{n}) =𝒜0(ω(𝒜1,,𝒜n))\displaystyle=\hskip 2.84526pt\mathcal{A}_{0}\otimes\left(\omega(\mathcal{A}_{1},...,\mathcal{A}_{n})\right)
II:\displaystyle II: ϑ(𝒜0,,𝒜n)\displaystyle\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{n}) =(ω(𝒜0,,𝒜n1))𝒜n\displaystyle=\hskip 2.84526pt\left(\omega(\mathcal{A}_{0},...,\mathcal{A}_{n-1})\right)\otimes\mathcal{A}_{n}
III:\displaystyle III: ϑ(𝒜0,,𝒜n)\displaystyle\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{n}) =(ω(𝒜0,,𝒜k1))(χ(𝒜k,,𝒜n))\displaystyle=\hskip 2.84526pt\left(\omega(\mathcal{A}_{0},...,\mathcal{A}_{k-1})\right)\otimes\left(\chi(\mathcal{A}_{k},...,\mathcal{A}_{n})\right)

where 1<k<n1<k<n and ω(𝒜1,,𝒜n)\omega(\mathcal{A}_{1},...,\mathcal{A}_{n}), ω(𝒜0,,𝒜n1)\omega(\mathcal{A}_{0},...,\mathcal{A}_{n-1}), ω(𝒜0,,𝒜k1)\omega(\mathcal{A}_{0},...,\mathcal{A}_{k-1}) and χ(𝒜k,,𝒜n)\chi(\mathcal{A}_{k},...,\mathcal{A}_{n}) are the parenthezations of the multi-matrix algebras 𝒜1𝒜n\mathcal{A}_{1}\otimes\cdots\otimes\mathcal{A}_{n}, 𝒜0𝒜n1\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n-1}, 𝒜0𝒜k1\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{k-1} and 𝒜k𝒜n\mathcal{A}_{k}\otimes\cdots\otimes\mathcal{A}_{n} for which the above equations hold. In each of the three cases we define the associated nn-blooms recursively as follows.

I:\displaystyle I: !ϑ(1,,n)\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n}) =!(!ω(2,,n)1)\displaystyle=\hskip 5.69054pt\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{2},...,\mathscr{E}_{n})\circ\hskip 2.13394pt\mathscr{E}_{1}\right)
II:\displaystyle II: !ϑ(1,,n)\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n}) =!(ntr)!ω(1,,n1)\displaystyle=\hskip 5.69054pt\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{n-1})
III:\displaystyle III: !ϑ(1,,n)\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n}) =!(!χ(k+1,,n)ktr)!ω(1,,k1)\displaystyle=\hskip 5.69054pt\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\chi}(\mathscr{E}_{k+1},...,\mathscr{E}_{n})\circ\hskip 2.13394pt\mathscr{E}_{k}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{k-1})

The next proposition shows that !ϑ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}, when viewed as a map on nn-chains, is an actual nn-bloom, i.e., that !ϑ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta} satisfies the condition (6.3) in Definition 6.2.

Proposition 6.8.

Let n>1n>1, let (𝒜0,,𝒜n)(\mathcal{A}_{0},...,\mathcal{A}_{n}) be an (n+1)(n+1)-tuple of multi-matrix algebras, let ϑ(𝒜0,,𝒜n)\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{n}) be a parenthezation of 𝒜0𝒜n\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}, and let !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} be a bloom map. Then !ϑ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta} is an nn-bloom, i.e.,

tri!ϑ(1,,n)={i1for i{1,,n}id𝒜0for i=0\emph{tr}_{i}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n})=\begin{cases}\mathscr{E}_{i}\circ\cdots\circ\mathscr{E}_{1}\quad\quad\text{for $i\in\{1,...,n\}$}\\ \mathrm{id}_{\mathcal{A}_{0}}\quad\quad\hskip 34.99677pt\text{for $i=0$}\end{cases} (6.9)

for all (1,,n)𝕋(𝒜0,,𝒜n)(\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathbb{TP}(\mathcal{A}_{0},...,\mathcal{A}_{n}).

Proof.

We use induction on nn. For n=2n=2 the statement follows from Proposition 5.10. Now suppose the result holds for n=m1>2n=m-1>2, and let ϑ(𝒜0,,𝒜m)\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{m}) be a parenthezation of 𝒜0𝒜m\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{m}. We then consider the three cases as for ϑ(𝒜0,,𝒜m)\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{m}) as defined above.

Case I: In this case we have ϑ(𝒜0,,𝒜m)=𝒜0(ω(𝒜1,,𝒜m))\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{m})=\mathcal{A}_{0}\otimes\left(\omega(\mathcal{A}_{1},...,\mathcal{A}_{m})\right) and

!ϑ(1,,m)=!(!ω(2,,m)1)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{m})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{2},...,\mathscr{E}_{m})\circ\hskip 2.13394pt\mathscr{E}_{1}\right)

for all mm-chains (1,,m)(\mathscr{E}_{1},...,\mathscr{E}_{m}). For i=0i=0 we then have

tr0!ϑ(1,,n)=tr0!(!ω(2,,m)1)=id𝒜0,\text{tr}_{0}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n})=\text{tr}_{0}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{2},...,\mathscr{E}_{m})\circ\hskip 2.13394pt\mathscr{E}_{1}\right)=\mathrm{id}_{\mathcal{A}_{0}},

where the last equality follows from the definition of a bloom map. As for 0<im0<i\leq m, we have

tri!ϑ(1,,n)\displaystyle\text{tr}_{i}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n}) =\displaystyle= tr𝒜0𝒜i^𝒜m!(!ω(2,,m)1)\displaystyle\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{m}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{2},...,\mathscr{E}_{m})\circ\hskip 2.13394pt\mathscr{E}_{1}\right)
=\displaystyle= tr𝒜1𝒜i^𝒜mtr𝒜0!(!ω(2,,m)1)\displaystyle\text{tr}_{\mathcal{A}_{1}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{m}}\circ\text{tr}_{\mathcal{A}_{0}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{2},...,\mathscr{E}_{m})\circ\hskip 2.13394pt\mathscr{E}_{1}\right)
=\displaystyle= tr𝒜1𝒜i^𝒜m!ω(2,,m)1\displaystyle\text{tr}_{\mathcal{A}_{1}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{m}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{2},...,\mathscr{E}_{m})\circ\hskip 2.13394pt\mathscr{E}_{1}
=induction\displaystyle\overset{\text{induction}}{=} (i2)1\displaystyle(\mathscr{E}_{i}\circ\cdots\circ\mathscr{E}_{2})\circ\mathscr{E}_{1}
=\displaystyle= i1,\displaystyle\mathscr{E}_{i}\circ\cdots\circ\mathscr{E}_{1},

thus !ϑ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta} is an mm-bloom.

Case II: In this case we have ϑ(𝒜0,,𝒜m)=(ω(𝒜0,,𝒜m1))𝒜m\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{m})=\left(\omega(\mathcal{A}_{0},...,\mathcal{A}_{m-1})\right)\otimes\mathcal{A}_{m} and

!ϑ(1,,m)=!(mtr)!ω(1,,m1)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{m})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{m}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{m-1})

for all mm-chains (1,,m)(\mathscr{E}_{1},...,\mathscr{E}_{m}), where tr=tr𝒜0𝒜m2:𝒜0𝒜m1𝒜m1\text{tr}=\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{m-2}}:\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{m-1}\longrightarrow\mathcal{A}_{m-1}. For i=mi=m we then have

trm!ϑ(1,,n)\displaystyle\text{tr}_{m}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n}) =\displaystyle= tr𝒜0𝒜m1!(mtr)!ω(1,,m1)\displaystyle\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{m-1}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{m}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{m-1})
=\displaystyle= m(tr!ω(1,,m1))\displaystyle\mathscr{E}_{m}\circ\left(\text{tr}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{m-1})\right)
=induction\displaystyle\overset{\text{induction}}{=} m(m11)\displaystyle\mathscr{E}_{m}\circ\left(\mathscr{E}_{m-1}\circ\cdots\circ\mathscr{E}_{1}\right)
=\displaystyle= m1.\displaystyle\mathscr{E}_{m}\circ\cdots\circ\mathscr{E}_{1}.

As for 0i<m0\leq i<m, we have

tri!ϑ(1,,n)\displaystyle\text{tr}_{i}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n}) =\displaystyle= tr𝒜0𝒜i^𝒜m1(tr𝒜m!(mtr))!ω(1,,m1)\displaystyle\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{m-1}}\circ\left(\text{tr}_{\mathcal{A}_{m}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{m}\circ\text{tr})\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{m-1})
=\displaystyle= tr𝒜0𝒜i^𝒜m1id𝒜0𝒜m1!ω(1,,m1)\displaystyle\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{m-1}}\circ\mathrm{id}_{\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{m-1}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{m-1})
=\displaystyle= tr𝒜0𝒜i^𝒜m1!ω(1,,m1)\displaystyle\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{m-1}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{m-1})
=induction\displaystyle\overset{\text{induction}}{=} {i1for i{1,,m1}id𝒜0for i=0,\displaystyle\begin{cases}\mathscr{E}_{i}\circ\cdots\circ\mathscr{E}_{1}\quad\quad\text{for $i\in\{1,...,m-1\}$}\\ \mathrm{id}_{\mathcal{A}_{0}}\quad\quad\hskip 34.99677pt\text{for $i=0$},\end{cases}

thus !ϑ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta} is an mm-bloom.

Case III: In this case we have ϑ(𝒜0,,𝒜m)=(ω(𝒜0,,𝒜k1))(χ(𝒜k,,𝒜m))\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{m})=\left(\omega(\mathcal{A}_{0},...,\mathcal{A}_{k-1})\right)\otimes\left(\chi(\mathcal{A}_{k},...,\mathcal{A}_{m})\right) for some 1<k<m1<k<m and

!ϑ(1,,m)=!(!χ(k+1,,m)ktr)!ω(1,,k1)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{m})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\chi}(\mathscr{E}_{k+1},...,\mathscr{E}_{m})\circ\hskip 2.13394pt\mathscr{E}_{k}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{k-1})

for all mm-chains (1,,m)(\mathscr{E}_{1},...,\mathscr{E}_{m}), where tr=tr𝒜0𝒜k2:𝒜0𝒜k1𝒜k1\text{tr}=\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{k-2}}:\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{k-1}\longrightarrow\mathcal{A}_{k-1}. Now let =tri!ϑ(1,,n)\aleph=\text{tr}_{i}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n}). Then for 0i<k0\leq i<k we have

\displaystyle\aleph =\displaystyle= tri!ϑ(1,,n)\displaystyle\text{tr}_{i}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n})
=\displaystyle= tr𝒜0𝒜i^𝒜m!(!χ(k+1,,m)ktr)!ω(1,,k1)\displaystyle\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{m}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\chi}(\mathscr{E}_{k+1},...,\mathscr{E}_{m})\circ\hskip 2.13394pt\mathscr{E}_{k}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{k-1})
=\displaystyle= tr𝒜0𝒜i^𝒜k1(tr𝒜k𝒜m!(!χ(k+1,,m))ktr)!ω(1,,k1)\displaystyle\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{k-1}}\circ\left(\text{tr}_{\mathcal{A}_{k}\otimes\cdots\mathcal{A}_{m}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\chi}(\mathscr{E}_{k+1},...,\mathscr{E}_{m})\right)\circ\hskip 2.13394pt\mathscr{E}_{k}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{k-1})
=\displaystyle= tr𝒜0𝒜i^𝒜k1id𝒜0𝒜k1!ω(1,,k1)\displaystyle\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{k-1}}\circ\mathrm{id}_{\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{k-1}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{k-1})
=\displaystyle= tr𝒜0𝒜i^𝒜k1!ω(1,,k1)\displaystyle\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{k-1}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{k-1})
=induction\displaystyle\overset{\text{induction}}{=} {i1for i{1,,k1}id𝒜0for i=0.\displaystyle\begin{cases}\mathscr{E}_{i}\circ\cdots\circ\mathscr{E}_{1}\quad\quad\text{for $i\in\{1,...,k-1\}$}\\ \mathrm{id}_{\mathcal{A}_{0}}\quad\quad\hskip 34.99677pt\text{for $i=0$}.\end{cases}

As for kimk\leq i\leq m, we have

\displaystyle\aleph =\displaystyle= tri!ϑ(1,,n)\displaystyle\text{tr}_{i}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n})
=\displaystyle= tr𝒜0𝒜i^𝒜m!(!χ(k+1,,m)ktr)!ω(1,,k1)\displaystyle\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{m}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\chi}(\mathscr{E}_{k+1},...,\mathscr{E}_{m})\circ\hskip 2.13394pt\mathscr{E}_{k}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{k-1})
=\displaystyle= tr𝒜k𝒜i^𝒜m(tr𝒜0𝒜k1!(!χ(k+1,,m)ktr))!ω(1,,k1)\displaystyle\text{tr}_{\mathcal{A}_{k}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{m}}\circ\left(\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{k-1}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\chi}(\mathscr{E}_{k+1},...,\mathscr{E}_{m})\circ\hskip 2.13394pt\mathscr{E}_{k}\circ\text{tr}\right)\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{k-1})
=\displaystyle= tr𝒜k𝒜i^𝒜m(!χ(k+1,,m)ktr)!ω(1,,k1)\displaystyle\text{tr}_{\mathcal{A}_{k}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{m}}\circ\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\chi}(\mathscr{E}_{k+1},...,\mathscr{E}_{m})\circ\hskip 2.13394pt\mathscr{E}_{k}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{k-1})
=\displaystyle= (tr𝒜k𝒜i^𝒜m!χ(k+1,,m))k(tr𝒜0𝒜k2!ω(1,,k1))\displaystyle\left(\text{tr}_{\mathcal{A}_{k}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{m}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\chi}(\mathscr{E}_{k+1},...,\mathscr{E}_{m})\right)\circ\hskip 2.13394pt\mathscr{E}_{k}\circ\left(\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{k-2}}\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{k-1})\right)
=induction\displaystyle\overset{\text{induction}}{=} ({ik+1for i{k+1,,m}id𝒜kfor i=k)kk11\displaystyle\left(\begin{cases}\mathscr{E}_{i}\circ\cdots\circ\mathscr{E}_{k+1}\quad\quad\text{for $i\in\{k+1,...,m\}$}\\ \mathrm{id}_{\mathcal{A}_{k}}\quad\quad\hskip 46.37813pt\text{for $i=k$}\end{cases}\right)\circ\mathscr{E}_{k}\circ\mathscr{E}_{k-1}\circ\cdots\circ\mathscr{E}_{1}
=\displaystyle= i1,\displaystyle\mathscr{E}_{i}\circ\cdots\circ\mathscr{E}_{1},

thus !ϑ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta} is an mm-bloom, as desired. ∎

Remark 6.10.

If ϑ(𝒜0,,𝒜n)\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{n}) and ϑ~(𝒜0,,𝒜n)\widetilde{\vartheta}(\mathcal{A}_{0},...,\mathcal{A}_{n}) are two parenthezations of 𝒜0𝒜n\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n} such that ϑ~(𝒜0,,𝒜n)\widetilde{\vartheta}(\mathcal{A}_{0},...,\mathcal{A}_{n}) is obtained from ϑ(𝒜0,,𝒜n)\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{n}) by an associator transformation of the form 𝒜(𝒞)(𝒜)𝒞\mathcal{A}\otimes(\mathcal{B}\otimes\mathcal{C})\longmapsto(\mathcal{A}\otimes\mathcal{B})\otimes\mathcal{C}, then there exists a 2-chain

𝒜𝒞\mathcal{A}\overset{\mathscr{E}}{\longrightarrow}\mathcal{B}\overset{\mathscr{F}}{\longrightarrow}\mathcal{C}

such that !ϑ~\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\widetilde{\vartheta}} is obtained from !ϑ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta} by an associator transformation at the level of maps associated with the 2-chain (,)(\mathscr{E},\mathscr{F}), i.e.,

!ϑ=!(!())!(tr)!()=!ϑ~.\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F})\circ\mathscr{E})\longmapsto\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\widetilde{\vartheta}}.

The previous remark motivates the following definition.

Definition 6.11.

Let n>1n>1 and let 0<k<ln0<k<l\leq n. The associator relation on the set of maps 𝕋(𝒜lk,𝒜lk𝒜l)\mathbb{TP}(\mathcal{A}_{l-k},\mathcal{A}_{l-k}\otimes\cdots\otimes\mathcal{A}_{l}) is the subset

𝒜𝕋(𝒜lk,𝒜lk𝒜l)×𝕋(𝒜lk,𝒜lk𝒜l)\mathscr{A}\subset\mathbb{TP}(\mathcal{A}_{l-k},\mathcal{A}_{l-k}\otimes\cdots\otimes\mathcal{A}_{l})\times\mathbb{TP}(\mathcal{A}_{l-k},\mathcal{A}_{l-k}\otimes\cdots\otimes\mathcal{A}_{l})

given by (ψ,φ)𝒜(\psi,\varphi)\in\mathscr{A} if and only if φ\varphi is obtained from ψ\psi from successive transformations of the form

!(!())!(tr)!().\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F})\circ\mathscr{E})\longmapsto\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}).

In such a case, we use the notation ψφ\psi\searrow\varphi to denote the fact that (ψ,φ)𝒜(\psi,\varphi)\in\mathscr{A}.

Lemma 6.12.

Let k>0k>0. Then !(!(1)!(k))!(1tr)!(ktr)!()\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}_{1})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}_{k})\circ\mathscr{E})\searrow\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}_{1}\circ\emph{tr})\circ\cdots\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}_{k}\circ\emph{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}).

Proof.

We use induction on kk. The case k=1k=1 holds by definition of the associator relation, so now assume the result holds for k=m1k=m-1. Then

!(!(1)!(m1)!(m))\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}_{1})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}_{m-1})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}_{m})\circ\mathscr{E}) induction\displaystyle\overset{\text{induction}}{\searrow} !(1tr)!(m1tr)!(!(m))\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}_{1}\circ\text{tr})\circ\cdots\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}_{m-1}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}_{m})\circ\mathscr{E})
associator\displaystyle\overset{\text{associator}}{\searrow} !(1tr)!(m1tr)!(mtr)!(),\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}_{1}\circ\text{tr})\circ\cdots\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}_{m-1}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{F}_{m}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}),

thus the result holds for k=mk=m, as desired. ∎

Proposition 6.13.

Let n>1n>1, and let ϑ(𝒜0,,𝒜n)\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{n}) be a parenthezation of 𝒜0𝒜n\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}. Then

!ϑ(1,,n)!(ntr)!(n1tr)!(2tr)!(1).\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n})\searrow\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n}\circ\emph{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n-1}\circ\emph{tr})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2}\circ\emph{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1}).
Proof.

We will consider each of the three cases given above for the definition of !ϑ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}, and use strong induction on nn. The case n=2n=2 follows by definition, so now assume the result holds for all mm with 1<m<n1<m<n.

Case I: Indeed,

!ϑ(1,,n)\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n}) =\displaystyle= !(!ω(2,,n)1)\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{2},...,\mathscr{E}_{n})\circ\hskip 2.13394pt\mathscr{E}_{1})
induction\displaystyle\overset{\text{induction}}{\searrow} !(!(ntr)!(3tr)!(2)1)\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n}\circ\text{tr})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{3}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2})\circ\mathscr{E}_{1}\right)
Lemma 6.12\displaystyle\overset{\text{Lemma~{}\ref{LXMMA8971}}}{\searrow} !(ntr)!(n1tr)!(2tr)!(1),\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n-1}\circ\text{tr})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1}),

where in the last line we note that we have repeatedly used the elision trtr=tr\text{tr}\circ\text{tr}=\text{tr}, thus the result holds.

Case II: Indeed,

!ϑ(1,,n)\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n}) =\displaystyle= !(ntr)!ω(1,,n1)\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{n-1})
induction\displaystyle\overset{\text{induction}}{\searrow} !(ntr)!(n1tr)!(2tr)!(1),\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n-1}\circ\text{tr})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1}),

as desired.

Case III: Let =!ϑ(1,,n)\aleph=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n}). Then

\displaystyle\aleph =\displaystyle= !ϑ(1,,n)\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n})
=\displaystyle= !(!χ(k+1,,n)ktr)!ω(1,,k1)\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\chi}(\mathscr{E}_{k+1},...,\mathscr{E}_{n})\circ\hskip 2.13394pt\mathscr{E}_{k}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\omega}(\mathscr{E}_{1},...,\mathscr{E}_{k-1})
induction\displaystyle\overset{\text{induction}}{\searrow} !(!(ntr)!(k+1)ktr)!(k1tr)!(2tr)!(1)\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}\left(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n}\circ\text{tr})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{k+1})\circ\mathscr{E}_{k}\circ\text{tr}\right)\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{k-1}\circ\text{tr})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1})
Lemma 6.12\displaystyle\overset{\text{Lemma~{}\ref{LXMMA8971}}}{\searrow} !(ntr)!(k+1tr)!(ktr)!(k1tr)!(2tr)!(1),\displaystyle\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n}\circ\text{tr})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{k+1}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{k}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{k-1}\circ\text{tr})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1}),

where again in the last line we note that we have repeatedly used the elision trtr=tr\text{tr}\circ\text{tr}=\text{tr}, thus the result holds. ∎

The following theorem follows directly from Proposition 6.13, and will form the basis for a definition of states over time associated with nn-step processes, as we will see in the next section.

Theorem 6.14.

If !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} is associative, then

!ϑ(1,,n)=!(ntr)!(n1tr)!(2tr)!(1)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n}\circ\emph{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n-1}\circ\emph{tr})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2}\circ\emph{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1})

for every parenthezation ϑ(𝒜0,,𝒜n)\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{n}) of 𝒜0𝒜n\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}.

7 States over time for nn-step processes

In this section we use bloom maps for nn-chains to define states over time for arbitrary finite-step quantum processes.

Definition 7.1.

Given an (n+1)(n+1)-tuple (𝒜0,,𝒜n)(\mathcal{A}_{0},...,\mathcal{A}_{n}) of multi-matrix algebras, an element

(ρ,1,,n)𝒮(𝒜0)×𝕋(𝒜0,𝒜1)××𝕋(𝒜n1,𝒜n)\left(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}\right)\in\mathcal{S}(\mathcal{A}_{0})\times\mathbb{CPTP}(\mathcal{A}_{0},\mathcal{A}_{1})\times\cdots\times\mathbb{CPTP}(\mathcal{A}_{n-1},\mathcal{A}_{n})

will be referred to as an nn-step process, and the set 𝒮(𝒜0)×𝕋(𝒜0,𝒜1)××𝕋(𝒜n1,𝒜n)\mathcal{S}(\mathcal{A}_{0})\times\mathbb{CPTP}(\mathcal{A}_{0},\mathcal{A}_{1})\times\cdots\times\mathbb{CPTP}(\mathcal{A}_{n-1},\mathcal{A}_{n}) of all such nn-step processes will be denoted by 𝒫(𝒜0,,𝒜n)\mathscr{P}(\mathcal{A}_{0},...,\mathcal{A}_{n}). Given an nn-step process (ρ,1,,n)\left(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}\right), we let ρ0=ρ\rho_{0}=\rho and ρi=(i1)(ρ)\rho_{i}=(\mathscr{E}_{i}\circ\cdots\circ\mathscr{E}_{1})(\rho) for all i{1,,n}i\in\{1,...,n\}.

Definition 7.2.

An nn-step state over time function associates every (n+1)(n+1)-tuple (𝒜0,,𝒜n)(\mathcal{A}_{0},\ldots,\mathcal{A}_{n}) of multi-matrix algebras with a map

ψ:𝒫(𝒜0,,𝒜n)𝒯(𝒜0𝒜n)\psi:\mathscr{P}(\mathcal{A}_{0},...,\mathcal{A}_{n})\longrightarrow\mathscr{T}\left(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}\right)

such that

tri(ψ(ρ,1,,n))=ρi,\text{tr}_{i}\left(\psi(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\right)=\rho_{i}, (7.3)

for all i{0,,n}i\in\{0,...,n\}. In such a case, the pseudo-density operator ψ(ρ,1,,n)\psi(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) will be referred to as the state over time associated with the nn-step process (ρ,1,,n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}). For n=1n=1 an nn-step state over time function will be referred to simply as a state over time function.

Theorem 7.4.

Let !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} be a bloom map which is hermitian, let n>0n>0, and let ψ\psi be the map on nn-step processes given by

ψ(ρ,1,,n)=(!(ntr)!(2tr)!(1))(ρ).\psi(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})=(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n}\circ\emph{tr})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2}\circ\emph{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1}))(\rho). (7.5)

Then the following statements hold.

  1. i.

    The map ψ\psi is an nn-step state over time function.

  2. ii.

    If !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} is also associative and (ρ,1,,n)𝒫(𝒜0,,𝒜n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathscr{P}(\mathcal{A}_{0},...,\mathcal{A}_{n}), then ψ(ρ,1,,n)=!ϑ(1,,n)(ρ)\psi(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}_{\vartheta}(\mathscr{E}_{1},...,\mathscr{E}_{n})(\rho) for every parenthezation ϑ(𝒜0,,𝒜n)\vartheta(\mathcal{A}_{0},...,\mathcal{A}_{n}) of 𝒜0𝒜n\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}.

Proof.

Item i: Let (𝒜0,,𝒜n)(\mathcal{A}_{0},...,\mathcal{A}_{n}) be an (n+1)(n+1)-tuple of multi-matrix algebras, let (ρ,1,,n)𝒫(𝒜0,,𝒜n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathscr{P}(\mathcal{A}_{0},...,\mathcal{A}_{n}) be an nn-step process. Then for all i{0,,n}i\in\{0,...,n\} we have

tri(ψ(ρ,1,,n))\displaystyle\text{tr}_{i}\left(\psi(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\right) =\displaystyle= tr𝒜0𝒜i^𝒜n((!(ntr)!(2tr)!(1))(ρ))\displaystyle\text{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\widehat{\mathcal{A}_{i}}\otimes\cdots\otimes\mathcal{A}_{n}}\left((\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{n}\circ\text{tr})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1}))(\rho)\right)
=(6.9)\displaystyle\overset{\eqref{BXCXDMA381}}{=} {(i1)(ρ)for i{1,,n}id𝒜0(ρ)for i=0\displaystyle\begin{cases}(\mathscr{E}_{i}\circ\cdots\circ\mathscr{E}_{1})(\rho)\quad\quad\text{for $i\in\{1,...,n\}$}\\ \mathrm{id}_{\mathcal{A}_{0}}(\rho)\quad\quad\hskip 44.38622pt\text{for $i=0$}\end{cases}
=\displaystyle= ρi,\displaystyle\rho_{i},

thus ψ\psi satisfies condition (7.3) of being a state over time function.

To conclude the proof, we must show that ψ(ρ,1,,n)\psi(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) is self-adjoint, for which we use induction on nn. For n=1n=1 we have ψ(ρ,1)=!(1)(ρ)\psi(\rho,\mathscr{E}_{1})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1})(\rho), and since 1\mathscr{E}_{1} is CPTP it is {\dagger}-preserving, thus !(1)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1}) is {\dagger}-preserving by the hermitian assumption on !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}. We then have

!(1)(ρ)=!(1)(ρ)=!(1)(ρ),\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1})(\rho)^{{\dagger}}=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1})(\rho^{{\dagger}})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1})(\rho),

where the last equality follows from the fact that ρ\rho is a state (and so self-adjoint), thus !(1)(ρ)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1})(\rho) is self-adjoint. Now assume the result holds for n=m1>1n=m-1>1, and let σ=(!(m1tr)!(2tr)!(1))(ρ)\sigma=(\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{m-1}\circ\text{tr})\circ\cdots\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{2}\circ\text{tr})\circ\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{1}))(\rho), so that

ψ(ρ,1,,m)=!(mtr)(σ).\psi(\rho,\mathscr{E}_{1},...,\mathscr{E}_{m})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{m}\circ\text{tr})(\sigma).

Now since m\mathscr{E}_{m} and tr are both completely positive, it follows that mtr\mathscr{E}_{m}\circ\text{tr} is {\dagger}-preserving, thus by the hermitian assumption on !\operatorname{\rotatebox[origin={c}]{180.0}{$!$}} it follows that !(mtr)\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{m}\circ\text{tr}) is {\dagger}-preserving as well. We then have

ψ(ρ,1,,m)=!(mtr)(σ)=!(mtr)(σ)=!(mtr)(σ)=ψ(ρ,1,,m),\psi(\rho,\mathscr{E}_{1},...,\mathscr{E}_{m})^{{\dagger}}=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{m}\circ\text{tr})(\sigma)^{{\dagger}}=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{m}\circ\text{tr})(\sigma^{{\dagger}})=\operatorname{\rotatebox[origin={c}]{180.0}{$!$}}(\mathscr{E}_{m}\circ\text{tr})(\sigma)=\psi(\rho,\mathscr{E}_{1},...,\mathscr{E}_{m}),

where the last equality follows from the fact that σ\sigma is self-adjoint by the inductive hypothesis, thus ψ(ρ,1,,m)\psi(\rho,\mathscr{E}_{1},...,\mathscr{E}_{m}) is self-adjoint.

Item ii: The statement follows from Theorem 6.14. ∎

8 The \Yinyang-function

Since the symmetric bloom \Yinyang is hermitian and associative, it follows from Theorem 7.4 that \Yinyang extends to a uniquely defined nn-step state over time function for all nn\in{{\mathbb{N}}}, which we refer to as the \Yinyang-function. In this section we will prove various properties the \Yinyang-function satisfies, including multi-linearity, classical reducibility and a multi-marginal property that will be used in the next section to solve an open problem from the literature. We also prove a general formula for the \Yinyang-function evaluated on an nn-step process (ρ,1,,n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) in terms of ρ\rho and the channel states 𝒥[1],,𝒥[n]\mathscr{J}[\mathscr{E}_{1}],...,\mathscr{J}[\mathscr{E}_{n}]. To conclude the section, we show that in the case of a closed system of dynamically evolving qubits, the \Yinyang-function recovers the pseudo-density matrix formalism first introduced by Fitzsimons, Jones and Vedral [FJV15].

Definition 8.1.

The \Yinyang-function is the function on all finite-step processes given by

\Yinyang(ρ,1,,n)=(\Yinyang(ntr)\Yinyang(2tr)\Yinyang(1))(ρ).\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})=(\text{\Yinyang}(\mathscr{E}_{n}\circ\text{tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{2}\circ\text{tr})\circ\text{\Yinyang}(\mathscr{E}_{1}))(\rho). (8.2)
Remark 8.3.

Note that the \Yinyang-function as given by (8.2) is well-defined even if ρ\rho is not a state and even if i\mathscr{E}_{i} is not CPTP for any i{1,,n}i\in\{1,...,n\}. As such, we may view the \Yinyang-function as being defined on the set

𝒜0×Hom(𝒜0,,𝒜n)\mathcal{A}_{0}\times\text{Hom}(\mathcal{A}_{0},...,\mathcal{A}_{n})

for any (n+1)(n+1)-tuple of multi-matrix algebras (𝒜0,,𝒜n)(\mathcal{A}_{0},...,\mathcal{A}_{n}). However, the element \Yinyang(ρ,1,,n)\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) will only be viewed as a state over time when (ρ,1,,n)𝒫(𝒜0,,𝒜n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathscr{P}(\mathcal{A}_{0},...,\mathcal{A}_{n}).

We now prove various properties the \Yinyang-function satisfies, but first we introduce some notation.

Notation 8.4.

Let 𝒜=𝒜0𝒜n\mathcal{A}=\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}. Given 0i1<<imn0\leq i_{1}<\cdots<i_{m}\leq n, we let

tri1im:𝒜𝒜i1𝒜im{\rm tr}_{i_{1}\cdots i_{m}}:\mathcal{A}\longrightarrow\mathcal{A}_{i_{1}}\otimes\cdots\otimes\mathcal{A}_{i_{m}}

denote the partial trace map.

Theorem 8.5.

The \Yinyang-function satisfies the following properties.

  1. i.

    (Hermiticity) \Yinyang(ρ,1,,n)\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) is self-adjoint of unit trace for all finite step processes (ρ,1,,n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}).

  2. ii.

    (Multi-Linearity) The \Yinyang-function is multi-linear, i.e.,

    \Yinyang(λρ+(1λ)σ,1,,n)=λ\Yinyang(ρ,1,,n)+(1λ)\Yinyang(σ,1,,n)\text{\Yinyang}(\lambda\rho+(1-\lambda)\sigma,\mathscr{E}_{1},...,\mathscr{E}_{n})=\lambda\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})+(1-\lambda)\text{\Yinyang}(\sigma,\mathscr{E}_{1},...,\mathscr{E}_{n}) (8.6)

    and

    \Yinyang(ρ,1,,λi+(1λ)i,,n)=λ\Yinyang(ρ,1,,i,,n)+(1λ)\Yinyang(ρ,n,,i,,n)\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\lambda\mathscr{E}_{i}+(1-\lambda)\mathscr{F}_{i},...,\mathscr{E}_{n})=\lambda\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{i},...,\mathscr{E}_{n})+(1-\lambda)\text{\Yinyang}(\rho,\mathscr{E}_{n},...,\mathscr{F}_{i},...,\mathscr{E}_{n}) (8.7)

    for all i{1,,n}i\in\{1,...,n\} and for all λ\lambda\in{{\mathbb{C}}}.

  3. iii.

    (Reduction) \Yinyang(ρ,1,,n)=\Yinyang(ρ,1,,\Yinyang(i+1)i,i+2tr,i+3,,n)\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})=\text{\Yinyang}\left(\rho,\mathscr{E}_{1},...,\text{\Yinyang}(\mathscr{E}_{i+1})\circ\mathscr{E}_{i}\hskip 0.99585pt,\hskip 0.99585pt\mathscr{E}_{i+2}\circ\emph{tr},\hskip 0.99585pt\mathscr{E}_{i+3},...,\mathscr{E}_{n}\right) for all i{1,,n1}i\in\{1,...,n-1\}.

  4. iv.

    (The Multi-Marginal Property) Let 0i1<<imn0\leq i_{1}<\cdots<i_{m}\leq n. Then

    tri1im(\Yinyang(ρ,1,,n))=\Yinyang(ρi1,i2i1+1,,imim1+1)\emph{tr}_{i_{1}\cdots i_{m}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\right)=\text{\Yinyang}(\rho_{i_{1}},\mathscr{E}_{i_{2}}\circ\cdots\circ\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{m}}\circ\cdots\circ\mathscr{E}_{i_{m-1}+1}) (8.8)
  5. v.

    (Classical Reducibility) If (ρ,1,,n)𝒫(X0,,Xn)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathscr{P}({{\mathbb{C}}}^{X_{0}},...,{{\mathbb{C}}}^{X_{n}}) is a classical process, then

    \Yinyang(ρ,1,,n)=(x0,,xn)X0××Xn(xn|xn1)(x1|x0)(x0),\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})=\bigoplus_{(x_{0},...,x_{n})\in X_{0}\times\cdots\times X_{n}}\mathbb{P}(x_{n}|x_{n-1})\cdots\mathbb{P}(x_{1}|x_{0})\mathbb{P}(x_{0}),

    where (x0)\mathbb{P}(x_{0}) is the classical distribution associated with ρ\rho and (xi|xi1)\mathbb{P}(x_{i}|x_{i-1}) are conditional probabilities associated with 𝒥[i]\mathscr{J}[\mathscr{E}_{i}].

Proof.

Item i: The statement follows from item i of Theorem 7.4.

Item ii: Equation (8.6) follows from the fact that \Yinyang(ntr)\Yinyang(1)\text{\Yinyang}(\mathscr{E}_{n}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{1}) is a linear map, while equation (8.7) follows from the fact that

\Yinyang((λi+(1λ)i)tr)=λ\Yinyang(itr)+(1λ)\Yinyang(itr)\text{\Yinyang}((\lambda\mathscr{E}_{i}+(1-\lambda)\mathscr{F}_{i})\circ{\rm tr})=\lambda\text{\Yinyang}(\mathscr{E}_{i}\circ{\rm tr})+(1-\lambda)\text{\Yinyang}(\mathscr{F}_{i}\circ{\rm tr})

for all i{1,,n}i\in\{1,...,n\} and all λ\lambda\in{{\mathbb{C}}}.

Item iii: Let i{1,,n1}i\in\{1,...,n-1\}, and let =\Yinyang(ρ,1,,\Yinyang(i+1)i,i+2tr,i+3,,n)\aleph=\text{\Yinyang}\left(\rho,\mathscr{E}_{1},...,\text{\Yinyang}(\mathscr{E}_{i+1})\circ\mathscr{E}_{i}\hskip 0.99585pt,\hskip 0.99585pt\mathscr{E}_{i+2}\circ\text{tr},\hskip 0.99585pt\mathscr{E}_{i+3},...,\mathscr{E}_{n}\right). Then

\displaystyle\aleph =\displaystyle= \Yinyang(ρ,1,,\Yinyang(i+1)i,i+2tr,i+3,,n)\displaystyle\text{\Yinyang}\left(\rho,\mathscr{E}_{1},...,\text{\Yinyang}(\mathscr{E}_{i+1})\circ\mathscr{E}_{i}\hskip 0.99585pt,\hskip 0.99585pt\mathscr{E}_{i+2}\circ\text{tr},\hskip 0.99585pt\mathscr{E}_{i+3},...,\mathscr{E}_{n}\right)
=\displaystyle= (\Yinyang(ntr)\Yinyang(\Yinyang(i+1)itr)\Yinyang(1))(ρ)\displaystyle\left(\text{\Yinyang}(\mathscr{E}_{n}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}\left(\text{\Yinyang}(\mathscr{E}_{i+1})\circ\mathscr{E}_{i}\circ{\rm tr}\right)\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{1})\right)(\rho)
=\displaystyle= (\Yinyang(ntr)\Yinyang(i+1tr)\Yinyang(itr)\Yinyang(1))(ρ)\displaystyle\left(\text{\Yinyang}(\mathscr{E}_{n}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{i+1}\circ{\rm tr})\circ\text{\Yinyang}(\mathscr{E}_{i}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{1})\right)(\rho)
=\displaystyle= \Yinyang(ρ,1,,n),\displaystyle\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}),

where the third equality follows from the associativity of the symmetric bloom.

Item iv: The proof relies on two lemmas, which we will prove after we finish proving the theorem. The case m=1m=1 follows from item i of Theorem 7.4, so now suppose m>1m>1. The partial trace map tri1im{\rm tr}_{i_{1}\cdots i_{m}} may be written as

tri1im=(tr𝒜i1+1𝒜i21tr𝒜im1+1𝒜im1)tr𝒜0𝒜i11𝒜im+1𝒜n,{\rm tr}_{i_{1}\cdots i_{m}}=\left({\rm tr}_{\mathcal{A}_{i_{1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{2}-1}}\circ\cdots\circ{\rm tr}_{\mathcal{A}_{i_{m-1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{m}-1}}\right)\circ{\rm tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{{i_{1}}-1}\otimes\mathcal{A}_{{i_{m}}+1}\otimes\cdots\otimes\mathcal{A}_{n}},

where if ijij1=1i_{j}-i_{j-1}=1, we set tr𝒜ij1+1𝒜ij1=id{\rm tr}_{\mathcal{A}_{i_{j-1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{j}-1}}=\mathrm{id}. Now by items i and ii Lemma 8.9 we have

tr𝒜0𝒜i11𝒜im+1𝒜n(\Yinyang(ρ,1,,n))=\Yinyang(ρi1,i1+1,,im)𝒜i1𝒜i1+1𝒜im,{\rm tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{{i_{1}}-1}\otimes\mathcal{A}_{{i_{m}}+1}\otimes\cdots\otimes\mathcal{A}_{n}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\right)=\text{\Yinyang}(\rho_{i_{1}},\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{m}})\in\mathcal{A}_{i_{1}}\otimes\mathcal{A}_{i_{1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{m}},

thus

tri1im(\Yinyang(ρ,1,,n))\displaystyle{\rm tr}_{i_{1}\cdots i_{m}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\right) =\displaystyle= tr𝒜i1+1𝒜i21tr𝒜im1+1𝒜im1(\Yinyang(ρi1,i1+1,,im))\displaystyle{\rm tr}_{\mathcal{A}_{i_{1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{2}-1}}\circ\cdots\circ{\rm tr}_{\mathcal{A}_{i_{m-1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{m}-1}}\left(\text{\Yinyang}(\rho_{i_{1}},\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{m}})\right)
=\displaystyle= \Yinyang(ρi1,i2i1+1,,imim1+1),\displaystyle\text{\Yinyang}(\rho_{i_{1}},\mathscr{E}_{i_{2}}\circ\cdots\circ\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{m}}\circ\cdots\circ\mathscr{E}_{i_{m-1}+1}),

where the last equality follows from Lemma 8.10.

Item v: In the proof of Proposition 3.10 we established that if :XY\mathscr{E}:{{\mathbb{C}}}^{X}\longrightarrow{{\mathbb{C}}}^{Y} is a classical channel, then

𝒥[]=(x,y)X×Yyx(δxδy),\mathscr{J}[\mathscr{E}]=\sum_{(x,y)\in X\times Y}\mathscr{E}_{yx}(\delta_{x}\otimes\delta_{y}),

where we recall yx\mathscr{E}_{yx} are the conditional probabilities associated with the classical channel \mathscr{E}. The statement then follows directly from the definition of the symmetric bloom. ∎

We now prove the lemmas needed for the proof of the multi-marginal property of the \Yinyang-function (item iv of Theorem 8.5).

Lemma 8.9.

Let (ρ,1,,n)𝒫(𝒜0,,𝒜n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathscr{P}(\mathcal{A}_{0},...,\mathcal{A}_{n}) be an nn-step process. Then the following statements hold.

  1. i.

    tr𝒜0𝒜i1(\Yinyang(ρ,1,,n))=\Yinyang(ρi,i+1,,n)\emph{tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{i-1}}(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}))=\text{\Yinyang}(\rho_{i},\mathscr{E}_{i+1},...,\mathscr{E}_{n}) for all i{1,,n}i\in\{1,...,n\}.

  2. ii.

    tr𝒜i+1𝒜n(\Yinyang(ρ,1,,n))=\Yinyang(ρ,1,,i)\emph{tr}_{\mathcal{A}_{i+1}\otimes\cdots\otimes\mathcal{A}_{n}}(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}))=\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{i}) for all i{0,,n1}i\in\{0,...,n-1\}.

Proof.

Item i: We use induction on ii. For i=0i=0 we have

tr𝒜0(\Yinyang(ρ,1,,n))\displaystyle{\rm tr}_{\mathcal{A}_{0}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\right) =\displaystyle= tr𝒜0((\Yinyang(ntr)\Yinyang(2tr)\Yinyang(1))(ρ))\displaystyle{\rm tr}_{\mathcal{A}_{0}}\left(\left(\text{\Yinyang}(\mathscr{E}_{n}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{2}\circ{\rm tr})\circ\text{\Yinyang}(\mathscr{E}_{1})\right)(\rho)\right)
=\displaystyle= tr𝒜0(\Yinyang(\Yinyang(ntr)\Yinyang(2)1)(ρ))\displaystyle{\rm tr}_{\mathcal{A}_{0}}\left(\text{\Yinyang}\left(\text{\Yinyang}(\mathscr{E}_{n}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{2})\circ\mathscr{E}_{1}\right)(\rho)\right)
=\displaystyle= (\Yinyang(ntr)\Yinyang(2))(1(ρ))\displaystyle\left(\text{\Yinyang}(\mathscr{E}_{n}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{2})\right)(\mathscr{E}_{1}(\rho))
=\displaystyle= \Yinyang(ρ1,2,,n),\displaystyle\text{\Yinyang}(\rho_{1},\mathscr{E}_{2},...,\mathscr{E}_{n}),

where the second equality follows from Lemma 6.12 and the third equality follows from the definition of a bloom map, thus the result holds for i=0i=0. Now suppose the result holds for i=m1i=m-1 with m1<nm-1<n. Then

tr𝒜0𝒜m1(\Yinyang(ρ,1,,n))\displaystyle{\rm tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{m-1}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\right) =\displaystyle= tr𝒜m1(tr𝒜0𝒜m2(\Yinyang(ρ,1,,n)))\displaystyle{\rm tr}_{\mathcal{A}_{m-1}}\left({\rm tr}_{\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{m-2}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\right)\right)
=\displaystyle= tr𝒜m1(\Yinyang(ρm1,m,n))\displaystyle{\rm tr}_{\mathcal{A}_{m-1}}\left(\text{\Yinyang}(\rho_{m-1},\mathscr{E}_{m},...\mathscr{E}_{n})\right)
=\displaystyle= \Yinyang(ρm,m+1,n),\displaystyle\text{\Yinyang}(\rho_{m},\mathscr{E}_{m+1},...\mathscr{E}_{n}),

where the last equality follows from a calculation similar to the i=0i=0 case. It then follows that the result holds for i=mi=m, as desired.

Item ii: We prove the equivalent statement tr𝒜ni𝒜n(\Yinyang(ρ,1,,n))=\Yinyang(ρ,1,,ni1){\rm tr}_{\mathcal{A}_{n-i}\otimes\cdots\otimes\mathcal{A}_{n}}(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}))=\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n-i-1}) for all i{0,,n1}i\in\{0,...,n-1\} by induction on ii. For i=0i=0 we have

tr𝒜n(\Yinyang(ρ,1,,n))=tr𝒜n(\Yinyang(ntr)(\Yinyang(ρ,1,,n1)))=\Yinyang(ρ,1,,n1),{\rm tr}_{\mathcal{A}_{n}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\right)={\rm tr}_{\mathcal{A}_{n}}\left(\text{\Yinyang}(\mathscr{E}_{n}\circ{\rm tr})\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n-1})\right)\right)=\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n-1}),

where the last equality follows from the definition of bloom map, thus the result holds for i=0i=0. Now suppose the result holds for i=m1i=m-1 with m1<n1m-1<n-1. Then

tr𝒜nm𝒜n(\Yinyang(ρ,1,,n))\displaystyle{\rm tr}_{\mathcal{A}_{n-m}\otimes\cdots\otimes\mathcal{A}_{n}}(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})) =\displaystyle= tr𝒜nm(tr𝒜nm+1𝒜n(\Yinyang(ρ,1,,n)))\displaystyle{\rm tr}_{\mathcal{A}_{n-m}}\left({\rm tr}_{\mathcal{A}_{n-m+1}\otimes\cdots\otimes\mathcal{A}_{n}}(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}))\right)
=\displaystyle= tr𝒜nm(\Yinyang(ρ,1,,n(m1)1))\displaystyle{\rm tr}_{\mathcal{A}_{n-m}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n-(m-1)-1})\right)
=\displaystyle= tr𝒜nm(\Yinyang(ρ,1,,nm))\displaystyle{\rm tr}_{\mathcal{A}_{n-m}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n-m})\right)
=\displaystyle= \Yinyang(ρ,1,,nm1),\displaystyle\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n-m-1}),

where the last equality follows from a calculation similar to the i=0i=0 case. It then follows that the result holds for i=mi=m, as desired. ∎

Lemma 8.10.

Let (1,,n)𝕋(𝒜0,,𝒜n)(\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathbb{TP}(\mathcal{A}_{0},...,\mathcal{A}_{n}) be an nn-chain, let 0=i1<<im=n0=i_{1}<\cdots<i_{m}=n, let ρ𝒜i1\rho\in\mathcal{A}_{i_{1}}, and let 𝒜i1𝒜i2𝒜im\aleph\in\mathcal{A}_{i_{1}}\otimes\mathcal{A}_{i_{2}}\otimes\cdots\otimes\mathcal{A}_{i_{m}} be the element given by

=tr𝒜im1+1𝒜im1tr𝒜i1+1𝒜i21(\Yinyang(ρ,i1+1,,im)),\aleph={\rm tr}_{\mathcal{A}_{i_{m-1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{m}-1}}\circ\cdots\circ{\rm tr}_{\mathcal{A}_{i_{1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{2}-1}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{m}})\right),

where we set tr𝒜ij1+1𝒜ij1=id{\rm tr}_{\mathcal{A}_{i_{j-1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{j}-1}}=\mathrm{id}   if   ijij1=1i_{j}-i_{j-1}=1. Then

=\Yinyang(ρ,i2i1+1,,imim1+1).\aleph=\text{\Yinyang}\left(\rho,\mathscr{E}_{i_{2}}\circ\cdots\circ\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{m}}\circ\cdots\circ\mathscr{E}_{i_{m-1}+1}\right). (8.11)
Proof.

We use induction on mm. The case m=1m=1 follows from item ii of Lemma 8.9. So assume the result holds for m=k1>1m=k-1>1, and let k\aleph_{k} be the element given by

k=tr𝒜ik1+1𝒜ik1tr𝒜i1+1𝒜i21(\Yinyang(ρ,i1+1,,ik)).\aleph_{k}={\rm tr}_{\mathcal{A}_{i_{k-1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{k}-1}}\circ\cdots\circ{\rm tr}_{\mathcal{A}_{i_{1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{2}-1}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{k}})\right).

The result then follows once we show

k=\Yinyang(ρ,i2i1+1,,ikik1+1).\aleph_{k}=\text{\Yinyang}\left(\rho,\mathscr{E}_{i_{2}}\circ\cdots\circ\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{k}}\circ\cdots\circ\mathscr{E}_{i_{k-1}+1}\right).

For this, we first consider the 2-chain

𝒜i1𝒜i1+1𝒜i21𝒜i2𝒜i2+1𝒜im,\mathcal{A}_{i_{1}}\overset{\mathscr{E}}{\longrightarrow}\mathcal{A}_{i_{1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{2}-1}\overset{\mathscr{F}}{\longrightarrow}\mathcal{A}_{i_{2}}\otimes\mathcal{A}_{i_{2}+1}\otimes\cdots\otimes\mathcal{A}_{i_{m}},

where =\Yinyang(i21tr)\Yinyang(i1+2)i1+1\mathscr{E}=\text{\Yinyang}(\mathscr{E}_{i_{2}-1}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{i_{1}+2})\circ\mathscr{E}_{i_{1}+1} and =\Yinyang(iktr)\Yinyang(i2+1)i2tr\mathscr{F}=\text{\Yinyang}(\mathscr{E}_{i_{k}}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{i_{2}+1})\circ\mathscr{E}_{i_{2}}\circ{\rm tr}. Then after letting χ=\Yinyang(ρ,,)\chi=\text{\Yinyang}(\rho,\mathscr{E},\mathscr{F}), we have

χ\displaystyle\chi =\displaystyle= \Yinyang(ρ,,)\displaystyle\text{\Yinyang}(\rho,\mathscr{E},\mathscr{F})
=\displaystyle= (\Yinyang(tr)\Yinyang())(ρ)\displaystyle\left(\text{\Yinyang}(\mathscr{F}\circ{\rm tr})\circ\text{\Yinyang}(\mathscr{E})\right)(\rho)
=\displaystyle= (\Yinyang(\Yinyang(iktr)\Yinyang(i2+1)i2tr)\Yinyang(\Yinyang(i21tr)\Yinyang(i1+2)i1+1))(ρ)\displaystyle\left(\text{\Yinyang}(\text{\Yinyang}(\mathscr{E}_{i_{k}}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{i_{2}+1})\circ\mathscr{E}_{i_{2}}\circ{\rm tr})\circ\text{\Yinyang}\left(\text{\Yinyang}(\mathscr{E}_{i_{2}-1}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{i_{1}+2})\circ\mathscr{E}_{i_{1}+1}\right)\right)(\rho)
=\displaystyle= (\Yinyang(i2tr)\Yinyang(i1+2tr)\Yinyang(i1+1))(ρ)\displaystyle\left(\text{\Yinyang}(\mathscr{E}_{i_{2}}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{i_{1}+2}\circ{\rm tr})\circ\text{\Yinyang}(\mathscr{E}_{i_{1}+1})\right)(\rho)
=\displaystyle= \Yinyang(ρ,i1+1,,ik),\displaystyle\text{\Yinyang}(\rho,\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{k}}),

where the fourth equality follows from the associativity of the symmetric bloom. Then after letting ξ=tr𝒜i1+1𝒜121(\Yinyang(ρ,i1+1,,ik))\xi={\rm tr}_{\mathcal{A}_{i_{1}+1}\otimes\cdots\otimes\mathcal{A}_{1_{2}-1}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{k}})\right), we then have

ξ\displaystyle\xi =\displaystyle= tr𝒜i1+1𝒜121(\Yinyang(ρ,i1+1,,ik))\displaystyle{\rm tr}_{\mathcal{A}_{i_{1}+1}\otimes\cdots\otimes\mathcal{A}_{1_{2}-1}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{k}})\right)
=\displaystyle= tr𝒜i+1𝒜j1(\Yinyang(ρ,,))\displaystyle{\rm tr}_{\mathcal{A}_{i+1}\otimes\cdots\otimes\mathcal{A}_{j-1}}\left(\text{\Yinyang}(\rho,\mathscr{E},\mathscr{F})\right)
=(5.16)\displaystyle\overset{\eqref{CMPXALTX67}}{=} \Yinyang(ρ,)\displaystyle\text{\Yinyang}(\rho,\mathscr{F}\circ\mathscr{E})
=\displaystyle= \Yinyang(ρ,\Yinyang(iktr)\Yinyang(i2+1)i2tr\Yinyang(i21tr)\Yinyang(i1+2)i1+1)\displaystyle\text{\Yinyang}\left(\rho,\text{\Yinyang}(\mathscr{E}_{i_{k}}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{i_{2}+1})\circ\mathscr{E}_{i_{2}}\circ{\rm tr}\circ\text{\Yinyang}(\mathscr{E}_{i_{2}-1}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{i_{1}+2})\circ\mathscr{E}_{i_{1}+1}\right)
=\displaystyle= \Yinyang(ρ,\Yinyang(iktr)\Yinyang(i2+1)i2i21i1+1)\displaystyle\text{\Yinyang}\left(\rho,\text{\Yinyang}(\mathscr{E}_{i_{k}}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{i_{2}+1})\circ\mathscr{E}_{i_{2}}\circ\mathscr{E}_{i_{2}-1}\circ\cdots\circ\mathscr{E}_{i_{1}+1}\right)
=\displaystyle= \Yinyang(\Yinyang(iktr)\Yinyang(i2+1)i2i21i1+1)(ρ)\displaystyle\text{\Yinyang}\left(\text{\Yinyang}(\mathscr{E}_{i_{k}}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{i_{2}+1})\circ\mathscr{E}_{i_{2}}\circ\mathscr{E}_{i_{2}-1}\circ\cdots\circ\mathscr{E}_{i_{1}+1}\right)(\rho)
=\displaystyle= (\Yinyang(iktr)\Yinyang(i2+1tr)\Yinyang(i2i21i1+1))(ρ)\displaystyle\left(\text{\Yinyang}(\mathscr{E}_{i_{k}}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{i_{2}+1}\circ{\rm tr})\circ\text{\Yinyang}\left(\mathscr{E}_{i_{2}}\circ\mathscr{E}_{i_{2}-1}\circ\cdots\circ\mathscr{E}_{i_{1}+1}\right)\right)(\rho)
=\displaystyle= \Yinyang(ρ,i2i21i1+1,i2+1,,ik),\displaystyle\text{\Yinyang}(\rho,\mathscr{E}_{i_{2}}\circ\mathscr{E}_{i_{2}-1}\circ\cdots\circ\mathscr{E}_{i_{1}+1},\mathscr{E}_{i_{2}+1},...,\mathscr{E}_{i_{k}}),

where the fifth equality follows from the fact that i2tr\Yinyang(i21tr)\Yinyang(i1+2)i1+1=i2i21i1+1\mathscr{E}_{i_{2}}\circ{\rm tr}\circ\text{\Yinyang}(\mathscr{E}_{i_{2}-1}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{i_{1}+2})\circ\mathscr{E}_{i_{1}+1}=\mathscr{E}_{i_{2}}\circ\mathscr{E}_{i_{2}-1}\circ\cdots\circ\mathscr{E}_{i_{1}+1}, and the seventh equality follows from the associativity of symmetric bloom. Now let σ=\Yinyang(ρ,i2i21i1+1)\sigma=\text{\Yinyang}(\rho,\mathscr{E}_{i_{2}}\circ\mathscr{E}_{i_{2}-1}\circ\cdots\circ\mathscr{E}_{i_{1}+1}), so that

\Yinyang(ρ,i2i21i1+1,i2+1,,ik)\displaystyle\text{\Yinyang}(\rho,\mathscr{E}_{i_{2}}\circ\mathscr{E}_{i_{2}-1}\circ\cdots\circ\mathscr{E}_{i_{1}+1},\mathscr{E}_{i_{2}+1},...,\mathscr{E}_{i_{k}}) =\displaystyle= (\Yinyang(iktr)\Yinyang(i2i21i1+1))(ρ)\displaystyle\left(\text{\Yinyang}(\mathscr{E}_{i_{k}}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}\left(\mathscr{E}_{i_{2}}\circ\mathscr{E}_{i_{2}-1}\circ\cdots\circ\mathscr{E}_{i_{1}+1}\right)\right)(\rho)
=\displaystyle= (\Yinyang(iktr)\Yinyang(i2+1tr))(σ)\displaystyle\left(\text{\Yinyang}(\mathscr{E}_{i_{k}}\circ{\rm tr})\circ\cdots\circ\text{\Yinyang}\left(\mathscr{E}_{i_{2}+1}\circ{\rm tr}\right)\right)(\sigma)
=\displaystyle= \Yinyang(σ,i2+1tr,i2+2,,ik).\displaystyle\text{\Yinyang}(\sigma,\mathscr{E}_{i_{2}+1}\circ{\rm tr},\mathscr{E}_{i_{2}+2},...,\mathscr{E}_{i_{k}}).

We then have

k\displaystyle\aleph_{k} =\displaystyle= tr𝒜ik1+1𝒜ik1tr𝒜i1+1𝒜i21(\Yinyang(ρ,i1+1,,ik))\displaystyle{\rm tr}_{\mathcal{A}_{i_{k-1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{k}-1}}\circ\cdots\circ{\rm tr}_{\mathcal{A}_{i_{1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{2}-1}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{k}})\right)
=\displaystyle= tr𝒜ik1+1𝒜ik1tr𝒜i2+1𝒜i31(tr𝒜i1+1𝒜121(\Yinyang(ρ,i1+1,,ik)))\displaystyle{\rm tr}_{\mathcal{A}_{i_{k-1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{k}-1}}\circ\cdots\circ{\rm tr}_{\mathcal{A}_{i_{2}+1}\otimes\cdots\otimes\mathcal{A}_{i_{3}-1}}\left({\rm tr}_{\mathcal{A}_{i_{1}+1}\otimes\cdots\otimes\mathcal{A}_{1_{2}-1}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{k}})\right)\right)
=\displaystyle= tr𝒜ik1+1𝒜ik1tr𝒜i2+1𝒜i31(\Yinyang(ρ,i2i21i1+1,i2+1,,ik))\displaystyle{\rm tr}_{\mathcal{A}_{i_{k-1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{k}-1}}\circ\cdots\circ{\rm tr}_{\mathcal{A}_{i_{2}+1}\otimes\cdots\otimes\mathcal{A}_{i_{3}-1}}\left(\text{\Yinyang}(\rho,\mathscr{E}_{i_{2}}\circ\mathscr{E}_{i_{2}-1}\circ\cdots\circ\mathscr{E}_{i_{1}+1},\mathscr{E}_{i_{2}+1},...,\mathscr{E}_{i_{k}})\right)
=\displaystyle= tr𝒜ik1+1𝒜ik1tr𝒜i2+1𝒜i31(\Yinyang(σ,i2+1tr,i2+2,,ik))\displaystyle{\rm tr}_{\mathcal{A}_{i_{k-1}+1}\otimes\cdots\otimes\mathcal{A}_{i_{k}-1}}\circ\cdots\circ{\rm tr}_{\mathcal{A}_{i_{2}+1}\otimes\cdots\otimes\mathcal{A}_{i_{3}-1}}\left(\text{\Yinyang}(\sigma,\mathscr{E}_{i_{2}+1}\circ{\rm tr},\mathscr{E}_{i_{2}+2},...,\mathscr{E}_{i_{k}})\right)
=induction\displaystyle\overset{\text{induction}}{=} \Yinyang(σ,i3i2+1tr,,ikik1+1)\displaystyle\text{\Yinyang}\left(\sigma,\mathscr{E}_{i_{3}}\circ\cdots\circ\mathscr{E}_{i_{2}+1}\circ{\rm tr},...,\mathscr{E}_{i_{k}}\circ\cdots\circ\mathscr{E}_{i_{k-1}+1}\right)
=\displaystyle= \Yinyang(\Yinyang(ρ,i2i21i1+1),i3i2+1tr,,ikik1+1)\displaystyle\text{\Yinyang}\left(\text{\Yinyang}(\rho,\mathscr{E}_{i_{2}}\circ\mathscr{E}_{i_{2}-1}\circ\cdots\circ\mathscr{E}_{i_{1}+1}),\mathscr{E}_{i_{3}}\circ\cdots\circ\mathscr{E}_{i_{2}+1}\circ{\rm tr},...,\mathscr{E}_{i_{k}}\circ\cdots\circ\mathscr{E}_{i_{k-1}+1}\right)
=\displaystyle= \Yinyang(ρ,i2i1+1,,ikik1+1),\displaystyle\text{\Yinyang}\left(\rho,\mathscr{E}_{i_{2}}\circ\cdots\circ\mathscr{E}_{i_{1}+1},...,\mathscr{E}_{i_{k}}\circ\cdots\circ\mathscr{E}_{i_{k-1}+1}\right),

where the final equality follows from item iii of Theorem ii, thus concluding the proof. ∎

We now prove a formula for \Yinyang(ρ,1,,n)\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) in terms of ρ\rho and the channel states 𝒥[1]\mathscr{J}[\mathscr{E}_{1}], …, 𝒥[n]\mathscr{J}[\mathscr{E}_{n}].

Notation 8.12.

Let [n]={1,,n}[n]=\{1,...,n\} for every natural number nn\in{{\mathbb{N}}}. For every subset {i1,,im}[n]\{i_{1},...,i_{m}\}\subset[n], the subscripts iji_{j} are chosen so that i1<<imi_{1}<\cdots<i_{m}. In such a case, we denote the complement [n]{i1,,im}[n]\setminus\{i_{1},...,i_{m}\} by {i¯1,,i¯nm}\{\underline{i}_{1},...,\underline{i}_{n-m}\}, with the same convention that i¯1<<i¯nm\underline{i}_{1}<\cdots<\underline{i}_{n-m}.

Notation 8.13.

Given an nn-chain (1,,n)𝕋(𝒜0,,𝒜n)(\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathbb{TP}(\mathcal{A}_{0},...,\mathcal{A}_{n}) and an element j[n]j\in[n], we let 𝒥~[j]𝒜0𝒜n\widetilde{\mathscr{J}}[\mathscr{E}_{j}]\in\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n} be the element given by

𝒥~[j]={𝒥[1]𝟙if j=1𝟙𝒥[j]𝟙if 1<j<n𝟙𝒥[n]if j=n.\widetilde{\mathscr{J}}[\mathscr{E}_{j}]=\begin{cases}\mathscr{J}[\mathscr{E}_{1}]\otimes\mathds{1}\quad\hskip 19.63246pt\text{if $j=1$}\\ \mathds{1}\otimes\mathscr{J}[\mathscr{E}_{j}]\otimes\mathds{1}\quad\text{if $1<j<n$}\\ \mathds{1}\otimes\mathscr{J}[\mathscr{E}_{n}]\quad\hskip 18.20973pt\text{if $j=n$}.\end{cases}
Theorem 8.14.

Let (ρ,1,,n)𝒫(𝒜0,,𝒜n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathscr{P}(\mathcal{A}_{0},...,\mathcal{A}_{n}) be an nn-step process. Then

\Yinyang(ρ,1,,n)=12n{i1,,im}[n]𝒥~[im]𝒥~[i1](ρ𝟙)𝒥~[i¯1]𝒥~[i¯nm]\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})=\frac{1}{2^{n}}\sum_{\{i_{1},...,i_{m}\}\subset[n]}\widetilde{\mathscr{J}}[\mathscr{E}_{i_{m}}]\cdots\widetilde{\mathscr{J}}[\mathscr{E}_{i_{1}}](\rho\otimes\mathds{1})\widetilde{\mathscr{J}}[\mathscr{E}_{\underline{i}_{1}}]\cdots\widetilde{\mathscr{J}}[\mathscr{E}_{\underline{i}_{n-m}}] (8.15)
Proof.

We use induction on nn. For n=1n=1 the result holds by definition of the \Yinyang-function. So now suppose the result holds for n=k1>1n=k-1>1. Then after letting =\Yinyang(ρ,1,,k)\aleph=\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{k}), we have

\displaystyle\aleph =\displaystyle= \Yinyang(ρ,1,,k)\displaystyle\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{k})
=\displaystyle= (\Yinyang(ktr)\Yinyang(2tr)\Yinyang(1))(ρ)\displaystyle(\text{\Yinyang}(\mathscr{E}_{k}\circ\text{tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{2}\circ\text{tr})\circ\text{\Yinyang}(\mathscr{E}_{1}))(\rho)
=\displaystyle= \Yinyang(ktr)(\Yinyang(ρ,1,,k1))\displaystyle\text{\Yinyang}(\mathscr{E}_{k}\circ\text{tr})\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{k-1})\right)
=\displaystyle= 𝕁𝕠𝕣(\Yinyang(ρ,1,,k1)𝟙,𝒥[ktr])\displaystyle\mathbb{Jor}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{k-1})\otimes\mathds{1},\mathscr{J}[\mathscr{E}_{k}\circ{\rm tr}]\right)
=\displaystyle= 𝕁𝕠𝕣(\Yinyang(ρ,1,,k1)𝟙,𝟙𝒥[k])\displaystyle\mathbb{Jor}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{k-1})\otimes\mathds{1},\mathds{1}\otimes\mathscr{J}[\mathscr{E}_{k}]\right)
=\displaystyle= 𝕁𝕠𝕣((12k1{i1,,im}[k1]𝒥~[im]𝒥~[i1](ρ𝟙)𝒥~[i¯1]𝒥~[i¯k1m])𝟙,𝒥~[k])\displaystyle\mathbb{Jor}\left(\left(\frac{1}{2^{k-1}}\sum_{\{i_{1},...,i_{m}\}\subset[k-1]}\widetilde{\mathscr{J}}[\mathscr{E}_{i_{m}}]\cdots\widetilde{\mathscr{J}}[\mathscr{E}_{i_{1}}](\rho\otimes\mathds{1})\widetilde{\mathscr{J}}[\mathscr{E}_{\underline{i}_{1}}]\cdots\widetilde{\mathscr{J}}[\mathscr{E}_{\underline{i}_{k-1-m}}]\right)\otimes\mathds{1},\widetilde{\mathscr{J}}[\mathscr{E}_{k}]\right)
=\displaystyle= 12k{i1,,im}[k]𝒥~[im]𝒥~[i1](ρ𝟙)𝒥~[i¯1]𝒥~[i¯km],\displaystyle\frac{1}{2^{k}}\sum_{\{i_{1},...,i_{m}\}\subset[k]}\widetilde{\mathscr{J}}[\mathscr{E}_{i_{m}}]\cdots\widetilde{\mathscr{J}}[\mathscr{E}_{i_{1}}](\rho\otimes\mathds{1})\widetilde{\mathscr{J}}[\mathscr{E}_{\underline{i}_{1}}]\cdots\widetilde{\mathscr{J}}[\mathscr{E}_{\underline{i}_{k-m}}],

where the second to last equality follows from the inductive assumption. It then follows that the result holds for n=kn=k, as desired. ∎

We now extend the normalized Jordan product to an nn-ary operation for all n>0n>0, which we will then use to derive a formula for the \Yinyang-function which may be viewed as the nn-step generalization of the 2-step formula (5.14).

Definition 8.16.

Give a multi-matrix algebra 𝒜\mathcal{A}, the extended Jordan product is the map

𝕁𝕠𝕣:𝒜××𝒜𝒜\mathbb{Jor}:\mathcal{A}\times\cdots\times\mathcal{A}\longrightarrow\mathcal{A}

given by

𝕁𝕠𝕣(A1,,An)=𝕁𝕠𝕣(A0,𝕁𝕠𝕣(A1,𝕁𝕠𝕣(,𝕁𝕠𝕣(An1,An)))).\mathbb{Jor}(A_{1},...,A_{n})=\mathbb{Jor}\left(A_{0},\mathbb{Jor}(A_{1},\mathbb{Jor}(\cdots,\mathbb{Jor}(A_{n-1},A_{n})))\right). (8.17)
Theorem 8.18.

Let (ρ,1,,n)𝒫(𝒜0,,𝒜n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathscr{P}(\mathcal{A}_{0},...,\mathcal{A}_{n}) be an nn-step process. Then

\Yinyang(ρ,1,,n)=𝕁𝕠𝕣(ρ𝟙,𝒥[1]𝟙,,𝟙𝒥[j]𝟙,,𝟙𝒥[n])\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})=\mathbb{Jor}\left(\rho\otimes\mathds{1},\mathscr{J}[\mathscr{E}_{1}]\otimes\mathds{1},...,\mathds{1}\otimes\mathscr{J}[\mathscr{E}_{j}]\otimes\mathds{1},...,\mathds{1}\otimes\mathscr{J}[\mathscr{E}_{n}]\right) (8.19)
Proof.

The statement follows from Theorem 8.14 together with the fact that (ρ𝟙)(\rho\otimes\mathds{1}) commutes with 𝒥~[j]\widetilde{\mathscr{J}}[\mathscr{E}_{j}] for all j[n]j\in[n] with j>1j>1. ∎

To conclude this section, we show how the \Yinyang-function recovers the pseudo-density matrix formalism for qubits.

Example 8.20 (The pseudo-density matrix formalism for systems of qubits).

Let ρi=1k𝕄2\rho\in\bigotimes_{i=1}^{k}\mathbb{M}_{2} be the initial state of an kk-qubit system, let {σ0,σ1,σ2,σ3}\{\sigma_{0},\sigma_{1},\sigma_{2},\sigma_{3}\} be the Pauli basis of 𝕄2\mathbb{M}_{2}, and let (ρ,1,,n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) be an nn-step process with initial state ρ\rho. In the formalism introduced by Fitzsimons, Jones and Vedral [FJV15], the pseudo-density matrix associated with (ρ,1,,n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) is the matrix Rnj=0n𝕄2kR_{n}\in\bigotimes_{j=0}^{n}\mathbb{M}_{2}^{\otimes k} given by

Rn=12k(n+1)i0=14kin=14k(σ~iα)α=0nα=0nσ~iα,R_{n}=\frac{1}{2^{k(n+1)}}\sum_{i_{0}=1}^{4^{k}}\cdots\sum_{i_{n}=1}^{4^{k}}\left\langle\left(\widetilde{\sigma}_{i_{\alpha}}\right)_{\alpha=0}^{n}\right\rangle\bigotimes_{\alpha=0}^{n}\widetilde{\sigma}_{i_{\alpha}},

where σ~iα{σ0,,σ3}k\widetilde{\sigma}_{i_{\alpha}}\in\{\sigma_{0},...,\sigma_{3}\}^{\otimes k} and {σ~iα}α=0n\left\langle\left\{\widetilde{\sigma}_{i_{\alpha}}\right\}_{\alpha=0}^{n}\right\rangle denotes the expectation value associated with the observable (σ~iα)α=0n\left(\widetilde{\sigma}_{i_{\alpha}}\right)_{\alpha=0}^{n}. In [LQDV], it was recently proved that

R1=𝕁𝕠𝕣(ρ𝟙,𝒥[1])\Yinyang(ρ,1)&Rn=𝕁𝕠𝕣(Rn1𝟙,𝟙𝒥[n]),R_{1}=\mathbb{Jor}(\rho\otimes\mathds{1},\mathscr{J}[\mathscr{E}_{1}])\equiv\text{\Yinyang}(\rho,\mathscr{E}_{1})\quad\&\quad R_{n}=\mathbb{Jor}\left(R_{n-1}\otimes\mathds{1},\mathds{1}\otimes\mathscr{J}[\mathscr{E}_{n}]\right),

and since

\Yinyang(ρ,1,,n)\displaystyle\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) =\displaystyle= (\Yinyang(ntr)\Yinyang(2tr)\Yinyang(1))(ρ)\displaystyle(\text{\Yinyang}(\mathscr{E}_{n}\circ\text{tr})\circ\cdots\circ\text{\Yinyang}(\mathscr{E}_{2}\circ\text{tr})\circ\text{\Yinyang}(\mathscr{E}_{1}))(\rho)
=\displaystyle= \Yinyang(ntr)(\Yinyang(ρ,1,,n1))\displaystyle\text{\Yinyang}(\mathscr{E}_{n}\circ\text{tr})\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n-1})\right)
=\displaystyle= 𝕁𝕠𝕣(\Yinyang(ρ,1,,n1)𝟙,𝒥[ntr])\displaystyle\mathbb{Jor}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n-1})\otimes\mathds{1},\mathscr{J}[\mathscr{E}_{n}\circ{\rm tr}]\right)
=\displaystyle= 𝕁𝕠𝕣(\Yinyang(ρ,1,,n1)𝟙,𝟙𝒥[n]),\displaystyle\mathbb{Jor}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n-1})\otimes\mathds{1},\mathds{1}\otimes\mathscr{J}[\mathscr{E}_{n}]\right),

it follows via induction that \Yinyang(ρ,1,,n)=Rn\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})=R_{n} (the final equality follows from item iii of Lemma 5.6). As such, the \Yinyang-function recovers the pseudo-density matrix formalism for systems of qubits first introduced in [FJV15], and moreover, Theorems 8.14 and 8.18 yield new formulas for RnR_{n}.

9 Extracting dynamics from a pseudo-density matrix

In this section, we derive an explicit expression for the inverse of the \Yinyang-function restricted to a suitably nice subset of nn-step processes. In particular, given a pseudo-density matrix τ\tau, we show how to identify τ\tau with an nn-step process (ρ,1,,n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) such that \Yinyang(ρ,1,,n)=τ\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})=\tau. This solves an open problem from [jia2023], where it is stated "Another interesting and closely relevant open question is, for a given PDO (pseudo-density matrix), how to find a quantum process to realize it.".

Notation 9.1.

Given an element τ𝒜0𝒜n\tau\in\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n} with tr0(τ)𝒜0{\rm tr}_{0}(\tau)\in\mathcal{A}_{0} invertible, we let Xτ𝒜0𝒜nX_{\tau}\in\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n} be the unique element such that τ=𝕁𝕠𝕣((tr(τ0)𝟙),Xτ)\tau=\mathbb{Jor}(({\rm tr}(\tau_{0})\otimes\mathds{1}),X_{\tau}).

Notation 9.2.

Given an (n+1)(n+1)-tuple (𝒜0,,𝒜n)(\mathcal{A}_{0},...,\mathcal{A}_{n}) of multi-matrix algebras, we let 𝒫+(𝒜0,,𝒜n)𝒫(𝒜0,,𝒜n)\mathscr{P}_{+}(\mathcal{A}_{0},...,\mathcal{A}_{n})\subset\mathscr{P}(\mathcal{A}_{0},...,\mathcal{A}_{n}) denote the subset given by

𝒫+(𝒜0,,𝒜n)={(ρ,1,,n)𝒫(𝒜0,,𝒜n)|ρiis invertible for i=0,,n1},\mathscr{P}_{+}(\mathcal{A}_{0},...,\mathcal{A}_{n})=\left\{(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathscr{P}(\mathcal{A}_{0},...,\mathcal{A}_{n})\hskip 2.84526pt|\hskip 2.84526pt\rho_{i}\hskip 2.84526pt\text{is invertible for $i=0,...,n-1$}\right\},

and we let 𝒯(𝒜0𝒜n)𝒯(𝒜0𝒜n)\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n})\subset\mathscr{T}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}) be the subset consisting of elements τ𝒯(𝒜0𝒜n)\tau\in\mathscr{T}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}) such that

  • τi𝒯(𝒜i1𝒜i)\tau_{i}\in\mathscr{T}_{*}(\mathcal{A}_{i-1}\otimes\mathcal{A}_{i}) for i=1,,ni=1,...,n, where 𝒯(𝒜i1𝒜i)\mathscr{T}_{*}(\mathcal{A}_{i-1}\otimes\mathcal{A}_{i}) is as defined by (4.15).

  • Xτ=𝕁𝕠𝕣(Xτ1𝟙,,𝟙Xτn)X_{\tau}=\mathbb{Jor}(X_{\tau_{1}}\otimes\mathds{1},...,\mathds{1}\otimes X_{\tau_{n}})

where τi=tri1,i(τ)\tau_{i}=\text{tr}_{i-1,i}(\tau) and tri1,i:𝒜0𝒜n𝒜i1𝒜i\text{tr}_{i-1,i}:\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}\longrightarrow\mathcal{A}_{i-1}\otimes\mathcal{A}_{i} is the partial trace map.

Remark 9.3.

We have reason to believe that the two conditions defining 𝒯(𝒜0𝒜n)\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}) are in fact equivalent, but we have yet to find a proof.

We now are going to prove that the \Yinyang-function restricted to 𝒫+(𝒜0,,𝒜n)\mathscr{P}_{+}(\mathcal{A}_{0},...,\mathcal{A}_{n}) defines a bijection onto 𝒯(𝒜0𝒜n)\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}), and we will also give a precise formula for the inverse. Before doing so however, we first prove that the image of the \Yinyang-function restricted to 𝒫+(𝒜0,,𝒜n)\mathscr{P}_{+}(\mathcal{A}_{0},...,\mathcal{A}_{n}) indeed lies in 𝒯(𝒜0𝒜n)\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}).

Proposition 9.4.

Let (ρ,1,,n)𝒫+(𝒜0,,𝒜n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathscr{P}_{+}(\mathcal{A}_{0},...,\mathcal{A}_{n}). Then \Yinyang(ρ,1,,n)𝒯(𝒜0𝒜n)\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}).

Proof.

Let (ρ,1,,n)𝒫+(𝒜0,,𝒜n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathscr{P}_{+}(\mathcal{A}_{0},...,\mathcal{A}_{n}), let τ=\Yinyang(ρ,1,,n)\tau=\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}), and let

τi=tri1,i(\Yinyang(ρ,1,,n))𝒜i1𝒜i\tau_{i}=\text{tr}_{i-1,i}\left(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\right)\in\mathcal{A}_{i-1}\otimes\mathcal{A}_{i}

for all i[n]i\in[n]. Then it follows from the multi-marginal property of the \Yinyang-function that τi=\Yinyang(ρi1,i)\tau_{i}=\text{\Yinyang}(\rho_{i-1},\mathscr{E}_{i}), and since ρi1\rho_{i-1} is invertible by the definition of 𝒫+(𝒜0,,𝒜n)\mathscr{P}_{+}(\mathcal{A}_{0},...,\mathcal{A}_{n}), it follows that (ρi1,i)𝒫+(𝒜i1,𝒜i)(\rho_{i-1},\mathscr{E}_{i})\in\mathscr{P}_{+}(\mathcal{A}_{i-1},\mathcal{A}_{i}), thus τi𝒯(𝒜i1𝒜i)\tau_{i}\in\mathscr{T}_{*}(\mathcal{A}_{i-1}\otimes\mathcal{A}_{i}) for i=1,,ni=1,...,n by Lemma 4.16, showing that τ\tau satisfies the first condition for being an element of 𝒯(𝒜0𝒜n)\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}). Moreover, it follows from Theorem 4.17 that i=𝒥1(Xτi)\mathscr{E}_{i}=\mathscr{J}^{-1}(X_{\tau_{i}}) for all i[n]i\in[n], thus

τ\displaystyle\tau =\displaystyle= \Yinyang(ρ,1,,n)\displaystyle\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})
=\displaystyle= \Yinyang(tr0(τ),𝒥1(Xτ1),,𝒥1(Xτn))\displaystyle\text{\Yinyang}({\rm tr}_{0}(\tau),\mathscr{J}^{-1}(X_{\tau_{1}}),...,\mathscr{J}^{-1}(X_{\tau_{n}}))
=(8.19)\displaystyle\overset{\eqref{CFX8771}}{=} 𝕁𝕠𝕣(tr0(τ)𝟙,𝒥(𝒥1(Xτ1))𝟙,,𝟙𝒥(𝒥1(Xτn)))\displaystyle\mathbb{Jor}\left({\rm tr}_{0}(\tau)\otimes\mathds{1},\mathscr{J}\left(\mathscr{J}^{-1}(X_{\tau_{1}})\right)\otimes\mathds{1},...,\mathds{1}\otimes\mathscr{J}\left(\mathscr{J}^{-1}(X_{\tau_{n}})\right)\right)
=\displaystyle= 𝕁𝕠𝕣(tr0(τ)𝟙,Xτ1𝟙,,𝟙Xτn)\displaystyle\mathbb{Jor}\left({\rm tr}_{0}(\tau)\otimes\mathds{1},X_{\tau_{1}}\otimes\mathds{1},...,\mathds{1}\otimes X_{\tau_{n}}\right)
=\displaystyle= 𝕁𝕠𝕣((tr0(τ)𝟙),𝕁𝕠𝕣(Xτ1𝟙,,𝟙Xτn))\displaystyle\mathbb{Jor}\left(({\rm tr}_{0}(\tau)\otimes\mathds{1}),\mathbb{Jor}\left(X_{\tau_{1}}\otimes\mathds{1},...,\mathds{1}\otimes X_{\tau_{n}}\right)\right)
\displaystyle\implies Xτ=𝕁𝕠𝕣(Xτ1𝟙,,𝟙Xτn),\displaystyle X_{\tau}=\mathbb{Jor}\left(X_{\tau_{1}}\otimes\mathds{1},...,\mathds{1}\otimes X_{\tau_{n}}\right),

thus τ\tau satisfies the second condition for being an element of 𝒯(𝒜0𝒜n)\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}). It then follows that τ=\Yinyang(ρ,1,,n)𝒯(𝒜0𝒜n)\tau=\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}), as desired.

Theorem 9.5.

Let (𝒜0,,𝒜n)(\mathcal{A}_{0},...,\mathcal{A}_{n}) be an (n+1)(n+1)-tuple of multi-matrix algebras, let

𝔖:𝒫+(𝒜0,,𝒜n)𝒯(𝒜0𝒜n)\mathfrak{S}:\mathscr{P}_{+}(\mathcal{A}_{0},...,\mathcal{A}_{n})\longrightarrow\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n})

be the restriction of the \Yinyang-function to 𝒫+(𝒜0,,𝒜n)\mathscr{P}_{+}(\mathcal{A}_{0},...,\mathcal{A}_{n}), and let

:𝒯(𝒜0𝒜n)𝒫+(𝒜0,,𝒜n)\mathfrak{C}:\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n})\longrightarrow\mathscr{P}_{+}(\mathcal{A}_{0},...,\mathcal{A}_{n})

be the map given by

(τ)=(tr0(τ),𝒥1(Xτ1),,𝒥1(Xτn)),\mathfrak{C}(\tau)=\left(\emph{tr}_{0}(\tau),\mathscr{J}^{-1}(X_{\tau_{1}}),...,\mathscr{J}^{-1}(X_{\tau_{n}})\right), (9.6)

where τi=tri1,i(τ)\tau_{i}=\emph{tr}_{i-1,i}\left(\tau\right) and Xτi𝒜i1𝒜iX_{\tau_{i}}\in\mathcal{A}_{i-1}\otimes\mathcal{A}_{i} is defined via (4.13) for all i{1,,n}i\in\{1,...,n\}. Then 𝔖\mathfrak{S} is a bijection, and =𝔖1\mathfrak{C}=\mathfrak{S}^{-1}.

Proof.

Let (ρ,1,,n)𝒫+(𝒜0,,𝒜n)(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\in\mathscr{P}_{+}(\mathcal{A}_{0},...,\mathcal{A}_{n}), let τ=𝔖(ρ,1,,1)\Yinyang(ρ,1,,n)\tau=\mathfrak{S}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{1})\equiv\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}), and let

τi=tri1,i(𝔖(ρ,1,,n))𝒜i1𝒜i\tau_{i}=\text{tr}_{i-1,i}\left(\mathfrak{S}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\right)\in\mathcal{A}_{i-1}\otimes\mathcal{A}_{i}

for all i{1,,n}i\in\{1,...,n\}. Then we have seen in the proof of Proposition 9.4 that 𝒥1(Xτi)=i\mathscr{J}^{-1}(X_{\tau_{i}})=\mathscr{E}_{i}, thus

(𝔖)(ρ,1,,n)\displaystyle(\mathfrak{C}\circ\mathfrak{S})(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}) =\displaystyle= (𝔖(ρ,1,,n))\displaystyle\mathfrak{C}\left(\mathfrak{S}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})\right)
=\displaystyle= (\Yinyang(ρ,1,,n))\displaystyle\mathfrak{C}(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}))
=\displaystyle= (tr0(\Yinyang(ρ,1,,n)),𝒥1(Xτ1),,𝒥1(Xτn))\displaystyle\left(\text{tr}_{0}(\text{\Yinyang}(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n})),\mathscr{J}^{-1}(X_{\tau_{1}}),...,\mathscr{J}^{-1}(X_{\tau_{n}})\right)
=\displaystyle= (ρ,1,,n).\displaystyle(\rho,\mathscr{E}_{1},...,\mathscr{E}_{n}).

It then follows that 𝔖=id\mathfrak{C}\circ\mathfrak{S}=\mathrm{id}, thus 𝔖\mathfrak{S} is injective.

Now let τ𝒯(𝒜0𝒜n)\tau\in\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}), so that

(τ)=(tr0(τ),𝒥1(Xτ1),,𝒥1(Xτn)).\mathfrak{C}(\tau)=\left({\rm tr}_{0}(\tau),\mathscr{J}^{-1}(X_{\tau_{1}}),...,\mathscr{J}^{-1}(X_{\tau_{n}})\right).

From the definition of 𝒯(𝒜0𝒜n)\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}) we have τi𝒯(𝒜i1,𝒜i)\tau_{i}\in\mathscr{T}_{*}(\mathcal{A}_{i-1},\mathcal{A}_{i}) for all i[n]i\in[n], from which it follows that 𝒥1(Xτi)\mathscr{J}^{-1}(X_{\tau_{i}}) is CPTP for all i[n]i\in[n] and also that tr0(τ){\rm tr}_{0}(\tau) is invertible, thus (τ)𝒫+(𝒜0,,𝒜n)\mathfrak{C}(\tau)\in\mathscr{P}_{+}(\mathcal{A}_{0},...,\mathcal{A}_{n}). We then have

(𝔖)(τ)\displaystyle(\mathfrak{S}\circ\mathfrak{C})(\tau) =\displaystyle= \Yinyang(tr0(τ),𝒥1(Xτ1),,𝒥1(Xτn))\displaystyle\text{\Yinyang}\left({\rm tr}_{0}(\tau),\mathscr{J}^{-1}(X_{\tau_{1}}),...,\mathscr{J}^{-1}(X_{\tau_{n}})\right)
=(8.19)\displaystyle\overset{\eqref{CFX8771}}{=} 𝕁𝕠𝕣(tr0(τ)𝟙,𝒥(𝒥1(Xτ1))𝟙,,𝟙𝒥(𝒥1(Xτn)))\displaystyle\mathbb{Jor}\left({\rm tr}_{0}(\tau)\otimes\mathds{1},\mathscr{J}\left(\mathscr{J}^{-1}(X_{\tau_{1}})\right)\otimes\mathds{1},...,\mathds{1}\otimes\mathscr{J}\left(\mathscr{J}^{-1}(X_{\tau_{n}})\right)\right)
=\displaystyle= 𝕁𝕠𝕣(tr0(τ)𝟙,Xτ1𝟙,,𝟙Xτn)\displaystyle\mathbb{Jor}\left({\rm tr}_{0}(\tau)\otimes\mathds{1},X_{\tau_{1}}\otimes\mathds{1},...,\mathds{1}\otimes X_{\tau_{n}}\right)
=\displaystyle= 𝕁𝕠𝕣((tr0(τ)𝟙),𝕁𝕠𝕣(Xτ1𝟙,,𝟙Xτn))\displaystyle\mathbb{Jor}\left(({\rm tr}_{0}(\tau)\otimes\mathds{1}),\mathbb{Jor}\left(X_{\tau_{1}}\otimes\mathds{1},...,\mathds{1}\otimes X_{\tau_{n}}\right)\right)
=\displaystyle= 𝕁𝕠𝕣((tr0(τ)𝟙),Xτ)\displaystyle\mathbb{Jor}\left(({\rm tr}_{0}(\tau)\otimes\mathds{1}),X_{\tau}\right)
=\displaystyle= τ,\displaystyle\tau,

where the second to last equality follows from the fact that τ𝒯(𝒜0𝒜n)\tau\in\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}). It then follows that 𝔖=id\mathfrak{S}\circ\mathfrak{C}=\mathrm{id}, thus concluding the proof. ∎

Remark 9.7.

As the inverse 𝒥1:𝒜Hom(𝒜,)\mathscr{J}^{-1}:\mathcal{A}\otimes\mathcal{B}\longrightarrow\text{Hom}(\mathcal{A},\mathcal{B}) of the Jamiołowski isomorphism admits the explicit formula

𝒥1(τ)(ρ)=tr𝒜((ρ𝟙)τ),\mathscr{J}^{-1}(\tau)(\rho)=\text{tr}_{\mathcal{A}}((\rho\otimes\mathds{1})\tau),

and since there are algorithms to compute XτX_{\tau} with τ𝒯(𝒜0𝒜n)\tau\in\mathscr{T}_{*}(\mathcal{A}_{0}\otimes\cdots\otimes\mathcal{A}_{n}), the formula (9.6) for the inverse of the \Yinyang-function is easily implementable in symbolic computing software.

References