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Unlimited One-Way Steering

Pavel Sekatski [email protected]    Florian Giraud    Roope Uola    Nicolas Brunner Department of Applied Physics, University of Geneva, Switzerland
Abstract

This work explores the asymmetry of quantum steering in a setup using high-dimensional entanglement. We construct entangled states with the following properties: (i)(i) one party (Alice) can never steer the state of the other party (Bob), considering the most general measurements, and (ii)(ii) Bob can strongly steer the state of Alice, demonstrating genuine high-dimensional steering. In other words, Bob can convince Alice that they share an entangled state of arbitrarily high Schmidt number, while Alice can never convince Bob that the state is even simply entangled. In this sense, one-way steering can become unlimited. A key result for our construction is a condition for the joint measurability of all high-dimensional measurements subjected to the combined effect of noise and loss, which is of independent interest.

I Introduction

Nonlocality is among the central features of quantum theory. This effect manifests at different levels. On the one hand, in the mathematical structure of the theory (i.e. the Hilbert space) via the notion of entanglement Horodecki et al. (2009); Gühne and Tóth (2009). On the other hand, the measurement statistics of local measurements performed on entangled states can feature strong nonlocal correlations, which are incompatible with any local model Brunner et al. (2014). Initially, believed to be two facets of the same phenomenon, it is now clear that entanglement and quantum Bell nonlocality are in fact inherently different, see e.g. Werner (1989); Barrett (2002).

Another perspective on quantum nonlocality is provided by the notion of quantum steering, see Cavalcanti and Skrzypczyk (2016); Uola et al. (2020) for reviews. This effect, formalized by Wiseman, Jones and Doherty Wiseman et al. (2007), takes its roots in the early works of Einstein-Podolsky-Rosen Einstein et al. (1935) and Schrödinger Schrödinger (1935, 1936). Here, an untrusted party (Bob) wants to convince another party (Alice) that they share an entangled state. By demonstrating his ability to remotely steer Alice’s local state in different measurement basis, Bob can convince Alice. Steering is thus an inherently asymmetrical task, contrary to entanglement and Bell nonlocality.

Interestingly, the asymmetry of steering is also observed at the level of quantum states. Specifically, there exist entangled states that lead to steering from Bob to Alice, but not the other way around Bowles et al. (2014); Bob can convince Alice that the shared state is entangled, while Alice can never convince Bob, even if she would use all possible local measurements. This effect, coined “one-way steering” has attracted considerable attention in recent years, with many examples in low-dimensional (mostly two-qubit) Bowles et al. (2014); Skrzypczyk et al. (2014); Quintino et al. (2015); Bowles et al. (2016); Zeng (2022); Márton et al. (2021) and continuous variable Gaussian systems Midgley et al. (2010); Olsen (2013) as well as experimental demonstrations, see e.g. Händchen et al. (2012); Wollmann et al. (2016); Sun et al. (2016); Tischler et al. (2018).

A relevant question is whether there exists different forms of one-way steering, and whether some are stronger than others. So far, this question has not been discussed, due to the lack of an appropriate measure for steering in this context. Here, we tackle this problem, taking advantage of the recently introduced notion of genuine high-dimensional steering Designolle et al. (2021), see also Designolle (2022); de Gois et al. (2022). This allows for a dimensional quantification of steering, specifically to lower bound the Schmidt number of an entangled state in a steering scenario. We then ask whether there exist entangled state with the following properties: (i) Alice cannot steer the state of Bob (even when allowing for all possible local measurements), and (ii) Bob can steer Alice’s state strongly, i.e. ensuring the presence of high-dimensional entanglement (Schmidt number at least dd). We answer this question in the affirmative by constructing a family of entangled states (of dimension d×(d+1)d\times(d+1)) with the above two properties for any finite dimension dd. To do so, we exploit the connection between steering and measurement incompatibility Quintino et al. (2014); Uola et al. (2014, 2015); Kiukas et al. (2017), and also its recent generalisation Jones et al. (2022) to high-dimensional steering and the concept of nn-simulability of measurements Ioannou et al. (2022). Notably, we give a sufficient condition for the joint measurability of the set of all POVMs subjected to the combined effect of noise and loss.

II Question and main result

We consider a scenario where two distant parties, Alice and Bob, share an entangled state ϱAB\varrho_{AB}. Each party performs local measurements, represented by sets of positive-operator-valued measures (POVMs). Specifically, Alice’s measurements are described by a set of POVMs {Aa|x}\{A_{a|x}\}, where xx denotes the measurement choice and aa the outcome, with the properties that Aa|x0A_{a|x}\geq 0 and aAa|x=𝟙\sum_{a}A_{a|x}=\openone for all aa and xx. Similarly, for Bob we define the set of POVMs {Bb|y}\{B_{b|y}\}.

As the shared state is entangled, the effect of each party’s local measurements is to remotely prepare (steer) the other party’s state. This is described via two so-called state assemblages: {σb|yA}\{\sigma^{A}_{b|y}\} describe the states of Alice’s system conditioned on Bob’s measurement while {σa|xB}\{\sigma^{B}_{a|x}\} are Bob’s states given Alice’s measurement. These are given by

σb|yA=trB[(𝟙𝔹𝕓|𝕪)ϱ𝔸𝔹],\displaystyle\sigma^{A}_{b|y}=\text{tr}_{B}[(\openone\otimes B_{b|y})\varrho_{AB}]\,, (1)
σa|xB=trA[(Aa|x𝟙)ϱ𝔸𝔹].\displaystyle\sigma^{B}_{a|x}=\text{tr}_{A}[(A_{a|x}\otimes\openone)\varrho_{AB}]\,. (2)

The main question we address here is how different these two state assemblages can become, in other words how asymmetric the effect of steering can be. Loosely speaking we are looking for an entangled state ϱAB\varrho_{AB} such that one of the state assemblages, say {σb|yA}\{\sigma^{A}_{b|y}\}, is classical (in the sense that it can never lead to steering), while the other assemblage {σa|xB}\{\sigma^{B}_{a|x}\} is highly non-classical (in the sense that it exhibits strong steering, witnessing high entanglement dimensionality).

Before defining the problem more precisely, let us first observe that we are looking for some entangled states that are high-dimensional and asymmetrical. Obviously, if the state ϱAB\varrho_{AB} would be symmetrical under the exchange of Alice and Bob, any assemblage obtainable in one direction can also be obtained the other way around. Moreover, the state ϱAB\varrho_{AB} should feature a high entanglement dimensionality, as quantified here via the Schmidt number Terhal and Horodecki (2000); Sanpera et al. (2001): the Schmidt number of ϱAB\varrho_{AB} is the minimum nn such that there exists a decomposition ϱAB=jpj|ψjψj|\varrho_{AB}=\sum_{j}p_{j}\ket{\psi_{j}}\bra{\psi_{j}} where all |ψj\ket{\psi_{j}} are pure entangled states of Schmidt rank at most nn.

More formally, we are looking for entangled states with the following two properties:

  1. 1.

    The assemblage {σb|yA}\{\sigma^{A}_{b|y}\} admits a local hidden state (LHS) model Wiseman et al. (2007):

    σb|yA=p(b|y)λp(λ|b,y)σλb,y.\sigma^{A}_{b|y}=p(b|y)\sum_{\lambda}p(\lambda|b,y)\sigma_{\lambda}\quad\forall b,y\,. (3)

    Importantly, this should hold for any set of measurements for Bob {Bb|y}\{B_{b|y}\}. Thus, no steering from Bob to Alice is possible, implying that Bob can never convince Alice that the shared state ϱAB\varrho_{AB} is entangled.

  2. 2.

    The assemblage {σa|xB}\{\sigma^{B}_{a|x}\} is not (d1)(d-1)-preparable Designolle et al. (2021), i.e.

    σa|xBλp(λ)trA[(Ma|xλ𝟙)ρ𝔸𝔹λ]\sigma^{B}_{a|x}\neq\sum_{\lambda}p(\lambda)\text{tr}_{A}[(M^{\lambda}_{a|x}\otimes\openone)\rho^{\lambda}_{AB}] (4)

    with p(λ)p(\lambda) an arbitrary probability distribution, Ma|xλM^{\lambda}_{a|x} arbitrary measurements (in d\mathbb{C}^{d}), and all ρABλ\rho^{\lambda}_{AB} being arbitrary states of Schmidt number (at most) d1d-1. Thus, genuine dd-dimensional steering is demonstrated, so that Alice can convince Bob that the shared state ϱAB\varrho_{AB} has large Schmidt number dd. For short, we say that ϱAB\varrho_{AB} is dd-steerable.

In the following, we construct a family of entangled states in dimension d×(d+1)d\times(d+1) with the above two properties, for any dd. This shows that one-way steering can become unlimited, in the sense of being maximally asymmetrical. Specifically, we start from a maximally entangled two-qudit state |ϕd+=1dk=0d1|k,k\ket{\phi^{+}_{d}}=\frac{1}{\sqrt{d}}\sum_{k=0}^{d-1}\ket{k,k} (with the notation Φd+=|ϕd+ϕd+|\Phi_{d}^{+}=\outerproduct{\phi^{+}_{d}}{\phi^{+}_{d}}) and apply successively a white noise (depolarizing) channel followed by a loss channel, defined by

𝒲p\displaystyle\mathcal{W}_{p} :ρpρ+(1p)tr[ρ]𝟙dd,\displaystyle:\rho\mapsto p\,\rho+(1-p)\text{tr}[\rho]\frac{\mathds{1}_{d}}{d}\,, (5)
η\displaystyle\mathcal{L}_{\eta} :ρηρ+(1η)tr[ρ]| ø  ø |.\displaystyle:\rho\mapsto\eta\,\rho+(1-\eta)\text{tr}[\rho]\outerproduct{\textup{\,\o\,}}{\textup{\,\o\,}}. (6)

The depolarizing (loss) channel replaces the original state with a maximally mixed state (vacuum state | ø  ø |\outerproduct{\textup{\,\o\,}}{\textup{\,\o\,}}) with probability 1p1-p (1η1-\eta). Note that the vacuum state represents an additional ”level”, orthogonal to the input Hilbert space, so that the output state has dimension d+1d+1. After the two channels, we obtain the final state

ϱAB(η,p)\displaystyle\varrho_{AB}^{(\eta,p)} =id(η𝒲p)[Φ+]\displaystyle=\mathrm{id}\otimes(\mathcal{L}_{\eta}\circ\mathcal{W}_{p})[\Phi^{+}] (7)
=ηpΦd++η(1p)𝟙d𝟙dd2+(1η)𝟙dd| ø  ø |,\displaystyle=\eta p\,\Phi^{+}_{d}+\eta(1-p)\frac{\mathds{1}_{d}\otimes\mathds{1}_{d}}{d^{2}}+(1-\eta)\frac{\mathds{1}_{d}}{d}\otimes\outerproduct{\textup{\,\o\,}}{\textup{\,\o\,}}\,,

which is of dimension d×(d+1)d\times(d+1).

Result 1. The states ϱAB(η,p)\varrho_{AB}^{(\eta,p)} defined in Eq. (7) satisfy the following properties

(i) ϱAB(η,p)\varrho_{AB}^{(\eta,p)} is dd-steerable from Alice to Bob if

p>ddd+11d1.p>\frac{d\sqrt{\frac{d}{d+1}}-1}{d-1}. (8)

(ii) ϱAB(η,p)\varrho_{AB}^{(\eta,p)} is unsteerable from Bob to Alice if

η(1p)d1.\eta\leq(1-p)^{d-1}. (9)

Therefore, for any dd there are parameter values (η,p)(\eta,p) such that conditions (i)(i) and (ii)(ii) are both satisfied, thus demonstrating unlimited one-way steering. This regime corresponds to low noise and high loss.

III Proof outline

III.1 Statement (i)

The first step is to show that the loss channel can be removed, because when acting on the trusted party it does not affect the steering properties of the underlying assemblage. This statement is formalized by the following lemma.

Lemma 1. Consider an assemblage {σa|x}a,x\{\sigma_{a|x}\}_{a,x} and a loss channel η\mathcal{L}_{\eta} for η>0\eta>0. The assemblage {σa|x=η[σa|x]}a,x\left\{\sigma^{\prime}_{a|x}=\mathcal{L}_{\eta}[\sigma_{a|x}]\right\}_{a,x} is nn-steerable if and only if {σa|x}a,x\{\sigma_{a|x}\}_{a,x} is.

The full proof is in Appendix A. The idea is to show that if the assemblage {σa|x}a,x\left\{\sigma^{\prime}_{a|x}\right\}_{a,x} is nn-preparable, one can construct an nn-preparable model for {σa|x}a,x\{\sigma_{a|x}\}_{a,x}. The converse holds trivially (via application of the loss channel).

Hence, to prove that ρAB(η,p)\rho^{(\eta,p)}_{AB} is dd-steerable from Bob to Alice it is sufficient to show that the isotropic state ρAB(1,p)=pΦd++(1p)𝟙d𝟙dd2\rho^{(1,p)}_{AB}=p\,\Phi^{+}_{d}+(1-p)\frac{\mathds{1}_{d}\otimes\mathds{1}_{d}}{d^{2}} is, which is to be expected for weak enough noise. Indeed, it was shown in Jones et al. (2022) that this is the case for pp satisfying Eq. (8), concluding the proof of statement (i)(i). Note that, in the above, dd-steerability is demonstrated when using the set of all measurements. Instead, one could use a simpler (and practical) witness based on a pair of MUBs, certifying dd-steerability of ρAB(1,p)\rho_{AB}^{(1,p)} Designolle et al. (2021) for

p(d+d1)d11(d1)(d1+1).p\geq\frac{(d+\sqrt{d}-1)\sqrt{d-1}-1}{(d-1)(\sqrt{d-1}+1)}. (10)

In turn, one shows dd-steerability of ρAB(η,p)\rho_{AB}^{(\eta,p)} for any η>0\eta>0, via the filtering procedure discussed in the proof of the lemma 2, which removes the vacuum component | ø \ket{\textup{\,\o\,}}.

III.2 Statement (ii)

Here we exploit the connection Uola et al. (2015) between the notions of steering and measurement compatibility, formally introduced in the next section. In particular, for any CPTP map \mathcal{E} it is known that a state assemblage given by the operators σb|y=tr[(𝟙Mb|y)(id[Φ+])]\sigma_{b|y}=\tr[(\mathds{1}\otimes M_{b|y})(\mathrm{id}\otimes\mathcal{E}[\Phi^{+}])] admits a LHS model iff the set of measurements {[Mb|y]}\{\mathcal{E}^{*}[M_{b|y}]\} is jointly measurable. Hence, the state (id[Φ+])(\mathrm{id}\otimes\mathcal{E}[\Phi^{+}]) is unsteerable if and only if the channel \mathcal{E} is incompatibility breaking Heinosaari et al. (2015), that is for any measurement assemblage {Mb|y}\{M_{b|y}\} the resulting assemblage {[Mb|y]}\{\mathcal{E}^{*}[M_{b|y}]\} is jointly measurable. Thus, statement (ii)(ii) is implied by the following lemma.

Lemma 2 The channel =η𝒲p\mathcal{E}=\mathcal{L}_{\eta}\circ\mathcal{W}_{p} is incompatibility breaking if η(1p)d1\eta\leq(1-p)^{d-1}.

The proof of the lemma is rather straightforward once we have solved the question of joint measurability of all measurements subject to white noise and loss in a given finite dimension dd. This is a question of independent interest, and will be exposed in the next section. Then these results are used in Appendix C to give a formal proof of Lemma 2.

IV Compatibility of all measurements with losses and noise

Measurement incompatibility is a property of a set of measurements which cannot be performed simultaneously on a single copy of a system. Formally, a set of measurements {Ma|x}\{M_{a|x}\} is called incompatible if there exists no ”parent” POVM {Eλ}\{E_{\lambda}\} and classical post-processings {p(a|x,λ)}\{p(a|x,\lambda)\} such that

Ma|x=λp(a|x,λ)Eλ.\displaystyle M_{a|x}=\sum_{\lambda}p(a|x,\lambda)E_{\lambda}. (11)

Sets of measurements allowing for such a model are called jointly measurable; see e.g. Heinosaari et al. (2016); Gühne et al. (2023) for reviews.

Our focus is on the (in)compatibility of measurements which are noisy and lossy. For any POVM {Ma}\{M_{a}\} with mm outputs a=1,,ma=1,\dots,m acting on a system of dimension dd, we define its imperfect version {M¯a(η,p)}\{\bar{M}_{a}^{(\eta,p)}\} as a POVM with m+1m+1 elements given by

M¯a(η,p)={ηpMa+η(1p)(trMa)𝟙dda=1,,m(1η)𝟙da= ø .\bar{M}_{a}^{(\eta,p)}=\begin{cases}\eta p\,M_{a}+\eta(1-p)(\tr M_{a})\frac{\mathds{1}_{d}}{d}&a=1,\dots,m\\ (1-\eta)\mathds{1}_{d}&a=\textup{\,\o\,}\end{cases}. (12)

The noisy measurement apparatus behaves like the ideal one with probability ηp\eta p, produces a random outcome with probability η(1p)\eta(1-p), and does not click with probability 1η1-\eta (formally this corresponds to the no-click outcome a= ø a=\textup{\,\o\,}). We are interested in knowing for which values of (η,p)(\eta,p) all measurements in a given dimension dd become compatible. The following result gives a sufficient condition.

Result 2. The set of all POVMs {M¯a|U(η,p)}U\{\bar{M}_{a|U}^{(\eta,p)}\}_{U} on d\mathds{C}^{d} is jointly measurable if

η(1p)d1.\eta\leq(1-p)^{d-1}. (13)

Proof: To prove the result we present an explicit construction, inspired by Hirsch et al. (2013), able to simulate all imperfect POVMs with noise parameters fulfilling Eq. (13). Note that it is sufficient to show that it can simulate all rank-one POVMs {Ma=αa|φaφa|}\{M_{a}=\alpha_{a}\outerproduct{\varphi_{a}}{\varphi_{a}}\} with αa=d\sum\alpha_{a}=d, since they can be post-processed to simulate all other measurements. The post-processing is done by mixing POVM elements, i.e. coarse-graining the corresponding outputs, by linearity it is consistent with the noisification of the measurements defined in Eq. (12).

We take the parent POVM to be the covariant one – the continuous-valued measurement with the density

E𝒛=d|𝒛𝒛|E_{\bm{z}}=d\outerproduct{\bm{z}}{\bm{z}} (14)

where |𝒛=k=0d1zk|k\ket{\bm{z}}=\sum_{k=0}^{d-1}z_{k}\ket{k} with 𝒛d\bm{z}\in\mathds{C}^{d} and |𝒛|2=1|\bm{z}|^{2}=1, and d𝒛\differential\bm{z} is the invariant (under unitary transformations) measure over pure quantum states |𝒛\ket{\bm{z}} such that d𝒛|𝒛𝒛|=1d𝟙d\int\differential\bm{z}\outerproduct{\bm{z}}{\bm{z}}=\frac{1}{d}\mathds{1}_{d}.

To define the response function p(a|𝒛)p(a|\bm{z}) simulating a notified rank-one POVM {αa|φaφa|}a\{\alpha_{a}\outerproduct{\varphi_{a}}{\varphi_{a}}\}_{a} we proceed in two steps. First one samples a possible output aa from the probability distribution αad\frac{\alpha_{a}}{d}. Given aa and the corresponding state |φa\ket{\varphi_{a}}, one simulates a two outcome POVM {Na(a),N ø (a)}\{N_{a}^{(a)},N_{\textup{\,\o\,}}^{(a)}\} by using the deterministic response functions

p(a)(a|𝒛)={1|φa|𝒛|2t0otherwisep^{(a)}(a|\bm{z})=\begin{cases}1&|\innerproduct{\varphi_{a}}{\bm{z}}|^{2}\geq t\\ 0&\text{otherwise}\end{cases} (15)

and p(a)( ø |𝒛)=1p(a)(a|𝒛)p^{(a)}(\textup{\,\o\,}|\bm{z})=1-p^{(a)}(a|\bm{z}), with a parameter t[0,1]t\in[0,1]. Let us now compute the resulting operators

Na(a)=d𝒛p(a)(a|𝒛)E𝒛N_{a}^{(a)}=\int\differential\bm{z}\,p^{(a)}(a|\bm{z})E_{\bm{z}} (16)

and N ø (a)=𝟙dNa(a)N_{\textup{\,\o\,}}^{(a)}=\mathds{1}_{d}-N_{a}^{(a)}. One notes that Na(a)=UNa(a)UN_{a}^{(a)}=UN_{a}^{(a)}U^{\dagger} is invariant under all unitary transformations UU which leave the state |φa\ket{\varphi_{a}} unchanged U|φa=|φaU\ket{\varphi_{a}}=\ket{\varphi_{a}}, since both the measure d𝒛\differential\bm{z} and the response function p(a)(a|𝒛)p^{(a)}(a|\bm{z}) are invariant under such transformations. In other words, any such unitary commutes with our operator [U,Na(a)]=0[U,N_{a}^{(a)}]=0. It follows that Na(a)N_{a}^{(a)} has |φa\ket{\varphi_{a}} for an eigenstate, and is furthermore proportional to identity on the orthogonal subspace. I.e. it is of the form

Na(a)=A¯d(t)|φaφa|+B¯d(t)𝟙d|φaφa|d1,N_{a}^{(a)}=\bar{A}_{d}(t)\outerproduct{\varphi_{a}}{\varphi_{a}}+\bar{B}_{d}(t)\frac{\mathds{1}_{d}-\outerproduct{\varphi_{a}}{\varphi_{a}}}{d-1}, (17)

with the scalar functions that can be computed as

A¯d(t)=trNa(a)|φaφa|=dd𝒛p(a)(a|𝒛)|φa|𝒛|2A¯d(t)+B¯d(t)=trNa(a)=dd𝒛p(a)(a|𝒛).\begin{split}\bar{A}_{d}(t)=\tr N_{a}^{(a)}\outerproduct{\varphi_{a}}{\varphi_{a}}&=d\int\differential\bm{z}\,p^{(a)}(a|\bm{z})|\innerproduct{\varphi_{a}}{\bm{z}}|^{2}\\ \bar{A}_{d}(t)+\bar{B}_{d}(t)=\tr N_{a}^{(a)}&=d\int\differential\bm{z}\,p^{(a)}(a|\bm{z}).\end{split} (18)

These integrals are straightforward to compute. In the Appendix B we show that they give A¯d(t)=(1t)d1((d1)t+1)\bar{A}_{d}(t)=(1-t)^{d-1}\left((d-1)t+1\right) and T¯d(t)A¯d(t)+B¯d(t)=d(1t)d1\bar{T}_{d}(t)\equiv\bar{A}_{d}(t)+\bar{B}_{d}(t)=d(1-t)^{d-1}.

Finally, averaging over the sampled value aa we see that this strategy simulates Na=αadNa(a)N_{a}=\frac{\alpha_{a}}{d}N_{a}^{(a)} and N ø =aαadN ø (a)N_{\textup{\,\o\,}}=\sum_{a}\frac{\alpha_{a}}{d}N_{\textup{\,\o\,}}^{(a)}, leading to

Na=αad(A¯d(t)|φaφa|+B¯d(t)𝟙d|φaφa|d1)N_{a}=\frac{\alpha_{a}}{d}\left(\bar{A}_{d}(t)\outerproduct{\varphi_{a}}{\varphi_{a}}+\bar{B}_{d}(t)\frac{\mathds{1}_{d}-\outerproduct{\varphi_{a}}{\varphi_{a}}}{d-1}\right) (19)

and N ø =𝟙daNaN_{\textup{\,\o\,}}=\mathds{1}_{d}-\sum_{a}N_{a}. On the other hand from Eq. (12) we obtain M¯a(η,p)=αdd(dηp|φaφa|+η(1p)𝟙d)\bar{M}_{a}^{(\eta,p)}=\frac{\alpha_{d}}{d}\big{(}d\,\eta p\outerproduct{\varphi_{a}}{\varphi_{a}}+\eta(1-p)\mathds{1}_{d}\big{)}. Comparing with Eq. (19) we conclude that for

η=T¯d(t)d=(1t)d1,p=dA¯d(t)T¯d(t)(d1)T¯d(t)=t\begin{split}\eta&=\frac{\bar{T}_{d}(t)}{d}=(1-t)^{d-1},\\ p&=\frac{d\bar{A}_{d}(t)-\bar{T}_{d}(t)}{(d-1)\bar{T}_{d}(t)}=t\end{split} (20)

and t[0,1]t\in[0,1] any POVM {M¯a(η,p)}\{\bar{M}_{a}^{(\eta,p)}\} can be simulated. Reproducing measurements that are even more noisy is straightforward by adding noise to the above construction. Hence, for any noise level p=tp=t our construction simulates all POVMs {M¯a(η,p)}\{\bar{M}_{a}^{(\eta,p)}\} if η(1t)d1=(1p)d1\eta\leq(1-t)^{d-1}=(1-p)^{d-1}. \square

V Discussion and conclusion

Considering a bipartite steering scenario based on high-dimensional entanglement, we have investigated how asymmetrical the effect of steering can become. We presented entangled states ϱAB\varrho_{AB} such that Alice can convince Bob that ϱAB\varrho_{AB} is of high Schmidt number, while Bob can never convince Alice that the state is even entangled. Thus, one-way steering can become unlimited.

Specifically, we constructed families of entangled states that exhibit genuine dd-dimensional steering in one direction, while remaining unsteerable (under the most general measurements) in the other direction. We note that this construction can be straightforwardly generalized to states with genuine nn-dimensional steering in one direction and unsteerable the other way around, for any 1<nd1<n\leq d 111For this, one should simply adapt the bounds in Eqs (8) and (10) to demand only nn-steerability, following the results of Jones et al. (2022) for (8) and Designolle et al. (2021) for (10).. A practical implementation of such states should be feasible, e.g. with the setups of Refs Zeng et al. (2018); Srivastav et al. (2022) for an experimental demonstration of unlimited one-way steering.

Finally, our work also contributed to the characterisation of joint measurability in high-dimensional measurements. In particular, we obtained a criterion for the compatibility of arbitrary measurements subjected to both noise and loss. This result is of independent interest and may find other applications.

Acknowledgments.— We acknowledge financial support from the Swiss National Science Foundation (projects 192244, Ambizione PZ00P2-202179, and NCCR SwissMAP).

References

Appendix A Proof of Lemma 1

Lemma 1: Consider an assemblage {σa|x}a,x\{\sigma_{a|x}\}_{a,x} and a loss channel η\mathcal{L}_{\eta} for η>0\eta>0. The assemblage {σa|x=η[σa|x]}a,x\left\{\sigma^{\prime}_{a|x}=\mathcal{L}_{\eta}[\sigma_{a|x}]\right\}_{a,x} is nn-steerable if and only if {σa|x}a,x\{\sigma_{a|x}\}_{a,x} is.

First let us prove that if {σa|x}a,x\{\sigma_{a|x}\}_{a,x} is not nn-steerable than {σa|x}a,x\{\sigma_{a|x}^{\prime}\}_{a,x} is neither. Here, {σa|x}a,x\{\sigma_{a|x}\}_{a,x} is not nn-steerable means that there exist a rank-nn hidden state (RnnHS) model which allows to simulate the assemblage. Such a model consists of a probability density p(λ)p(\lambda) over states ψAB(λ)\psi_{AB}(\lambda) that all have Schmidt rank at most nn, and a collection of ”response POVMs” Aa|x,λA_{a|x,\lambda} performed by Alice, such that

σa|x=dλp(λ)trAψAB(λ)Aa|x,λ.\sigma_{a|x}=\int\differential\lambda p(\lambda)\tr_{A}\psi_{AB}(\lambda)A_{a|x,\lambda}. (21)

It is straightforward to define RnnHS model for the assemblage {σa|x}a,x\{\sigma_{a|x}^{\prime}\}_{a,x} by simply applying the loss channel to each state ψAB(λ)=idAη[ψAB(λ)]\psi^{\prime}_{AB}(\lambda)=\mathrm{id}_{A}\otimes\mathcal{L}_{\eta}[\psi_{AB}(\lambda)]. Note that the loss channel (as any local map) cannot increase the Schmidt rank of a state and

dλp(λ)trA(idAη[ψAB(λ)])Aa|x,λ=η[dλp(λ)trAψAB(λ)Aa|x,λ]=[σa|x]=σa|x.\begin{split}\int\differential\lambda p(\lambda)\tr_{A}(\mathrm{id}_{A}\otimes\mathcal{L}_{\eta}[\psi_{AB}(\lambda)])A_{a|x,\lambda}=\mathcal{L}_{\eta}\left[\int\differential\lambda p(\lambda)\tr_{A}\psi_{AB}(\lambda)A_{a|x,\lambda}\right]=\mathcal{L}[\sigma_{a|x}]=\sigma_{a|x}^{\prime}.\end{split} (22)

Next, we prove the other direction – if {σa|x}a,x\{\sigma_{a|x}^{\prime}\}_{a,x} is not nn-steerable then {σa|x}a,x\{\sigma_{a|x}\}_{a,x} is neither. Now we start with a RnnHS model for the assemblage model p(λ)p^{\prime}(\lambda), ψAB(λ)\psi_{AB}^{\prime}(\lambda) and Aa|x,λA_{a|x,\lambda} for the assemblage

σa|x=dλp(λ)trAψAB(λ)Aa|x,λ.\sigma_{a|x}^{\prime}=\int\differential\lambda p^{\prime}(\lambda)\tr_{A}\psi^{\prime}_{AB}(\lambda)A_{a|x,\lambda}. (23)

The states σa|x\sigma_{a|x}^{\prime} and ϕAB(λ)\phi_{AB}^{\prime}(\lambda) are here defined on a (d+1)(d+1)-dimensional system involving an additional level | ø \ket{\textup{\,\o\,}} introduced by the loss channel. Let us define the restriction of these states on the subspace orthogonal to |ψ\ket{\psi} projected by Π[1,d]\Pi_{[1,d]}. For the hidden state we obtain

ψAB(λ)(idAΠ[1,d])ψAB(λ)(idAΠ[1,d])q[1,d](λ)withq[1,d](λ)=tr(idAΠ[1,d])ψAB(λ)\psi_{AB}(\lambda)\equiv\frac{(\mathrm{id}_{A}\otimes\Pi_{[1,d]})\psi_{AB}^{\prime}(\lambda)(\mathrm{id}_{A}\otimes\Pi_{[1,d]})}{q_{[1,d]}(\lambda)}\qquad\text{with}\qquad q_{[1,d]}(\lambda)=\tr(\mathrm{id}_{A}\otimes\Pi_{[1,d]})\psi_{AB}^{\prime}(\lambda) (24)

that also have Schmidt rank at most nn, and define new probability density

p(λ)p(λ)q[1,d](λ)dλp(λ)q[1,d](λ).p(\lambda)\equiv\frac{p^{\prime}(\lambda)q_{[1,d]}(\lambda)}{\int\differential\lambda p^{\prime}(\lambda)q_{[1,d]}(\lambda)}. (25)

Here the denominator is

dλp(λ)q[1,d](λ)=tr(idAΠ[1,d])dλp(λ)ψAB(λ)=trΠ[1,d]aσa|x=η\int\differential\lambda p^{\prime}(\lambda)q_{[1,d]}(\lambda)=\tr(\mathrm{id}_{A}\otimes\Pi_{[1,d]})\int\differential\lambda p^{\prime}(\lambda)\psi^{\prime}_{AB}(\lambda)=\tr\Pi_{[1,d]}\sum_{a}\sigma_{a|x}^{\prime}=\eta (26)

Now compute which assemblage the novel RnnHS model (p(λ),ψAB(λ),Aa|x,λ)(p(\lambda),\psi_{AB}(\lambda),A_{a|x,\lambda}) simulates

σ~a|x=dλp(λ)trAψAB(λ)(idAΠ[1,d])Aa|x,λ=dλp(λ)trA(idAΠ[1,d])ψAB(λ)(idAΠ[1,d])Aa|x,λdλp(λ)q[1,d](λ)=Π[1,d](dλp(λ)trAψAB(λ)Aa|x,λ)Π[1,d]η=1ηΠ[1,d]σa|xΠ[1,d]=σa|x\begin{split}\tilde{\sigma}_{a|x}&=\int\differential\lambda p(\lambda)tr_{A}\psi_{AB}(\lambda)(\mathrm{id}_{A}\otimes\Pi_{[1,d]})A_{a|x,\lambda}\\ &=\frac{\int\differential\lambda p^{\prime}(\lambda)tr_{A}(\mathrm{id}_{A}\otimes\Pi_{[1,d]})\psi^{\prime}_{AB}(\lambda)(\mathrm{id}_{A}\otimes\Pi_{[1,d]})A_{a|x,\lambda}}{\int\differential\lambda p^{\prime}(\lambda)q_{[1,d]}(\lambda)}\\ &=\frac{\Pi_{[1,d]}\left(\int\differential\lambda p^{\prime}(\lambda)tr_{A}\psi^{\prime}_{AB}(\lambda)A_{a|x,\lambda}\right)\Pi_{[1,d]}}{\eta}\\ &=\frac{1}{\eta}\Pi_{[1,d]}\sigma_{a|x}^{\prime}\ \Pi_{[1,d]}\\ &=\sigma_{a|x}\end{split} (27)

To show the last equality consider the action of loss channel on σa|x\sigma_{a|x}

σa|x=η(σa|x)=ησa|x+(1η)| ø  ø |σa|x=1ηΠ[1,d]σa|xΠ[1,d].\sigma_{a|x}^{\prime}=\mathcal{L}_{\eta}(\sigma_{a|x})=\eta\,\sigma_{a|x}+(1-\eta)\outerproduct{\textup{\,\o\,}}{\textup{\,\o\,}}\implies\sigma_{a|x}=\frac{1}{\eta}\Pi_{[1,d]}\sigma_{a|x}^{\prime}\Pi_{[1,d]}. (28)

This concludes the proof. \square

Appendix B Computing the integrals in the proof of the Result 2

Here we compute that integrals

A¯d(t)=dd𝒛pt(a)(a|𝒛)|φa|𝒛|2T¯d(t)=dd𝒛pt(a)(a|𝒛),\begin{split}\bar{A}_{d}(t)&=d\int\differential\bm{z}\,p_{t}^{(a)}(a|\bm{z})\,\left|\innerproduct{\varphi_{a}}{\bm{z}}\right|^{2}\\ \bar{T}_{d}(t)&=d\int\differential\bm{z}\,p_{t}^{(a)}(a|\bm{z}),\end{split} (29)

where d𝒛\differential\bm{z} is the unitary-invariant measure over quantum state |𝒛\ket{\bm{z}} and pt(a)(a|𝒛)={1|φa|𝒛|2t0otherwisep_{t}^{(a)}(a|\bm{z})=\begin{cases}1&|\innerproduct{\varphi_{a}}{\bm{z}}|^{2}\geq t\\ 0&\text{otherwise}\end{cases}. It will be convening to work in the basis where |φa=|0\ket{\varphi_{a}}=\ket{0}, since the measure is invariant we directly get

A¯d(t)=dd𝒛Ξ[𝒛]|0|𝒛|2T¯d(t)=dd𝒛Ξ[𝒛],\begin{split}\bar{A}_{d}(t)&=d\int\differential\bm{z}\,\Xi[\bm{z}]\,\left|\innerproduct{0}{\bm{z}}\right|^{2}\\ \bar{T}_{d}(t)&=d\int\differential\bm{z}\,\Xi[\bm{z}],\end{split} (30)

and Ξt(𝒛)={1|0|𝒛|2t0otherwise\Xi_{t}(\bm{z})=\begin{cases}1&|\innerproduct{0}{\bm{z}}|^{2}\geq t\\ 0&\text{otherwise}\end{cases}. To compute these integrals we parametrize the quantum states |𝒛=k=0d1zk|k\ket{\bm{z}}=\sum_{k=0}^{d-1}z_{k}\ket{k} with

𝒛=(z0z1zd2zd1)=(cos(θ1)eiϕ2sin(θ1)cos(θ2)eiϕd1sin(θ1)sin(θd2)cos(θd1)eiϕdsin(θ1)sin(θd2)sin(θd1))\bm{z}=\left(\begin{array}[]{c}z_{0}\\ z_{1}\\ \vdots\\ z_{d-2}\\ z_{d-1}\end{array}\right)=\left(\begin{array}[]{c}\cos(\theta_{1})\\ e^{\mathrm{i}\phi_{2}}\sin(\theta_{1})\cos(\theta_{2})\\ \vdots\\ e^{\mathrm{i}\phi_{d-1}}\sin(\theta_{1})\dots\sin(\theta_{d-2})\cos(\theta_{d-1})\\ e^{\mathrm{i}\phi_{d}}\sin(\theta_{1})\dots\sin(\theta_{d-2})\sin(\theta_{d-1})\end{array}\right) (31)

for ϕi[0,2π]\phi_{i}\in[0,2\pi] and θi[0,π2]\theta_{i}\in[0,\frac{\pi}{2}], for which the invariant measure takes the form Hayashi (1998)

d𝒛=(d1)!πd1Πi=1d1sin2d2i1(θi)cos(θi)dθiΠj=2ddϕj.\differential\bm{z}=\frac{(d-1)!}{\pi^{d-1}}\Pi_{i=1}^{d-1}\sin^{2d-2i-1}(\theta_{i})\cos(\theta_{i})\differential\theta_{i}\,\Pi_{j=2}^{d}\differential\phi_{j}. (32)

For such parametrization we obtain |0|𝒛|2=|z0|2=cos2(θ1)|\innerproduct{0}{\bm{z}}|^{2}=|z_{0}|^{2}=\cos^{2}(\theta_{1}) and

Ξt(𝒛)Ξt(θ1)={1cos2(θ1)t0otherwise.\Xi_{t}(\bm{z})\equiv\Xi_{t}(\theta_{1})=\begin{cases}1&\cos^{2}(\theta_{1})\geq t\\ 0&\text{otherwise}.\end{cases} (33)

The integrand only depends on the angle θ1\theta_{1}, which allow us to write

T¯d(t)=dN0π/2sin2d3(θ1)cos(θ1)dθ1Ξt(θ1)A¯d(t)=dN0π/2sin2d3(θ1)cos(θ1)dθ1Ξt(θ1)cos2(θ1).\begin{split}\bar{T}_{d}(t)&=\frac{d}{N}\int_{0}^{\pi/2}\sin^{2d-3}(\theta_{1})\cos(\theta_{1})\differential\theta_{1}\Xi_{t}(\theta_{1})\\ \bar{A}_{d}(t)&=\frac{d}{N}\int_{0}^{\pi/2}\sin^{2d-3}(\theta_{1})\cos(\theta_{1})\differential\theta_{1}\Xi_{t}(\theta_{1})\cos^{2}(\theta_{1}).\end{split} (34)

with the normalization constant N=0π/2sin2d3(θ1)cos(θ1)dθ1N=\int_{0}^{\pi/2}\sin^{2d-3}(\theta_{1})\cos(\theta_{1})\differential\theta_{1}. Perform the variable change x=sin2(θ1)x=\sin^{2}(\theta_{1}) with dθ1=dx2sin(θ1)cos(θ1){\differential\theta_{1}}=\frac{\differential x}{2\sin(\theta_{1})\cos(\theta_{1})}, we obtain the expressions used in the main text

N=1201xd2dx=12(d1)T¯d(t)=d2N01txd2dx=d(1t)d1A¯d(t)=d2N01txd2(1x)dx=(1t)d1((d1)t+1).\begin{split}N&=\frac{1}{2}\int_{0}^{1}x^{d-2}\differential x=\frac{1}{2(d-1)}\\ \bar{T}_{d}(t)&=\frac{d}{2N}\int_{0}^{1-t}x^{d-2}\differential x=d(1-t)^{d-1}\\ \bar{A}_{d}(t)&=\frac{d}{2N}\int_{0}^{1-t}x^{d-2}(1-x)\differential x=(1-t)^{d-1}((d-1)t+1).\end{split} (35)

Appendix C Proof of Lemma 2

Lemma 2 The channel =η𝒲p\mathcal{E}=\mathcal{L}_{\eta}\circ\mathcal{W}_{p} is incompatibility breaking if η(1p)d1\eta\leq(1-p)^{d-1}.

Proof: By Result 3 we know that all POVMs {M¯a|U(η,p)}U\{\bar{M}_{a|U}^{(\eta,p)}\}_{U} in dimension dd subject to white noise and loss are jointly measurable if η(1p)d1\eta\leq(1-p)^{d-1}. Lemma 2 is thus implied by the following result.

Lemma 2.1 The channel =η𝒲p\mathcal{E}=\mathcal{L}_{\eta}\circ\mathcal{W}_{p} is incompatibility breaking if and only if the set of all POVMs {M¯a|x(η,p)}\{\bar{M}_{a|x}^{(\eta,p)}\} subject to loss and white noise is jointly measurable.

Proof: This claim might seem a tautology at first glance, but at closer inspection one realises that the channel \mathcal{E} outputs (d+1)(d+1)-dimensional, while M¯a|x\bar{M}_{a|x} describe measurements of a dd-dimensional system. So the predicate does not refer to the same sets of POVMs.

More precisely, let the set {Ma|x}\{M_{a|x}^{\prime}\} denote all POVMs acting in dimension d+1d+1. Then \mathcal{E} is incompatibility breaking iff the measurement assemblage

{M^a|x(η,p)[Ma|x]}\{\hat{M}_{a|x}^{(\eta,p)}\equiv\mathcal{E}^{*}[M^{\prime}_{a|x}]\} (36)

is jointly measurable. While the set {M¯a|x(η,p)}\{\bar{M}_{a|x}^{(\eta,p)}\} contains the POVMs

M¯a(η,p)={ηpMa+η(1p)(trMa)𝟙dda=1,,m(1η)𝟙da= ø \bar{M}_{a}^{(\eta,p)}=\begin{cases}\eta p\,M_{a}+\eta(1-p)(\tr M_{a})\frac{\mathds{1}_{d}}{d}&a=1,\dots,m\\ (1-\eta)\mathds{1}_{d}&a=\textup{\,\o\,}\end{cases} (37)

where Ma|xM_{a|x} runs through all POVMs in dimension dd. We want to show that the assemblage {M^a|x(η,p)}\{\hat{M}_{a|x}^{(\eta,p)}\} is compatible if and only if {M¯a|x(η,p)}\{\bar{M}_{a|x}^{(\eta,p)}\} is, where both sets contain measurements acting on a dd-dimensional quantum system.

To start let us have a closer look on the assemblage {M^a|x}\{\hat{M}_{a|x}\}. The loss channel η\mathcal{L}_{\eta} has d+1d+1 Kraus operators

K=ηk=1d|kk|ΠKk=1η| ø k|k=0,,d1\begin{split}K_{\checkmark}&=\sqrt{\eta}\underbrace{\sum_{k=1}^{d}\outerproduct{k}{k}}_{\equiv\Pi_{\checkmark}}\\ K_{k}&=\sqrt{1-\eta}\outerproduct{\textup{\,\o\,}}{k}\qquad k=0,\dots,d-1\end{split} (38)

Thus, for any measurement MaM_{a}^{\prime} on the Hilbert space d+1=span{|0,,|d1,| ø }\mathcal{H}_{d+1}=\text{span}\{\ket{0},\dots,\ket{d-1},\ket{\textup{\,\o\,}}\} we have [Ma]=𝒲pη[Ma]\mathcal{E}^{*}[M_{a}^{\prime}]=\mathcal{W}_{p}^{*}\circ\mathcal{L}_{\eta}^{*}[M_{a}^{\prime}] and

η[Ma]=KMaK+k=0d1KkMaKk=ηΠMaΠ+(1η)𝟙d ø |Ma| ø q(a),\begin{split}\mathcal{L}_{\eta}^{*}[M_{a}^{\prime}]&=K_{\checkmark}^{\dagger}M_{a}^{\prime}K_{\checkmark}+\sum_{k=0}^{d-1}K_{k}^{\dagger}M_{a}^{\prime}K_{k}\\ &=\eta\,\Pi_{\checkmark}M_{a}^{\prime}\Pi_{\checkmark}+(1-\eta)\mathds{1}_{d}\underbrace{\bra{\textup{\,\o\,}}M_{a}^{\prime}\ket{\textup{\,\o\,}}}_{\equiv q(a)},\end{split} (39)

where q(a)q(a) is a probability distribution since q(a)=tr| ø  ø |Ma0q(a)=\tr\outerproduct{\textup{\,\o\,}}{\textup{\,\o\,}}M_{a}^{\prime}\geq 0 and aq(a)=tr| ø  ø |aMa=1\sum_{a}q(a)=\tr\outerproduct{\textup{\,\o\,}}{\textup{\,\o\,}}\sum_{a}M_{a}^{\prime}=1. In addition, the operators MaΠMaΠM_{a}\equiv\Pi_{\checkmark}M_{a}^{\prime}\Pi_{\checkmark} define a valid POVM on the dd-dimensional subspace d=span{|0,,|d1}\mathcal{H}_{d}=\text{span}\{\ket{0},\dots,\ket{d-1}\} prior to losses. We thus have

η[Ma]=ηMa+(1η)q(a)𝟙d.\begin{split}\mathcal{L}_{\eta}^{*}[M_{a}^{\prime}]=\eta M_{a}+(1-\eta)q(a)\mathds{1}_{d}.\end{split} (40)

Finally adding the noise channel we get

M^a(η,p)=𝒲pη[Ma]=η𝒲p[Ma]+(1η)q(a)𝒲p[𝟙d]=ηMa(1,p)+(1η)q(a)𝟙d=ηpMa+η(1p)(trMa)𝟙dd+(1η)q(a)𝟙d=M¯a(η,p)+q(a)M¯ ø (η,p).\begin{split}\hat{M}_{a}^{(\eta,p)}&=\mathcal{W}_{p}^{*}\circ\mathcal{L}_{\eta}^{*}[M_{a}^{\prime}]\\ &=\eta\mathcal{W}_{p}^{*}[M_{a}]+(1-\eta)q(a)\mathcal{W}_{p}^{*}[\mathds{1}_{d}]\\ &=\eta M_{a}^{(1,p)}+(1-\eta)q(a)\mathds{1}_{d}\\ &=\eta pM_{a}+\eta(1-p)(\tr M_{a})\frac{\mathds{1}_{d}}{d}+(1-\eta)q(a)\mathds{1}_{d}\\ &=\bar{M}_{a}^{(\eta,p)}+q(a)\bar{M}_{\textup{\,\o\,}}^{(\eta,p)}.\end{split} (41)

We see that if the POVM M¯a(η,p)\bar{M}_{a}^{(\eta,p)} can be simulated, one can also use it to simulate M^a(η,p)\hat{M}_{a}^{(\eta,p)} by simply relabeling the output  ø  to aa with probability q(a)q(a). Hence if the set of all POVMs {M¯a|x(η,p)}\{\bar{M}_{a|x}^{(\eta,p)}\} is compatible this is also the case of {M^a|x(η,p)}\{\hat{M}_{a|x}^{(\eta,p)}\}.

To show the converse consider all POVMs on the (d+1)(d+1)-dimensional space with the elements of the form {M1,,Ma,,M ø =| ø  ø |}\{M_{1},\dots,M_{a},\dots,M_{\textup{\,\o\,}}=\outerproduct{\textup{\,\o\,}}{\textup{\,\o\,}}\} with  ø |Ma| ø =0\bra{\textup{\,\o\,}}M_{a}\ket{\textup{\,\o\,}}=0, and {M1,,Ma,}\{M_{1},\dots,M_{a},\dots\} (without the M ø M_{\textup{\,\o\,}} element) running through all POVMs on the dd-dimensional subspace d=span{|0,,|d1}\mathcal{H}_{d}=\text{span}\{\ket{0},\dots,\ket{d-1}\}. For a POVM of such form we have q(a)=0q(a)=0, q( ø )=1q(\textup{\,\o\,})=1 and by virtue of Eq.

M^a(η,p)=M¯a(η,p)M^ ø (η,p)=(1η)M¯ ø (η,p).\begin{split}\hat{M}_{a}^{(\eta,p)}&=\bar{M}_{a}^{(\eta,p)}\\ \hat{M}_{\textup{\,\o\,}}^{(\eta,p)}&=(1-\eta)\bar{M}_{\textup{\,\o\,}}^{(\eta,p)}.\end{split} (42)

This if any measuremnt M^a(η,p)\hat{M}_{a}^{(\eta,p)} can be simulated any measurement M¯a(η,p)\bar{M}_{a}^{(\eta,p)} can be as well. \square