Unveiling NPT bound problem: From Distillability Sets to Inequalities and Multivariable Insights

Entanglement [1] is what distinguished the quantum world from classical worlds. Unfortunately, our understanding of it remains quite limited to this day. Entanglement exhibits varying degrees of strength, the highest of which falls into a subclass called “maximal entanglement”. Bell state is one such example. No consensus has yet been reached about what qualifies as a “weak entanglement”. Quantum states with small Schmidt numbers are generally seen as weakly entangled, since quantum states with Schmidt number 1 are separable. Undistillable entanglement is another kind of widely accepted weak entanglement. Relationships between these two kinds of weak entanglements are discussed in [2].

Entanglement distillation [3] is the process of obtaining pure Bell states only by Local Operations and Classical Communication (LOCC) from multiple copies of an entangled bipartite state. If N copies of a given entangled bipartite state can be transformed into pure Bell states through LOCC, then the entangled bipartite state is referred to as N-distillable. If not, then it’s N-undistillable. If a particular entangled bipartite state is N-undistillable for arbitrary N, it is considered undistillable and is also referred to as a bound entangled state.

Another property concerned here is “Positive Partial Transpose” (PPT). PPT refers to a bipartite quantum state whose partial transpose is positive semidefinite. It is established in [4] that Positive Partial Transpose (PPT) entanglement implies bound entanglement, and therefore weakly entangled in terms of distillability. However, the question of whether the converse is true remains an open problem [5]. This problem is usually referred to as the NPT bound problem, or the distillability problem. The popular conjecture regarding this problem is that the converse is not true, and that NPT bound entanglement exists. A number of special circumstances have been worked out in [6, 7, 8, 9, 10]. Indirect approaches to this problem include expansion of distillation operations to k-extendible operations [11], catalysis-assisted distillation [12], and dually non-entangling and PPT-preserving channels [13], linking the problem with squashed entanglement [14] and linear preserver [15], studying the transformation of bound entangled states under certain dynamic process [16], or considering the problem in a broader scenario such as hyperquantum states [17].

In three separate attempts to directly tackle the distillability problem [18], [19] and [20], they each had a subset of quantum states singled out, and proved that for any finite number of copies N, there exist undistillable states in this subset. However, as N approaches infinity, the subset shrinks to emptiness. It has been shown in [21] that N-undistillability does not imply (N+1)-undistillability, by demonstrating a set of distillable-only-by-arbitrarily-large-number-of-copies state, thus ruling out the easy route of concentrating on few-copies distillation.

For the Werner states [22], a family of single-parameter states, the existence of NPT bound states implies existence of NPT bound Werner states [23]. Therefore, it is sufficient to only study Werner states. Due to the inherent complexity of addressing N-copy Werner state distillability, over the last decade, direct attacks on the problem have been primarily fixating on the seemingly straightforward problem of 2-copy Werner state distillability, but so far only limited progress has been made. In an attempt to attack the 2-copy Werner state distillability problem, the special case of $4\times 4$ bipartite states is considered in [24], in which case the distillability condition can be reduced to a convenient form of “half-property”, and subsequently reformulated the distillability problem into a matrix analysis problem. For a special case where the matrices are normal, a proof was given.

In [25] the problem is reformulated and established that for a $d\times d$ bipartite Werner state, the 2-copy Werner state undistillability problem is equivalent to a certain $2d^{2}\times 2d^{2}$ matrix being positive semidefinite. Positive definiteness of the block matrices in the upper-left and lower-right corners has been proved.

In this work, we introduce a method that decomposes bound entanglement verification into repeated steps of 1-undistillability verification, before giving a new bound for N-undistillability that is independent of quantum state dimensions $d$, and therefore different from that given in [18], [19] and [20]. However, similar to the problem encountered in [18], [19] and [20], our set of N-undistillable states also shrinks to emptiness for $N\to\infty$. Also, an equivalent matrix analysis inequality form is presented, which is applicable to all finite dimensionalities, as opposed to the case in [24] when only the $d=4$ case is considered. Another equivalent version of inequalities regarding only rank-one matrices is also included, which takes on a form reminiscent of Cauchy-Schwartz inequality. Finally, we take on a brand-new perspective of considering it as a multi-variable function problem, finding critical points, proving non-convexity and conjecturing Hessian positivity.

We will be using $\mathcal{H}_{A}$ and $\mathcal{H}_{B}$ to denote Hilbert spaces pertaining to A and B subsystems, and $X_{i}$ to denote matrices of rank less than or equal to $i$. Throughout the paper we will be considering multi-partite quantum states with identical dimensions, i.e. quantum states on $\mathcal{H}_{A}\otimes\mathcal{H}_{B}$, where $dim(\mathcal{H}_{A})=dim(\mathcal{H}_{B})=d$. This is justified by noticing that any quantum state of $d_{1}\times d_{2}$ can be equivalently transformed into a quantum state of identical dimensions $max\{d_{1},d_{2}\}\times max\{d_{1},d_{2}\}$, with all unnecessary elements set to zero.

The structure of this paper is as follows: In Section II, we introduce a matrix transformation that takes N-distillability problem into a higher dimensional 1-distillability problem. In Section III, the process of $N$-undistillability verification is transformed into $log(N)$ repeated steps of 1-undistillability verification. In Section IV, upper bounds on $N$-undistillability are given, yielding a result similar to that of [18], [19] and [20], effectively finding shrinking families of N-undistillable quantum states. In Section V, equivalent partial trace inequalities are presented before turning them into a matrix analysis problem. The rank-two matrix analysis problem is also reduced to a rank-one matrix analysis problem, resulting in a Cauchy-Schwartz inequality look-alike. In Section VI, we take on a new perspective by treating the problem as a multi-variable function, proving critical points, non-convexity and conjecturing positive-semidefinite Hessians.

The open problem of equivalence between PPT entanglement and bound entanglement can be resolved by finding bound entangled NPT states. Since Peres [26] has already shown that NPT is a sufficient condition for entanglement, attention can be limited to the state being both “bound”, which is, N-undistillable for arbitrarily large N, and NPT, which is, having a negative partial transpose. Compared to ascertaining positive semidefiniteness of a certain matrix, the major difficulty usually lies in finding a concise mathematical interpretation for undistillability, and asserting it for two types of arbitrariness: arbitrary LOCC operation and arbitrarily finitely many copies of the entangled state. This will be the main focus of this section, and the analysis of mathematical structure will eventually lead to a transformation that reduces N-undistillability verification to 1-undistillability verification of a transformed state.

Let’s recall the concept of Schmidt decomposition, Schmidt coefficients and Schmidt rank [27]. Schmidt decomposition of a pure state $|\psi\rangle$ in $\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ is:

$$|\psi\rangle=\sum_{i=1}^{d}\alpha_{i}|u_{i}v_{i}\rangle,$$

(1)

where $\mathcal{H}_{A}$ and $\mathcal{H}_{B}$ are both d-dimensional Hilbert spaces, with $\{|u_{1}\rangle,\cdots,|u_{d}\rangle\}$ and $\{|v_{1}\rangle,\cdots,|v_{d}\rangle\}$ as their respective orthonormal bases. Schmidt coefficients $\alpha_{i}$ are real, nonnegative and unique up to re-ordering. The number of non-zero Schmidt coefficients is called Schmidt rank of pure state $|\psi\rangle$. $SR(|\psi\rangle)$ denotes the Schmidt rank of $|\psi\rangle$.

A bipartite quantum state is called N-distillable if pure Bell states can be obtained using LOCC, from N-copies of the state. It has been proven in [28] that all entangled $2\times 2$ bipartite quantum states are distillable, therefore, all we need to do is transform the original quantum state into a $2\times 2$ entangled state by LOCC. This is demonstrated in the following result: A bipartite quantum state $\rho$ on $\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ is N-undistillable iff for any $d^{N}\times d^{N}$ bipartite pure state $|\psi^{SR\leq 2}\rangle$ on $\mathcal{H}_{A}^{\otimes N}\otimes\mathcal{H}_{B}^{\otimes N}$ with Schmidt rank $\leq$ 2, the following holds:

$$\langle\psi^{SR\leq 2}|(\rho^{T_{A}})^{\otimes N}|\psi^{SR\leq 2}\rangle\geq 0,$$

(2)

where $\rho^{T_{A}}=(T\otimes I)\rho$ is the partial transpose of $\rho$, with $T$ being the transpose operator. The above expression overlooks the pairing between different subsystems and so an additional operator $M_{N}$ from $(\mathcal{H}_{A}\otimes\mathcal{H}_{B})^{\otimes N}$ to $\mathcal{H}_{A}^{\otimes N}\otimes\mathcal{H}_{B}^{\otimes N}$ is introduced to make it meaningful:

Operator $M_{N}$ then takes the $A_{1}B_{1}A_{2}B_{2}\cdots A_{N}B_{N}$ structure of $(\rho^{T_{A}})^{\otimes N}$ and merges their A-subsystems and B-subsystems, resulting in a structure of $A_{1}A_{2}\cdots A_{N}B_{1}B_{2}\cdots B_{N}$ that suits the pure quantum state, thus explicitly demonstrating the process of system pairing which is essential to any further calculations derived from this representation of undistillability.

For better representation of undistillability and further use in following sections, we use a certain state-operator isomorphism introduced in [24]. We notice that it is in essence a Choi-Jamiokowski isomorphism followed by a transposition on the resulting matrix. For convenience and consistency with previous works, we stick to this “transposed Choi-Jamiokowski isomorphism” when presenting our findings.

Note that Schmidt decomposition of a pure quantum state $|\psi\rangle$ is related to the singular value decomposition of $\Psi(|\psi\rangle)$ through state-operator isomorphism. In fact, Schmidt coefficients of $|\psi\rangle$ is the same as the singular values of $\Psi(|\psi\rangle)$, and Schmidt rank of $|\psi\rangle$ is equal to the rank of $\Psi(|\psi\rangle)$.

The following is a more concise mathematical representation of N-undistillability: A bipartite quantum state $\rho$ is N-undistillable iff:

$$[\Psi^{-1}(X_{2})]^{\dagger}M_{N}(\rho^{T_{A}})^{\otimes N}M_{N}^{\dagger}\Psi^{-1}(X_{2})\geq 0,$$

(6)

where $X_{2}$ is any $d^{N}\times d^{N}$ matrix of rank $\leq 2$. Note that $\Psi^{-1}(X_{2})$ is equivalent to $|\psi^{SR\leq 2}\rangle$, a pure bipartite quantum state with Schmidt rank no larger than 2. For the case of $N=1$, 1-undistillability is:

$$[\Psi^{-1}(X_{2})]^{\dagger}\rho^{T_{A}}\Psi^{-1}(X_{2})=\langle\psi^{SR\leq 2}|\rho^{T_{A}}|\psi^{SR\leq 2}\rangle\geq 0.$$

(7)

Effectively, the problem of $\rho$ being N-copy undistillable is equivalent to the problem of $M_{N}(\rho^{\otimes N})M_{N}^{\dagger}$ being 1-undistillable, but on a Hilbert space with higher dimensions $\mathcal{H^{\prime}}_{A}\otimes\mathcal{H^{\prime}}_{B}$, $dim(\mathcal{H^{\prime}}_{A})=dim(\mathcal{H^{\prime}}_{B})=d^{N}$.

Since discussions of 1-copy distillability problem rarely involves specification of finite dimensional Hilbert space dimensionality, and most of the theorems and methods are applicable to arbitrary dimensions, this transformation poses an advantage. In the next section, a specific symmetry of $M_{2^{N}}$ transformations is utilized for decomposing the increasingly complicated structure of $M_{2^{N}}$ into identical steps repeated for $N$ times. In fact, verification of $2^{N}$-undistillability is enough to assert bound entanglement.

From the observation that (N+1)-undistillability implies N-undistillability, we can get the following result which relaxes the N-undistillability requirement for all N to only an infinite sequence of N diverging to infinity:

In the 2-copy case, $M_{2}$ transformation is simply:

$$E=M_{2}=\sum|i_{A_{1}}i_{A_{2}}i_{B_{1}}i_{B_{2}}\rangle\langle i_{A_{1}}i_{B_{1}}i_{A_{2}}i_{B_{2}}|,$$

(8)

which is merely an exchange of the second and third parts of $B_{1}$ and $A_{2}$. From now on we will always refer to the 2-copy case of $M_{2}$ as $E$, emphasizing that it is only a simple exchange as opposed to complicated $M_{N}$ of larger copy numbers.

According to Eq.6, the quantum state $\rho$ is 2-undistillable iff $E(\rho\otimes\rho)E^{\dagger}$ is 1-undistillable.

In the 4-copy case, $M_{4}$ can actually be decomposed into two steps:

		$\displaystyle A_{1}B_{1}A_{2}B_{2}A_{3}B_{3}A_{4}B_{4}$		(9)
	$\displaystyle\to$	$\displaystyle A_{1}A_{2}B_{1}B_{2}A_{3}A_{4}B_{3}B_{4}$
	$\displaystyle\to$	$\displaystyle A_{1}A_{2}A_{3}A_{4}B_{1}B_{2}B_{3}B_{4}.$

Step 1: do the 2-copy exchange on subsystems 1,2 ($A_{1}B_{1}A_{2}B_{2}\to A_{1}A_{2}B_{1}B_{2}$) and subsystems 3,4 ($A_{3}B_{3}A_{4}B_{4}\to A_{3}A_{4}B_{3}B_{4}$). This step is no different than in the 2-copy case, except that it is done on two systems simultaneously. Group every neighbouring 2 subsystems(in terms of $A,B$), so that the eight parts are now seen as four parts ($A_{1}A_{2},B_{1}B_{2},A_{3}A_{4},B_{3}B_{4}$ to $ABAB$). In order to separate this newly obtained four-part state from 2-copy Werner states, we denote this state $\rho(e)\otimes\rho(e)$, where $\rho(e)=E(\rho\otimes\rho)E^{\dagger}$.

Step 2: Do the 2-copy exchange $E$ again, obtaining $\rho(e^{2})=E(\rho(e)\otimes\rho(e))E^{\dagger}$, where $E$ is taken with regard to the “merged” parts. This step takes once-transformed Werner states into twice-transformed Werner states.

The above steps of repeating the $E$ transformation twice, is equivalent to doing the $M_{4}$ translations directly. The original state $\rho$ is 4-copy undistillable iff $\rho(e^{2})$ is one-copy undistillable.

Similarly, in the $2^{N}$-copy case, E is decomposed into N steps, with each step taking state $\rho(e^{i-1})\otimes\rho(e^{i-1})$, producing a state $\rho(e^{i})=E(\rho(e^{i-1})\otimes\rho(e^{i-1}))E^{\dagger}$, verifying its positivity on Schmidt rank $\leq 2$ states and effectively ascertaining $2^{i}$-copy undistillability, before passing two copies of this state on, as a starting state for the next step. The following theorem demonstrates the process.

In other words, we are essentially searching for an infinite sequence of points connected by the operation $E$, with the first point being $\rho=\rho(e^{0})$, second point being $\rho(e^{1})=E(\rho(e^{0})\otimes\rho(e^{0}))E^{\dagger}$, and the $(i+2)$-th point being $\rho(e^{i+1})=E(\rho(e^{i})\otimes\rho(e^{i}))E^{\dagger}$. If points of this infinite sequence always fall within the set of 1-undistillability, then we can safely say that the starting point $\rho$ is N-undistillable for arbitrary N, or that it is a bound state.

The problem of Werner state N-undistillability can be equivalently written as inequalities regarding the Frobenius norm of partial traces of a rank-2 matrix.

Take 2-undistillability as an example. $\langle\psi^{SR\leq 2}|E(I\otimes G+G\otimes I)E^{\dagger}|\psi^{SR\leq 2}\rangle$ can be written with regard to the Frobenius norms of two partial traces:

$$\langle\psi^{SR\leq 2}|E(I\otimes G)E^{\dagger}|\psi^{SR\leq 2}\rangle=||Tr_{2}(X_{2}(\psi))||_{F}^{2},$$

(29)

$$\langle\psi^{SR\leq 2}|E(G\otimes I)E^{\dagger}|\psi^{SR\leq 2}\rangle=||Tr_{1}(X_{2}(\psi))||_{F}^{2},$$

(30)

where $||\cdot||_{F}$ denotes Frobenius norm:

$$||X||_{F}=\sqrt{\sum_{ij}|X_{ij}|^{2}}=\sqrt{Tr(X^{\dagger}X)}.$$

(31)

$X_{2}(\psi)$ is a $d^{2}\times d^{2}$ matrix of rank $\leq$ 2, obtained by state-operator isomorphism, from $|\psi^{SR\leq 2}\rangle$, a pure bipartite $d^{2}\times d^{2}$ state with Schmidt rank $\leq$ 2. For simplicity, we will be writing it as $X_{2}$ from now on.

The normalization of $|\psi^{SR\leq 2}\rangle$ requires that $||X_{2}||_{F}^{2}=Tr(X_{2}^{\dagger}X_{2})=1$. Now the problem of Werner state 2-undistillability is equivalent to: Find a range of $\beta$ that makes the following inequality always holds for arbitrary $X_{2}$ with rank no larger than 2 (The requirement of $||X_{2}||_{F}^{2}=1$ can be lifted due to homogeneity):

		$\displaystyle||X_{2}||_{F}^{2}+\beta^{2}|Tr(X_{2})|^{2}$		(32)
		$\displaystyle+\beta(||Tr_{1}(X_{2})||_{F}^{2}+||Tr_{2}(X_{2})||_{F}^{2})\geq 0.$

N-undistillability has a similar form:

The proof of this is presented in Appendix C. A similar result is obtained by [29].

The partial trace inequalities are with regard to a matrix $X_{2}$ of rank one or two. From now on we consider the case where $\beta$ is taken to be $-\frac{1}{2}$ according to popular guess. For 2-undistillability problem, The rank one case can be trivially proved by making use of the following lemma:

Proof of this lemma is presented in Appendix E. For $X_{2}$ of rank two, the inequality of concern when $\beta=-\frac{1}{2}$ is:

$$||Tr_{2}(X_{2})||_{F}^{2}+||Tr_{1}(X_{2})||_{F}^{2}-\frac{1}{2}|Tr(X_{2})|^{2}\leq 2||X_{2}||_{F}^{2}=2.$$

(35)

For any square matrix $X_{2}$ of rank two, the singular decomposition can be applied, decomposing the matrix as the sum of two rank one matrices:

$$X_{2}=\sigma_{1}X_{2}^{1}+\sigma_{2}X_{2}^{2}=\sigma_{1}\mathbf{u}^{1}\mathbf{v}^{1\dagger}+\sigma_{2}\mathbf{u}^{2}\mathbf{v}^{2\dagger},$$

(36)

where $\sigma_{1}^{2}+\sigma_{2}^{2}=1$, and both $\sigma_{1}$ and $\sigma_{2}$ are positive. We introduce $d\times d$ matrices $U_{1},U_{2},V_{1},V_{2}$ that are the result of state-operator isomorphisms, corresponding to $\mathbf{u}^{1},\mathbf{u}^{2},\mathbf{v}^{1},\mathbf{v}^{2}$ respectively.

Eq.35 then becomes:

		$\displaystyle\sigma_{1}^{2}(Tr(V_{1}U_{1}^{\dagger}U_{1}V_{1}^{\dagger})+Tr(U_{1}^{\dagger}V_{1}V_{1}^{\dagger}U_{1}))$		(37)
		$\displaystyle+\sigma_{2}^{2}(Tr(V_{2}U_{2}^{\dagger}U_{2}V_{2}^{\dagger})+Tr(U_{2}^{\dagger}V_{2}V_{2}^{\dagger}U_{2}))$
		$\displaystyle+\sigma_{1}\sigma_{2}2Re[Tr(V_{1}U_{1}^{\dagger}U_{2}V_{2}^{\dagger})+Tr(U_{1}^{\dagger}V_{1}V_{2}^{\dagger}U_{2})]$
		$\displaystyle-\frac{|\sigma_{1}Tr(U_{1}V_{1}^{\dagger})+\sigma_{2}Tr(U_{2}V_{2}^{\dagger})|^{2}}{2}\leq 2.$

For simplicity the above is rewritten as:

$$\sigma_{1}^{2}P+\sigma_{2}^{2}Q+\sigma_{1}\sigma_{2}R\leq 2,$$

(38)

where $P,Q,R$ are:

$$P=Tr(U_{1}^{\dagger}V_{1}V_{1}^{\dagger}U_{1})+Tr(V_{1}U_{1}^{\dagger}U_{1}V_{1}^{\dagger})-\frac{|Tr(U_{1}V_{1}^{\dagger})|^{2}}{2},$$

(39)

$$Q=Tr(U_{2}^{\dagger}V_{2}V_{2}^{\dagger}U_{2})+Tr(V_{2}U_{2}^{\dagger}U_{2}V_{2}^{\dagger})-\frac{|Tr(U_{2}V_{2}^{\dagger})|^{2}}{2},$$

(40)

		$\displaystyle R=2Re[Tr(V_{1}U_{1}^{\dagger}U_{2}V_{2}^{\dagger})+Tr(U_{1}^{\dagger}V_{1}V_{2}^{\dagger}U_{2})$		(41)
		$\displaystyle-\frac{Tr^{*}(U_{1}V_{1}^{\dagger})Tr(U_{2}V_{2}^{\dagger})}{2}].$

Maximization regarding variables $\sigma_{1},\sigma_{2}$ reduces the question to proving:

$$R^{2}\leq 4(2-P)(2-Q),$$

(42)

with normalization and orthogonality conditions requiring:

$$||U_{1}||_{F}^{2}=||V_{1}||_{F}^{2}=||U_{2}||_{F}^{2}=||V_{2}||_{F}^{2}=1,$$

(43)

and

$$tr(V_{2}^{\dagger}V_{1})=tr(U_{2}^{\dagger}U_{1})=0.$$

(44)

The special case of $U_{1}=V_{1}$, in other words, $X_{2}$ being the sum of a normal matrix and a rank-1 matrix, has been proved in [29].

In fact, by a slight change of representation, we can also get an equivalent expression of Werner state N-undistillability inequality, regarding only rank one matrices:

Proof of this equivalence is presented in Appendix D.

It is then established that N-undistillability problem for arbitrary N can be written as inequalities concerning rank one matrices. Although the above form looks like a Cauchy-Schwartz inequality, $f_{N}(X_{1},X^{\prime}_{1})$ cannot be seen as an inner product and enable the direct application of Cauchy-Schwartz inequality. $f_{N}(X_{1},X^{\prime}_{1})$ is obviously conjugate symmetric, and for rank one matrices $X_{1}$ and $X^{\prime}_{1}$, positivity of $f_{N}(X_{1},X_{1})$ and $f_{N}(X^{\prime}_{1},X^{\prime}_{1})$ can be ascertained. However, rank-one matrices do not compose a vector space, since the sum of two rank-one matrices can be a rank-two matrix. Although arbitrary finite rank matrices do compose a finite dimensional vector space, proving positivity of $f_{N}(A,A)$ for arbitrary rank matrix $A$ is both beyond our ability and our need, since proving positivity of $f_{N}(X_{2},X_{2})$ for arbitrary rank two matrix $X_{2}$ is, in fact, equivalent to Eq.33, and is enough for proof of N-undistillability.

We make another attempt at this problem by seeing it as a multi-variable function. For simplicity we consider only the real case of the 2-distillability problem, that is, $X_{1}$, $X^{\prime}_{1}$ are $d^{2}\times d^{2}$ real matrices. For convenience of expression we use $f(C,D)=f_{2}(X_{1},X^{\prime}_{1})$ to denote the function concerned. Thus, the 2-undistillability problem is equivalent to proving the following inequality for all $d^{2}\times d^{2}$-dimensional rank one matrices $C,D$:

$$f^{2}(C,D)\leq f(C,C)f(D,D),$$

(47)

where

		$\displaystyle f(C,D)=Tr(C^{T}D)$		(48)
		$\displaystyle+\beta(Tr(C_{1}^{T}D_{1})+Tr(C_{2}^{T}D_{2}))+\beta^{2}Tr(C^{T})Tr(D),$

where we have used $Tr_{1}(C)=C_{2},Tr_{2}(C)=C_{1},Tr_{1}(D)=D_{2},Tr_{2}(D)=D_{1}$ for simplicity. A multi-variable function $g(C)$ is defined if we see $D$ as a constant $D_{0}$:

$$g(C)_{D_{0}}=f(C,C)f(D_{0},D_{0})-f^{2}(C,D_{0}),$$

(49)

where $D_{0}$ is an arbitrary $d^{2}\times d^{2}$-dimensional real matrix of rank 1. Both $C$ and $D_{0}$ can be written as outer products of $d^{2}$-dimensional real vectors $\mathbf{w}$, $\mathbf{x},\mathbf{y},\mathbf{z}$:

$$C=\mathbf{w}\mathbf{x}^{T},D_{0}=\mathbf{y}\mathbf{z}^{T}.$$

(50)

The multi variables of function $g(C)$ are taken to be the components of vector $\mathbf{w},\mathbf{x}$, represented by $w_{ij},x_{ij}$. Indices i and j take their values in $[0,1,\cdots,d-1]$, and are combined together to represent the $(i\times d+j)$-th component of the vector. The problem is therefore transformed into proving that multi-variable function $g(C)_{D_{0}}$ is always non-negative, for all variables $w_{ij},x_{ij}$ and all possible parameters $y_{ij},z_{ij}$. It is clear that when $C=D_{0}$, $g(D_{0})_{D_{0}}=f^{2}(D_{0},D_{0})-f^{2}(D_{0},D_{0})=0$, and we conjecture that these are global minimums.