Blog > What the Operator-Schmidt Decomposition Tells Us About Entanglement

## What the Operator-Schmidt Decomposition Tells Us About Entanglement

June 27th, 2014

In quantum information theory, the well-known Schmidt decomposition theorem tells us that every pure quantum state $|\phi\rangle\in\mathbb{C}^d\otimes\mathbb{C}^d$ can be written in the form

$|\phi\rangle =\displaystyle\sum_{i=1}^d\gamma_i|a_i\rangle\otimes|b_i\rangle,$

where each $\gamma_i \geq 0$ is a real scalar and the sets $\{|a_i\rangle\}$ and $\{|b_i\rangle\}$ form orthonormal bases of $\mathbb{C}^d$.

The Schmidt decomposition theorem isn’t anything fancy: it is just the singular value decomposition in disguise (the $\gamma_i$‘s are singular values of some matrix and the sets $\{|a_i\rangle\}$ and $\{|b_i\rangle\}$ are its left and right singular vectors). However, it tells us everything we could ever want to know about the entanglement of $|\phi\rangle$: it is entangled if and only if it has more than one non-zero $\gamma_i$, and in this case the question of “how much” entanglement is contained within $|\phi\rangle$ is answered by a certain function of the $\gamma_i$‘s.

Well, we can find a similar decomposition of mixed quantum states. If $\rho\in M_d\otimes M_d$ is a mixed quantum state then it can be written in its operator-Schmidt decomposition:

$\rho=\displaystyle\sum_{i=1}^{d^2}\gamma_iA_i\otimes B_i,$

where each $\gamma_i \geq 0$ is a real scalar and the sets $\{A_i\}$ and $\{B_i\}$ form orthonormal bases of Hermitian matrices in $M_d$ (under the Hilbert–Schmidt inner product $\langle A,B\rangle :=\mathrm{Tr}(A^\dagger B)$).

Once again, we haven’t really done anything fancy: the operator-Schmidt decomposition is also just the singular value decomposition in disguise in almost the exact same way as the regular Schmidt decomposition. However, its relationship with entanglement of mixed states is much weaker (as we might expect from the fact that the singular value decomposition can be computed in polynomial time, but determining whether a mixed state is entangled or separable (i.e., not entangled) is expected to be hard [1]). In this post, we’ll investigate some cases when the operator-Schmidt decomposition does let us conclude that $\rho$ is separable or entangled.

### Proving a State is Entangled: The Realignment Criterion

One reasonably well-known method for proving that a mixed state is entangled is the realignment criterion [2,3]. What is slightly less well-known is that the realignment criterion can be phrased in terms of the coefficients $\{\gamma_i\}$ in the operator-Schmidt decomposition of $\rho$.

Theorem 1 (realignment criterion). Let $\rho\in M_d\otimes M_d$ have operator-Schmidt decomposition

$\rho=\displaystyle\sum_{i=1}^{d^2}\gamma_iA_i\otimes B_i.$

If $\sum_{i=1}^{d^2}\gamma_i>1$ then $\rho$ is entangled.

Proof. The idea is to construct a specific entanglement witness that detects the entanglement in $\rho$. In particular, the entanglement witness that we will use is $W:=I-\sum_{i=1}^{d^2}A_i\otimes B_i$. To see that $W$ is indeed an entanglement witness, we must show that $(\langle a|\otimes\langle b|)W(|a\rangle\otimes|b\rangle)\geq 0$ for all $|a\rangle\in\mathbb{C}^m$ and $|b\rangle\in\mathbb{C}^n$. Well, some algebra shows that

$(\langle a|\otimes\langle b|)W(|a\rangle\otimes|b\rangle) = 1 - \displaystyle\sum_{i=1}^{d^2}\langle a|A_i|a\rangle\langle b|B_i|b\rangle,$

so it suffices to show that $\sum_{i=1}^{d^2}\langle a|A_i|a\rangle\langle b|B_i|b\rangle\leq 1$. To see this notice that

$\displaystyle\sum_{i=1}^{d^2}\langle a|A_i|a\rangle\langle b|B_i|b\rangle\leq \sqrt{\sum_{i=1}^{d^2}(\langle a|A_i|a\rangle)^2}\sqrt{\sum_{i=1}^{d^2}(\langle b|B_i|b\rangle)^2}= 1,$

where the inequality is the Cauchy–Schwarz inequality and the equality comes from the fact that the sets $\{A_i\}$ and $\{B_i\}$ are orthonormal bases, so $\sum_{i=1}^{d^2}(\langle a|A_i|a\rangle)^2=\big\langle |a\rangle\langle a|,|a\rangle\langle a|\big\rangle=1$ (and similarly for $|b\rangle$).

Now that we know that $W$ is an entanglement witness, we must check that it detects the entanglement in $\rho$ (that is, we want to show that $\mathrm{Tr}(W\rho)<0$). This is straightforward to show by making use of the fact that the sets $\{A_i\}$ and $\{B_i\}$ are orthonormal:

$\mathrm{Tr}(W\rho) = \displaystyle\mathrm{Tr}\left( \Big(I - \sum_{i=1}^{d^2}A_i\otimes B_i\Big)\Big(\sum_{i=1}^{d^2}\gamma_iA_i\otimes B_i\Big)\right)=1-\mathrm{Tr}\left(\sum_{i,j=1}^{d^2}\gamma_iA_iA_j\otimes B_iB_j\right)$

${}_{{}_{{}_{.}}} \phantom{\mathrm{Tr}(W\rho)}=\displaystyle 1-\sum_{i,j=1}^{d^2}\gamma_i\mathrm{Tr}(A_iA_j)\mathrm{Tr}(B_iB_j)=1-\sum_{i=1}^{d^2}\gamma_i<0.$

It follows that $\rho$ is entangled, which completes the proof.

A more popular formulation of the realignment criterion says that if we define the realignment map $R$ by $R(|i\rangle\langle j|\otimes|k\rangle\langle\ell|)=|i\rangle\langle k|\otimes|j\rangle\langle\ell|$ and extending by linearity, and let $\|\cdot\|_{tr}$ denote the trace norm (i.e., the sum of the singular values), then $\|R(\rho)\|_{tr}>1$ implies that $\rho$ is entangled. The equivalence of these two formulations of the realignment criterion comes from the fact that the singular values of $R(\rho)$ are exactly the coefficients $\{\gamma_i\}$ in its operator-Schmidt decomposition.

### Proving a State is Entangled: Beyond the Realignment Criterion

We might naturally wonder whether we can prove that even more states are entangled based on their operator-Schmidt decomposition than those detected by the realignment criterion. The following theorem gives one sense in which the answer to this question is “no”: if we only look at “nice” functions of the coefficients $\{\gamma_i\}$ then the realignment criterion gives the best method of entanglement detection possible.

Theorem 2. Let $f : \mathbb{R}^{d^2}\rightarrow\mathbb{R}_{\geq 0}$ be a symmetric gauge function (i.e., a norm that is invariant under permutations and sign changes of the $d^2$ entries of the input vector). If we can conclude that $\rho\in M_d\otimes M_d$ is entangled based on the value of $f(\gamma_1,\gamma_2,\ldots,\gamma_{d^2})$ then it must be the case that $\sum_{i=1}^{d^2}\gamma_i>1$.

Proof. Without loss of generality, we scale $f$ so that $f(1,0,0,\ldots,0) = 1$. We first prove two facts about $f$.

Claim 1: $f(\gamma_1,\gamma_2,\ldots,\gamma_{d^2})\geq \frac{1}{d}$ for all mixed states $\rho\in M_d\otimes M_d$. This follows from the fact that $\max_i\gamma_i\geq\frac{1}{d}$ (which itself is kind of a pain to prove: it follows from the fact that the $1\rightarrow\infty$ Schatten norm of the realignment map $R$ is $d$, but if anyone knows of a more direct and/or simpler way to prove this, I’d love to see it). If we assume without loss of generality that $\gamma_1\geq\frac{1}{d}$ then

$f(\gamma_1,\gamma_2,\ldots,\gamma_{d^2}) = \frac{1}{2}\big(f(\gamma_1,\gamma_2,\ldots,\gamma_{d^2})+f(\gamma_1,-\gamma_2,\ldots,-\gamma_{d^2})\big)$

${}_{{}_{.}} \phantom{f(\gamma_1,\gamma_2,\ldots,\gamma_{d^2})} \geq f(\gamma_1,0,0,\ldots,0) = \gamma_1f(1,0,0,\ldots,0) = \gamma_1 \geq \frac{1}{d},$

as desired.

Claim 2: There exists a separable state $\rho\in M_d\otimes M_d$ for which $f(\gamma_1,\gamma_2,\ldots,\gamma_{d^2})$ equals any given value in the interval $[\frac{1}{d},1]$. To see why this is the case, first notice that there exists a separable state $\rho\in M_d\otimes M_d$ with $\gamma_1=1$ and $\gamma_i=0$ for all $i\geq 2$: the state $\rho=|0\rangle\langle 0|\otimes|0\rangle\langle 0|$ is one such example. Similarly, there is a separable state with $\gamma_1=\frac{1}{d}$ and $\gamma_i=0$ for all $i\geq 2$: the state $\rho=\frac{1}{d^2}I\otimes I$ is one such example. Furthermore, it is straightforward to interpolate between these two extremes to find separable states (even product states) with $\gamma_i = 0$ for all $i\geq 2$ and any value of $\gamma_1 \in[\frac{1}{d},1]$. For such states we have

$f(\gamma_1,\gamma_2,\ldots,\gamma_{d^2}) = f(\gamma_1,0,0,\ldots,0)=\gamma_1f(1,0,0,\ldots,0)=\gamma_1,$

which can take any value in the interval $[\frac{1}{d},1]$ as claimed.

By combining claims 1 and 2, we see that we could only ever use the value of $f(\gamma_1,\gamma_2,\ldots,\gamma_{d^2})$ to conclude that $\rho$ is entangled if $f(\gamma_1,\gamma_2,\ldots,\gamma_{d^2})>1$. However, in this case we have

$\displaystyle\sum_{i=1}^{d^2}\gamma_i = f(\gamma_1,0,0,\ldots,0) + f(0,\gamma_2,0,\ldots,0) + \cdots + f(0,0,0,\ldots,\gamma_{d^2})$

${}_{{}_{.}}\displaystyle\phantom{\sum_{i=1}^{d^2}\gamma_i}\geq f(\gamma_1,\gamma_2,\ldots,\gamma_{d^2}) >1,$

which completes the proof.

Theorem 2 can be phrased naturally in terms of the other formulation of the realignment criterion as well: it says that there is no unitarily-invariant matrix norm $\|\cdot\|_{u}$ with the property that we can use the value of $\|R(\rho)\|_{u}$ to conclude that $\rho$ is entangled, except in those cases where the trace norm (i.e., the realignment criterion) itself already tells us that $\rho$ is entangled.

Nonetheless, we can certainly imagine using functions of the coefficients $\{\gamma_i\}$ that are not symmetric gauge functions. Alternatively, we could take into account some (hopefully easily-computable) properties of the matrices $\{A_i\}$ and $\{B_i\}$. One such method for detecting entanglement that depends on the coefficients $\{\gamma_i\}$ and the trace of each $\{A_i\}$ and $\{B_i\}$ is as follows.

Theorem 3 [4,5]. Let $\rho\in M_d\otimes M_d$ have operator-Schmidt decomposition

$\rho=\displaystyle\sum_{i=1}^{d^2}\gamma_iA_i\otimes B_i.$

If

$\displaystyle\sum_{i=1}^{d^2}\big|\gamma_i - \gamma_i^2\mathrm{Tr}(A_i)\mathrm{Tr}(B_i)\big| + \frac{1}{2}\sum_{i=1}^{d^2}\gamma_i^2\big(\mathrm{Tr}(A_i)^2+\mathrm{Tr}(B_i)^2\big) > 1$

then $\rho$ is entangled.

I won’t prove Theorem 3 here, but I will note that it is strictly stronger than the realignment criterion, which can be seen by showing that the left-hand side of Theorem 3 is at least as large as the left-hand side of Theorem 1. To show this, observe that

$\displaystyle\sum_{i=1}^{d^2}\big|\gamma_i - \gamma_i^2\mathrm{Tr}(A_i)\mathrm{Tr}(B_i)\big| \geq \sum_{i=1}^{d^2}\gamma_i - \sum_{i=1}^{d^2}\gamma_i^2\mathrm{Tr}(A_i)\mathrm{Tr}(B_i)$

and

$\displaystyle\frac{1}{2}\sum_{i=1}^{d^2}\gamma_i^2\big(\mathrm{Tr}(A_i)^2+\mathrm{Tr}(B_i)^2\big) - \sum_{i=1}^{d^2}\gamma_i^2\mathrm{Tr}(A_i)\mathrm{Tr}(B_i) = \frac{1}{2}\sum_{i=1}^{d^2}\big(\gamma_i\mathrm{Tr}(A_i)-\gamma_i\mathrm{Tr}(B_i)\big)^2,$

which is nonnegative.

### Proving a State is Separable

Much like we can use the operator-Schmidt decomposition to sometimes prove that a state is entangled, we can also use it to sometimes prove that a state is separable. To this end, we will use the operator-Schmidt rank of $\rho$, which is the number of non-zero coefficients $\{\gamma_i\}$ in its operator-Schmidt decomposition. One trivial observation is as follows:

Proposition 4. If the operator-Schmidt rank of $\rho$ is $1$ then $\rho$ is separable.

Proof. If the operator-Schmidt rank of $\rho$ is $1$ then we can write $\rho=A\otimes B$ for some $A,B\in M_d$. Since $\rho$ is positive semidefinite, it follows that either $A$ and $B$ are both positive semidefinite or both negative semidefinite. If they are both positive semidefinite, we are done. If they are both negative semidefinite then we can write $\rho=(-A)\otimes(-B)$ and then we are done.

Somewhat surprisingly, however, we can go further than this: it turns out that all states with operator-Schmidt rank $2$ are also separable, as was shown in [6].

Theorem 5 [6]. If the operator-Schmidt rank of $\rho$ is $2$ then $\rho$ is separable.

Proof. If $\rho$ has operator-Schmidt rank $2$ then it can be written in the form $\rho=A\otimes B + C\otimes D$ for some $A,B,C,D\in M_d$. Throughout this proof, we use the notation $a:=\mathrm{Tr}(A), b:=\mathrm{Tr}(B)$, and so on.

Since $\rho$ is positive semidefinite, so are each of its partial traces. Thus $bA+dC$ and $aB+cD$ are both positive semidefinite operators. It is then straightforward to verify that

$(bA+dC)\otimes(aB+cD) + (cA-aC)\otimes(dB-bD)= A\otimes B + C\otimes D =\rho.$

What is important here is that we have found a rank-$2$ tensor decomposition of $\rho$ in which one of the terms is positive semidefinite. Now we define

$X := \big((bA+dC)^{-1/2}\otimes(aB+cD)^{-1/2}\big) \rho \big((bA+dC)^{-1/2}\otimes(aB+cD)^{-1/2}\big)$

and notice that $X = I\otimes I + E\otimes F$ for some $E,F\in M_d$ (in order to do this, we actually need the partial traces of $\rho$ to be nonsingular, but this is easily taken care of by standard continuity arguments, so we’ll sweep it under the rug). Furthermore, $X$ is also positive semidefinite, and it is separable if and only if $\rho$ is separable. Since $X$ is positive semidefinite, we know that $1 + \lambda_i(E)\lambda_j(F) \geq 0$ for all eigenvalues $\lambda_i(E)$ of $E$ and $\lambda_j(F)$ of $F$. If we absorb scalars between $E$ and $F$ so that $\max_i\lambda_i(E) = 1$ then this implies that $\lambda_j(F) \geq -1$ for all $j$. Thus $I-E$ and $I+F$ are both positive semidefinite. Furthermore, a straightforward calculation shows that

$(I-E)\otimes I + E \otimes (I+F) = I\otimes I + E\otimes F = X.$

We now play a similar game as before: we define a new matrix

$Y := \big((I-E)^{-1/2}\otimes(I+F)^{-1/2}\big)X\big((I-E)^{-1/2}\otimes(I+F)^{-1/2}\big)$

and notice that $Y = I \otimes G + H \otimes I$ for some $G,H\in M_d$ (similar to before, we note that there is a standard continuity argument that can be used to handle the fact that $I-E$ and $I+F$ might be singluar). The minimum eigenvalue of $Y$ is then $\lambda_{\textup{min}}(G)+\lambda_{\textup{min}}(H)$, which is non-negative as a result of $Y$ being positive semidefinite. It then follows that

$Y = I \otimes (G - \lambda_{\textup{min}}(G)I) + (H - \lambda_{\textup{min}}(H)I) \otimes I + \big(\lambda_{\textup{min}}(G)+\lambda_{\textup{min}}(H)\big)I \otimes I.$

Since each term in the above decomposition is positive semidefinite, it follows that $Y$ is separable, which implies that $X$ is separable, which finally implies that $\rho$ is separable.

In light of Theorem 6, it seems somewhat natural to ask how far we can push things: what values of the operator-Schmidt rank imply that a state is separable? Certainly we cannot expect all states with an operator-Schmidt rank of $4$ to be separable, since every state in $M_2 \otimes M_2$ has operator-Schmidt rank $4$ or less, and there are entangled states in this space (more concretely, it’s easy to check that the maximally-entangled pure state has operator-Schmidt rank $4$).

This left the case of operator-Schmidt rank $3$ open. Very recently, it was shown in [7] that a mixed state in $M_2 \otimes M_d$ with operator-Schmidt rank $3$ is indeed separable, yet there are entangled states with operator-Schmidt rank $3$ in $M_3 \otimes M_3$.

References

1. L. Gurvits. Classical deterministic complexity of Edmonds’ problem and quantum entanglement. In Proceedings of the Thirty-Fifth Annual ACM Symposium on Theory of Computing, pages 10–19, 2003. E-print: arXiv:quant-ph/0303055
2. K. Chen and L.-A. Wu. A matrix realignment method for recognizing entanglement. Quantum Inf. Comput., 3:193–202, 2003. E-print: arXiv:quant-ph/0205017
3. O. Rudolph. Some properties of the computable cross norm criterion for separability. Phys. Rev. A, 67:032312, 2003. E-print:  E-print: arXiv:quant-ph/0212047
4. C.-J. Zhang, Y.-S. Zhang, S. Zhang, and G.-C. Guo. Entanglement detection beyond the cross-norm or realignment criterion. Phys. Rev. A, 77:060301(R), 2008. E-print: arXiv:0709.3766 [quant-ph]
5. O. Gittsovich, O. Gühne, P. Hyllus, and J. Eisert. Unifying several separability conditions using the covariance matrix criterion. Phys. Rev. A, 78:052319, 2008. E-print: arXiv:0803.0757 [quant-ph]
6. D. Cariello. Separability for weak irreducible matrices. E-print: arXiv:1311.7275 [quant-ph]
7. D. Cariello. Does symmetry imply PPT property?. E-print: arXiv:1405.3634 [math-ph]
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1. July 3rd, 2014 at 11:41 | #1

Hi Nathaniel,
You discuss testing whether a given state is entangled and mention the question of “how much” entanglement is contained within the state. My question is related: Given two entangled states, there are many different measures of entanglement used to compare them in order to say something like “this state is more entangled than that one” (the entropy of entanglement that you linked is one such measure, but there is for example the geometric measure of entanglement). Is it just personal preference which one to use? It seems if there were a “best” measure of entanglement, then everyone would use that one. So, why is this not the case?

2. July 4th, 2014 at 16:37 | #2

Sarah :
Is it just personal preference which one to use?

Personal preference is part of it, but much more important is what exactly you want the entanglement measure to tell you. The entropy of entanglement can really be thought of as “how much” entanglement is within a state if you’re thinking of entanglement as a resource. For example, if you have 1 bit of entanglement (i.e., 1 ebit) according to the entropy of entanglement then you can teleport 1 qubit.

The geometric measure of entanglement, on the other hand, is a measure of how close a state is to the set of separable states. While this quantity certainly is interesting mathematically, I’m admittedly not aware of any statements of the form “you can do task XYZ with a given quantum state if and only if its geometric measure of entanglement is at least c”.

If you look at entanglement measures on mixed states, things get even more diverse:

– The distillable entanglement tells you how much entanglement you can extract out of a state (if you are restricted to certain natural operations).

– The entanglement cost tells you how much entanglement has to be used to create a state (if you are restricted to that same set of operations).

– The negativity is an entanglement measure that doesn’t really have a great physical interpretation, but it’s much easier to compute than other entanglement measures.

Etc. It depends on context, what exactly what you want your entanglement measure to tell you, and how hard you are willing to work mathematically to get your hands on it.

It’s quite analogous to norms, really. There are tons of different norms on vectors/matrices/whatever that are frequently used. Which one you use depends a bit on the context of the problem, a bit on whichever one happens to be nicest to work with mathematically, and a bit on personal preference.

3. July 4th, 2014 at 21:05 | #3

Ah, thank you!

4. August 21st, 2014 at 11:53 | #4

Hi Nathaniel,
Nice to see your article on separability tests. However, I would like to know that whether the operator-Schmidt decomposition for a mixed density operator \rho will be unique. For a pure density operator \sigma it will be unique I feel.

5. August 22nd, 2014 at 10:59 | #5

@Nirman Ganguly – The operator-Schmidt decomposition is as unique as the singular value decomposition is. Specifically:

– The operator-Schmidt coefficients $\{\gamma_i\}$ are uniquely determined by $\rho$.

– If all of the operator-Schmidt coefficients are distinct, then the operators $\{A_i\}$ and $\{B_i\}$ in the operator-Schmidt decomposition are uniquely determined as well (up to multiplication by a complex phase $e^{i\theta}$).

– If there are degenerate (i.e., repeated) operator-Schmidt coefficients, then we can replace the $A_i$‘s by any linear combination of the $A_i$‘s corresponding to the same operator-Schmidt coefficient.

6. July 5th, 2016 at 01:52 | #6

Very clear and very helpful, just want to say thanks.

1. October 11th, 2018 at 20:20 | #1