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## How to compute hard-to-compute matrix norms

January 11th, 2016

There are a wide variety of different norms of matrices and operators that are useful in many different contexts. Some matrix norms, such as the Schatten norms and Ky Fan norms, are easy to compute thanks to the singular value decomposition. However, the computation of many other norms, such as the induced p-norms (when p ≠ 1, 2, ∞), is NP-hard. In this post, we will look at a general method for getting quite good estimates of almost any matrix norm.

The basic idea is that every norm can be written as a maximization of a convex function over a convex set (in particular, every norm can be written as a maximization over the unit ball of the dual norm). However, this maximization is often difficult to deal with or solve analytically, so instead it can help to write the norm as a maximization over two or more simpler sets, each of which can be solved individually. To illustrate how this works, let’s start with the induced matrix norms.

### Induced matrix norms

The induced p → q norm of a matrix B is defined as follows:

$\displaystyle\|B\|_{p\rightarrow q}:=\max\big\{\|B\mathbf{x}\|_q : \|\mathbf{x}\|_p = 1\big\},$

where

$\displaystyle\|\mathbf{x}\|_p := \left(\sum_{i}|x_i|^p\right)^{1/p}$

is the vector p-norm. There are three special cases of these norms that are easy to compute:

1. When p = q = 2, this is the usual operator norm of B (i.e., its largest singular value).
2. When p = q = 1, this is the maximum absolute column sum: $\|B\|_{1\rightarrow 1} = \max_j\sum_i|b_{ij}|$.
3. When p = q = ∞, this is the maximum absolute row sum: $\|B\|_{\infty\rightarrow \infty} = \max_i\sum_j|b_{ij}|$.

However, outside of these three special cases (and some other special cases, such as when B only has real entries that are non-negative [1]), this norm is much messier. In general, its computation is NP-hard [2], so how can we get a good idea of its value? Well, we rewrite the norm as the following double maximization:

$\displaystyle\|B\|_{p\rightarrow q}=\max\big\{|\mathbf{y}^*B\mathbf{x}| : \|\mathbf{x}\|_p = 1, \|\mathbf{y}\|_{q^\prime} = 1\big\},$

where $q^\prime$ is the positive real number such that $1/q + 1/q^\prime = 1$ (and we take $q^\prime = 1$ if $q = \infty$, and vice-versa). The idea is then to maximize over $\mathbf{x}$ and $\mathbf{y}$ one at a time, alternately.

1. Start by setting $j = 1$ and fixing a randomly-chosen vector $\mathbf{x}_0$, scaled so that $\|\mathbf{x}_0\|_{p} = 1$.
2. Compute

$\max\big\{|\mathbf{y}^*B\mathbf{x}_{j-1}| : \|\mathbf{y}\|_{q^\prime} = 1\big\},$

keeping $\mathbf{x}_{j-1}$ fixed, and let $\mathbf{y}_j$ be the vector attaining this maximum. By Hölder’s inequality, we know that this maximum value is exactly equal to $\|B\mathbf{x}_{j-1}\|_{q}$. Furthermore, the equality condition of Hölder’s inequality tells us that the vector $\mathbf{y}_j$ attaining this maximum is the one with complex phases that are the same as those of $B\mathbf{x}_{j-1}$, and whose magnitudes are such that $|\mathbf{y}_j|^{q^\prime}$ is a multiple of $|B\mathbf{x}_{j-1}|^q$ (here the notation $|\cdot|^q$ means we take the absolute value and the q-th power of every entry of the vector).

3. Compute

$\max\big\{|\mathbf{y}_j^*B\mathbf{x}| : \|\mathbf{x}\|_{p} = 1\big\},$

keeping $\mathbf{y}_j$ fixed, and let $\mathbf{x}_j$ be the vector attaining this maximum. By an argument almost identical to that of step 2, this maximum is equal to $\|B^*\mathbf{y}_j\|_{p^\prime}$, where $p^\prime$ is the positive real number such that $1/p + 1/p^\prime = 1$. Furthermore, the vector $\mathbf{x}_j$ attaining this maximum is the one with complex phases that are the same as those of $B^*\mathbf{y}_j$, and whose magnitudes are such that $|\mathbf{x}_j|^p$ is a multiple of $|B^*\mathbf{y}_j|^{p^\prime}$.

4. Increment $j$ by 1 and return to step 2. Repeat until negligible gains are made after each iteration.

This algorithm is extremely quick to run, since Hölder’s inequality tells us exactly how to solve each of the two maximizations separately, so we’re left only performing simple vector calculations at each step. The downside of this algorithm is that, even though it will always converge to some local maximum, it might converge to a value that is smaller than the true induced p → q norm. However, in practice this algorithm is fast enough that it can be run several thousand times with different (randomly-chosen) starting vectors $\mathbf{x}_0$ to get an extremely good idea of the value of $\|B\|_{p\rightarrow q}$.

It is worth noting that this algorithm is essentially the same as the one presented in [3], and reduces to the power method for finding the largest singular value when p = q = 2. This algorithm has been implemented in the QETLAB package for MATLAB as the InducedMatrixNorm function.

### Induced Schatten superoperator norms

There is a natural family of induced norms on superoperators (i.e., linear maps $\Phi : M_n \rightarrow M_n$) as well. First, for a matrix $X \in M_n$, we define its Schatten p-norm to be the p-norm of its vector of singular values:

$\|X\|_p := \left(\sum_{i=1}^n \sigma_i(X)^p\right)^{1/p}.$

Three special cases of the Schatten p-norms include:

• p = 1, which is often called the “trace norm” or “nuclear norm”,
• p = 2, which is often called the “Frobenius norm” or “Hilbert–Schmidt norm”, and
• p = ∞, which is the usual operator norm.

The Schatten norms themselves are easy to compute (since singular values are easy to compute), but their induced counter-parts are not.

Given a superoperator $\Phi : M_n \rightarrow M_n$, its induced Schatten p → q norm is defined as follows:

$\|\Phi\|_{p\rightarrow q} := \max\big\{ \|\Phi(X)\|_q : \|X\|_p = 1 \big\}.$

These induced Schatten norms were studied in some depth in [4], and crop up fairly frequently in quantum information theory (especially when p = q = 1) and operator theory (especially when p = q = ∞). The fact that they are NP-hard to compute in general is not surprising, since they reduce to the induced matrix norms (discussed earlier) in the case when $\Phi$ only acts on the diagonal entries of $X$ and just zeros out the off-diagonal entries. However, it seems likely that this norm’s computation is also difficult even in the special cases p = q = 1 and p = q = ∞ (however, it is straightforward to compute when p = q = 2).

Nevertheless, we can obtain good estimates of this norm’s value numerically using essentially the same method as discussed in the previous section. We start by rewriting the norm as a double maximization, where each maximization individually is easy to deal with:

$\|\Phi\|_{p\rightarrow q} = \max\big\{ |\mathrm{Tr}(Y^*\Phi(X))| : \|X\|_p = 1, \|Y\|_{q^\prime} = 1\big\},$

where $q^\prime$ is again the positive real number (or infinity) satisfying $1/q + 1/q^\prime = 1$. We now maximize over $X$ and $Y$, one at a time, alternately, just as before:

1. Start by setting $j = 1$ and fixing a randomly-chosen matrix $X_0$, scaled so that $\|X_0\|_p = 1$.
2. Compute

$\max\big\{|\mathrm{Tr}(Y^*\Phi(X_{j-1})| : \|Y\|_{q^\prime} = 1\big\},$

keeping $X_{j-1}$ fixed, and let $Y_j$ be the matrix attaining this maximum. By the Hölder inequality for Schatten norms, we know that this maximum value is exactly equal to $\|\Phi(X_{j-1})\|_{q}$. Furthermore, the matrix $Y_j$ attaining this maximum is the one with the same left and right singular vectors as $\Phi(X_{j-1})$, and whose singular values are such that there is a constant $c$ so that $\sigma_i(Y_j)^{q^\prime} = c\sigma_i(\Phi(X_{j-1}))^q$ for all $i$ (i.e., the vector of singular values of $Y_j$, raised to the $q^\prime$ power, is a multiple of the vector of singular values of $\Phi(X_{j-1})$, raised to the $q$ power).

3. Compute

$\max\big\{|\mathrm{Tr}(Y_j^*\Phi(X)| : \|X\|_{p} = 1\big\},$

keeping $Y_j$ fixed, and let $X_j$ be the matrix attaining this maximum. By essentially the same argument as in step 2, we know that this maximum value is exactly equal to $\|\Phi^*(Y_j)\|_{p^\prime}$, where $\Phi^*$ is the map that is dual to $\Phi$ in the Hilbert–Schmidt inner product. Furthermore, the matrix $X_j$ attaining this maximum is the one with the same left and right singular vectors as $\Phi^*(Y_j)$, and whose singular values are such that there is a constant $c$ so that $\sigma_i(X_j)^{p} = c\sigma_i(\Phi^*(Y_j))^{p^\prime}$ for all $i$.

4. Increment $j$ by 1 and return to step 2. Repeat until negligible gains are made after each iteration.

The above algorithm is almost identical to the algorithm presented for induced matrix norms, but with absolute values and complex phases of the vectors $\mathbf{x}$ and $\mathbf{y}$ replaced by the singular values and singular vectors of the matrices $X$ and $Y$, respectively. The entire algorithm is still extremely quick to run, since each step just involves computing one singular value decomposition.

The downside of this algorithm, as with the induced matrix norm algorithm, is that we have no guarantee that this method will actually converge to the induced Schatten p → q norm; only that it will converge to some lower bound of it. However, the algorithm works pretty well in practice, and is fast enough that we can simply run it a few thousand times to get a very good idea of what the norm actually is. If you’re interested in making use of this algorithm, it has been implemented in QETLAB as the InducedSchattenNorm function.

### Entanglement Norms

The central idea used for the previous two families of norms can also be used to get lower bounds on the following norm on $M_m \otimes M_n$ that comes up from time to time when dealing with quantum entanglement:

$\|X\|_{S(1)} := \max\Big\{\big|(\mathbf{v}\otimes\mathbf{w})^*X(\mathbf{x} \otimes \mathbf{y})\big| : \|\mathbf{v}\| = \|\mathbf{w}\| = \|\mathbf{x}\| = \|\mathbf{y}\| = 1\Big\}.$

(As a side note: this norm, and some other ones like it, were the central focus on my thesis.) This norm is already written for us as a double maximization, so the idea presented in the previous two sections is somewhat clearer from the start: we fix randomly-generated vectors $\mathbf{v}$ and $\mathbf{x}$ and then maximize over all vectors $\mathbf{w}$ and $\mathbf{y}$, which can be done simply by computing the left and right singular vectors associated with the maximum singular value of the operator

$(\mathbf{v} \otimes I)^*X(\mathbf{x} \otimes I) \in M_n.$

We then fix $\mathbf{w}$ and $\mathbf{y}$ as those singular vectors and then maximize over all vectors $\mathbf{v}$ and $\mathbf{x}$ (which is again a singular value problem), and we iterate back and forth until we converge to some value.

As with the previously-discussed norms, this algorithm always converges, and it converges to a lower bound of $\|X\|_{S(1)}$, but perhaps not its exact value. If you want to take this algorithm out for a spin, it has been implemented in QETLAB as the sk_iterate function.

It’s also worth mentioning that this algorithm generalizes straightforwardly in several different directions. For example, it can be used to find lower bounds on the norms $\|\cdot\|_{S(k)}$ where we maximize on the left and right by pure states with Schmidt rank not larger than k rather than separable pure states, and it can be used to find lower bounds on the geometric measure of entanglement [5].

References:

1. D. Steinberg. Computation of matrix norms with applications to robust optimization. Research thesis. Technion – Israel University of Technology, 2005.
2. J. M. Hendrickx and A. Olshevsky. Matrix p-norms are NP-hard to approximate if p ≠ 1,2,∞. 2009. E-print: arXiv:0908.1397
3. D. W. Boyd. The power method for ℓp norms. Linear Algebra and Its Applications, 9:95–101, 1974.
4. J. Watrous. Notes on super-operator norms induced by Schatten norms. Quantum Information & Computation, 5(1):58–68, 2005. E-print: arXiv:quant-ph/0411077
5. T.-C. Wei and P. M. Goldbart. Geometric measure of entanglement and applications to bipartite and multipartite quantum states. Physical Review A, 68:042307, 2003. E-print: arXiv:quant-ph/0212030
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## 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]
Tags:

## Counting the Possible Orderings of Pairwise Multiplication

February 12th, 2014

Suppose we are given n distinct positive real numbers $a_1 > a_2 > \cdots > a_n > 0$. The question we are going to consider in this post is as follows:

Question. How many different possible orderings are there of the $n(n+1)/2$ numbers $\{a_ia_j\}_{1\leq i\leq j\leq n}$?

To help illustrate what we mean by this question, consider the n = 2 case, where $a_1 > a_2 > 0$. Then the 3 possible products of $a_1$ and $a_2$ are $a_1^2, a_2^2, a_1a_2$, and it is straightforward to see that we must have $a_1^2 > a_1a_2> a_2^2$, so there is only one possible ordering in the n = 2 case.

In the n = 3 case, we have $a_1 > a_2 > a_3 > 0$ and 6 possible products: $a_1^2, a_2^2, a_3^2, a_1a_2, a_1a_3, a_2a_3$. Some relationships between these 6 numbers are immediate, such as $a_1^2 > a_1a_2 > a_1a_3 > a_2a_3 > a_3^2$. However, it could be the case that either $a_2^2 > a_1a_3$ or $a_1a_3 > a_2^2$ (we ignore the degenerate cases where two products are equal to each other), so there are two different possible orderings in this case:

$a_1^2 > a_1a_2 > a_2^2 > a_1a_3 > a_2a_3 > a_3^2\quad\text{ or }\\ a_1^2 > a_1a_2 > a_1a_3 > a_2^2 > a_2a_3 > a_3^2.$

In this post, we will consider the problem of how many such orderings exist for larger values of n. This problem arises naturally from a problem in quantum entanglement: the number of such orderings is exactly the minimum number of linear matrix inequalities needed to characterize the eigenvalues of quantum states that are “PPT from spectrum” [1].

### A Rough Upper Bound

We now begin constructing upper bounds on the number of possible orderings of $\{a_ia_j\}_{1\leq i\leq j\leq n}$. Since we are counting orderings between $n(n+1)/2$ numbers, a trivial upper bound is given by $(n(n+1)/2)!$, since that is the number of possible orderings of $n(n+1)/2$ arbitrary numbers. However, this quantity is a gross overestimate.

We can improve this upper bound by creating an $n \times n$ matrix whose $(i,j)$-entry is $a_ia_j$ (note that this matrix is symmetric, positive semidefinite, and has rank 1, which is roughly how the connection to quantum entanglement arises). For example, in the n = 4 case, this matrix is as follows:

$\begin{bmatrix}a_1^2 & a_1a_2 & a_1a_3 & a_1a_4 \\ * & a_2^2 & a_2a_3 & a_2a_4 \\ * & * & a_3^2 & a_3a_4 \\ * & * & * & a_4^2\end{bmatrix},$

where we have used asterisks (*) to indicate entries that are determined by symmetry. The fact that $a_1 > a_2 > \cdots > a_n > 0$ implies that the rows and columns of the upper-triangular part of this matrix are decreasing. Thus we can get an upper bound to the solution to our problem by counting the number of ways that we can place the numbers $1, 2, \ldots, n(n+1)/2$ (exactly once each) in the upper-triangular part of a matrix in such a way that the rows and columns of that upper-triangular part are decreasing. For example, this can be done in 2 different ways in the n = 3 case:

$\begin{bmatrix}6 & 5 & 4 \\ * & 3 & 2 \\ * & * & 1\end{bmatrix} \quad \text{and} \quad \begin{bmatrix}6 & 5 & 3\\ * & 4 & 2\\ * & * & 1\end{bmatrix}.$

The matrix above on the left corresponds to the case $a_1a_3 > a_2^2$ discussed earlier, while the matrix above on the right corresponds to the case $a_2^2 > a_1a_3$.

A formula for the number of such ways to place the integers $1, 2, \ldots, n(n+1)/2$ in a matrix was derived in [2] (see also A003121 in the OEIS), which immediately gives us the following upper bound on the number of orderings of the products $\{a_ia_j\}_{1\leq i\leq j\leq n}$:

$\displaystyle(n(n+1)/2)! \frac{1! 2! \cdots (n-1)!}{1! 3! \cdots (2n-1)!}.$

For n = 1, 2, 3, …, this formula gives the values 1, 1, 2, 12, 286, 33592, 23178480, …

### A Better Upper Bound

Before improving the upper bound that we just presented, let’s first discuss why it is not actually a solution to the original question. In the n = 4 case, our best upper bound so far is 12, since there are 12 different ways to place the integers $1,2,\ldots,10$ in the upper-triangular part of a $4 \times 4$ matrix such that the rows and columns of that upper-triangular part are decreasing. However, one such placement is as follows:

$\begin{bmatrix}10 & 9 & 7 & 6 \\ * & 8 & 5 & 3 \\ * & * & 4 & 2 \\ * & * & * & 1\end{bmatrix}.$

The above matrix corresponds to the following inequalities in terms of $\{a_ia_j\}_{1\leq i\leq j\leq n}$:

$a_1^2 > a_1a_2 > a_2^2 > a_1a_3 > a_1a_4 > a_2a_3 > a_3^2 > a_2a_4 > a_3a_4 > a_4^2.$

The problem here is that there actually do not exist real numbers $a_1 > a_2 > a_3 > a_4 > 0$ that satisfy the above string of inequalities. To see this, notice in particular that we have the following three inequalities: $a_2^2 > a_1a_3$, $a_1a_4 > a_2a_3$, and $a_3^2 > a_2a_4$. However, multiplying the first two inequalities together gives $a_1a_2^2a_4 > a_1a_2a_3^2$, so $a_2a_4 > a_3^2$, which contradicts the third inequality.

More generally, there can not be indices $i,j,k,\ell,m,n$ such that we simultaneously have the following three inequalities:

$a_ia_j > a_ka_\ell$, $a_\ell a_m > a_j a_n$, and $a_i a_m < a_k a_n$.

I am not aware of a general formula for the number integer matrices that do not lead to these types of “bad” inequalities, but I have computed this quantity for n ≤ 7 (C code is available here), which gives the following better upper bound on the number of possible orderings of the products $\{a_ia_j\}_{1\leq i\leq j\leq n}$ for n = 1, 2, 3, …: 1,1,2,10,114,2612,108664, …, which we see is significantly smaller than the upper bound found in the previous section for n ≥ 5.

### This Bound is Not Tight

It is straightforward to write a script that generates random numbers $a_1 > a_2 > \cdots > a_n > 0$ and determines the resulting ordering of the pairwise products $\{a_ia_j\}_{1\leq i\leq j\leq n}$. By doing this, we can verify that the upper bounds from the previous section are in fact tight when n ≤ 5. However, when n = 6, we find that 4 of the 2612 potential orderings do not seem to actually be attained by any choice of $a_1 > a_2 > \cdots > a_6 > 0$. One of these “problematic” orderings is the one that arises from the following matrix:

$\begin{bmatrix}21 & 20 & 19 & 18 & 17 & 11\\ * & 16 & 15 & 14 & 10 & 6\\ * & * & 13 & 12 & 8 & 5\\ * & * & * & 9 & 7 & 3\\ * & * & * & * & 4 & 2\\ * & * & * & * & * & 1\end{bmatrix}$

The problem here is that the above matrix implies the following 5 inequalities:

$a_1a_5 > a_2^2, \quad \ \ a_2a_4 > a_3^2, \quad \ \ a_2a_5 > a_4^2, \quad \ \ a_3a_4 > a_1a_6, \quad \text{and }\ \ \ a_3a_6 > a_5^2.$

However, multiplying the first four inequalities gives $a_1a_2^2a_3a_4^2a_5^2 > a_1a_2^2a_3^2a_4^2a_6$, so $a_5^2 > a_3a_6$, which contradicts the fifth inequality above. We can similarly prove that the other 3 seemingly problematic orderings are in fact not attainable, so there are exactly 2608 possible orderings in the n = 6 case.

I haven’t been able to compute the number of orderings when n ≥ 7, as my methods for obtaining upper and lower bounds are both much too slow in these cases. The best bounds that I have in the n = 7 case say that the number of orderings is between 50900 and 108664, inclusive.

Update [Feb. 13, 2014]: Giovanni Resta has improved the lower bound in the n = 7 case to 107498, which narrows the n = 7 region down considerably. I’ve also improved the upper bound to 108146 (see this improved version of the C script). In all likelihood, 107498 is the correct number of orderings in this case, and it’s the upper bound 108146 that needs to be further improved.

Update [Feb. 14, 2014]: This sequence is now in the OEIS. See A237749.

Update [Feb. 18, 2014]: Hans Havermann has found a couple of references that talk about this problem (in the language of Golomb rulers) and compute all values for n ≤ 7. See [3] and [4].

References

1. R. Hildebrand. Positive partial transpose from spectra. Phys. Rev. A, 76:052325, 2007. E-print: arXiv:quant-ph/0502170
2. R. M. Thrall. A combinatorial problem. Michigan Math. J., 1:81–88, 1952.
3. M. Beck, T. Bogart, and T. Pham. Enumeration of Golomb rulers and acyclic orientations of mixed graphs. Electron. J. Combin., 19:42, 2012. E-print: arXiv:1110.6154 [math.CO]
4. T. Pham. Enumeration of Golomb rulers. Master’s Thesis, San Francisco State University, 2011.

## In Search of a 4-by-11 Matrix

October 1st, 2013

IMPORTANT UPDATE [January 30, 2014]: I have managed to solve the 4-by-11 case: there is no such matrix! Details of the computation that led to this result, as well as several other related results, are given in [4]. See Table 3 in that paper for an updated list of which cases still remain open (the smallest open cases are now 5-by-11 and 6-by-10).

After spinning my wheels on a problem for far too long, I’ve decided that it’s time to enlist the help of the mathematical and programming geniuses of the world wide web. The problem I’m interested in asks for a 4-by-11 matrix whose columns satisfy certain relationships. While the conditions are relatively easy to state, the problem size seems to be just slightly too large for me to solve myself.

### The Problem

The question I’m interested in (for reasons that are explained later in this blog post) is, given positive integers p and s, whether or not there exists a p-by-s matrix M with the following three properties:

1. Every entry of M is a nonzero integer;
2. The sum of any two columns of M contains a 0 entry; and
3. There is no way to append a (s+1)th column to M so that M still has property 2.

In particular, I’m interested in whether or not such a matrix M exists when p = 4 and s = 11. But to help illustrate the above three properties, let’s consider the p = 3, s = 4 case first, where one such matrix M is:

$M = \begin{bmatrix}1 & -1 & 2 & -2 \\ 1 & -2 & -1 & 2 \\ 1 & 2 & -2 & -1\end{bmatrix}.$

The fact that M satisfies condition 2 can be checked by hand easily enough. For example, the sum of the first two columns of M is [0, -1, 3]T which contains a 0 entry, and it is similarly straightforward to check that the other 5 sums of two columns of M each contain a 0 entry as well.

Checking property 3 is slightly more technical (NP-hard, even), but is still doable in small cases such as this one. For the above example, suppose that we could add a 5th column (which we will call z = [z1, z2, z3]T) to M such that its sum with any of the first 4 columns has a 0 entry. By looking at M’s first column, we see that one of z’s entries must be -1 (and by the cyclic symmetry of the entries of the last 3 columns of M, we can assume without loss of generality that z1 = -1). By looking at the last 3 columns of M, we then see that either z2 = 2 or z3 = -2, either z2 = 1 or z3 = 2, and either z2 = -2 or z3 = 1. Since there is no way to simultaneously satisfy all 3 of these requirements, no such column z exists.

### What’s Known (and What Isn’t)

As I mentioned earlier, the instance of this problem that I’m really interested in is when p = 4 and s = 11. Let’s first back up and briefly discuss what is known for different values of p and s:

• If s ≤ p then M does not exist. To see this, simply note that property 3 can never be satisfied since you can always append one more column. If we denote the (i,j)-entry of M by mij and the i-th entry of the new column z by zi, then you can choose zi = -mii for i = 1, 2, …, s.
• Given p, the smallest value of s for which M exists is: (a) s = p+1 if p is odd, (b) s = p+2 if p = 4 or p ≡ 2 (mod 4), (c) s = p+3 if p = 8, and (d) s = p+4 otherwise. This result was proved in [1] (the connection between that paper and this blog post will be explained in the “Motivation” section below).
• If s > 2p then M does not exist. In this case, there is no way to satisfy property 2. This fact is trivial when p = 1 and can be proved for all p by induction (an exercise left to the reader?).
• If s = 2p then M exists. To see this claim, let the columns of M be the 2p different columns consisting only of the entries 1 and -1. To see that property 2 is satisfied, simply notice that each column is different, so for any pair of columns, there is a row in which one column is 1 and the other column is -1. To see that property 3 is satisfied, observe that any new column must also consist entirely of 1’s and -1’s. However, every such column is already a column of M itself, and the sum of a column with itself will not have any 0 entries.
• If s = 2p – 4 (and p ≥ 3) then M exists. There is an inductive construction (with the p = 3, s = 4 example from the previous section as the base case) that works here. More specifically, if we let Mp denote a matrix M that works for a given value of p and s = 2p – 4, we let Bp be the matrix from the s = 2p case above, and 1k denotes the row vector with k ones, then
$M_{p+1} = \begin{bmatrix}M_p & B_p \\ 1_{2^p-4} & -1_{2^p}\end{bmatrix}$

is a solution to the problem for p’ = p+1 and s’ = 2p+1 – 4.
• If 2p – 3 ≤ s ≤ 2p – 1 then M does not exist. This is a non-trivial result that follows from [2].

Given p, the above results essentially tell us the largest and smallest values of s for which a solution M to the problem exists. However, we still don’t really know much about when solutions exist for intermediate values of s – we just have scattered results that say a solution does or does not exist in certain specific cases, without really illuminating what is going on. The following table summarizes what we know about when solutions do and do not exist for small values of p and s (a check mark ✓ means that a solution exists, a dash – means no solution exists, and ? means we don’t know).

s \ p 1 2 3 4 5
1
2
3
4
5
6
7
8
9 ?
10 ?
11 ? ?
12
13
14
15
16
17 – 26
27 ?
28
29
30
31
32

The table above shows why I am interested in the p = 4, s = 11 case: it is the only case when p ≤ 4 whose solution still is not known. The other unknown cases (i.e., p = 5 and s ∈ {9,10,11,27}, and far too many to list when p ≥ 6) would be interesting to solve as well, but are a bit lower-priority.

### Some Simplifications

Some assumptions about the matrix M can be made without loss of generality, in order to reduce the search space a little bit. For example, since the values of the entries of M don’t really matter (other than the fact that they come in positive/negative pairs), the first column of M can always be chosen to consist entirely of ones (or any other value). Similarly, permuting the rows or columns of M does not affect whether or not it satisfies the three desired properties, so you can assume (for example) that the first row is in non-decreasing order.

Finally, since there is no advantage to having the integer k present in M unless -k is also present somewhere in M (i.e., if M does not contain any -k entries, you could always just replace every instance of k by 1 without affecting any of the three properties we want), we can assume that the entries of M are between -floor(s/2) and floor(s/2), inclusive.

### Motivation

The given problem arises from unextendible product bases (UPBs) in quantum information theory. A set of pure quantum states $|v_1\rangle, \ldots, |v_s\rangle \in \mathbb{C}^{d_1} \otimes \cdots \otimes \mathbb{C}^{d_p}$ forms a UPB if and only if the following three properties hold:

1. (product) Each state $|v_j\rangle$ is a product state (i.e., can be written in the form $|v_j\rangle = |v_j^{(1)}\rangle \otimes \cdots \otimes |v_j^{(p)}\rangle$, where $|v_j^{(i)}\rangle \in \mathbb{C}^{d_i}$ for all i);
2. (basis) The states are mutually orthogonal (i.e., $\langle v_i | v_j \rangle = 0$ for all i ≠ j); and
3. (unextendible) There does not exist a product state $|z\rangle$ with the property that $\langle z | v_j \rangle = 0$ for all j.

UPBs are useful because they can be used to construct quantum states with very strange entanglement properties [3], but their mathematical structure still isn’t very well-understood. While we can’t really expect an answer to the question of what sizes of UPBs are possible when the local dimensions $d_1, \ldots, d_p$ are arbitrary (even just the minimum size of a UPB is still not known in full generality!), we might be able to hope for an answer if we focus on multi-qubit systems (i.e., the case when $d_1 = \cdots = d_p = 2$).

In this case, the 3 properties above are isomorphic in a sense to the 3 properties listed at the start of this post. We associate each state $|v_j\rangle$ with the j-th column of the matrix M. To each state in the product state decomposition of $|v_j\rangle$, we associate a unique integer in such a way that orthogonal states are associated with negatives of each other. The fact that $\langle v_i | v_j \rangle = 0$ for all i ≠ j is then equivalent to the requirement that te sum of any two columns of M has a 0 entry, and unextendibility of the product basis corresponds to not being able to add a new column to M without destroying property 2.

Thus this blog post is really asking whether or not there exists an 11-state UPB on 4 qubits. In order to illustrate this connection more explicitly, we return to the p = 3, s = 4 example from earlier. If we associate the matrix entries 1 and -1 with the orthogonal standard basis states $|0\rangle, |1\rangle \in \mathbb{C}^2$ and the entries 2 and -2 with the orthogonal states $|\pm\rangle := (|0\rangle \pm |1\rangle)/\sqrt{2}$, then the matrix M corresponds to the following set of s = 4 product states in $\mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \mathbb{C}^2$:

$|0\rangle|0\rangle|0\rangle, \quad |1\rangle|-\rangle|+\rangle, \quad |+\rangle|1\rangle|-\rangle, \quad|-\rangle|+\rangle|1\rangle.$

The fact that these states form a UPB is well-known – this is the “Shifts” UPB from [3], and was one of the first UPBs found.

References

1. N. Johnston. The minimum size of qubit unextendible product bases. In Proceedings of the 8th Conference on the Theory of Quantum Computation, Communication and Cryptography (TQC), 2013. E-print: arXiv:1302.1604 [quant-ph], 2013.
2. L. Chen and D. Ž. Ðjoković. Separability problem for multipartite states of rank at most four. J. Phys. A: Math. Theor., 46:275304, 2013. E-print: arXiv:1301.2372 [quant-ph]
3. C. H. Bennett, D. P. DiVincenzo, T. Mor, P. W. Shor, J. A. Smolin, and B. M. Terhal. Unextendible product bases and bound entanglement. Phys. Rev. Lett., 82:5385–5388, 1999. E-print: arXiv:quant-ph/9808030
4. N. Johnston. The structure of qubit unextendible product basesJournal of Physics A: Mathematical and Theoretical, 47:424034, 2014. E-print: arXiv:1401.7920 [quant-ph], 2014.

## The Spectrum of the Partial Transpose of a Density Matrix

July 3rd, 2013

It is a simple fact that, given any density matrix (i.e., quantum state) $\rho\in M_n$, the eigenvalues of $\rho$ are the same as the eigenvalues of $\rho^T$ (the transpose of $\rho$). However, strange things can happen if we instead only apply the transpose to one half of a quantum state. That is, if $\rho\in M_m \otimes M_n$ then its eigenvalues in general will be very different from the eigenvalues of $(id_m\otimes T)(\rho)$, where $id_m$ is the identity map on $M_m$ and $T$ is the transpose map on $M_n$ (the map $id_m\otimes T$ is called the partial transpose).

In fact, even though $\rho$ is positive semidefinite (since it is a density matrix), the matrix $(id_m\otimes T)(\rho)$ in general can have negative eigenvalues. To see this, define $p:={\rm min}\{m,n\}$ and let $\rho=|\psi\rangle\langle\psi|$, where

$|\psi\rangle=\displaystyle\frac{1}{\sqrt{p}}\sum_{j=1}^{p}|j\rangle\otimes|j\rangle$

is the standard maximally-entangled pure state. It then follows that

$(id_m\otimes T)(\rho)=\displaystyle\frac{1}{p}\sum_{i,j=1}^{p}|i\rangle\langle j|\otimes|j\rangle\langle i|$,

which has $p(p+1)/2$ eigenvalues equal to $1/p$$p(p-1)/2$ eigenvalues equal to $-1/p$, and $p|m-n|$ eigenvalues equal to $0$.

The fact that $(id_m\otimes T)(\rho)$ can have negative eigenvalues is another way of saying that the transpose map is positive but not completely positive, and thus plays a big role in entanglement theory. In this post we consider the question of how exactly the partial transpose map can transform the eigenvalues of $\rho$:

Question. For which ordered lists $\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_{mn}\in\mathbb{R}$ does there exist a density matrix $\rho$ such that $(id_m\otimes T)(\rho)$ has eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_{mn}$?

### The Answer for Pure States

In the case when $\rho$ is a pure state (i.e., has rank 1), we can completely characterize the eigenvalues of $(id_m\otimes T)(\rho)$ by making use of the Schmidt decomposition. In particular, we have the following:

Theorem 1. Let $|\phi\rangle$ have Schmidt rank $r$ and Schmidt coefficients $\alpha_1\geq\alpha_2\geq\cdots\geq\alpha_r>0$. Then the spectrum of $(id_m\otimes T)(|\phi\rangle\langle\phi|)$ is

$\{\alpha_i^2 : 1\leq i\leq r\}\cup\{\pm\alpha_i\alpha_j:1\leq i,

together with the eigenvalue $0$ with multiplicity $p|n-m|+p^2-r^2$.

Proof. If $|\phi\rangle$ has Schmidt decomposition

$\displaystyle|\phi\rangle=\sum_{i=1}^r\alpha_i|a_i\rangle\otimes|b_i\rangle$

then

$\displaystyle(id_m\otimes T)(|\phi\rangle\langle\phi|)=\sum_{i,j=1}^r\alpha_i\alpha_j|a_i\rangle\langle a_j|\otimes|b_j\rangle\langle b_i|.$

It is then straightforward to verify, for all $1\leq i, that:

• $|a_i\rangle\otimes|b_i\rangle$ is an eigenvector with eigenvalue $\alpha_i^2$;
• $|a_i\rangle\otimes|b_j\rangle\pm|a_j\rangle\otimes|b_i\rangle$ is an eigenvector with eigenvalue $\pm\alpha_i\alpha_j$; and
• ${\rm rank}\big((id_m\otimes T)(|\phi\rangle\langle\phi|)\big)= r^2$, from which it follows that the remaining $p|n-m|+p^2-r^2$ eigenvalues are $0$.

Despite such a simple characterization in the case of rank-1 density matrices, there is no known characterization for general density matrices, since eigenvalues aren’t well-behaved under convex combinations.

### The Number of Negative Eigenvalues

Instead of asking for a complete characterization of the possible spectra of $(id_m\otimes T)(\rho)$, for now we focus on the simpler question that asks how many of the eigenvalues of $(id_m\otimes T)(\rho)$ can be negative. Theorem 1 answers this question when $\rho=|\phi\rangle\langle\phi|$ is a pure state: the number of negative eigenvalues is $r(r-1)/2$, where r is the Schmidt rank of $|\phi\rangle$. Since $r\leq p$, it follows that $(id_m\otimes T)(\rho)$ has at most $p(p-1)/2$ negative eigenvalues when $\rho$ is a pure state.

It was conjectured in [1] that a similar fact holds for general (not necessarily pure) density matrices $\rho$ as well. In particular, they conjectured that if $\rho\in M_n\otimes M_n$ then $(id_n\otimes T)(\rho)$ has at most $n(n-1)/2$ negative eigenvalues. However, this conjecture is easily shown to be false just by randomly-generating many density matrices $\rho$ and then counting the number of negative eigenvalues of $(id_n\otimes T)(\rho)$; density matrices whose partial transposes have more than $n(n-1)/2$ negative eigenvalues are very common.

In [2,3], it was shown that if $\rho\in M_m\otimes M_n$ then $(id_m\otimes T)(\rho)$ can not have more than $(m-1)(n-1)$ negative eigenvalues. In [4], this bound was shown to be tight when ${\rm min}\{m,n\}=2$ by explicitly constructing density matrices $\rho\in M_2\otimes M_n$ such that $(id_2\otimes T)(\rho)$ has $n-1$ negative eigenvalues. Similarly, this bound was shown to be tight via explicit construction when $m=n=3$ in [3]. Finally, it was shown in [5] that this bound is tight in general. That is, we have the following result:

Theorem 2. The maximum number of negative eigenvalues that $(id_m\otimes T)(\rho)$ can have when $\rho\in M_m\otimes M_n$ is $(m-1)(n-1)$.

It is worth pointing out that the method used in [5] to prove that this bound is tight is not completely analytic. Instead, a numerical method was presented that is proved to always generate a density matrix $\rho\in M_m\otimes M_n$ such that $(id_m\otimes T)(\rho)$ has $(m-1)(n-1)$ negative eigenvalues. Code that implements this numerical procedure in MATLAB is available here, but no general analytic form for such density matrices is known.

### Other Bounds on the Spectrum

Unfortunately, not a whole lot more is known about the spectrum of $(id_m\otimes T)(\rho)$. Here are some miscellaneous other results that impose certain restrictions on its maximal and minimal eigenvalues (which we denote by $\lambda_\textup{max}$ and $\lambda_\textup{min}$, respectively):

Theorem 3 [3]. $1\geq\lambda_\textup{max}\geq\lambda_\textup{min}\geq -1/2$.

Theorem 4 [2]. $\lambda_\textup{min}\geq\lambda_\textup{max}(1-{\rm min}\{m,n\})$.

Theorem 5 [6]. If $(id_m\otimes T)(\rho)$ has $q$ negative eigenvalues then

$\displaystyle\lambda_\textup{min}\geq\lambda_\textup{max}\Big(1-\big\lceil\tfrac{1}{2}\big(m+n-\sqrt{(m-n)^2+4q-4}\big)\big\rceil\Big)$ and

$\displaystyle\lambda_\textup{min}\geq\lambda_\textup{max}\Big(1-\frac{mn\sqrt{mn-1}}{q\sqrt{mn-1}+\sqrt{mnq-q^2}}\Big)$.

However, these bounds in general are fairly weak and the question of what the possible spectra of $(id_m\otimes T)(\rho)$ are is still far beyond our grasp.

References

1. R. Xi-Jun, H. Yong-Jian, W. Yu-Chun, and G. Guang-Can. Partial transposition on bipartite system. Chinese Phys. Lett., 25:35, 2008.
2. N. Johnston and D. W. Kribs. A family of norms with applications in quantum information theory. J. Math. Phys., 51:082202, 2010. E-print: arXiv:0909.3907 [quant-ph]
3. S. Rana. Negative eigenvalues of partial transposition of arbitrary bipartite states. Phys. Rev. A, 87:054301, 2013. E-print: arXiv:1304.6775 [quant-ph]
4. L. Chen, D. Z. Djokovic. Qubit-qudit states with positive partial transpose. Phys. Rev. A, 86:062332, 2012. E-print: arXiv:1210.0111 [quant-ph]
5. N. Johnston. Non-positive-partial-transpose subspaces can be as large as any entangled subspace. Phys. Rev. A, 87:064302, 2013. E-print: arXiv:1305.0257 [quant-ph]
6. N. Johnston. Norms and Cones in the Theory of Quantum Entanglement. PhD thesis, University of Guelph, 2012.

## The Minimal Superpermutation Problem

April 10th, 2013

Imagine that there is a TV series that you want to watch. The series consists of n episodes, with each episode on a single DVD. Unfortunately, however, the DVDs have become mixed up and the order of the episodes is in no way marked (and furthermore, the episodes of the TV show are not connected by any continuous storyline – there is no way to determine the order of the episodes just from watching them).

Suppose that you want to watch the episodes of the TV series, consecutively, in the correct order. The question is: how many episodes must you watch in order to do this?

To illustrate what we mean by this question, suppose for now that n = 2 (i.e., the show was so terrible that it was cancelled after only 2 episodes). If we arbitrarily label one of the episodes “1” and the other episode “2”, then we could watch the episodes in the order “1”, “2”, and then “1” again. Then, regardless of which episode is really the first episode, we’ve seen the two episodes consecutively in the correct order. Furthermore, this is clearly minimal – there is no way to watch fewer than 3 episodes while ensuring that you see both episodes in the correct order.

So what is the minimal number of episodes we must watch for a TV show consisting of n episodes? Somewhat surprisingly, no one knows. So let’s discuss what is known.

### Minimal Superpermutations

Rephrased a bit more mathematically, we are interested in finding a shortest possible string on the symbols “1”, “2”, …, “n” that contains every permutation of those symbols as a contiguous substring. We call a string that contains every permutation in this way a superpermutation, and one of minimal length is called a minimal superpermutation. Minimal superpermutations when n = 1, 2, 3, 4 are easily found via brute-force computer search, and are presented here:

n Minimal Superpermutation Length
1 1 1
2 121 3
3 123121321 9
4 123412314231243121342132413214321 33

By the time n = 5, the strings we are looking for are much too long to find via brute-force. However, the strings in the n ≤ 4 cases provide some insight that we can hope might generalize to larger n. For example, there is a natural construction that allows us to construct a short superpermutation on n+1 symbols from a short superpermutation on n symbols (which we will describe in the next section), and this construction gives the minimal superpermutations presented in the above table when n ≤ 4.

Similarly, the minimal superpermutations in the above table can be shown via brute-force to be unique (up the relabeling the characters – for example, we don’t count the string “213212312” as distinct from “123121321”, since they are related to each other simply by interchanging the roles of “1” and “2”). Are minimal superpermutations unique for all n?

### Minimal Length

A trivial lower bound on the length of a superpermutation on n symbols is n! + n – 1, since it must contain each of the n! permutations as a substring – the first permutation contributes a length of n characters to the string, and each of the remaining n! – 1 permutations contributes a length of at least 1 character more.

It is not difficult to improve this lower bound to n! + (n-1)! + n – 2 (I won’t provide a proof here, but the idea is to note that when building the superpermutation, you can not add more than n-1 permutations by appending just 1 character each to the string – you eventually have to add 2 or more characters to add a permutation that is not already present). In fact, this argument can be stretched further to show that n! + (n-1)! + (n-2)! + n – 3 is a lower bound as well (a rough proof is provided here). However, the same arguments do not seem to extend to lower bounds like n! + (n-1)! + (n-2)! + (n-3)! + n – 4 and so on.

There is also a trivial upper bound on the length of a minimal superpermutation: n×n!, since this is the length of the string obtained by writing out the n! permutations in order without overlapping. However, there is a well-known construction of small superpermutations that provides a much better upper bound, which we now describe.

Suppose we know a small superpermutation on n symbols (such as one of the superpermutations provided in the table in the previous section) and we want to construct a small superpermutation on n+1 symbols. To do so, simply replace each permutation in the n-symbol superpermutation by: (1) that permutation, (2) the symbol n+1, and (3) that permutation again. For example, if we start with the 2-symbol superpermutation “121”, we replace the permutation “12” by “12312” and we replace the permutation “21” by “21321”, which results in the 3-symbol superpermutation “123121321”. The procedure for constructing a 4-symbol superpermutation from this 3-symbol superpermutation is illustrated in the following diagram:

A diagram that demonstrates how to construct a small superpermutation on 4 symbols from a small superpermutation on 3 symbols.

It is a straightforward inductive argument to show that the above method produces n-symbol superpermutations of length $\sum_{k=1}^nk!$ for all n. Although it has been conjectured that this superpermutation is minimal [1], this is only known to be true when n ≤ 4.

### Uniqueness

As a result of minimal superpermutations being unique when n ≤ 4, it has been conjectured that they are unique for all n [1]. However, it turns out that there are in fact many superpermutations of the conjectured minimal length – the main result of [2] shows that there are at least

$\displaystyle\prod_{k=1}^{n-4}(n-k-2)^{k\cdot k!}$

distinct n-symbol superpermutations of the conjectured minimal length. For n ≤ 4, this formula gives the empty product (and thus a value of 1), which agrees with the fact that minimal superpermutations are unique in these cases. However, the number of distinct superpermutations then grows extremely quickly with n: for n  = 5, 6, 7, 8, there are at least 2, 96, 8153726976, and approximately 3×1050 superpermutations of the conjectured minimal length. The 2 such superpermutations in the n = 5 case are as follows (each superpermutation has length 153 and is written on two lines):

12345123415234125341235412314523142531423514231542312453124351243152431254312
1345213425134215342135421324513241532413524132541321453214352143251432154321

and

12345123415234125341235412314523142531423514231542312453124351243152431254312
1354213524135214352134521325413251432513425132451321543215342153241532145321

Similarly, a text file containing all 96 known superpermutations of the expected minimal length 873 in the n = 6 case can be viewed here. It is unknown, however, whether or not these superpermutations are indeed minimal or if there are even more superpermutations of the conjectured minimal length.

Update [Aug. 13, 2014]: Ben Chaffin has shown that minimal superpermutations in the n = 5 case have length 153, and he has also shown that there are exactly 8 (not just 2) distinct minimal superpermutations in this case. See the write up here.

IMPORTANT UPDATE [August 22, 2014]: Robin Houston has disproved the minimal superpermutation conjecture for all n ≥ 6. See here.

References

1. D. Ashlock and J. Tillotson. Construction of small superpermutations and minimal injective superstrings. Congressus Numerantium, 93:91–98, 1993.
2. N. Johnston. Non-uniqueness of minimal superpermutations. Discrete Mathematics, 313:1553–1557, 2013.

Other Random Links Related to This Problem

1. A180632 – the main OEIS entry for this problem