Is the square of a separable state again separable?

Question

Famously, a bipartite quantum state $\rho\in\mathbb C^{n\times n}\otimes\mathbb C^{m\times m}$ is defined to be separable if and only if $$ \rho\in{\rm conv}\{ \omega\otimes\sigma : \omega\in\mathbb C^{n\times n},\sigma\in\mathbb C^{m\times m}\text{ density matrices} \}\,, $$ i.e., if and only if $\rho=\sum_{j=1}^k\lambda_j\omega_j\otimes\sigma_j$ with states $\omega_j,\sigma_j$, and $\lambda_j\geq 0$, $\sum_{j=1}^k\lambda_j=1$. This notion of separability can be generalized to positive semi-definite matrices $A\in\mathbb C^{n\times n}\otimes\mathbb C^{m\times m}$ by saying that $A$ is separable if there exist positive semi-definite matrices $B_1,\ldots,B_k\in\mathbb C^{n\times n}$ and $C_1,\ldots,C_k\in\mathbb C^{m\times m}$ such that $$ A=\sum_{j=1}^k B_j\otimes C_j\,. $$

One interesting question is which classes of maps preserve this separability structure. Such separability-preserving maps include: entanglement-breaking channels, separable operations (e.g., LOCC operations), or the partial transpose. Going one step further, but one could relax this notion to include non-linear operations, as well, eg., squaring the density matrix, i.e., applying the map $A \mapsto A^2$. This—obviously non-linear—map is natural to consider as it is positive and unital. This raises the question:

If $\rho$ is separable, is $\rho^2$ necessarily separable?

Based on some easy observations, this claim holds in specific cases:

• The square of a product state is trivially still a product state; in particular, the claim holds for all pure states $\rho$ (i.e., all extreme points of the separable states).
• If $\rho$ is diagonal in a product basis, then $\rho^2$ is also diagonal and hence separable.

However, does squaring preserve separability in general?

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(This is a Q&A style question meant as a contribution to the list of counterexamples in quantum information)

Answer 1

It turns out that just because $\rho$ is separable, $\rho^2$ need not be separable. For a counterexample, consider $$ \rho = \frac{1}{2} ( |00\rangle \langle 00|\,+ \,|\!+\!+\rangle \langle+\! +\!|), $$ where $|+\rangle = \frac{1}{\sqrt{2}} (1, 1)^\top$. In matrix form, this state is given by $$ \rho = \frac{1}{8} \begin{pmatrix} 5 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}, $$ and it is separable by construction (or by PPT). Now consider its square $$ \rho^2 = \frac{1}{16} \begin{pmatrix} 7 & 2 & 2 & 2 \\ 2 & 1 & 1 & 1 \\ 2 & 1 & 1 & 1 \\ 2 & 1 & 1 & 1 \end{pmatrix}, $$ and notice that the partial transpose of $\rho^2$ features a negative block $$ \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} $$ (up to a global factor) which has a negative determinant. Hence $\rho^2$ is entangled due to PPT.

Answer 2

Let me give an alternative answer, with three goals: First, to show that this behavior (i.e. that for a separable $\rho$, $\rho^2$ is entangled) is very likely (and if one also asks for higher powers, generic -- that is, any random separable state will have entangled powers); second, that this also happens in the multi-partite setting; and third, to give a physics-based intuition why this is the case.

Let me start with the physics-based intuition, as this is what led me here: We can think of $\rho$ as the Gibbs state of some Hamiltonian, $\rho=\exp[-\beta H]$.$^1$ Then, $\rho^2$ is the Gibbs state at $\beta'=2\beta$, or half the temperature.

This is saying that when we compare $\rho^2$ and $\rho$, we compare the entanglement of a state at some temperature ($\rho^2$) and at twice the temperature ($\rho$). Now, if we go to very high temperature, the state is getting closer and closer (at an exponential rate) to a maximally mixed state. On the other hand, it is also well known that if a state is sufficiently close to a maximally mixed state, it is separable.

We can use this to "reverse engineer" examples:

1. Start from an entangled state $\sigma$, and take its eigenvalue decomposition. Then, replace any eigenvalue $\lambda$ by $\lambda^{1/n}$. For a sufficiently large $n$, the resulting state $\rho$ will become separable (as it approaches the maximally mixed state); conversely, $\rho$ then forms an example for which $\rho^n$ is separable. (From there, we can also find a pair of states $\tau$ where $\tau$ is separable while $\tau^2$ is entangled, since the transition must happen at some power $n$).
2. (Genericness:) Take an arbitrary separable state $\rho$, with the only constraint that its largest eigenvalue is unique, and the corresponding eigenvector is entangled.$^2$ Then, for increasing $n$, $\rho^n$ approaches the projector onto that largest eigenvector, and thus, for large enough $n$, $\rho^n$ will be entangled.

Finally, the very same arguments also apply in the multipartite setting, since a state which is close enough to the identity is also separable in a multipartite sense.

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$^1$ There's some caveats if $\rho$ does not have full rank, but this is a subclass of measure zero. Indeed, two-qubit states with Bell rank two form a class of counterexamples, as they are all entangled independent of the weight (unless both weights are equal).

$^2$ Any random separable $\rho$ will have this property. One way to get a subclass of random separable states $\rho$ (with that property) is to start with a random $\rho$ and mix it with the maximally mixed state until it becomes separable.