Unitary over bipartite states that can turn a non-product state into a product state

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Consider a bipartite quantum state $\rho_{AB}$ over a product of finite-dimensional Hilbert spaces $\mathcal{H}_A \otimes \mathcal{H}_B$. Does there exists a unitary $U$ over $\mathcal{H}_A \otimes \mathcal{H}_B$ such that:

  • For any bipartite $\rho_{AB}$, the reduced density matrix over $A$ is preserved $\mathrm{tr}_B{[U(\rho_{AB})U^*]} = \rho_A$.
  • For some a priori given $\sigma_{AB}$, the output is a product state: $U(\sigma_{AB})U^* = \sigma_A \otimes \omega_B$ (where $\omega_B$ can be anything).

I'm also interested in the case where such properties might only hold up to some arbitrary $\epsilon >0$ accuracy.

Artemy

Posted 2020-12-19T20:14:30.060

Reputation: 173

The way you currently write the question you can just take any unitary that acts trivially on system A. For example the identity on both systems. Then the first property is immediate and the second property holds for any product state. – Rammus – 2020-12-19T21:06:23.173

@Rammus I clarified the question a bit -- $\sigma_{AB}$ is some fixed a priori specified state (which will in general not be a product state) – Artemy – 2020-12-19T21:48:43.817

Answers

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The first condition is satisfied for example by unitaries of the form $U = e^{i\theta}I_A \otimes U_B$ where $I_A$ is identity on subsystem $A$, $U_B$ is any unitary on subsystem $B$ and the phase factor $e^{i\theta}$ is irrelevant.


Let us consider the second condition. It turns out that the condition cannot be guaranteed for all states $\sigma_{AB}$. More precisely, there are states $\sigma_{AB}$ such that for every unitary $U$ and every state $\omega_B$ of subsystem $B$ we have $U\sigma_{AB}U^* \ne \sigma_A \otimes \omega_B$. This is a consequence of two facts: that unitary transformations preserve eigenvalues of density matrices and that spectra (sets of eigenvalues) of generic density matrices cannot be reproduced by spectra of product states.

More formally, we can state the first fact by saying that for every unitary $U$, $\lambda$ is an eigenvalue of $\rho$ if and only if it is an eigenvalue of $U\rho U^*$.

In order to show the second fact, first note that the eigenvalues of an $n \times n$ density matrix lie in an $(n-1)$-simplex. Let $n_A = \dim \mathcal{H}$ and $n_B = \dim \mathcal{H}_B$. If $\lambda^A_i$ denotes the eigenvalues of $\sigma_A$ and $\lambda^B_j$ denotes the eigenvalues of $\omega_B$ then the eigenvalues of $\sigma_A \otimes \omega_B$ are the products $\lambda^{AB}_{ij} = \lambda^A_i \lambda^B_j$. Thus, the eigenvalues of $\sigma_A \otimes \omega_B$ lie in the Cartesian product of two simplices that can be described using $(n_A - 1) (n_B - 1)$ real parameters. On the other hand, the eigenvalues of an arbitrary joint density matrix on systems $A$ and $B$ lie in an $(n_An_B - 1)$-simplex. Thus, by a simple parameter counting argument we see that the set of spectra of product states is a measure zero subset of the spectra of arbitrary states.

For a concrete example, suppose that $A$ and $B$ are qubits and that $\sigma_{AB}$ has eigenvalues $0, \frac{1}{4}, \frac{1}{4}, \frac{1}{2}$. Note that there do not exist two sets of numbers $\{\lambda^A_1, \lambda^A_2\}$ and $\{\lambda^B_1, \lambda^B_2\}$ such that

$$ \{\lambda^A_1\lambda^B_1, \lambda^A_1\lambda^B_2, \lambda^A_2\lambda^B_1, \lambda^A_2\lambda^B_2\} = \{0, \frac{1}{4}, \frac{1}{4}, \frac{1}{2}\}. $$

Consequently, there is no unitary $U$ such that $U\sigma_{AB}U^* = \sigma_A \otimes \omega_B$.

Adam Zalcman

Posted 2020-12-19T20:14:30.060

Reputation: 4 307

Still working through your post, thanks. However, I don't think its true that "by the cyclic property of the trace the first property is satisfied by every unitary". The cyclic property of trace holds for the full trace, not for the partial trace. Consider for example the unitary that simply swaps the states of $A$ and $B$ (assuming those spaces have the same dimension). Clearly this doesn't obey the first property in general. – Artemy – 2020-12-20T00:25:52.850

Oops, you are right! I'll edit my answer. – Adam Zalcman – 2020-12-20T00:29:16.703