Let $\rho^{AB}$ be a bipartite state, and let $\rho^A$ denote the partial trace. Suppose $$ \lVert \rho^A - |\sigma\rangle\langle\sigma|^A \rVert_1 \leq \varepsilon $$ for some pure state $|\sigma\rangle$. Here $\lVert\cdot\rVert_1$ denotes the trace norm. Is it true that $$ \lVert \rho^{AB} - |\sigma\rangle\langle\sigma|^A\otimes \rho^B \rVert_1 = O(\varepsilon) $$ This holds in the exact case, that is if the partial trace of a bipartite state is pure, then it is a product state (see reference). I am wondering if this approximate version is true as well. Any thoughts are appreciated!

  • $\begingroup$ @NorbertSchuch If $\rho^{AB}$ entangled, then $\rho^A$ cannot be pure. I must have mistaken "mixed" and "not pure". I changed the title. $\endgroup$ – user114158 Feb 10 at 21:38
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    $\begingroup$ This is still not correct. What you mean (I guess - and what the question asks) is not separable, but product. $\endgroup$ – Norbert Schuch Feb 11 at 7:38
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    $\begingroup$ @NorbertSchuch I see, sorry about this imprecision! Thanks :) $\endgroup$ – user114158 Feb 11 at 13:13

Let us purify $\rho_{AB}$ to $\psi_{ABR} \in H_A\otimes H_B\otimes H_R$. You are given that

$$ \lVert \rho_A - |\sigma\rangle\langle\sigma|_A \rVert_1 \leq \epsilon $$

By a tight version of Fannes inequality, we have

$$S(\rho_A) \leq \epsilon\log d + H(\epsilon, 1-\epsilon), $$

where $S(A)$ is the von Neumann entropy, $H(X)$ is the binary entropy and $d$ is the dimension of $H_A$. This gives us a lower bound on the largest eigenvalue of $\rho_A$ i.e. $\lambda_1 > 1 - \delta$. I have not worked out $\delta = \delta(\epsilon)$ here but I think this should be possible.

Meanwhile, the Schmidt decomposition of $\psi_{ABR}$ is

$$\psi_{ABR} = \sum_i \sqrt{\lambda_i}\vert i\rangle_A\otimes \vert \tilde{i}\rangle_{BR}$$

Taking the trace over $R$ of $\vert\psi\rangle\langle\psi\vert_{ABR}$ and denoting $\text{Tr}_R \vert\tilde{i}\rangle\langle\tilde{i}\vert = \omega^i_B$, we have

\begin{align} \rho_{AB} &= \lambda_1\vert i\rangle\langle i\vert\otimes\omega^i_B + \sum_{(i,j)\neq (1,1)} \sqrt{\lambda_i\lambda_j} \vert i\rangle\langle j\vert\otimes\text{Tr}_R(\vert\tilde{i}\rangle\langle\tilde{j}\vert) \\ &= \lambda_1\vert i\rangle\langle i\vert\otimes\omega^i_B + O(\delta) \end{align}

Thus, we have $$\lVert \rho_{AB} - \vert i\rangle\langle i\vert\otimes\omega^i_B \rVert_1 \leq O(\delta)$$

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