If $\text{Tr}_B \rho^{AB}$ is almost pure, then $\rho^{AB}$ is almost a product state?

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!

• @NorbertSchuch If $\rho^{AB}$ entangled, then $\rho^A$ cannot be pure. I must have mistaken "mixed" and "not pure". I changed the title. – user114158 Feb 10 '20 at 21:38
• This is still not correct. What you mean (I guess - and what the question asks) is not separable, but product. – Norbert Schuch Feb 11 '20 at 7:38
• @NorbertSchuch I see, sorry about this imprecision! Thanks :) – user114158 Feb 11 '20 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 1\rangle\langle 1\vert\otimes\omega^1_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 1\rangle\langle 1\vert\otimes\omega^1_B + O(\delta) \end{align}

Thus, we have $$\lVert \rho_{AB} - \vert 1\rangle\langle 1\vert\otimes\omega^1_B \rVert_1 \leq O(\delta)$$