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

Question

Let $rho^{AB}$ be a bipartite state, and let $rho^A$ denote the partial trace. Suppose 
$$ lVert rho^A - |sigmaranglelanglesigma|^A rVert_1 leq varepsilon $$ 
for some pure state $|sigmarangle$. Here $lVertcdotrVert_1$ denotes the trace norm. Is it true that
$$ lVert rho^{AB} - |sigmaranglelanglesigma|^Aotimes rho^B rVert_1 = O(varepsilon) $$
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!

rnva · Answer

Let us purify $rho_{AB}$ to $psi_{ABR} in H_Aotimes H_Botimes H_R$. You are given that
$$ lVert rho_A - |sigmaranglelanglesigma|_A rVert_1 leq epsilon $$
By a tight version of Fannes inequality, we have
$$S(rho_A) leq epsilonlog d + H(epsilon, 1-epsilon), $$
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
$$psi_{ABR} = sum_i sqrt{lambda_i}vert irangle_Aotimes vert tilde{i}rangle_{BR}$$
Taking the trace over $R$ of $vertpsiranglelanglepsivert_{ABR}$ and denoting $text{Tr}_R verttilde{i}ranglelangletilde{i}vert = omega^i_B$, we have
begin{align}
rho_{AB} &= lambda_1vert 1ranglelangle 1vertotimesomega^1_B + sum_{(i,j)neq (1,1)} sqrt{lambda_ilambda_j} vert iranglelangle jvertotimestext{Tr}_R(verttilde{i}ranglelangletilde{j}vert) 
&= lambda_1vert 1ranglelangle 1vertotimesomega^1_B + O(delta)
end{align}
Thus, we have
$$lVert rho_{AB} -  vert 1ranglelangle 1vertotimesomega^1_B rVert_1 leq O(delta)$$

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

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