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Consider two quantum states $\rho_A$ and $\sigma_A$, and define the classical-quantum state over a classical binary system $B$ and $A$, $$\omega_{AB}^\epsilon :=\epsilon \vert 0 \rangle \langle 0 \vert \rho_A + (1-\epsilon) \vert 1 \rangle \langle 1 \vert \sigma_A.$$ I am interested in the conditional von Neumann entropy $S(B\vert A)_{\omega_{AB}^\epsilon} = S(\omega_{AB}^\epsilon)- S(\omega_{A}^\epsilon)$. This is related to quantum state discrimination.

If $\rho_A$ and $\sigma_A$ are orthogonal, it is easy to show that $S(B\vert A)=0$. In case they are not orthogonal, I seek a lower bound of the form $$S(B\vert A) \ge -c_1\epsilon \ln \epsilon + c_2 \epsilon,$$ where $c_1>0$ and $c_2$ is any constant. I am particularly interested in the case where $c_1$ does not depend on the dimensionality of $A$, and applies even in the countably-infinite dimensional case.

One can derive a bound of this type for a classical system in the following way. Define $c = \max_a [\min\{\rho_A(a), \sigma_A(a)\}]$, and let $\hat{a}$ indicate the state of $A$ such that $c = \min\{\rho_A(\hat{a}), \sigma_A(\hat{a})\}$. Then, write \begin{align} S(B|A)&=\sum_a\left[\epsilon \rho_A(a)\ln\frac{\epsilon \rho_A(a)+(1-\epsilon)\sigma_A(a)}{\epsilon \rho_A(a)} + (1- \epsilon) \sigma_A(a)\ln\frac{\epsilon \rho_A(a)+(1-\epsilon)\sigma_A(a)}{(1- \epsilon)\sigma_A(a)}\right]\\ &\ge \epsilon \rho_A(\hat{a})\ln\frac{\epsilon \rho_A(\hat{a})+(1-\epsilon)\sigma_A(\hat{a})}{\epsilon \rho_A(\hat{a})}\\ &\ge \epsilon \rho_A(\hat{a})\ln\frac{c}{\epsilon \rho_A(\hat{a})}\\ &= -\epsilon \rho_A(\hat{a})\ln\epsilon +\epsilon \rho_A(\hat{a})\ln \frac{c}{\rho_A(\hat{a})} \end{align}

However, I have not seen something similar for the quantum case.

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I found a very nice result in Audenaert, "Quantum skew divergence", J of Mathematical Physics, 2014 (also cited in Kim and Ruskai, "Bounds on the concavity of quantum entropy", J of Mathematical Physics, 2014). Theorem 14 in Audenaert bounds the Holevo information between $A$ and $B$ as $$\chi(B;A) \le h(\epsilon) T(\rho_A, \sigma_A) $$ where $h(\epsilon)=-\epsilon \ln \epsilon - (1-\epsilon) \ln (1-\epsilon)$ is the binary Shannon entropy and $T(\rho_A, \sigma_A) = \frac{1}{2} \lVert \rho_A-\sigma_A\rVert_1$ is the trace distance. For our classical-quantum state, $S(B|A)= S(B) - \chi(B;A)$, which gives $$S(B|A) \ge (1 - T(\rho_A, \sigma_A)) h(\epsilon),$$ which is exactly the kind of bound I was looking for.

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