Let $\rho_{XA}$ be a classical-quantum state, i.e., $\rho_{XA} = \sum_{x} p(x) |x\rangle \langle x| \otimes \rho_A^x$.

How to prove that the conditional von Neumann entropy $S(X|A) = S(\rho_{XA}) - S(\rho_A)$ is non-negative?


We will use the upper bound on the entropy of a mixture (for proof see for example theorem 11.10 on p.518 in Nielsen & Chuang)

$$ S\left(\sum_k p_k \rho_k\right) \leq H(p) + \sum_k p_k S(\rho_k)\tag1 $$

where $H(p) = -\sum_k p_k \log p_k$.

Set $p_x := p(x)$. Note that if $|\psi_y^x\rangle$ is an eigenvector of $\rho_A^x$ associated to eigenvalue $q_y^x$ then $|x\rangle\otimes|\psi_y^x\rangle$ is an eigenvector of $\rho_{XA}$ associated to eigenvalue $p_x q_y^x$. Therefore

$$ \begin{align} S(\rho_{XA}) &= -\sum_x \sum_y p_x q_y^x \log(p_x q_y^x) \\ &= -\sum_x p_x \log p_x - \sum_x p_x \sum_y q_y^x \log q_y^x \\ &= H(p) + \sum_x p_x S(\rho_A^x)\tag2. \end{align} $$

Next, compute $\rho_A$

$$ \rho_A = {\rm tr}_X \rho_{XA} = \sum_x p_x \rho_A^x $$

and so

$$ S(\rho_A) = S\left(\sum_x p_x \rho_A^x\right)\tag3. $$

Finally, recognize $(3)$ and $(2)$ as the left and right hand sides of $(1)$ with the appropriate substitution to get

$$ S(\rho_A) \leq H(p) + \sum_x p_x S(\rho_A^x) \\ 0 \leq S(\rho_{XA}) - S(\rho_A) $$

which is the desired inequality.

  • 1
    $\begingroup$ Nice answer! Could you please recheck equation 3? It seems there is a typo. $\endgroup$ – Hasan Iqbal Dec 17 '20 at 6:15
  • $\begingroup$ Thanks for catching! Fixed. $\endgroup$ – Adam Zalcman Dec 17 '20 at 15:22

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