5
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Nice answer! Could you please recheck equation 3? It seems there is a typo. $\endgroup$ Dec 17 '20 at 6:15
  • $\begingroup$ Thanks for catching! Fixed. $\endgroup$ Dec 17 '20 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.