# Prove that the conditional entropy of a classical-quantum state is non-negative

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.

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