### Question 8.

Regarding Q3 of the 2009 exam: I don’t know how to show part b), that \[ S(A|B) \geq - \log \dim \mathcal{H}_A. \]

This one’s a little tricky. This question is Theorem 11.5.1 of Mark Wilde’s book, and he uses a purification to solve it, which seems like a good way to go. The way that comes more naturally to me is as follows. Since the quantum conditional entropy is concave, it’s minimum over the set of states must occur on a pure state. To see this, consider \(\rho_{AB} = \sum_i p_i | \psi_i \rangle\langle \psi_i |_{AB}\). Then \[ \min_i S(A|B)_{\psi_i} \leq \sum_i p_i S(A|B)_{\psi_i} \leq S(A|B)_\rho \] so the conditional entropy of some pure state \(\psi_i\) is less than the conditional entropy of \(\rho\). But the conditional entropy of a pure state is simple to compute: \[ S(A|B)_\psi = S(AB)_\psi - S(B)_\psi = -S(B)_\psi = - S(A)_\psi \] using the definition of the conditional entropy in the first equality, that the von Neumann entropy of any pure state is zero (i.e. \(S(AB)_\psi = 0\)) in the second, and in the third that the entropy of either reduced state of a pure state is the same. Note \(S(A)_\psi\) is notation for the von Neumann entropy of the reduced state \(\operatorname{tr}_B \psi_{AB}\), which is in general a mixed state. To minimize \(S(A|B)_\psi\), therefore, we must maximize \(S(A)_\psi\). But we know the von Neumann entropy of any state on \(A\) is bounded by \(\log d_A\). Thus, \(S(A)_\psi\leq \log d_A\) and \[ S(A|B)_\psi \geq - \log d_A. \] Note we actually can find \(S(A|B)_\psi \geq - \log (\min\{d_A,d_B\})\) for \(d_B = \dim \mathcal{H}_B\) by also bounding \(S(B)_\psi \leq \log d_B\).