# Exercise 11

## from Example Sheet 3

### Exercise 11.

Projective measurements do not decrease von Neumann entropy

Suppose a projective measurement described by a set of projection operators $$\{P_i\}$$ is performed on a quantum system, but we never learn the result of the measurement. If the state of the system before the measurement was $$\rho$$ then the state after the measurement is given by $\rho^\prime = \sum_i P_i \rho P_i.$ Prove that the entropy of this final state is at least as great as the original entropy: $S(\rho^\prime) \ge S(\rho),$ with equality if and only if $$\rho = \rho^\prime$$.

We consider $D(\rho\|\rho') = \operatorname{tr}[ \rho (\log \rho - \log \rho')].$(1) We have that $\rho' P_j = \left(\sum_i P_i \rho P_i\right) P_j = P_j \rho P_j^2= P_j^2 \rho P_j = P_j \left(\sum_i P_i \rho P_i\right) = P_j \rho'.$ Thus, $$[P_i, \rho'] =0$$, which implies $$[P_i, \log \rho']=0$$. Now, since $$\sum_i P_i = I$$, \begin{aligned} \operatorname{tr}[\rho \log (\rho')] &= \sum_i\operatorname{tr}[ P_i \rho \log (\rho')] = \sum_i\operatorname{tr}[ P_i^2 \rho \log (\rho')] \\ &= \sum_i\operatorname{tr}[ P_i \rho \log (\rho') P_i] \\ &= \sum_i\operatorname{tr}[ P_i \rho P_i \log (\rho')] \\ &= \operatorname{tr}[ \rho' \log (\rho')]. \end{aligned} Thus, (1) becomes $D(\rho\|\rho') = -S(\rho) + S(\rho').$ Klein’s inequality gives $$D(\rho\|\rho')\geq 0$$. So we have $S(\rho) \leq S(\rho').$