Exercise 11

from Example Sheet 4

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 \[ S(\rho) \leq S(\rho'). \]