Given a set of measurement operators $\{M_i\}$ that sum to unity, consider the post-measurement states on some $\rho$ as $\rho_i:=(\sqrt{M_i}\rho\sqrt{M_i})/p_i$ and $p_i:=\mathrm{Tr}(M_i\rho)$.

It's known that $S(\rho)\ge \sum_i p_i S(\rho_i)$ and the entropy difference is referred to as the information gain. Is the stronger version $S(\rho)\ge S(\rho_i), \forall i$ true? If not, any counterexample?


1 Answer 1


The stronger inequality holds if $\rho$ is a pure state. You can see it from the fact that $\sqrt M \rho \sqrt M$ has unit rank if $\rho$ does. Thus in all such cases you have $S(\rho)=S(\rho_i)=0$ (which is also obvious from the fact that you must have $0=S(\rho)\ge \sum_i p_i S(\rho_i)$.

It doesn't hold in general though. Consider for example a single-qubit state of the form $$\rho=\begin{pmatrix}\epsilon&0\\0&1-\epsilon\end{pmatrix}$$ for some small $\epsilon>0$. This state is "close" to pure and therefore has a very small entropy. Consider now the POVM with elements $$M_1 = \begin{pmatrix}1&0\\0&\frac{\epsilon}{1-\epsilon}\end{pmatrix}, \qquad M_2=I-M_1.$$ Then $$\rho_1 = \frac{\sqrt M_1 \rho \sqrt M_1}{p_1} =\frac12\begin{pmatrix}1&0\\0&1\end{pmatrix}, \quad p_1=2\epsilon.$$ It follows that in this case $S(\rho)\simeq0$ but $S(\rho_1)=1$.

If you want to know what this would look like in practice, you can implement this POVM as the two-qubit circuit $(I\otimes U)\operatorname{CNOT}(I\otimes U^\dagger)$, with input at the first register $\rho$, and input at the second register $|0\rangle$, and $U$ a single-qubit unitary such that $$UXU^\dagger=\begin{pmatrix}\sqrt{\frac{\epsilon}{1-\epsilon}} & \square \\ \sqrt{\frac{1-2\epsilon}{1-\epsilon}} & \square\end{pmatrix},$$ where you can put whatever you want in place of the squares as long as it makes the overall matrix unitary. Then, finding the outcome $|0\rangle$ at the second register "projects" the first register on $\sqrt M_1\rho\sqrt M_1/p_1=I/2$. Of course, the probability of finding this outcome is also proportional to $\epsilon$, thus very small.

Note that here I used a POVM which displays the "entropy increasing" effect only for a specific input state. I'm not sure it's possible to have a POVM that gives $S(\rho)< S(\rho_i)$ for some $i$ for all $\rho$.


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