# Can post-measurement states have entropy larger than the original state?

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?

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 $$\rho$$.