Conceptual problem with post-measurement density matrix

Question

I'm asked to derive what the density matrix would be after a measure of $|\phi \rangle$ ($\rho \rightarrow \rho'$). At first I thought that, since $\rho$ has an statistical meaning on an ensemble of mixed states, it should remains the same after, say, some measure $A |\phi \rangle \rightarrow |a_k \rangle$, even if this one state "collapses". As far as I know, even if we measured all the states in the ensemble, $\rho$ would still be the same, since it carries only statistical information about the distribution of values in it. I've also found a simple example here in the quantum computing community that seems to say just that.

Even though, After some digging, I found out that we should have something as $$\rho' = \frac{P_n \rho P_n}{Tr(P_n \rho)}$$
Which I can kind of see from where it comes, but breaks my confidence in my first thinking process, since I am not being able to relate these 2.

This also gets me thinking about what exactly means to work with the density operator and what do we mean by saying "$|\phi\rangle$ is in a mixed-state". Is this last expression some kind of notation abuse? Since it is not like $\phi$ is a superposition or anything like that.

Note: This is my first question here, so I would really appreciate any tips on how to make this question better (or less worse)

Answer 1

A measurement, generally speaking, does change the density matrix because it destroys coherences, or, in other words, the probability distribution over the eigenvalues of observables that do not commute with the observable being measured is altered after the measurement. This is intuitively clear: let's say you have a spin$-1/2$ particle in $\vert \uparrow\rangle_z$ state. If you measure its $x$ spin, the post-measurement ensemble has $\langle \sigma_z\rangle=0$ whereas the pre-measurement ensemble has $\langle \sigma_z\rangle=1/2$.

This essentially answers your question, however, I will derive the post-measurement density matrix to show that the density matrix does indeed change. There is also a bit of nuance about what is meant by the "post-measurement" density matrix, so this is worthwhile to discuss.

Let's say you have a density matrix $\rho=\sum p_j\vert \psi_j\rangle\langle \psi_j\vert$. If you measure some observable $A=\sum_ka_k\vert a_k\rangle\langle a_k\vert$, the resultant ensemble is an ensemble wherein the system is in state $\vert a_k\rangle$ with probability $\langle a_k\vert \rho\vert a_k\rangle$, i.e., $\rho'=\sum_k \langle a_k\vert \rho\vert a_k\rangle \vert a_k\rangle\langle a_k\vert =\sum_k P_k\rho P_k$ where $P_k=\vert a_k\rangle\langle a_k\vert$. We have simply used the Born rule, the "state-reduction" postulate, and the definition of a density matrix to arrive at this conclusion. Notice that we are summing over the probabilities corresponding to the different possible outcomes of the measurement of $A$. If, however, we know that the outcome of the measurement of $A$ was $a_k$ then the post-measurement density matrix would be $\rho''=\vert a_k\rangle\langle a_k\vert = \frac{\langle a_k\vert \rho\vert a_k\rangle \vert a_k\rangle\langle a_k\vert }{\langle a_k\vert \rho\vert a_k\rangle}=\frac{P_k\rho P_k}{{\rm Tr}(\rho P_k)}$. The reason behind this last bit of seemingly pointless algebraic manipulation is that the expression in terms of projection operators applies to all projective measurements -- not just measurements of non-degenerate observables. Similarly, the result $\rho'=\sum_k P_k \rho P_k$ also generalizes to all projective measurements, i.e., also when $P_k\neq \vert a_k\rangle \langle a_k\vert$. In my experience, people can mean either $\rho'$ or $\rho''$ when they say "post-measurement" density matrix -- depending on the context.

PS: The reason the density matrix does not change after measurement in the example you linked is that the example talks about a maximally mixed density matrix, to begin with, i.e., a density matrix where the coherences are already destroyed (in all bases).