Measurement on density operator

Question:

A system in a mixed state $\rho$ is measured with the measurement described by a projection operator $P$.

1. What is the probability of the outcome?
2. What is the density operator of the system after a measurement?
3. State and explain whether $\rho$ is necessarily still in a mixed state after the measurement.

I have checked many textbooks, and basically all of them only discuss the the expectation value $\langle A \rangle =Tr(\rho A)$. What about the actual state after measurement? In wikipedia it says $\rho'=P\rho P$, but how to justify this expression? For (3), I only get $$Tr(\rho'^2)=Tr(P\rho PP\rho P)=Tr(P\rho P\rho P)=Tr(P\rho P\rho)=Tr(P\rho P\rho),$$ but then I don't know how to proceed. Any idea?

• What exactly is $P$ here? is it a projector? – ZeroTheHero Mar 31 '17 at 18:30
• Yes, $P$ is a projector – Physicist Mar 31 '17 at 18:40
• Please see section 2.5 of this note. In particular, the answer by @ZeroTheHero is wrong! And the wikipedia entry has been corrected to $\rho_i' = \frac{P_i \rho P_i}{\operatorname{tr}[\rho P_i]}$, if the measurement outcome is $i$. – taper Aug 12 '18 at 21:44
• @taper Maybe I'm misunderstanding something in your comment as my answer collapses to the expression $\vert\psi^\prime_j\rangle:= \frac{M_i\vert\psi_j\rangle}{\sqrt{p_{ji}}}$ for a pure state given in the section you refer to. Indeed the projector $P$ is written in those notes as $M_i^\dagger M_i$. – ZeroTheHero Aug 13 '18 at 14:40

After a measurement, the system should be in the eigenstate of $\hat A$ with eigenvalue $a_i$ (assuming no degeneracy in the spectrum) since you know the outcome was $a_i$. Normally, the measurement does not preserve the norm so physically you know that your density matrix after the measurement should be $$\hat \rho' \propto \vert a_i\rangle\langle a_i\vert\, .$$ where $\vert a_i\rangle$ is the eigenvector of $\hat A$ corresponding to the eigenvalue $a_i$.
Now, start with the projector $P$ and the density matrix $\rho$ written explicitly as \begin{align} P&=\vert a_i\rangle\langle a_i\vert \, ,\\ \hat \rho &= \sum_j p_j \vert \psi_j\rangle \langle \psi_j\vert \end{align} and examine $P\hat \rho P$. Inserting the expressions yields explicitly \begin{align} \hat \rho'&= \sum_j p_j \vert a_i\rangle\langle a_i\vert \psi_j\rangle \langle \psi_j \vert a_i\rangle\langle a_i\vert \\ &= \vert a_i\rangle\langle a_i\vert \left(\sum_j p_j \vert \langle a_i\vert \psi_j\rangle\vert^2\right)\, ,\\ &=\vert a_i\rangle \langle a_i\vert \beta_i\, , \qquad \beta_i> 0 \end{align} which is what you expect. Of course, as mentioned above, $\hat \rho'$ is no longer normalized - but that's to be expected because projectors are no norm-preserving operators.
In this example where there is no degeneracy, $\hat \rho'$ is a pure state. You can scratch your head as to whether or not this remains true if the eigenvalue $a_i$ occurs multiple times in the spectrum and, in particular, if this degeneracy changes the projector $P$.