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?

  • $\begingroup$ What exactly is $P$ here? is it a projector? $\endgroup$ – ZeroTheHero Mar 31 '17 at 18:30
  • $\begingroup$ Yes, $P$ is a projector $\endgroup$ – Physicist Mar 31 '17 at 18:40
  • $\begingroup$ 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 $i$. $\endgroup$ – taper Aug 12 '18 at 21:44
  • 1
    $\begingroup$ @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 $M_i^\dagger M_i$. $\endgroup$ – 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 $P$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.