I am a bit confused by the collapse of the density operator. Consider a system described by the density operator $$ \hat{\rho}=\sum_{m}P_{m}|\psi_{m}\rangle\langle\psi_{m}| $$ and a measurement of the operator $$ \hat{A}=\sum_{i}a_{i}\hat{\Pi}_{i} $$ where $\hat{\Pi}_{i}$ represents the projector onto eigenstates of $\hat{A}$ with eigenvalue $a_{i}$. Suppose that the measurement yields an outcome of $a_{j}$. I think that each state in the ensemble should collapse individually, as follows $$|\psi_{m}\rangle\rightarrow|\psi'_{m}\rangle=\frac{\hat{\Pi}_{j}|\psi_{m}\rangle}{\sqrt{\langle\psi_{m}|\hat{\Pi}_{j}|\psi_{m}\rangle }}$$ such that the density operator is given by $$ \hat{\rho}\rightarrow\hat{\rho}'=\sum_{m}P_{m}|\psi'_{m}\rangle\langle\psi'_{m}|=\sum_{m}P_{m}\frac{\hat{\Pi}_{j}|\psi_{m}\rangle\langle\psi_{m}|\hat{\Pi}_{j}}{\langle\psi_{m}|\hat{\Pi}_{j}|\psi_{m}\rangle } $$ However, the formula is in fact given by $$\hat{\rho}'=\frac{\hat{\Pi}_{j}\hat{\rho}\hat{\Pi}_{j}}{\mathrm{Tr}(\hat{\rho}\hat{\Pi}_{j})}$$ which as far as I can tell isn't equivalent to what I obtain. Is there any flaw in the reasoning behind the term I derived?


1 Answer 1


That's a very interesting observation you made. The issue boils down to the fact that your first method of computing a post-measurement state normalizes "too early". Effectively, in your first method you lose the fact that some of the $|\psi_m\rangle$ may have extremely small probability to produce outcome $j$ and others may have much larger probability. Once you normalize, you forget about this probability. It's probably best explained with an example.

Consider the qutrit states $$ |\psi_0\rangle = |0\rangle \qquad |\psi_1\rangle = \epsilon|1\rangle + \sqrt{1-\epsilon^2} |2\rangle, $$ where $\epsilon > 0$ is we taken to be small for the purpose of the example. Let $\Pi = |0\rangle\langle 0| + |1\rangle \langle 1|$ be a projector which forms a measurement $\{\Pi, I-\Pi\}$. For simplicity, we take $P_0 = P_1 = \frac12$. The initial density matrix is $$ \rho = \frac12 \begin{pmatrix} 1 & 0 & 0 \\ 0 & \epsilon^2 & \epsilon \sqrt{1-\epsilon^2} \\ 0 & \epsilon\sqrt{1-\epsilon^2} & 1- \epsilon^2 \end{pmatrix}\,. $$ Note that we observe the first outcome of the measurement with probability $$ \mathrm{Pr}[\text{observe }\Pi] = \frac{1 + \epsilon^2}{2}. $$ Finally, the usual way to compute the post-measurement state with density matrices says that the state after measurement will be $$ \rho_{\mathrm{final}} = \frac{1}{1 + \epsilon^2}\begin{pmatrix} 1 & 0 & 0 \\ 0 & \epsilon^2 & 0 \\ 0 & 0 &0 \end{pmatrix}\,. $$ Notice the probability that the final state is in the state $|1\rangle\langle1|$ is very small if $\epsilon$ is small. This makes sense as the only way that the state could end up in the state $|1\rangle\langle 1|$ is if we measured $|\psi_1\rangle$ and received outcome $\Pi$, which only happens with probability $\frac12 \epsilon^2$. Thus, we'd expect the final density matrix to reflect this.

What goes wrong?

So let's try to do it the way that you mentioned in the question body. The post-measurement state when we measure $|\psi_0\rangle$ and receive the first outcome is $$ \rho_0 = |0\rangle \langle 0| $$ and the post-measurement state when we measure $|\psi_1\rangle$ and receive the first outcome is $$ \rho_1 = |1\rangle \langle 1|. $$ Notice the problem: where has the $\epsilon$ gone? Well, it disappeared when we normalized, and so we lost information that this second event was actually very unlikely to occur. Thus, when we compute our final density matrix in the way you specified, we get $$ \rho_{\mathrm{final2}} = \frac12 \rho_0 + \frac12 \rho_{1} = \frac12 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}\, , $$ which is clearly not what we'd expect.

  • 1
    $\begingroup$ Hey thanks for this answer it's super clear and well thought out $\endgroup$ Commented Apr 17, 2023 at 18:18
  • $\begingroup$ No problem, glad it helped $\endgroup$
    – Rammus
    Commented Apr 18, 2023 at 9:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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