measurement probability from density operator?

Question

I've been through this before but I can't fully get my head round this upon review. So the density operator $\hat{\rho}=\sum_j p_j|\psi_j\rangle\!\langle \psi_{j}|$ for pure states $|\psi_{j}>$ at probabilities $p_j$. Suppose we wanted to measure a non degenerative operator, $\hat{A}=\sum_{l}\lambda_{l}|a_{l}><a_{l}|$ for eigenvalues $\lambda_{l}$ associated to eigenstates $|a_{l}>$. Probability of measuring $\lambda_{l}$ is as follows: $$P(measured\ value=\lambda_{l})=\sum_{j}p_{j}|\langle\psi_{j}|a_{l}\rangle|^{2}=\sum_{j}p_{j}<\psi_{j}|a_{l}\rangle\langle a_{l}|\psi_{j}>$$ We simplify this expression to $$\operatorname{Tr}(\hat{\rho}|a_{l}\rangle\langle a_{l}|)$$ However when I multiply out this second expression I get: $$Tr(\sum_{j}p_{j}<\psi_{j}|a_{l}>|\psi_{j}><a_{l}|)$$ $$=\sum_{j}p_{j}<\psi_{j}|a_{l}><\psi_{j}|a_{l}>$$ Which will not be the same as the first expression unless $<\psi_{j}|a_{l}>$ is real. This number is not necessarily real, for example if $|\psi_{l}>=|0>+i|1>$ and $|a_{l}>=|1>$. I have a feeling that I've made some stupid error somewhere, can anyone see where? Also sorry for the bra/ket formatting, i couldn't get the latex package to work.

Answer 1

You're doing the trace wrong :)

Let's look at it explicitly.

\begin{align} {\rm Tr}\Big(\sum_j p_j\langle\psi_j\vert a_l\rangle\vert\psi_j\rangle\langle a_l\vert\Big) &= \sum_i\bigg\langle a_i\bigg\vert \Big(\sum_j p_j \langle \psi_j\vert a_l\rangle\vert \psi_j\rangle\langle a_l\vert\Big)\bigg\vert a_i\bigg\rangle \\ &=\sum_j p_j\langle\psi_j \vert a_l\rangle\sum_i\langle a_i\vert \psi_j\rangle\langle a_l\vert a_i\rangle \\ &= \sum_j p_j\langle\psi_j\vert a_l\rangle \sum_i \langle a_i\vert \psi_j\rangle\delta_{li} \\ &= \sum_j p_j\langle\psi_j\vert a_l\rangle\langle a_l\vert \psi_j\rangle \\ &= \sum_j p_j\vert \langle \psi_j\vert a_l\rangle\vert^2\ \end{align} which is the expected result. We used the fact that the operator that you were trying to measure had a non-degenerate spectrum. More generally, you'd use the projection operators onto the distinct eigensubspaces of an operator, however, you can perform the same calculation because these projection operators would also be complete.