0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ $\operatorname{Tr}(\lambda |\psi\rangle\!\langle\phi| ) = \lambda \langle\phi|\psi\rangle$ $\endgroup$
    – glS
    Aug 5, 2022 at 12:23
  • $\begingroup$ So the ket remains a ket and the bra remains a bra? $\endgroup$ Aug 5, 2022 at 12:26
  • $\begingroup$ See circular property of trace. $\endgroup$
    – kludg
    Aug 5, 2022 at 14:55

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ This is great thanks! $\endgroup$ Aug 6, 2022 at 14:03
  • $\begingroup$ @AdrienAmour Glad it was helpful! Notice that the takehome message is what glS mentioned in the comment on your post. :) We implicitly derived the same result in this calculation but you can try arriving at it more explicitly using the same method of calculating trace that is used here. $\endgroup$ Aug 6, 2022 at 14:46

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.