I have a question regarding this exercise:

Let O be an observable on V. Show that $\langle v,O(v)\rangle= \mathrm{tr}(O|v\rangle\langle v|)$ for $v \in V$.

I thought that this exercise is quite easy because I thought I can use the fact that we define $|v\rangle\langle v|$ as density operator. Any other ideas or is my idea the right one?

  • $\begingroup$ Here are two hints, you can use either to find the desired conclusion. 1. The trace is cyclic; 2. You can take the trace in any orthonormal basis you want. Note that you should really try to use mathjax to write your equations though: here's a guide $\endgroup$
    – Rammus
    Oct 27, 2021 at 16:41

1 Answer 1


Extend $|v\rangle$ to an orthonormal basis $|u_1\rangle, |u_2\rangle, \dots, |u_n\rangle$ so that $|u_1\rangle=|v\rangle$. Then we have

$$ \mathrm{tr}(O|v\rangle\langle v|) = \sum_k\langle u_k|O|v\rangle\langle v|u_k\rangle = \langle v|O|v\rangle $$

which is the desired equality.


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