Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Does the expectation value of an observable must be equal to an eigenvalue of the corresponding operator? I already know that 0 is not an eigenvalue, but is there any other examples?

share|improve this question
add comment

2 Answers

A specific quantum mechanical example to show the contrary is spin for spin 1/2 systems. If you are in an eigenstate of the Sz operator, the expectation value of Sx is 0, but it has eigenvalues h/2, -h/2, where those are h-bars. Not exactly sure how to do latex here..

share|improve this answer
add comment

I would actually expect this to be rare, and only generically true when the state of the system corresponds to an eigenstate. This simply because, for a state $\psi = \sum a_{n}\lvert\phi_{n}\rangle$ with eigenvalues $V_{n}$, you would have $\langle V\rangle = \sum V_{n}\lvert a_{n}\rvert^{2}$, which is not constrained to be equal to one of the $V_{n}$. It's easy to check this for a two state system with the two values of $V_{n}$ different.

share|improve this answer
Right, for a classical experiment to get the basic idea, consider the expected number of eyes when throwing a die, namely $\frac{7}{2}$. –  Qmechanic Sep 19 '12 at 19:13
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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