# Proof: Operators with same expectation value

1. Jun 23, 2011

Given some state $\left|\psi\right\rangle$, and two operators $\hat{A}$ and $\hat{B}$, how do you prove that if $\langle\psi|\hat{A}|\psi\rangle = \langle\psi|\hat{B}| \psi\rangle$ then $\hat{A} = \hat{B}$ ?

2. Jun 23, 2011

### mathman

Is it true for every state?

3. Jun 23, 2011

### Goldstone1

Proving $$\hat{A}$$ is the same as $$\hat{B}$$ depends on the state of $$\psi$$. You've made some notation $$<\psi|\psi>$$ is a probability amplitude. Your operators are fine, but if they are equal, it depends on the state of $$|\psi>$$. Note though $$\mathcal{O} \cdot x \ne x \cdot \mathcal{O}$$.

4. Jun 23, 2011

### dextercioby

Well, you can't; you have the equality of the images for only one vector in the common domain. You must have much more than that, of course.