Proof: Operators with same expectation value

[FaradayCage]

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 $\hat{A} = \hat{B}$ ?

 

[mathman]

Is it true for every state?

 

[Goldstone1]

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 $\hat{A} = \hat{B}$ ?

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 $$\mathcal{O} \cdot x \ne x \cdot \mathcal{O}$$.

 

[dextercioby]

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 $\hat{A} = \hat{B}$ ?

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.