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.