Are all selfadjoint operators in quantum mechanics bounded?

[hitche]

Hi, I'd like to know if the following statement is true:
Let $\hat{A}, \hat{B}$ be operators for any two observables $A, B$. Then $\langle \hat{A} \rangle_{\psi} = \langle \hat{B} \rangle_{\psi} \forall \psi$ implies $\hat{A} = \hat{B}$.
Here, $\langle \hat{A} \rangle_{\psi} = \int_\mathbb{R^3} d^3x \psi^* \hat{A} \psi$ is standard definition of expectation value of an operator $\hat{A}$.
If this doesn't hold, could you provide any counterexample? Thank you!

 

[Meir Achuz]

Any two operators with odd parity would give zero for any wave function.

 

[cattlecattle]

yes - because the expected value of A-B will be zero for all psi, then by taking psi to be A-B's eigenvector, you reached the conclusion that A-B has only zero eigenvalues, which means A-B must be a zero operator.

 

[dextercioby]

Of course, not. The general solution of $\langle\psi, A\psi \rangle= \langle\psi, B\psi\rangle$ is that either one is an extension of the other. Equality follows iff both operators are bounded. As a counterexample would be any symmetric operator which is not self-adjoint.

 

[WannabeNewton]

yes - because the expected value of A-B will be zero for all psi, then by taking psi to be A-B's eigenvector, you reached the conclusion that A-B has only zero eigenvalues, which means A-B must be a zero operator.

Not that's wrong. If either A or B is unbounded then clearly your argument won't apply.

 

[cattlecattle]

Not that's wrong. If either A or B is unbounded then clearly your argument won't apply.

sorry, I failed to see which step was not applicable due to unboundedness

 

[hitche]

sorry, I failed to see which step was not applicable due to unboundedness

I'm new to the theory, but I think the reason is that spectrum of operator is defined for bounded operators. There exists also some extended definition of spectrum for unbounded operators, but it can be empty (unlike the spectrum of a bounded operator).
BUT I read that in quantum mechanics every observable is represented by selfadjoint linear operator and according to another resource every selfadjoint everywhere defined operator is bounded. On the other hand bounded self adjoint operator may have no eigenvalue. So.. I don't know!