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Considering a sandwich of the form $\langle{\psi}|{\hat{O}}|{\phi}\rangle$, where $\hat{O}$ is an operator and $\psi, \phi$ are state vectors, I know that if $\psi=\phi$ then the sandwich represents the expectation value of that operator $\langle \hat{O} \rangle$ i.e. the average value of all the possible outcomes of a measurement under identical conditions. But if $\psi \ne \phi$ does the sandwich have a physical meaning?

Moreover if the operator $\hat{O}$ is not hermitian, then the expectation value does not represent an obserable, so does $\langle{\psi}|{\hat{O}}|{\psi}\rangle$ still have a physical interpretation?

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    $\begingroup$ You are asking about the "physical meaning" of an arbitrary matrix element of an arbitrary matrix? $\endgroup$ Dec 7, 2022 at 15:23
  • $\begingroup$ Do people really call such matrix elements, such inner products, not only overlaps but also sandwiches? As for interpreting them, they're fixed by more easily-understood diagonal ($\psi=\phi$) elements if $\hat{O}$ is positive-definite; indeed, $\hat{O}$ is the difference of two such operators. $\endgroup$
    – J.G.
    Dec 7, 2022 at 16:26
  • $\begingroup$ @J.G. I've heard people use the word sandwich for them a lot in the US. Also as a verb e.g. "to sandwich the operator between $\langle \phi |$ and $|\psi \rangle$ $\endgroup$ Dec 7, 2022 at 17:42
  • $\begingroup$ @doublefelix Thanks for identifying a relevant nation. My experience with "overlap" is UK-based. $\endgroup$
    – J.G.
    Dec 7, 2022 at 19:15

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In general there is no physical meaning I am aware of even if $O$ is hermitian. Even when $O$ is the momentum operator, for example, taking $\langle \psi | \hat{p} | \phi \rangle$ has no specific interpretation for arbitrary $\psi, \phi$.

There is a specific set of rules at the foundation of QM which lead to interpretable connections to reality. Some examples which have a similar flavor to your question:

  • The probability of transitioning from an initial state $|\psi \rangle$ at $t=0$ to an eigenstate $|\lambda \rangle$ of some operator $O$ such that $O |\lambda \rangle = \lambda |\lambda \rangle$, is (if $\lambda$ has only 1 orthonormal eigenvector) $$|\langle \lambda | e^{-itH/\hbar }|\psi \rangle|^2$$
  • Letting $P_\lambda$ be a projector to the eigenspace corresponding to eigenvalue $\lambda$, the probability of getting outcome $\lambda$ when measuring a state $|\psi \rangle$ without evolution in between is $$\langle \psi | P_\lambda | \psi \rangle$$
  • The expectation value of an operator $O$ when measuring a particle in state $|\psi \rangle$ is $$\langle \psi | O | \psi \rangle$$

But aside from the specific connections to reality specified in the postulates of QM (see for example chapter 4 in Shankar's textbook principles of QM) there is no reason for an arbitrary mathematical expression to have a clear physical interpretation.

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  • $\begingroup$ "there is no reason for an arbitrary mathematical expression to have a clear physical interpretation". Especially in quantum mechanics, where the inferential gap between the mathematical formalism and the interpretational apparatus is huge. $\endgroup$
    – march
    Dec 7, 2022 at 17:28

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