Physics Asked on August 14, 2021
The closed-path Berry phase can have measurable effects and, if I am understanding correctly, is a measurable quantity in and of itself. If that is so, is there a Hermitian operator with Berry phases as its eigenvalues? What would it be, and what would its eigenstates be?
There is no single operator corresponding to the Berry phase because the Berry phase is not a property of a single quantum state, but of a full trajectory through state space. "Observables are Hermitian operators" does not mean "anything you can measure is a Hermitian operator", it means measurable properties of states are Hermitian operators acting on those states.
It easy to construct plenty of other things you can "observe" that are not actually observables/Hermitian operators, e.g. if you do repeated measurements on an identically realized state, you can compute (an approximation for) the standard deviation of the observable being measured, but the standard deviation itself is not a linear operator on states, either.
Correct answer by ACuriousMind on August 14, 2021
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