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Message: [QUOTE="vanhees71, post: 6875841, member: 260864"] For a pure state, represented by a normalized vector ##|\Psi \rangle## expectation value is $$\langle f(\hat{A}) = \langle \Psi|f(\hat{A}) \Psi \rangle=\langle f(\hat{A})^{\dagger} \Psi|\Psi \rangle,$$ for an arbitrary function ##f(\hat{A})##. It doesn't matter whether the operator is self-adjoint or not for the identity of the two expressions. Of course, such an operator cannot represent an observable to begin with, and you might argue that it doesn't make sense to call this expression an "expectation value" in the first place. [/QUOTE]