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So let's say we have an operator $\hat{A}$, and now I want to calculate the expectation value $\langle \psi|\hat{A}|\psi\rangle$ with an arbitrary ket state $|\psi\rangle$. Is it then always true that $\langle \psi|\hat{A}|\psi\rangle = \langle\psi|\hat{A}^{\dagger}|\psi\rangle$, assuming that $\langle \psi|\hat{A}|\psi\rangle$ is real?

If so, is there a way to prove this?

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3 Answers 3

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  1. The product in Hilbert space is conjugate symmetric $\langle\psi|\phi\rangle = \langle\phi|\psi\rangle^*$.

  2. The Hermitian conjugate of the operator statisfy $\langle\psi| A^\dagger \phi\rangle = \langle A\psi| \phi\rangle$.

  3. Then you get $\langle\psi| A^\dagger \psi\rangle = \langle \psi | A\psi \rangle^*$. So these two quantities are complex conjugates of each other. If one is real then the other one is real. The operators of physical observables in QM are Hermitian (self-adjoin), hence expectation values are real (which makes sense).

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It's easier to do if you don't use Dirac notation: Write $\langle \psi |A|\psi\rangle$ as $\langle \psi, A\psi\rangle$ then $$ \langle \psi, A\psi\rangle= \langle A^\dagger \psi, \psi\rangle, \quad \hbox{(definition of $A^\dagger$)}\\ = \langle \psi,A^\dagger \psi\rangle^*,\quad \hbox{(because $\langle \chi|\psi\rangle=\langle \psi|\chi\rangle^* $)}\\ =\langle \psi, A^\dagger \psi\rangle,\quad \hbox{(because expectation is real )}\\= \langle\psi|A^\dagger|\psi\rangle,\quad \hbox{(Back to Dirac)} $$

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One of the postulates of QM is that observables are given by the eigenvalues of hermitian operators (which must be real), so assuming the operator in question is an observable, you could argue it has to be so.

Edit: to put this in terms of expectation values specifically though, if the operator isn’t an observable I’m not sure the expectation value has any physical significance.

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