# Quantum Mechanics: Generic Question Concerning Expectation Value of Operator And its Adjoint

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?

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).
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 $$\langle \psi, A\psi\rangle= \langle A^\dagger \psi, \psi\rangle, \quad \hbox{(definition of A^\dagger)}\\ = \langle \psi,A^\dagger \psi\rangle^*,\quad \hbox{(because \langle \chi|\psi\rangle=\langle \psi|\chi\rangle^* )}\\ =\langle \psi, A^\dagger \psi\rangle,\quad \hbox{(because expectation is real )}\\= \langle\psi|A^\dagger|\psi\rangle,\quad \hbox{(Back to Dirac)}$$