Inner product of hermitian conjugate identity in QM

Question

Could someone kindly show me why

$\left \langle f \mid \hat{A} \mid g\right \rangle=\left \langle f \mid \hat{A}g \right \rangle =\int dxf^{*}\left ( \hat{A}g \right ) =\int dx \left ( \hat{A^{\dagger}f}^{} \right )^{*}g =\left \langle \hat{A^{\dagger}f} \mid g \right \rangle$

I do not understand the second last equivalence despite looking at it for a while. It doesn't make sense to me.

Any help is appreciated.

Answer 1

This is the necessary relationship that defines the hermitian conjugate operator $\hat{A^{\dagger}}$.

In other words the question is, if I operate on $g$ with $\hat{A}$ then what do I have to operate on $f$ with to get the same integral? This is new operator is given the symbol $\hat{A^{\dagger}}$ and named hermitian conjugate.

If it turns out that $\hat{A^{\dagger}} = \hat{A}$ then $\hat{A}$ is self-adjoint or hermitian.

For further reassurance see Equation (3.2) here