$\hat{p}$ is Hermitian and Hermitian operators $O$ satisfy, by definition,

$$\hat{O} = \hat{O}^\dagger$$

Adjoint is not a synonym for complex conjugate. $\hat{p} = -i\hbar \nabla \rightarrow +i\hbar \nabla^\dagger \rightarrow -i\hbar \nabla =\hat{p}^\dagger$, but $\hat{p} \neq \hat{p}^*$.

You're not getting your facts right at all.

How do we know from this $\langle W \rangle = \int_{-\infty}^{\infty} \bar{\Psi}\left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + W_p \right) \Psi dx$ or **this** $\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + W_p$ that we have an eigenfunctiuion and eigenvalue.

Answer: we don't.

All I know about operator $\bar{H}$ so far is this equation where $\langle W \rangle$ is an energy expected value:
\begin{align}
\langle W \rangle = \int_{-\infty}^{\infty} \bar{\Psi}\left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + W_p \right) \Psi dx
\end{align}

No, you don't.

Here's the mathematical side of what an eigenfunction and eigenvalue is:

Given a linear transformation $T : V \to V$, where $V$ is an infinite dimensional Hilbert or Banach space, then a scalar $\lambda$ is an eigenvalue if and only if there is some non-zero vector $v$ such that $T(v) = \lambda v$.

Here's the physics side (i.e. QM):

We postulate that the state of a system is described by some abstract vector (called a ket) $|\Psi\rangle$ that belongs to some abstract Hilbert space $\mathcal{H}$.

Next we postulate that this state evolves in time by some Hermitian operator $H$, which we call the Hamiltonian, via the Schrodinger equation. What is $H$? you guess and compare to experimental results (that's what physics is anyway).

Next we postulate for any measurable quantity, there exists some Hermitian operator $O$, and we further postulate that the average of many measurements of $O$ is given by $ \langle O \rangle = \langle \Psi | O | \Psi \rangle$.

Connection to wavefunctions: we pick the Hilbert space $L^2(\mathbb{R}^3)$ to work in, so $\Psi(x) = \langle x | \Psi \rangle$, and $\langle O \rangle = \int_{-\infty}^{\infty} \Psi^*(x) O(x) \Psi(x) dx$.

Ok, that's the end. The form of $H$ doesn't follow from the energy expected value.

Wait! I haven't even talked about eigenvalues and eigenfunctions. This is a useless post!

Answer: well you don't have to. But it is useful to find the eigenvalues and eigenfunctions of $H$, because the eigenfunctions of $H$ form a basis of the Hilbert space, and certain expressions become diagonal/more easily manipulated when we do whatever calculations we want to do.

So to find the eigenvalues of $H$, we simply solve the eigenvalue equation as stated above:
Solve
\begin{align}
H | \Psi_n \rangle = E_n | \Psi_n \rangle.
\end{align}
This is in the form $T(v) = \lambda v$.

So as Alfred Centauri says, we simply **want** to find the eigenfunctions of $H$. A more subtle question would be, how do we know they exist? The answer lies in spectral theory and Sturm-Liouville theory but nevermind for now, as physicists we assume they always exist.

So your additional question:

$\hat{a} \psi$ is an eigenfunction of operator$\hat{H}$ with
eigenvalue $(W-\hbar \omega)$.

Well.... that just follows straightaway. You said you already proved that $H a^\dagger \psi = (W - \hbar \omega) a^\dagger \psi$. So here $T$ = $H$, $a^\dagger \psi = v$, and $\lambda = (W - \hbar \omega)$. which is an eigenvalue equation $T(v) = \lambda v$. Thus, $a^\dagger \psi$ is an eigenfunction of $H$ with eigenvalue $(W-\hbar \omega)$.

## Best Answer

I'm getting the impression that a good part of this question (and perhaps also this physics.SE question?) arises from a wrong presumption that time and position should be on equal footing in quantum mechanics. They are not. The position $\hat{\bf r}$ is an

operator, while time $t$ is aparameter.(Notation: In the following boldface denotes a vector quantity, and a hat denotes an operator quantity.) In fact, Pauli's Theorem shows, under mild assumptions, that time cannot be an operator in quantum mechanics, cf. Ref. (1). See also this and this physics.SE questions.In the position space Schroedinger representation of the Schroedinger picture, we have a wave function $\Psi({\bf r},t)=\left<{\bf r}|\Psi(t)\right>$, the position operator $\hat{\bf r}$ becomes multiplication with ${\bf r}$, and the momentum operator $\hat{\bf p}= \frac{\hbar}{i}{\bf\nabla}$ is differentiation wrt. position. Both $\hat{\bf r}$ and $\hat{\bf p}$ are Hermitian operators. The Hamiltonian $\hat{H}=H(\hat{\bf r},\hat{\bf p})$ is composed out of the operators $\hat{\bf r}$ and $\hat{\bf p}$ in such a way that it is Hermitian.

It is wrong to claim that the Hamiltonian $\hat{H}$ is $i\hbar \partial_{t}$. (If it was, for starters, the Schroedinger equation $i\hbar \partial_{t}\Psi({\bf r},t)=\hat{ H}\Psi({\bf r},t)$ would become an empty statement without any content.) This should be compared with the fact that (in the position space Schroedinger representation) the momentum operator $\hat{\bf p}= \frac{\hbar}{i}{\bf\nabla}$ is a differentiation wrt. position. In particular, time and position differentiations are, in this sense, not on equal footing.

Finally, the frequency operator may be defined as $\frac{1}{h}\hat{H}=\frac{1}{h}{H}(\hat{\bf r},\hat{\bf p})$, as David Zaslavsky observes in a comment to lurscher's answer. The corresponding expectation value is

$$\left<\Psi(t) \left|\frac{1}{h}\hat{H} \right| \Psi(t)\right> ~=~ \int\! d^{3}r \ \Psi^{*}({\bf r},t)\ \frac{1}{h}\hat{H}\Psi({\bf r},t). $$

References:

(1) Wolfgang Pauli 1933, in

Handbuch der Physik (Encyclopedia of Physics)(ed. S. Fluegge), vol. 5, pp.1-168, Springer Verlag, 1958.