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 $(W-\hbar \omega)$.
$\newcommand{\ket}[1]{\lvert #1 \rangle}$There is no such thing as "looking like a collapsed wavefunction", even if you believe there's collapse.
Let's go to the finite dimensional case and have a simple two-level spin system, that is, our Hilbert space is spanned by, e.g., the definite spin states in the $z$-direction $\ket{\uparrow_z}$,$\ket{\downarrow_z}$.
Now, the state $\ket{\psi} = \frac{1}{\sqrt{2}}(\ket{\uparrow_z} + \ket{\downarrow_z})$ is not an eigenstate of $S_z$, and measurement of $S_z$ will collapse it with equal probability into $\ket{\uparrow_z}$ or $\ket{\downarrow_z}$. Yet, this is an eigenstate of the spin in $y$-direction, i.e. $\ket{\psi} = \ket{\uparrow_y}$ (or down, we'd have to check that by computation, but it doesn't matter for this argument). So, although you can "collapse" $\ket{\psi}$ into other states, it already looks like a collapsed state by your logic. This shows that the notion of "looking like a collapsed state" is not very useful to begin with.
Furthermore, you seem to be confused about the difference between a measurement (inducing collapse in some dictions) and the application of an operator. You say
So when the hamiltonian acts on a system in one of those eigenfunctions, that system always collapses to the same point in space?
but this is non-sensical. The action of the Hamiltonian is an infinitesimal time step, as the Schrödinger equation tells you:
$$ \mathrm{i}\hbar\partial_t \ket{\psi} = H\ket{\psi} $$
and, for an eigenstate $\ket{\psi_n}$, which is a solution to the time-independent equation with energy $E_n$, you have by definition $H\ket{\psi_n} = E_n\ket{\psi_n}$, that is, the Hamiltonian is a "do nothing" operation on eigenstates since multiplication by a number does not change the quantum state. That is, after all, why we are interested in the solutions to the time-independent equation - because these are the stationary states that do not evolve in time. This has nothing to do with collapse, or measurement.
Lastly, exactly determinate states of position are not, strictly speaking, quantum states, since the "eigenfunctions" of the position operator "multiplication by x" are Dirac deltas $\psi(x) = \delta(x-x_0)$, which are not proper square-integrable functions $L^2(\mathbb{R})$ as quantum states are usually required to be. But yes, this is "a spike", and conversely, the determinate momentum states are plane waves $\psi(x) = \mathrm{e}^{\frac{\mathrm{i}}{\hbar}px}$.
Best Answer
It means you can use the eigenvalue as a label for the state since (in your example) $\Delta p=0$. Not only is the average value $p$, but there is no fluctuation in this outcome.
Hence for instance we can label hydrogen states by energy, angular momentum, and projection because the eigenstates have “unique” (in the sense they do not fluctuate) values of these quantities. We can speak of a state having this energy, with this angular momentum etc because if you mak a measurement of energy on such an eigenstate you get a single answer.