How do we know that $\psi$ is the eigenfunction of an operator $\hat{H}$ with eigenvalue $W$?

Question

I am kind of new to this eigenvalue, eigenfunction and operator things, but I have come across this quote many times:

$\psi$ is the eigenfunction of an operator $\hat{H}$ with eigenvalue $W$.

First I need some explanation on how do we know this? All I know about operator $\hat{H}$ so far is this equation where $\langle W \rangle$ is an energy expected value:

\begin{align} \langle W \rangle &= \int \limits_{-\infty}^{\infty} \overline{\Psi}\, \left(- \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p\right) \Psi \, d x \end{align}

From which it follows that $\hat{H} = - \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p$.

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Additional question:

I know how to derive relation $\hat{H}\hat{a} = (W - \hbar \omega)\hat{a} \psi$ for which they state that:

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

I also know how to derive relation $\hat{H}\hat{a}^\dagger = (W + \hbar \omega)\hat{a}^\dagger \psi$ for which they state that:

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

How do we know this?

Answer 1

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 $(W-\hbar \omega)$.

Answer 2

First i need some explaination on how do we know this?

It's stipulated.

Maybe it will help your understanding if we phrase it this way:

Let $\psi$ be an eigenfunction of an operator $\hat{H}$ with eigenvalue $W$.

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(Update to address OPs comment).

Spectral Theorem:

Theorem. There exists an orthonormal basis of V consisting of eigenvectors of A. Each eigenvalue is real.

In the above, A is a Hermitian operator. In QM, The Hamiltonian operator $\hat{H}$, is a Hermitian operator corresponding to the classical total energy observable.

The spectral theorem essentially guarantees that not only are there eigenfunctions (eigenvectors, eigenstates) with real eigenvalues associated with Hermitian operators, but that the set of these eigenstates is complete, i.e., any possible state of the system can be expressed as a weighted sum of the eigenstates of the operator.

So, we know that there are eigenstates and eigenvalues associated with $\hat{H}$. $\psi$ is just a label for one in particular and $W$ is just a label for the associated eigenvalue.