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

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 $\hat{H} = - \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p$.

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?

-
Your additional question applies to the harmonic oscillator only. You should mention that. – Lagerbaer May 10 '13 at 16:49

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.

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.

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

-
Thank you for this explaination. It was brief and provided lots of good info. There is only one more thing. I don't quite understand this equation: $\langle O \rangle = \langle \Psi | O | \Psi \rangle$. Is this a scalar product with itself? And then an operator acts on this scalar product? I know that if we use a $\dagger$ on a ket we get a bra, so it must hold that: $\langle O\rangle = \langle \psi |O| \psi \rangle = |\psi\rangle^\dagger O|\psi\rangle$... But where is the integral? Shouldnt it be: $\langle O\rangle = \int |\psi\rangle^\dagger O|\psi\rangle d x$ ? – 71GA May 10 '13 at 17:28
$\langle \psi | O | \psi \rangle$ is an example of Dirac's Bra-ket notation. $| \psi \rangle$ is a ket, a vector in Hilbert space. $\langle \psi |$ is a bra, the covector in Hilbert space, such that $\langle \psi | \psi \rangle$ is an inner product. Sounds tricky, but actually it becomes much much easier than position representations $\psi(x)$. – innisfree May 10 '13 at 22:49
@71GA The expectation value of an operator $O$ is defined to be the 'sandwich' $\langle\Psi|O|\Psi\rangle$ which really means taking $O$ to act to the right on $|\Psi\rangle$ then taking the inner product of the two kets $|\Psi\rangle$ and $|O\Psi \rangle$ which is the same as taking a (different kind of) product of the bra $\langle\Psi|$ and the ket $|O\Psi\rangle$. At this stage, we haven't specified kind of inner product at all yet except that it obeys inner product properties. – nervxxx May 10 '13 at 23:35
@71GA When we take the position basis, we are specifying the kind of inner product by the use of the identity operator $1 = \int dx |x\rangle\langle x|$. Then $\langle O \rangle \equiv \langle\Psi|O|\Psi\rangle = \int dxdx' \langle\Psi|x\rangle\langle x|O|x'\rangle\langle x'|\Psi\rangle = \int dxdx' \Psi(x)^* O(x)\delta(x-x') \Psi(x') = \int dx \Psi(x)^* O(x) \Psi(x)$. See en.wikipedia.org/wiki/Bra-ket_notation – nervxxx May 10 '13 at 23:38
This Dirac notation is confusing for a starters ... I tried reading Zetilli and got lost ... there is so much of this stuff / rules ... – 71GA May 11 '13 at 7:11

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

It's stipulated.

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

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.
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.
I dont know if you understood my question right. How do we know from this $\langle W \rangle = \int \limits_{-\infty}^{\infty} \overline{\Psi}\, \left(- \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p\right) \Psi \, d x$ or **this** $\hat{H} = - \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p$ that we have an eigenfunctiuion and eigenvalue. – 71GA May 10 '13 at 15:49