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

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?

  • $\begingroup$ Your additional question applies to the harmonic oscillator only. You should mention that. $\endgroup$
    – Lagerbaer
    May 10, 2013 at 16:49
  • $\begingroup$ We find eigenfunctions and eigenvalues by solving $\hat{H} \Psi = <W> \Psi$, in your notation. $\endgroup$
    – my2cts
    Apr 30, 2018 at 21:33

2 Answers 2


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 $(W-\hbar \omega)$.

  • $\begingroup$ 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: $\langle O\rangle = \int |\psi\rangle^\dagger O|\psi\rangle d x$ ? $\endgroup$
    – 71GA
    May 10, 2013 at 17:28
  • $\begingroup$ $\langle \psi | O | \psi \rangle$ 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 $\psi(x)$. $\endgroup$
    – innisfree
    May 10, 2013 at 22:49
  • $\begingroup$ @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. $\endgroup$
    – nervxxx
    May 10, 2013 at 23:35
  • 1
    $\begingroup$ @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 $\endgroup$
    – nervxxx
    May 10, 2013 at 23:38
  • $\begingroup$ This Dirac notation is confusing for a starters ... I tried reading Zetilli and got lost ... there is so much of this stuff / rules ... $\endgroup$
    – 71GA
    May 11, 2013 at 7:11

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 $W$.

(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.

  • $\begingroup$ 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. $\endgroup$
    – 71GA
    May 10, 2013 at 15:49

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