# Do the eigenfunctions for the position operator form an orthogonal set?

Starting with,

$\hat{X}\psi = x\psi$

then,

$x\psi = x\psi$

$\psi = \psi$

So the eigenfunctions for this operator can equal anything (as long as they keep $\hat{X}$ linear and Hermitian), right?

Well, McQuarrie says that "the eigenfunctions of a Hermitian operator are orthogonal", which can be checked with:

$\int_{-\infty}^{\infty}\psi^*_m \psi_n\, dx = \langle m | n \rangle = 0$

But if the eigenfunctions can be anything, then that integral won't always equal zero. What am I missing here?

Thanks

## Answers and Replies

Matterwave
Science Advisor
Gold Member
So, you're messing up because you're using x as both the eigenvalue and the operator X. The correct equation should look like:

$$\hat{X}\psi=y\psi$$

We can't know a priori that the eigenfunction $\psi$ has eigenvalue x, so we must use y as the eigenvalue.

Therefore

$$x\psi=y\psi$$

And

$$(x-y)\psi=0$$

Let's make the notation better still:

$$(x-y)\psi_y(x)=0$$

So, our wave function is a function of x, and it has eigenvalue y for the x-operator.

What "function" can make this true for arbitrary x and y? Well, actually there is no function that can make this true for arbitrary x and y other than the trivial $\psi=0$ function but this is the trivial case. In fact, x has no "eigenfunctions" in the strict sense of the word "function". The only thing that could make that above equality work is:

$$\psi_y(x)=\delta(x-y)$$

Now you notice that this is STILL not square integrable, however, it IS orthogonal in some sense:

$$\int_{-\infty}^{\infty}\psi_y(x)\psi_z(x)dx=\int_{-\infty}^{\infty}\delta(x-y)\delta(x-z)dx=\delta(y-z)$$

This is the best that we can do. The eigenfunctions of x (and p) are not actually within the Hilbert space.

dextercioby
Science Advisor
Homework Helper
If X acts on wavefunctions of L2(R) as a multiplication through the real number x, then one can show that X is self-adjoint on his domain of definition. Then by the spectral theorem of Gelfand-Kostyuchenko-Maurin, an extension of it has a complete set of eigenfunctions is a rigged Hilbert space built over L2(R). The eigenfunctions are the Dirac delta distributions.

If X acts on wavefunctions of L2(R) as a multiplication through the real number x, then one can show that X is self-adjoint on his domain of definition. Then by the spectral theorem of Gelfand-Kostyuchenko-Maurin, an extension of it has a complete set of eigenfunctions is a rigged Hilbert space built over L2(R). The eigenfunctions are the Dirac delta distributions.
Thanks, but...

Your explanation

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My head

I think I'm going to have to find a more mathematically rigorous textbook on the subject as mine doesn't discuss Hilbert spaces, self-adjointness, or that spectral theorem.