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

[Runner 1]

Starting with,

[itex]\hat{X}\psi = x\psi[/itex]

then,

[itex]x\psi = x\psi[/itex]

[itex]\psi = \psi[/itex]

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

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

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

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

Thanks

 

[Matterwave]

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

[tex]\hat{X}\psi=y\psi[/tex]

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

Therefore

[tex]x\psi=y\psi[/tex]

And

[tex](x-y)\psi=0[/tex]

Let's make the notation better still:

[tex](x-y)\psi_y(x)=0[/tex]

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 [itex]\psi=0[/itex] 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:

[tex]\psi_y(x)=\delta(x-y)[/tex]

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

[tex]\int_{-\infty}^{\infty}\psi_y(x)\psi_z(x)dx=\int_{-\infty}^{\infty}\delta(x-y)\delta(x-z)dx=\delta(y-z)[/tex]

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

 

[dextercioby]

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.

 

[Runner 1]

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 headI 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.