# Does position operator have eigen wave function?

1. Dec 13, 2013

### nosafeway

I am learning quantum mechics. The hypothesis is:

In the quantum mechanics, all operators representing observables are Hermitian, and their eigen functions constitute complete systems. For a system in a state described by wave function ψ(x,t), a measurement of observable F is certain to return one of the eigenvalues of the operator F.

In the 1D infinite well, the position x should be an observable, its operator x is Hermitian, and its eigen function satisfies xg(x)=x'g(x), leading to g(x)=δ(x-x'), the eigenvalue x' is any real number. But g(x)=δ(x-x') cannot be normalized, and it cannot be a wave function since a wave function must be normalizable. Then my question is:

1. Is the position x a Hermitian operator?
2. If it is, why there is no eigen wave function?

2. Dec 13, 2013

### rubi

Yes, the position operator is hermitian. But for the spectral theorem, a slightly stronger condition is needed: self-adjointness. These conditions agree only in the finite dimensional case. In the infinite-dimensional case, the domain of the operator matters. The position operator is self-adjoint on $L^2(\mathbb R)$ if you restrict its domain to the space of Schwarz functions $\mathcal S(\mathbb R)$. (You can look up the all the definitions in Wikipedia.)

In infinite dimensions, the eigenvectors don't need to form a basis anymore. The position operator is an extreme example, since it doesn't have eigenvectors at all.

For the full spectral theorem, have a look at this: http://en.wikipedia.org/wiki/Spectral_theorem

3. Dec 13, 2013

### WannabeNewton

You're right that the eigenfunctions of the position operator cannot be normalized in the usual sense of the $L^2$ norm as we're dealing with distributions now i.e. they don't live in the usual Hilbert space of square Lebesgue integrable functions. The way to get around it is to use what's known as a Rigged Hilbert space: http://en.wikipedia.org/wiki/Rigged_hilbert_space

4. Dec 13, 2013

### rubi

One thing needs to be noted here, though: Rigged Hilbert spaces indeed allow you to define the notion of generalized eigenvectors, but there is still no sense in which these generalized eigenvectors would constitute a basis.

5. Dec 13, 2013

### kith

Don't they constitute a basis for the extended space Ωx?

6. Dec 13, 2013

### George Jones

Staff Emeritus
Some may have noticed that the original poster has been banned. nosafeway was the reincarnation of a previously banned member, who was the reincarnation of a previously banned member, who was ...