# Wavefunction Dimensionality

1. Apr 22, 2010

### shaiguy6

Hey Guys,

I'm getting a bit confused on the dimensionality of the wavefunction. I've seen the wavefunction described as:

(1) A vector of norm 1 in a finite dimensional Hilbert Space
(2) A vector of norm 1 in an infinite dimensional Hilbert Space
(3) A continuos function (it is my understanding that there can be equivalences between infinite dimensional spaces and functions ie. fourier series etc., is this what is happening here?)

Also, I'd like to make some statements and verify that they are correct:

* Observables are self-adjoint operators that act on the wavefunction
* The effect of a measurement is to force the state of the system to an eigenvector of the operator and the result of the measurement is the associated eigenvalue of that eigenvector
* Do self-adjoint operators always have n distinct eigenvalues? (where n is the dimensionality of the Hilbert space?)

Thanks!

2. Apr 22, 2010

### LostConjugate

In matrix notation a continuous function can be written as a vector where the index represents different possible values of the function's argument. This gives you a continuous index or infinite dimensional space. We find that functions have similar properties as vectors in this way and we can work with them easier. The wave function is just a function, we often times write it as a vector however. Hilbert space is the space of integrable functions. This is all mathematical notion keep in mind, not an observable space.

Observables are self-adjoint (hermition) operators that act on the wavefunction. In matrix notation they are nxn matrices.

Measurement does force the state into an eigenstate (vector and value) of the operator and yes the result of the measurement will be identically the eigenvalue of that operator.

I think your last question is correct, though I am not sure.

The wave function is always normalized to 1 in QM, mathematically the vector moves around on an infinite dimensional sphere.

Last edited: Apr 22, 2010
3. Apr 22, 2010

### fermi

Yes, because observables are assumed to be real valued (I cannot think of any exceptions) and the eigenvalues of self-adjoint operators are real. Operators that are not self-adjoint may have complex eigenvalues, and it is not possible in a straightforward way to interpret these as observables. So far so good. But the converse is not true. Some non-self-adjoint operators also have only real eigenvalues. Why do we exclude these from corresponding to physical observables? I am not sure that we do, and if indeed they are excluded, I do not know why. Maybe someone else will explain that, I would like to know.

No, the eigenvalues do not have to be distinct. Degenerate (i.e. repeated) eigenvalues may occur. For a finite n, you have precisely n eigenvalues, all of which are real, but not all of which are distinct. For an infinite dimensional case, the same result holds but it requires some supplementary assumptions about the Hilbert space.

4. Apr 22, 2010

### genneth

Actually, it's a bit of a lie that observables are real-valued. After all, a complex number is just two real ones. The amplitude of electric fields, for example, has a phase. It is represented as a non-Hermitian operator, and its expectation values are in general complex. This is fine.

5. Apr 22, 2010

### LostConjugate

This is not an example of a Quantum State though. All observables must be real-valued in QM otherwise they are not observable.

In classical mechanics such as EM waves, the complex number sometimes represents another function out of phase. Like sin and cos.