1. May 15, 2012

### O.J.

Theorem: For every Hermitian operator, there exists at least one basis consisting of its orthonormal eigen vectors. It is diagonal in this basis and has its eigenvalues as its diagonal entries.

The theory is apparently making an assumption that every Hermitian operator must have eigen values/vectors. Am I missing something here? Should ALL hermitian operators have eigen values/vectors?

2. May 15, 2012

### dextercioby

Yes, that is the essential content of the spectral theorem for selfadjoint operators in a Hilbert space which is one of the most widely known results of functional analysis.

3. May 15, 2012

### O.J.

Is there a proof? Can you link me to one? =)

4. May 15, 2012

### Fredrik

Staff Emeritus
There's no proof that can be fully understood by a typical physics student in less than a year. However, most linear algebra books contains a proof of the finite-dimensional case. (Look for the words "spectral theorem" in any of them).

There's a spectral theorem for compact normal operators that has a proof that can be understood by someone who's good at linear algebra and topology, and only covers a few pages. You might be interested in that, but you will need to learn a non-negligible amount of topology.

The spectral theorem for bounded normal operators is the really hard one. You need lots of functional analysis, topology and measure theory for that one. The two simpler theorems mentioned above are special cases of this one.

However, the one for bounded normal operators is a special case of the one for arbitrary (not necessarily bounded) normal operators, and even that one can't handle things like momentum eigenstates. For that you need an even more general spectral theorem.