# Problem with changing basis in Hilbert space

1. Jan 3, 2013

### Chain

The expansion theorem in quantum mechanics states that a general state of a system can be represented by a unique linear combination of the eigenstates of any Hermitian operator.

If that's the case then that would imply we would be able to represent the spin state of a particle in terms of it's energy eigenstates for example however surely it's impossible to do this for a particle in an infinite square well for instance, since the energy eigenstates are independent of spin.

2. Jan 3, 2013

### ModusPwnd

I dont see how that implication comes from the statement... You can represent the spin with a unique linear combination of the spin eigenstates, you can represent the energy state as a linear combination of energy eigenstates. Is this not right?

3. Jan 3, 2013

### Jano L.

There are two difficulties.

This is not necessarily always true. In quantum theory, it can be proven only for certain operators. I think perhaps the Hamiltonian of a harmonic oscillator is a good example in this respect. But I have read somewhere that it is difficult to prove generally that eigenfunctions of every Hermitian operator will form a basis. Often this has to be just assumed.

No, of course this is not so. The above theorem or assumption applies to single Hilbert space. Spin state and coordinate wave functions belong to different Hilbert spaces.

4. Jan 3, 2013

### dextercioby

Indeed, the nuclear spectral theorem is hard to prove, but nonetheless correct.

It's always assumed, if you minimize the mathematical rigurosity of your treatment.

The product of the two (i.e. the wavefunction of the system) belongs to the direct product (rigged) Hilbert space.

5. Jan 4, 2013

### Chain

Ah okay fair enough guys, I thought it literally meant a wavefunction could be expanded as the eigenstates of any Hermitian operator, so basically the theorem is true only if the observables belong to the same Hilbert space?

Thanks for the replies :)