# My gratiude/thankfulness for PF

1. Oct 18, 2005

### Neitrino

Dear PF,
Let me once again express you all my gratiude/thankfulness for PF. Many years ago I finished my bachelors courses (physics), but I could not continue Master, I had to work to earn money. But I want to learn it and now the PF remains the major source for the answers to my questions.

Gents,
How do i begin diagonalisation of Hamoltonians (in QFT ) expressed in terms of creation/annihilation operators?

Thks alot

2. Oct 18, 2005

### vanesch

Staff Emeritus
As far as I know, we only know how to do that for free field theories (and some toy models - but I don't know much about those myself).

The answer is Fock space. It is worked out in, for instance, Peskin and Schroeder (and some other books). Fock space is defined by a basis that diagonalises the Hamiltonian and the momentum operators Px, Py and Pz.
It is "sliced up" in 0 - particle (vacuum), 1 - particle, 2 - particle , ... etc... subspaces.

The 1-particle states have the peculiarity of having a relationship between the momentum eigenvalues (kx, ky, kz) and the energy eigenvalue (omega), given by the relativistic dispersion relation: E^2 = P^2 + m^2 (c = 1, hbar = 1).
Each individual state in that subspace corresponds to a particle of definite momentum and energy. It is interesting that this FOLLOWS from the diagonalisation procedure (essentially hopping around with creation and annihilition operators). It is this feature which makes people claim that second quantisation makes particles come out of quantum fields. It has not been put in hand in there. That there are states corresponding to momentum and energy eigenvalues which correspond to this E^2 = P^2 + m^2 relationship is what remains, in the quantum realm, of what we intuitively call "a particle".

3. Oct 18, 2005

### Neitrino

Thank you Vanesch,
Could advise me in PS where is that mentioned?

4. Oct 18, 2005

### vanesch

Staff Emeritus
I think, off the top of my head, that this must be in chapter 2 or 3. I don't know if they mention explicitly Fock space, however.

cheers,
Patrick.