Operators: how to get rid of the two extra terms?

Re: Operators

When we use an operator like this, we do so by taking its expectation value against some state.

[tex]\langle \psi | P | \psi \rangle[/tex]

But in this formalism, states are constructed out of momentum eigenstates.

[tex]|\psi\rangle = a^\dagger_p|0\rangle[/tex]

That means that terms like [itex]a^\dagger_p a^\dagger_{-p}[/itex] in the momentum operator will lead to things like

[tex]\langle 0 | a_p a^\dagger_p a^\dagger_{-p} a^\dagger_p|0\rangle[/tex]

Terms like this will always vanish, because there aren't enough [itex]a_p[/itex]'s to annihilate all of the [itex]a^\dagger_p[/itex]'s (and there aren't any at all to annihilate the [itex]a^\dagger_{-p}[/itex]'s), so some of them will hit the vacuum on the left and vanish. A similar argument holds for terms like [itex]a_p a_p[/itex].

I'm not sure whether that's a good answer, because you could say the same thing about the Hamiltonian too, and there those terms explicitly cancel without needing this argument, but I know that when you use it as an expectation value like this, those terms won't contribute.

 

I think Chopin's argument fails because the general state in Fock space is a superposition of 0-particle states, 1-particle states, 2-particle states, etc etc. It might very well survive after we hit it with 2 annihilation operators.

However if you do the x-integration carefully, I think you might find that the resulting integrand is odd in p, and so the whole thing does vanish anyway. If memory serves me correctly.

 

Oxvillian, you're right: http://physics.stackexchange.com/questions/24716/derivation-of-total-momentum-operator-qft

That's a much more satisfying answer than the one I gave--sorry for the red herring!