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

[latentcorpse]

I'm trying to work out equation 2.45 in these notes:

http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf

Anyway $$\phi(\vec{x})$$ and $$\pi(\vec{x})$$ are given in equations 2.18 and 2.19

When I multiply out $\pi(\vec{x}) \vec{\div} \phi(\vec{x})$ I get four terms.
After normal ordering I can combine two of them to get the $$a_{\vec{p}}^\dagger a_{\vec{p}}$$ term I am meant to but I also have a $$a_{\vec{-p}} a_{\vec{p}}$$ and a $$a_{\vec{-p}}^\dagger a_{\vec{p}}^\dagger$$ term that I cannot get rid of.

Does anyone know how to get rid of the two extra terms?

Cheers

 

[Chopin]

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

$$\langle \psi | P | \psi \rangle$$

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

$$|\psi\rangle = a^\dagger_p|0\rangle$$

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

$$\langle 0 | a_p a^\dagger_p a^\dagger_{-p} a^\dagger_p|0\rangle$$

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

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.

 

[latentcorpse]

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

$$\langle \psi | P | \psi \rangle$$

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

$$|\psi\rangle = a^\dagger_p|0\rangle$$

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

$$\langle 0 | a_p a^\dagger_p a^\dagger_{-p} a^\dagger_p|0\rangle$$

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

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.

so basically they've just looked ahead and said that those terms won't be relevant?

i agree that they won't be if you take an expectation value it just bugs me because what he's written as an equality doesn't appear to actually be an equality!

 

[Oxvillian]

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.

 

[Chopin]

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!