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

  • #1
1,444
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I'm trying to work out equation 2.45 in these notes:

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

Anyway [tex]\phi(\vec{x})[/tex] and [tex]\pi(\vec{x})[/tex] are given in equations 2.18 and 2.19

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

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

Cheers
 

Answers and Replies

  • #2
368
12


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.
 
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  • #3
1,444
0


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.
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!
 
  • #4
196
22
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. :frown:

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.
 

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