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The expansion of the Klein Gordon field and conjugate momentum field are

$\hat{\phi}(x) = \int \frac{d^3k}{(2 \pi)^3} \, \frac{1}{ \sqrt{2 E_{k}}} \left( \hat{a}_{k} + \hat{a}^{\dagger}_{-k} \right) e^{i k \cdot x}$


$\hat{\pi}(x) = \int \frac{d^3k}{(2 \pi)^3} \, \sqrt{\frac{E_{k}}{2}} \left( \hat{a}_{k} - \hat{a}^{\dagger}_{-k} \right) e^{i k \cdot x}$

The total momentum in the classical field is given by

$P^{i} = -\int d^3x \, \pi(x) \, \partial_{i} \phi(x)$

which is promoted to an operator

$\hat{P}^{i} = -\int d^3x \, \hat{\pi}(x) \, \partial_{i} \hat{\phi}(x)$

We are told that putting the expansion into this should give

$\int \frac{d^3p}{(2\pi)^3} \, p^i \left( \hat{a}_{p}^{\dagger} \hat{a}_{p} + \frac12 (2\pi)^3 \delta^{(3)}(0) \right)$

which I do get, however I also get terms involving $\hat{a}_{p}\hat{a}_{-p}$ and $\hat{a}^{\dagger}_{p} \hat{a}^{\dagger}_{-p}$. How in the world can these terms possibly go away? Have a made a mistake in the math (I am prone to mistakes at the moment.. quite tired) or is there some way that those terms cancel?

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up vote 3 down vote accepted

The two integrands $p^i\hat{a}_{{\bf p}}\hat{a}_{-{\bf p}}$ and $p^i\hat{a}^{\dagger}_{{\bf p}}\hat{a}^{\dagger}_{-{\bf p}}$ are antisymmetric wrt. ${\bf p} \leftrightarrow -{\bf p}$. Hence the corresponding integrals $\int d^3p(\ldots )$ are zero.

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Right, +1. The OP could have seen that these terms cancel and have to cancel. A simple way to see it is that $P$ commutes with the total number of excitations $N$, because it just moves these excitations to a different place. So it can't contain terms that change $N$ by plus or minus two. – Luboš Motl May 2 '12 at 7:49
ugh.... obviously. Thanks, I could have just thought a little bit before asking the question, but sometimes its just nice to be told the answer. Thanks! – Kyle May 2 '12 at 9:41
Apologies for being nosey, but are you a theoretical physicist in your main job or spare time? – Larry Harson May 2 '12 at 18:40

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