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

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

and

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

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?

Abhimanyu Pallavi Sudhir
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Kyle
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  • How did you get the first term. I am unable to get that. Please consider answering: https://physics.stackexchange.com/q/375585/ – Sidd Dec 21 '17 at 05:56

1 Answers1

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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.

Qmechanic
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