Total Momentum Operator for Free Scalar Field

[nicksauce]

Homework Statement

I want to show that
\mathbf{P} = -\int d^{3}x}\pi(x)\nabla\phi(x) = \int{\frac{d^{3}p}{(2\pi)^3}\mathbf{p}a_{p}^{\dagger}a_p

for the KG field.

Homework Equations

\phi(x) = \int{\frac{d^{3}p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p + a_{-p}^{\dagger})e^{ipx}

\pi(x) = -i\int{\frac{d^{3}p}{(2\pi)^3}\sqrt{\frac{\omega_k}{2}}(a_p - a_{-p}^{\dagger})e^{ipx}

The Attempt at a Solution

I'm having trouble seeing why this is true. What happens to the a_pa_k[/tex] and a_p^{\dagger}a_k^{\dagger}-like cross terms?

 

[gabbagabbahey]

I'd write \pi(x) and \phi(x) as

\pi(\textbf{x}) = -i\int{\frac{d^{3}\textbf{p}'}{(2\pi)^3}\sqrt{\frac{\omega_{p'}} {2}}(a_{\textbf{p}'} - a_{-\textbf{p}'}^{\dagger})e^{i\textbf{p}'\cdot\textbf{x}}

\phi(\textbf{x}) = \int{\frac{d^{3}\textbf{p}}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\textbf{p}}}}(a_{\textbf{p}} + a_{-\textbf{p}}^{\dagger})e^{i\textbf{p}\cdot\textbf{x}}

Then just calculate \mathbf{\nabla}\phi, substitute everything in and do the integration over \textbf{x} first. You should get something with a delta function like \delta^3(\textbf{p}+\textbf{p}'), which allows you to get rid of all the a_{\textbf{p}}a^{\dagger}_{-\textbf{p}'}[/itex]-like cross terms just by integrating over \textbf{p}'.