Homework Helper

## 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 $a_pa_k[/tex] and [itex]a_p^{\dagger}a_k^{\dagger}$-like cross terms?

gabbagabbahey
Homework Helper
Gold Member
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 [tex]a_{\textbf{p}}a^{\dagger}_{-\textbf{p}'}[/itex]-like cross terms just by integrating over $\textbf{p}'$.