# Charge and normal ordering in QFT

1. Nov 30, 2017

### Milsomonk

1. The problem statement, all variables and given/known data
Express the charge Q in terms of the creation and annihilation operators.

2. Relevant equations
$$\phi_{(x)}=\int \dfrac {d^3 p} {(2\pi)^3} \dfrac {1} {2 \omega_p} (a_p e^{i x \cdot p} + b^{\dagger}_{p} e^{-i x \cdot p})$$
$$\pi_{(x)}=\dfrac {-i} {2}\int \dfrac {d^3 p} {(2\pi)^3} (b_p e^{i x \cdot p} - a^{\dagger}_{p} e^{-i x \cdot p})$$
$$Q=-i\int d^3 x(\pi\phi - \phi^* \pi^*)$$

3. The attempt at a solution

Hey guys, little bit stuck with the normal ordering procedure. So i've basically plugged in my expressions for pi and phi into the expression for Q and arrived at the following:
$$Q=-i \int d^3x \int \dfrac {d^3 p} {(2\pi)^3} \dfrac {-i} {2\omega_p} (b_p b^{\dagger}_p - a^{\dagger}_p a_p)$$
So I know that I shouldn't have the spatial integral, but i'm not sure how to get rid of it and I know I need to normal order the operators but I'm stuck there to :/ any guidance would be massively appreciated :)

2. Nov 30, 2017

### TSny

The $p$ in the integral for $\pi$ should be distinguished from the $p$ in the integral for $\phi$. Use different symbols for these $p$'s. Thus, the product $\pi \phi$ will be a double integral over the two different $p$'s. The integration over $x$ is used to eliminate the exponentials and to introduce delta functions involving the two different $p$'s. The delta functions allow you to integrate over one of the $p$'s so that you end up with a single integral over the remaining $p$.

3. Nov 30, 2017

### Milsomonk

Oh ok, thanks very much :) Looks like I have it now.