# Explanation of Eq of Peskin's Intro to QFT Book

1. Feb 7, 2006

### beta3

Hi

Can someone please explain to me Eq 2.30 in Schroeder's and Peskin's book? (page 21)

how does he simplify the long equation of the commutator to this delta distribution?
$$\imath\delta^(^3^)\ (x - x')$$

Thanks

2. Feb 7, 2006

### SpaceTiger

Staff Emeritus
He starts with:

$$[a_{\textbf p},a_{\textbf p'}^{\sp\dagger}]=(2\pi)^3\delta^{(3)}(\textbf p-\textbf p')$$

This means that:

$$[a_{-\textbf p}^{\sp\dagger},a_{\textbf p'}]-[a_{\textbf p},a_{-\textbf p'}^{\sp\dagger}]=-2(2\pi)^3\delta^{(3)}(\textbf p+\textbf p')$$

Substituting this in and integrating over the delta function (i.e. replacing p with -p'), you get:

$$i\int \frac{d^3p'}{(2\pi)^3}e^{i\textbf p'(\textbf x'-\textbf x)}$$

which is just i times the inverse Fourier transform of $e^{i\textbf p'\textbf x'}$:

$$[\phi(\textbf x),\pi(\textbf x')]=i\delta^{(3)}(\textbf x-\textbf x')$$

Last edited: Feb 7, 2006
3. Feb 8, 2006

### beta3

thanks, but there's one thing i still don't understand

after substituting it in and then if you integrate, what happens to the term
$$\sqrt{\frac{\omega_p_'}{\omega_p}}$$

4. Feb 8, 2006

### nrqed

since p = p' (where by p I mean the magnituide of the three-vector), the two omegas are equal.

Pat

5. Feb 8, 2006

### beta3

ah, yeah, now everything makes sense

Thank you two both soo much