Explanation of Eq of Peskin's Intro to QFT Book

[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

 

[SpaceTiger]

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

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')

 

[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}}

 

[nrqed]

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}}

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

Pat

 

[beta3]

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

Pat

ah, yeah, now everything makes sense

Thank you two both soo much