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Explanation of Eq of Peskin's Intro to QFT Book

  1. Feb 7, 2006 #1
    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?
    [tex]\imath\delta^(^3^)\ (x - x') [/tex]



    Thanks
     
  2. jcsd
  3. Feb 7, 2006 #2

    SpaceTiger

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    He starts with:

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

    This means that:

    [tex][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')[/tex]

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

    [tex]i\int \frac{d^3p'}{(2\pi)^3}e^{i\textbf p'(\textbf x'-\textbf x)}[/tex]

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

    [tex][\phi(\textbf x),\pi(\textbf x')]=i\delta^{(3)}(\textbf x-\textbf x')[/tex]
     
    Last edited: Feb 7, 2006
  4. Feb 8, 2006 #3
    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
    [tex] \sqrt{\frac{\omega_p_'}{\omega_p}} [/tex]
     
  5. Feb 8, 2006 #4

    nrqed

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

    Pat
     
  6. Feb 8, 2006 #5
    ah, yeah, now everything makes sense

    Thank you two both soo much :wink:
     
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