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Homework Help: Qft peskin eqn 2.54

  1. Jan 6, 2009 #1
    QFT Peskin p.30 eqn 2.54

    1. The problem statement, all variables and given/known data

    i am perplexed with eqn 2.54 peskins introductory qft. just cant make out how to arrive at it from the previous step. i think that there are dirac delta funtions involved but simply cant make it out. can somebody help? provide some hint? thanks in advance for ur time

    3. The attempt at a solution
    [tex]\int\ \frac{d^3p} {(2\pi)^3}\ \{ \frac {1}{2E_p}\ e^{-ip.(x-y)}\left|_{p^0 = E_p}\ +\ \frac {1}{-2E_p}\ e^{-ip.(x-y)}\left|_{p^0 = -E_p}\ \}= \int\ \frac{d^3p} {(2\pi)^3}\ \int\ \frac{dp^0} {2p^0}\ e^{-ip.(x-y)}\ \{ \delta (p_0-E_p) +\delta (p_0+E_p)\ \}[/tex] [tex]= \int\ \frac{d^3p} {(2\pi)^3}\ \int\ dp^0\ e^{-ip.(x-y)}\ \delta(p^2-m^2)[/tex]

    dont know if iam on the right track.pls correct me if am wrong.
    Last edited: Jan 6, 2009
  2. jcsd
  3. Jan 7, 2009 #2
    It would be good, if you would write down the equation you want to prove, since I dont have the mentioned book...
  4. Jan 7, 2009 #3
    What P&S are doing in Eqn. 2.54 is re-writing a three-dimensional integral as a four-dimensional integral:
    [tex]\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}[\exp(-ip\cdot(x - y)) - \exp(ip\cdot (x - y))] = \int\frac{d^3p}{(2\pi)^3}\int\frac{dp^0}{2\pi i}\ \frac{-1}{p^2 - m^2}\exp(-ip\cdot(x - y)),[/tex]
    where [tex]x^0 > y^0[/tex].

    What I would do to understand this is start from the latter form and perform the [tex]p^0[/tex] integral. Break up the denominator into
    [tex]p^2 - m^2 = (p^0)^2 - \textbf p^2 - m^2,[/tex]
    which has poles at
    [tex]p^0 = \pm \sqrt{\textbf p^2 + m^2} = \pm E_p.[/tex]
    Contour integration should produce the first expression in the original post [which P&S give as an intermediate step] without too much trouble.
  5. Jan 7, 2009 #4
    i jumped into conclusions before reading the text further. sorry. anyways thanks so much for ur time & help.:smile:
    Last edited: Jan 7, 2009
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