Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: QFT in a nutshell: From field to particle

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data

    I don't understand how Zee gets Eq. (2) on p. 24:

    [itex]W(J) = - \frac{1}{2}\int \frac{d^4k}{(2\pi)^4} J(k)^\ast\frac{1}{k^2-m^2+i\varepsilon}J(k)[/itex]

    2. Relevant equations

    [itex]W(J) := - \frac{1}{2}\int d^4x\int d^4y J(x)D(x-y)J(y)[/itex]

    3. The attempt at a solution

    I don't see where the [itex]d^4k[/itex] and factors of 2pi are coming from. Manipulating the definition of W(J) I was able to show that

    [itex]W(J) = -\frac{1}{2}\int d^4 x e^{ikx}J(x)\frac{1}{k^2-m^2+i\varepsilon} J(k)[/itex]
    [itex]W(J) = -\frac{1}{2}J(-k)\frac{1}{k^2-m^2+i\varepsilon} J(k)[/itex]

    but this obviously doesn't agree.

    Last edited: Nov 19, 2007
  2. jcsd
  3. Nov 19, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    I see you have a power E^{ikx} but I don't see a k integral... the idea is to substitute J by its Fourier transform:
    [tex]J(x) = \int \frac{d^4 k}{(2\pi)^4} e^{i k x} J(k)[/tex]
    [tex]J(y)^* = \int \frac{d^4 k'}{(2\pi)^4} e^{-i k' x} J(k')^*[/tex]
    Now combine the exponentials and perform the x-integration, this will cancel out one of the k integrals
    [tex]\int dx e^{i (k - k') x} = \delta(k - k')[/tex]
    and then do the manipulations you did to get the [tex]\left(k^2 - m^2 + i\epsilon\right)^{-1}[/tex] and the other stuff right.
  4. Nov 19, 2007 #3
    Hi CompuChip,

    Thanks for replying.

    Why are you taking the complex conjugate of J(x)? The definition of W(J) is

    [itex]W(J) := - \frac{1}{2}\int d^4x\int d^4y J(x)D(x-y)J(y)[/itex]

    No complex conjugates there. My approach was to separate the [itex]e^{ik(x-y)}[/itex] term in the expression for the propagation D(x - y) giving

    [itex]W(J) := - \frac{1}{2}\int d^4x\int d^4y J(x)D(x-y)J(y)[/itex]
    [itex]W(J) := - \frac{1}{2}\int d^4x\int d^4y J(x)\frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon}J(y)[/itex]
    [itex]W(J) := - \frac{1}{2}\int d^4x\int d^4y e^{ikx}J(x)\frac{1}{k^2-m^2+i\epsilon}e^{-iky}J(y)[/itex]
    [itex]W(J) := - \frac{1}{2}\left(\int d^4x e^{ikx}J(x)\right)\frac{1}{k^2-m^2+i\epsilon}\left(\int d^4ye^{-iky}J(y)\right)[/itex]
    [itex]W(J) := - \frac{1}{2}J(-k)\frac{1}{k^2-m^2+i\epsilon}J(k)[/itex].

    Note quite right.
  5. Nov 19, 2007 #4
    Omg I'm so dumb! I forgot to put the integral in the propagator. Problem solved now (I think). If J is complex, why is it [itex]J^\ast(k)[/itex] and not J(-k)?
  6. Nov 19, 2007 #5


    User Avatar
    Science Advisor
    Homework Helper

    Ah, my mistake. You get a [tex]\delta(k + k')[/tex] from the (real)space-integral hence J(k) and J(-k).
    I might still be a little off here, but the general idea was as follows: J(x) is real. This implies that
    [tex]J(x) = \sum_{k = -\infty}^\infty e^{ikx} J(k)
    \sum_{k = -\infty}^\infty e^{-ikx} J(k)^* = J(x)^*
    so comparing the coefficients of the (basis) exponentials, we conclude that [tex]J(k) = J(-k)^*[/tex], therefore we can write J(k)* instead of J(-k).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook