1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

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

    CompuChip

    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]
    and
    [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

    CompuChip

    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)
    \stackrel{!}{=}
    \sum_{k = -\infty}^\infty e^{-ikx} J(k)^* = J(x)^*
    [/tex]
    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).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: QFT in a nutshell: From field to particle
  1. Particle in a field (Replies: 2)

Loading...