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

Fourier transorm problem has me stumped

  1. Jan 4, 2005 #1
    The ground state wave functional for the photon theory is given as

    [tex] \Psi_0[\tilde{a}] = \eta \exp \left(-\frac{1}{2} \int \frac{d^3k}{(2\pi)^3} \frac{(\vec{k}\times\tilde{a}(\vec{k}))\cdot(\vec{k}\times\tilde{a}(-\vec{k}))}{|\vec{k}|}\right)[/tex]

    where [tex]\tilde{a}[/tex] is given as the Fourier transform of [tex]a[/tex], that is,

    [tex] a_i(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\tilde{a}_i(\vec{k})e^{i\vec{k}\cdot\vec{x}} [/tex]

    Transforming back to [tex]a[/tex], the book now says that (10.81) is equivalent to

    [tex] \Psi_0[a] = \eta \exp \left(-\frac{1}{(2\pi)^2} \int d^3x d^3y \frac{(\nabla\times\vec{a}(\vec{x}))\cdot(\nabla\times\vec{a}(\vec{y}))}{|\vec{x}-\vec{y}|^2}\right) [/tex]

    I've had to think about this for a long time, and I'm still not sure I understand it exactly.
  2. jcsd
  3. Jan 5, 2005 #2
    Starting with 10.83,substituting a_i from 10.67, get to 10.81
  4. Jan 5, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    Yes,usually the other way around is more simple.Passing from the coordinate representation to momentum representation.I think you struggled a great deal proving the first part of the equivalence.
    Basically u really don't have to,usually in QFT,we like more the momentum representation,because we build the Feynman rules in the momentum representation.
    So probably the author went backwards.He should have given it first in the coordinate representation and then pass it to momentum.After all,once u have it in the momentum representation,what's the point of passing it into the coordinate representation??


    PS.I hope u see that the coordinate->momentum passing is immediate.
  5. Jan 5, 2005 #4
    It takes at least three steps

    I don't see it immediately.

    [tex]-\frac{1}{(2\pi)^2} \int d^3x d^3y \frac{(\nabla\times\vec{a}(\vec{x}))\cdot(\nabla\times\vec{a}(\vec{y}))}{|\vec{x}-\vec{y}|^2}

    = -\frac{1}{(2\pi)^2}\int\frac{d^3xd^3yd^3kd^3k'}{(2\pi)^3(2\pi)^3}\frac{-(\vec{k}\times\tilde{a}(\vec{k})e^{i\vec{k}\cdot\vec{x}})\cdot(\vec{k'}\times\tilde{a}(\vec{k'})e^{i\vec{k'}\cdot\vec{y}})}{|\vec{x}-\vec{y}|^2}[/tex]

    [tex]= -\frac{1}{2}\int\frac{d^3xd^3y}{(2\pi)^3}\frac{-(\vec{k}\times\tilde{a}(\vec{k}))\cdot(\vec{k'}\times\tilde{a}(\vec{k'}))\delta^3(\vec{k}+\vec{k'})}{|\vec{k}|}

    = -\frac{1}{2} \int \frac{d^3k}{(2\pi)^3} \frac{(\vec{k}\times\tilde{a}(\vec{k}))\cdot(\vec{ k}\times\tilde{a}(-\vec{k}))}{|\vec{k}|}[/tex]

    For the second step, one needs [tex]\int\frac{4\pi}{(2\pi)^3}d^3x\frac{e^{i\vec{k}\cdot\vec{x}}}{|\vec{x}-\vec{y}|^2} = \frac{e^{i\vec{k}\cdot\vec{y}}}{|\vec{k}|}[/tex], which to me is not immediately obvious. It seems to have something to do with calculating the anti-Laplacian by reasoning in analogy with the inverse square law of gravitation.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?