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QFT: Solving the integral for the Wightman function in Minkowski spacetime.

  1. Aug 18, 2012 #1
    1. The problem statement, all variables and given/known data

    How does one actually solve the integral for the Wightman function for a massless quantum scalar field in 4D Minkowski spacetime? That is, what is the integration technique to go from:

    [itex] \langle \hat{\phi}(x) \hat{\phi}(y) \rangle = \int_c d^4k \, \frac{1}{(2 \pi )^4}\frac{e^{ik(x-y)}}{ k^2} [/itex]

    to:

    [itex] \langle \hat{\phi}(x) \hat{\phi}(y) \rangle = \frac{1}{4 \pi^2 (x-y)^2} [/itex]?

    2. Relevant equations

    See above.

    3. The attempt at a solution

    The [itex]k_0[/itex] term is dealt with by contour integration. That part's easy. The problem I'm having is figuring out how to deal with the [itex]k_i[/itex] terms. It should just be a 3D Fourier transform, but I can't figure it out.

    I have to admit that it's embarrassing for me to post this, since I'm a postdoc and should know this off the top of my head. I know I have the answer buried somewhere in my notes, but I just finished an oversees move and all of my notes are still in transit. I haven't been able to find it online, either; seems most online lecture notes I've found either just show the final equation or leave the derivation as an exercise to the student. (Ha!)
     
    Last edited: Aug 18, 2012
  2. jcsd
  3. Aug 19, 2012 #2

    gabbagabbahey

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    Can you show your [itex]k_0[/itex] integration? I may have some ideas for the Fourier integral, but first I want to make sure we are talking about the same integral.
     
  4. Aug 19, 2012 #3
    If you consider the [itex]k_0[/itex] integral first, then you get:

    [itex]\int_c d^4k \, \frac{i}{(2\pi)^4}\frac{e^{ik(x-y)}}{k^2} = \int_c dk_0 \, \int_{-\infty}^{\infty} d^3\vec{k} \, \frac{i}{(2\pi)^4}\frac{e^{i\vec{k}(\vec{x}-\vec{y})-i k_0 (t_x-t_y)}}{-k_0^2+\vec{k}^2}[/itex]

    where the contour for [itex]G^+(x,y)[/itex] is a counterclockwise circle of radius epsilon around the positive root [itex]k_0 = |\vec{k}|[/itex]. (Note that I was missing a factor of [itex]i[/itex] in my first post.)

    The Laurent series for [itex]\frac{e^{iux}}{-u^2+a^2}[/itex] around the pole at [itex]u=a[/itex] is [itex]\frac{e^{iax}}{2a(-u+a)}+O(1)[/itex], so by method of residues the [itex]k_0[/itex] integral evaluates to:

    [itex]\int_c dk_0 \, \int_{-\infty}^{\infty} d^3\vec{k} \, \frac{i}{(2\pi)^4}\frac{e^{i\vec{k}(\vec{x}-\vec{y})-i k_0 (t_x-t_y)}}{-k_0^2+\vec{k}^2} = \int_{-\infty}^{\infty} d^3\vec{k} \, \frac{1}{(2\pi)^3}\frac{e^{i\vec{k}(\vec{x}-\vec{y})-i \omega_k (t_x-t_y)}}{2 \omega_k} [/itex]

    with [itex]\omega_k = |\vec{k}|[/itex].

    This is the same integral one gets from evaluation of [itex]\langle \hat{\phi}(x) \hat{\phi}(y) \rangle[/itex] with [itex]\hat{\phi}(x) \equiv \int \frac{d^3\vec{k}}{(2\pi)^{3/2}\sqrt{2\omega_k}} (\hat{a}_\vec{k} e^{i\vec{k}\vec{x}-i\omega_kt}+\hat{a}^\dagger_\vec{k} e^{-i\vec{k}\vec{x}+i\omega_kt}) [/itex], so I don't think it's helpful to evaluate the [itex]k_0[/itex] integral first. I'm not sure, though. The Fourier transform seems more difficult this way, but maybe I'm just not seeing the trick.
     
    Last edited: Aug 19, 2012
  5. Aug 19, 2012 #4

    gabbagabbahey

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    Right, so let's look at the Fourier transform first. It is sufficient to look at [itex]\int_{ \text{all } \vec{k} \text{-space} } d^3 \vec{k} \frac{e^{i\vec{k} \cdot (\vec{x} - \vec{y})}}{\omega_k^2-k_0^2} [/itex] , yes?

    I think this is best done in Spherical coordinates [itex](\omega_k, \theta, \phi)[/itex] with the polar axis aligned with [itex](\vec{x} - \vec{y})[/itex], which gives you something like

    [tex]2\pi \int_0^{\infty} \int_0^{\pi} \frac{e^{i \omega_k |\vec{x} - \vec{y}| \cos \theta}}{\omega_k^2-k_0^2}\omega_k^2\sin\theta d \omega_k d\theta[/tex]

    Do the angular integral first, and then I think you can just use the Residue theorem on the [itex]\omega_k[/itex] integral
     
  6. Aug 19, 2012 #5
    Ah ha! I think you've got it; this seems to stir some memories from my coursework. Let me sit down with pencil and paper and work through it in detail, then I'll post the full derivation.

    I'm surprised I didn't come up with that myself. Guess I got stuck on doing the integral in Cartesian coords.
     
  7. Aug 20, 2012 #6
    So, spherical coords. is indeed the right way to solve the integral, although it turns out to be easier to solve if you do the contour first. You get:

    [itex]
    \langle \hat{\phi}(x) \hat{\phi}(y) \rangle = \int_{-\infty}^{\infty} d^3\vec{k} \, \frac{1}{(2\pi)^3}\frac{e^{i\vec{k}(\vec{x}-\vec{y})-i \omega_k (t_x-t_y)}}{2 \omega_k} \\
    = \frac{1}{2(2\pi)^3} \int_0^\infty dk_r \int_{-1}^1 d\cos{k_\theta} \int_0^{2\pi} dk_\phi \, k_r e^{i k_r |\vec{x}-\vec{y}|\cos{k_\theta} - i k_r (t_x-t_y)} \\
    = -\frac{1}{4\pi^2} \frac{1}{|\vec{x}-\vec{y}|} \int_0^\infty dk_r \, \sin(k_r |\vec{x}-\vec{y}|)e^{- i k_r (t_x-t_y)}
    [/itex]

    To evaluate the final integral, you have to displace [itex](t_x-t_y)[/itex] slightly into the complex plane, [itex](t_x-t_y) \to (t_x-t_y -i\epsilon)[/itex], giving:
    [itex]-\frac{1}{4\pi^2} \frac{1}{|\vec{x}-\vec{y}|} \int_0^\infty dk_r \, \sin(k_r |\vec{x}-\vec{y}|)e^{- i k_r (t_x-t_y-i\epsilon)} = -\frac{1}{4\pi^2}\frac{1}{|\vec{x}-\vec{y}|^2 - (t_x-t_y)^2} = -\frac{1}{4\pi^2}\frac{1}{(x-y)^2-i\epsilon}[/itex]

    This is still a slightly unsatisfying way to do things, since it relies on the identity [itex]\vec{k} \cdot (\vec{x}-\vec{y}) = |\vec{k}||\vec{x}-\vec{y}|\cos \theta[/itex], and that's not guaranteed to hold if I move away from Minkowski spacetime. Oh well; no one ever said QFT was easy.

    Thanks for the help, gabba.
     
    Last edited: Aug 20, 2012
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