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Zee, Quantum Field Theory in a Nutshell, problem 1.3.1

  1. Jul 2, 2016 #1
    1. The problem statement, all variables and given/known data
    I'm working through Zee for some self study and I'm trying to do all the problems, which is understandably challenging. Problem 1.3.1 is where I'm currently stuck: Verify that D(x) decays exponentially for spacelike separation.

    2. Relevant equations
    The propagator in question is
    $$ D(x) = -i \int \frac{d^3k}{2(2\pi)^3} \frac{e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}}{\sqrt{\boldsymbol{k}^2+m^2}} $$

    3. The attempt at a solution
    Presumably, I would have to solve the integral and show that it decays exponentially (the spacelike aspect has already been taken into account for the above integral) and what I did was switch to spherical coordinates and integrated over the azimuthal:

    $$ D = \frac{-i}{8\pi^2}\int dr\,d\theta\frac{e^{-irx\cos\theta}}{\sqrt{r^2+m^2}}r\cos\theta $$

    This is where I'm stuck. The square root in the denominator suggests this is a branch cut integral but I haven't been able to find a source that explains it sufficiently. If anyone could help me figure this out I'd appreciate it. Thanks.
     
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  3. Jul 3, 2016 #2

    stevendaryl

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    I feel that there is something wrong with this equation. Does [itex]x[/itex] mean a spatial vector, or a 4-vector? If it's a 4-vector, then there is a problem, because on the right-hand side of the equals sign, there is no mention of the time component of [itex]x[/itex]. If [itex]x[/itex] is a spatial vector vector, then it doesn't make any sense to talk about spacelike separations.
     
  4. Jul 3, 2016 #3
    That is curious, though I wrote exactly what he has in the book. Initially, we have
    $$ D(x) = -i \int \frac{d^3k}{(2\pi)^32\omega_k}[e^{-i(\omega_kt - \boldsymbol{k}\cdot\boldsymbol{k})}\theta(x^0)+e^{i(\omega_kt - \boldsymbol{k}\cdot\boldsymbol{k})}\theta(-x^0)] $$
    which he reduces to the above equation in consideration of spacelike separation where ##x^0= 0##. So I suppose the notation on the left is a four vector, it just happens to be ##x = (0, x^1, x^2, x^3) \equiv \boldsymbol{x}##.
     
  5. Jul 8, 2016 #4
    I now realize I made a mistake in the spherical coordinate substitution, the integral should be
    $$ D(x) = \frac{-i}{8\pi^2}
    int dr\,d\theta \frac{e^{-irx\cos\theta}}{\sqrt{r^2+m^2}} r^2\sin\theta $$
    The integral in ##\theta## is fairly straight forward at this point, and I got
    $$ D(x) = \frac{-i}{4\pi x}\int dr \frac{r\sin(rx)}{\sqrt{r^2+m^2}} $$
    which i now have no idea how to solve
     
  6. Jul 9, 2016 #5

    malawi_glenn

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    That is because that integral has no solution in terms of elementary functions.
     
  7. Jul 26, 2016 #6
    If you assume that the magnitude of the radial velocity is much less than the speed of light, then you can expand the denominator in the integral to second order in ## p_r ##.
    ##\sqrt{p^2 + m^2} = m\sqrt{1 + \frac {p^2} {m^2}} \approx m(1 + \frac {p^2} {m^2}) = \frac {(p - im)(p + im)} {m}##
    The integral becomes
    ##D(x) \approx \frac {1} {2(2\pi)^2} \frac {m} {|x|} \int_{0}^\infty dr \frac {r[(-i) \sin(r|x|)]} {(r - im)(r + im)} = \frac {1} {2(2\pi)^2} \frac {m} {|x|} \mathcal {Im} \int_{0}^\infty dr \frac {r e^{-i|x|r}} {(r - im)(r + im)}##
    We have simple poles at ##z_0 = \pm im##. We use the Cauchy Integral theorem and the residue theorem, closing a semi-circular contour in the lower half-plane, and noting that the integral along the arc tends to zero as the radius tends to ## \infty## and the integral along the real axis from ##-\infty ## to ##+\infty## is twice the integral from 0 to ## \infty##. We find,
    ##D(x) \approx \frac {1} {16\pi}\frac{m} {|x|}e^{-|x|m}##
     
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