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Zee QFT In a Nutshell Propagator Question

  1. Apr 26, 2010 #1
    I've started working through Zee's book and have got to question I.3.2 - calculation of D(x) in 1+1 dimensions for t=0. The expression to evaluate becomes (omitting constant multipliers for simplicity)

    [tex]\int^{\infty}_{-\infty} dk \frac{e^{ikx}}{\sqrt{k^2+m^2}} [/tex]


    This is singular at k=+im and k=-im. I'll take the square root to have a branch cut on the imaginary axis from [itex]k=+im\rightarrow+i\infty[/itex] and [itex]k=-im\rightarrow-i\infty[/itex]. If I substitute z=-ik, I can put the branch cut on the real axis from [itex]z=+m\rightarrow+\infty[/itex] and [itex]z=-m\rightarrow-\infty[/itex] and the integral becomes

    [tex]\int^{+i\infty}_{-i\infty} dz \frac{e^{zx}}{\sqrt{m^2-z^2}} [/tex]

    For positive x, I close the contour in the half of the z plane with negative real value, so that the contribution of the semicircle at infinity vanishes. The answer I want then becomes the integral inwards and outwards along the branch cut. Since with the square root, the integral along the cut an infinitesimal distance above the cut is the negative of the integral an infinitesimal below it, I'm left with (again ignoring factors of 2 etc)

    [tex]\int^{-\infty}_{-m} dz \frac{e^{zx}}{\sqrt{m^2-z^2}} [/tex]

    ie

    [tex]e^{-mx}\int^{-\infty}_{-m} dz \frac{e^{(z+m)x}}{\sqrt{z-m}\sqrt{z+m}} [/tex]

    This is as far as I can get. If I expand the exponential, I can separate the integrand into a part which has a singular contribution at z=-m and the rest which doesn't, but beyond this I'm not sure where to go. Since it's an elementary question at the beginning of an introductory book, I expected it to have some simple closed form solution, so I think I must have made a mistake somewhere... Can anybody help ?

    (Question I.3.1 is pretty similar, but I have an integral in spherical polars in k space, which changes things slightly. I thought I'd tackle I.3.2 first since it should be simpler).
     
  2. jcsd
  3. Apr 26, 2010 #2
    The real part of your integral is the modified bessel function K_0(x) ... have a look at Mathworld [1], eq. (6), or some classical text about Bessel functions.

    The imaginary part vanishes, because it is antisymmetric in k.

    To my knowledge that's the best you can do --- the Bessel functions cannot be expressed simply in terms of elementary functions. It's frustrating but they keep popping up in field theory because of this kind of integral... I guess we just have to work with them!

    Hope that helps,

    Dave

    [1] http://mathworld.wolfram.com/ModifiedBesselFunctionoftheSecondKind.html
     
  4. Apr 26, 2010 #3
    The previous answer probably answered your question already, but are you even allowed to use contour integration techniques here? I mean, the integrand should satisfy

    [tex]
    \lim_{|Re \ z|\rightarrow \infty} zf(z) = 0,
    [/tex]

    if you want to close the contour in the upper or lower half-plane. But perhaps some convergence factor ei|0+| is implicitly assumed...
     
  5. Apr 26, 2010 #4

    Ah right thanks, I see.
    (There was a typo in the first integral, should have been exp(-ikx), not exp(ikx), but it doesn't change the conclusion).
    I'd never have recognized that as a Bessel function.

    thanks very much !!
     
  6. Apr 26, 2010 #5
    I missed a minus sign out of the numerator of the integrand in the original integral, should have been exp(-ikx). So for positive x, my contour in the k plane would be :

    Start at k = [itex]-\infty[/itex]
    Quarter circle round to k = [itex]-i\infty[/itex]
    In along the cut to k = -im
    Out along the cut to k = [itex]-i\infty[/itex]
    Quarter clrcle round to k = [itex]+\infty[/itex]

    Since the contour misses the singularities and the branch cut I was thinking that's valid. Is that right ?
     
  7. Apr 26, 2010 #6
    Forget what I said, the exponential factor takes care of the convergence in the appropriate half-plane :)
     
  8. May 5, 2010 #7
    Agh I've got stuck again! This time, looking at Zee question 1.4.1 - working out the analog of the inverse square law in (2+1) dimensions. Zee works it out in (3+1) dimensions, by evaluating the integral (omitting constant multipliers), note - he's already dealt with the time part here -

    [tex]\int^{\infty}_{-\infty} d^3k \frac{e^{i\mathbf{k}\cdot\mathbf{x}}}{\mathbf{k}^2+m^2} [/tex]

    which he does in an appendix. I understand that OK, but when I try to apply the same technique in (2+1) dimensions, my working goes like this:

    Introduce polars (k, [itex]\theta[/itex]) in k space in 2 dimensions. [itex]\mathbf{k}\cdot\mathbf{x}[/itex] becomes [itex]krcos\theta[/itex], where r is the length of [itex]\mathbf{x}[/itex]. The integral then becomes

    [tex]\int^{\infty}_{0} \int^{2\pi}_{0} \frac{e^{ikrcos\theta}}{k^2+m^2} d{\theta} k dk [/tex]

    Looking at the behaviour of the exponential over various parts of the range 0->2[itex]\pi[/itex], I ended up with

    [tex]\int^{\infty}_{0} \int^{\pi}_{0} \frac{cos(krcos\theta)}{k^2+m^2} d{\theta} kd k [/tex]

    Doing the [itex]\theta[/itex] integral, I get

    [tex]2\int^{\infty}_{0} \frac{Jo(kr)}{k^2+m^2} k dk [/tex]

    where Jo is a Bessel function of the first kind. By symmetry I can write:

    [tex]\int^{\infty}_{-\infty} \frac{Jo(kr)}{k^2+m^2} k dk [/tex]

    This is where I'm not sure how to proceed further. If Jo() went to zero as |k|->[itex]\infty[/itex] in the complex k plane, I could do the usual trick of contour integration using the appropriate pole of the denominator. However, I think Jo vanishes for large real arguments, but not imaginary ones, so that's out.

    I'm expecting the answer to be a potential which goes as log(r) (at least for m=0), since that gives me a force which goes as 1/r, which I'm guessing is what we would get with 2 spatial dimensions.

    Can somebody tell me where my working has gone wrong ?
     
  9. May 6, 2010 #8
    In case anyone is interested (LOL!) I may have made some progress. I had ended up with

    [tex]2\int^{\infty}_{0} \frac{Jo(kr)}{k^2+m^2} k dk [/tex]

    I found http://en.wikipedia.org/wiki/Hankel_transform" [Broken] on wikipedia, which seems to suggest that the integral is the Hankel Transform of

    [tex] \frac{1}{k^2+m^2} [/tex]

    which, according to the table is

    [tex] K_{0}(mr) [/tex]

    where [itex] K_{0} [/itex] is a modified Bessel function of the second kind. According to http://www.math.auckland.ac.nz/~hcoh001/as/page_375.htm" [Broken],

    [tex] K_{0}(mr) \approx -ln(mr) [/tex]

    so it's the right kind of behaviour. Only problem is it doesn't work for m=0 though, where I would expect the ln(r) potential, which I would differentiate to get a (1/r) force law. So I think I'm not quite there yet....
     
    Last edited by a moderator: May 4, 2017
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