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I am working with the propagator associated to the Klein-Gordon equation, as derived in "Quantum Physics a functional integral point of view", James Glimm, Arthur Jaffe or as derived here: http://www.wiese.itp.unibe.ch/lectures/fieldtheory.pdf § 5.4

It turns out that the propagator can be evaluated, and a close-form expression for it can be given, namely:

$$ C \left( m; \mathbf{x} - \mathbf{y} \right) = \left(\frac{1}{2 \pi}\right)^{-\frac{d}{2}} \left(\frac{m}{\left| \mathbf{x} - \mathbf{y} \right|}\right)^{\frac{d-2}{2}} K_{\frac{d-2}{2}} \left( m \left| \mathbf{x} - \mathbf{y} \right| \right) $$

where $K$ is the modified Bessel function of the second kind. I'd want to take the massless limit in two dimensions; when setting $d=2$ and $m=0$ one of the terms in the r.h.s. of the equation evaluates to $0^0$ while the modified Bessel function goes to infinity. How do I calculate the massless limit for the Klein-Gordon propagator in 2D?

Thank you!

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A nice way to see how the correlation function behaves is described here where it is shown that the propagator goes as $$C(r)=\frac{1}{2\pi}\log(r)$$ which can also be seen as given by Qmechanics hint. Now the interesting thing is not that it diverges at $r=0$ (this happens even in 4D where $C(r)=1/4 \pi^2 r^2$) but that it also diverges as $r\to \infty$. This is an infrared divergence I had not come across before. The Wikipedia article linked above states that this makes a two dimensional massless scalar field slightly tricky to define mathematically and also that you cannot have spontaneous breaking of a continuous symmetry in two dimensions. Very interesting!

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Hint: Use e.g. that the modified Bessel function $K_0$ behaves as minus the logarithm for small arguments near zero.

Reference:

  1. Abramowitz & Stegun, Handbook of Mathematical Functions, p. 375, eq. (9.6.8). For an online version see e.g. here.
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I would suggest you first set $d=2$ giving $$C \left( m; \mathbf{x} - \mathbf{y} \right) = \frac{1}{2 \pi} K_0 \left( m \left| \mathbf{x} - \mathbf{y} \right| \right)$$ and then take the massless limit.

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  • $\begingroup$ Maybe I should've mentioned that, I've tried this approach... When taking the $m \longrightarrow 0$ limit the Bessel function goes to $\infty$ so that the propagator is infinite everywhere... $\endgroup$ – zakk Aug 30 '12 at 10:10

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