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Why does propagator converge?

  1. Sep 5, 2013 #1
    The propagator $$\frac{1}{k^2+m^2} $$ diverges in the ultraviolet when integrated over all dkn, with n>=2. However, when you throw in an exponential,

    $$\Delta(x)=\int d^nk \frac{e^{ikx}}{k^2+m^2} $$ is convergent for x≠0

    Intuitively adding the oscillating exponential decreases the ultraviolet growth by alternating it with + and - that cancel when added.

    But consider 1 dimension, and the expression $$ \int_1^\infty \frac{dx}{x^n} $$. This expression is convergent for n>1. Now add an exponential: $$ \int_1^\infty dx \frac{ e^{ix}}{x^n} $$. This improves the convergence: now the expression is convergent for n>0. But for n<0, it doesn't converge.

    So intuitively, $$\Delta(x)=\int d^nk \frac{e^{ikx}}{k^2+m^2} $$ should not be convergent for n>=2. But it obviously is, since the propagator is convergent in dimensions greater than 2, I believe.

    What exactly is happening with these oscillations at infinity? Also I'm a bit bewildered at why $$\int_{-\infty}^{\infty} \cos(x^2) dx $$ is convergent. I can show it mathematically, but intuitively why does that expression converge and not $$\int_{-\infty}^{\infty} \cos(x) dx $$?
     
  2. jcsd
  3. Sep 5, 2013 #2
    Well, consider the functions ##f(M) = \int_{-M}^{M} \cos(x^2) dx## and ##g(M) = \int_{-M}^{M} \cos(x) dx##. Plot the two functions as ##M \to \infty## and it will be clear that ##f## goes to a limit, while ##g## doesn't. Intuitively, the area under each positive and negative oscillation of ##\cos(x^2)## shrinks as ##x \to \infty##, so the integral ##f## oscillates less and less. But the oscillations of ##\cos(x)## don't have smaller areas as ##x \to \infty##, so the integral ##g## oscillates with a constant amplitude and never settles down to a limit.
     
  4. Sep 6, 2013 #3
    Hi Geoduck,

    You are right when you say the Fourier transform to the propagator $$\frac{1}{k^2+m^2} $$ only diverges for x=0 and that's quite logical because the contribution of a position to itself must be a delta function and not only an infinitesimal contribution.

    From what I have read, the divergence problem arises when introducing the self-interacting lagrangian that depends on λ[itex]ψ^{4}[/itex], this new term modifies the propagator causing a mass renormalization, this mass renormalization turns to be divergent for high energy contributions. The Renormalization Group tries to solve these problems. However, I must recognise this issues related with the interacting lagrangian and its fluctuations scape quite from my comprehension.

    Best regards,
     
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