Renormalization with hard cutoff of a loop diagram with single vertex

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theittsco
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TL;DR Summary
Confusion on how to renormalize a loop diagram with only a single vertex. Feynman parameters don't work.
Trying to solve for the loop contribution when renormalizing a one loop ##\frac{\lambda}{4!}\phi^4## diagram with two external lines in ##d=4## dimensions. After writing down the Feynman rule I see that:

$$\frac{(-i\lambda)}{2}\int d^4q \frac{i}{q^2-m_{\phi}^2+i\epsilon} $$

But I see no way to convert this to Feynman parameters like in ##1/AB##. When I Wick rotate I get:

$$\frac{(\lambda)}{2}\int dq_E \frac{q_E^3}{q_E^2-m_{\phi}^2} $$

Which when integrated from ##0,\Lambda## (a hard cut off) via the great Mathematica yields:

$$\lambda/2 (1/2)(\Lambda^2-m^2\log(-m^2)+m^2\log(\Lambda^2-m^2))$$

My renormalization conditions are ##\Pi(m^2)=\Pi'(m^2)=0##, so when solving for counter terms:

$$i\Pi(p^2) = \lambda/2 (1/2)(\Lambda^2-m^2\log(-m^2)+m^2\log(\Lambda^2-m^2)) + i(\delta_Zp^2-\delta_m)$$

my counter terms still end up with ##\Lambda## dependence. So I messed up somewhere but I can't figure out where. Help please!
 
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Answers and Replies

  • #2
HomogenousCow
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Your counter terms should absolutely depend on the cut-off, that's how they cancel the divergences. There's no momentum dependence so ##\delta_Z = 0## and ##\delta_m## cancels out everything. Also I'm pretty sure you made a mistake when you did your wick-rotation, the sign in the denominator should be positive.

The whole idea behind renormalized perturbation theory is that you deform your theory with some parameter ##\Lambda## that makes the observables finite, then you ascribe ##\Lambda##-dependence to your constants which are selected to cancel the infinites and satisfy the renormalization conditions.
 
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