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I'm currently on the chapter of regularization on Zee's QFT book. For the $\phi^4$ theory, an amplitude for a single loop correction to order $\lambda^2$ is given by a diagram enter image description here

Following the Feynman rules, we obtain the amplitude $$ \mathcal{M}=\frac{i^2}{2}(-i\lambda)^2\int\frac{d^4k}{(2\pi)^4}\frac{1}{k^2-m^2+i\epsilon}\frac{1}{(K-k)^2-m^2+i\epsilon} $$ where $K=k_1+k_2$. Assuming that $m<<k$ and having the cut off $\Lambda$, he said that this is equal to $$ \mathcal{M}=iC\lambda^2log\bigg(\frac{\Lambda^2}{K^2}\bigg) $$ Where $C$ is just some constant after evaluation. I was trying to evaluate this integral to obtain the same answer as he got, but I am getting nowhere close. I don't even understand what the bounds are in the 4 integrals if we replace them by $\Lambda$. Are all the bounds $\Lambda$?

How did he get that answer?

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For the regulation of the amplitude $M = \frac{i^2}{2}(-i\lambda)^2 I$ with

$$I:=\int \frac{d^4k}{(2\pi)^4} \frac{1}{k^2-m^2}\frac{1}{(k-K)^2-m^2}$$

the following the strategy is applied (according to R.Feynman). With $A=(k-K)^2-m^2$ and $B=k^2-m^2$

$$I = \int \frac{d^4k}{(2\pi)^4} \frac{1}{A}\frac{1}{B}=\int_0^1 dx\int \frac{d^4k}{(2\pi)^4}\frac{1}{[xA + (1-x) B]^2}$$

Expanding the denominator we get:

$$I=\int_0^1 dx\int \frac{d^4k}{(2\pi)^4}\frac{1}{[k^2 - 2xK\cdot k + xK^2 -m^2]^2}$$

Next step is a substitution $l =k-xK$: ($l$ is a 4-vector with the components $(l^0,l^1,l^2,l^3)$):

$$I=\int_0^1 dx\int \frac{d^4l}{(2\pi)^4}\frac{1}{[l^2+x(1-x)K^2 -m^2]^2}$$

followed by a Wick rotation: $l^0_E = -il^0$ and $\vec{l_E} = \vec{l}$

$$I = i\int_0^1 dx\int \frac{d^4l_E}{(2\pi)^4}\frac{1}{[l_E^2-x(1-x)K^2 +m^2]^2}= i\int_0^1 dx\int \frac{d^4l_E}{(2\pi)^4}\frac{1}{[l_E^2+\Delta]^2}$$ with $\Delta = m^2-x(1-x)K^2$ Apparently Zee applied Pauli-Villars regularisation which consists of replacing

$$ \frac{1}{(l_E^2 +\Delta)^2} \longrightarrow \frac{1}{(l_E^2 +\Delta)^2}-\frac{1}{(l_E^2 +\Lambda^2)^2}$$ Applying it on our integral $I$ it becomes $I_\Lambda$:

$$I_\Lambda=i\int_0^1 dx\int \frac{d^4l_E}{(2\pi)^4}(\frac{1}{[l_E^2+\Delta]^2}-\frac{1}{[l_E^2+\Lambda^2]^2}) = i\int_0^1 dx\int d\Omega_4 \int_0^{\infty} \frac{dl_E }{(2\pi)^4} \left( \frac{l_E^3}{[l_E^2+\Delta]^2} - \frac{l_E^3}{[l_E^2+\Lambda^2]^2}\right)$$

A final substitution $z= l_E^2$ with $dz = 2l_E dl_E $ yields ($\int d\Omega_4 = 2\pi^2$):

$$I_\Lambda=i\int_0^1 dx\int_0^{\infty}\frac{dz}{(4\pi)^2} \left( \frac{z}{[z+\Delta]^2} - \frac{z}{[z+\Lambda^2]^2}\right)=\frac{i}{(4\pi)^2}\int_0^1 dx\,\, \log(\frac{\Lambda^2}{\Delta}) $$

Therefore we get for the regularized amplitude $M_\Lambda$:

$$ M_\Lambda = \frac{i^2}{2}(-i\lambda)^2 I_\Lambda =\frac{i\lambda^2}{32\pi^2} \int_0^1 dx\,\, \log\left(\frac{\Lambda^2}{m^2-x(1-x)K^2}\right)$$

which corresponds to Zee's formula (14) on page 152. In the computation the terms $+i\epsilon$ were omitted for simplicity.

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  • $\begingroup$ If $B=k^2-m^2$ then why is it only $k^2$ in this solution (and in Zee's solution)? Also, after doing this, how do you finally get $\mathcal{M}=iC\lambda^2log\bigg(\frac{\Lambda^2}{K^2}\bigg)$? I don't think you can get this by doing the integral over $x$. $\endgroup$ – Gradient137 Oct 26 '18 at 2:20
  • $\begingroup$ Note that $$\int_0^1 dx\,\, \log\left(\frac{\Lambda^2}{m^2-x(1-x)K^2}\right)=\log\frac{\Lambda^2}{m^2}+\int_0^1 dx\,\, \log\left(\frac{m^2}{m^2-x(1-x)K^2}\right)$$ $\endgroup$ – AccidentalFourierTransform Oct 26 '18 at 2:26
  • $\begingroup$ Zee just used a simplified expression in the text as he did not want to write a long formula which would require an empty space around it. Above all because in the same section he shows the correcter formula. For the understanding of the text this difference is not important, in particular in the eyes of Zee. BTW this calculation can be found again upon computation of the self-energy in QED, so it rather standard. Finally I don't understand the comment: If $B=k^2-m^2$, then why is $k^2$ in this solution ? It's not $k^2$ that is in the solution but $K^2$. $k^2$ is integrated over and disappears $\endgroup$ – Frederic Thomas Oct 26 '18 at 8:25
  • $\begingroup$ What I meant by the that is you said plug $B=k^2-m^2$ into the Feynman reparametrization. You did plug in $k^2-m^2$ but $k^2$ is the only one multiplied to $(1-x)$ therefore what you did is actually just plugged in $B=k^2$ and for some reason just added $-m^2$ without multiplying it in to the $(1-x)$ term. $\endgroup$ – Gradient137 Oct 26 '18 at 15:49
  • $\begingroup$ $xA+(1-x)B= x[(k-K)^2-m^2] + (1-x)(k^2-m^2) = x k^2 -2xk\cdot K +xK^2 -xm^2 + k^2-m^2 -xk^2+xm^2 = -2xk\cdot K +xK^2 +k^2-m^2 = k^2 -2xk\cdot K +xK^2 -m^2$. Therefore I do not leave out any term neither I add any further term. The term $xm^2$ just cancels out with another $-xm^2$. $\endgroup$ – Frederic Thomas Oct 26 '18 at 17:03
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If you do not need to know the constant $C$ then all you have to do is observe that the integral is Log divergent at large momentum, so it must contain a $\ln \Lambda^2$, but the argument of the log has to be dimensionless and the only dimensionful quantity that the integral dependens on is $K^2$ so it contains a $\ln (K^2/\Lambda^2)$. To get the factor of $i$ I assume that a Wick rotation is being made so that the $dk_0$ is being rotated to $i dk_0$.

If you need the actual number $C$ use the Feynman parameter method. If $m=0$ it is much easier to work in Euclidan signature and in configuation space.

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  • $\begingroup$ Oh I see, he never formally introduced the Feynman parametrization, except in an appendix. I'll try that, thanks. $\endgroup$ – Gradient137 Oct 25 '18 at 19:49
  • $\begingroup$ I was reading through Zee's usage of Feynman parametrization and saw what he did is that $\frac{1}{xy}=\int^1_0 d\alpha \frac{1}{(\alpha x+(1-\alpha)y)^2}$. In this case it should be $\int^1_0 d\alpha \frac{1}{(\alpha (K-k)^2+(1-\alpha)(k^2-m^2+i\epsilon))^2}$, but he wrote and continued writing $\int^1_0 d\alpha \frac{1}{(\alpha (K-k)^2+(1-\alpha)k^2-m^2+i\epsilon)^2}$, i.e. only the $k^2$ is multiplied with $(1-\alpha)$. Why did he do that? $\endgroup$ – Gradient137 Oct 25 '18 at 20:33

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