## Monday, June 20, 2016

### Loop correction to the propagator in $\phi^3$ theory

The loop correction to the propagator in the $\phi^3$ theory in six dimensions is given by the Feynman diagrams
Thus \begin{equation*} i \Pi (p) = \frac{g^2}{2} \int_0^1\!\!\! dx \int\!\! \frac{d^6k}{(2 \pi)^6} \frac{1}{(k^2 - \Delta(x,p) )^2} + i (p^2 \delta_{\phi} - \delta_m) \end{equation*} Here, $g$ is the coupling constant, $\delta_{\phi}$, $\delta_m$ are the counterterms and $\Delta(x,p) = m^2 - x(1-x) p^2$ with $m$ the mass of the particle.

In chapter 14 in his book, Srednicki calculates as follows:
1. calculate the integral in general dimension $d$
2. expand the result in small $\epsilon$ with $d = 6 - \epsilon$
3. fix the counterterms $\delta_{\phi}$ and $\delta_m$ using on-shell renormalization conditions
4. take the limit $\epsilon \to 0$
I find the calculation less messy if I change the order of the calculation:
1. calculate the integral in general dimension $d$
2. fix the counterterms $\delta_{\phi}$ and $\delta_m$ using on-shell renormalization conditions
3. take the limit $d\to 6$
Step 1 gives \begin{equation*} \Pi(p) = A \int_0^1\!\!\! dx\ \Delta(x,p)^{d/2 - 2} + p^2 \delta_{\phi} - \delta_m \end{equation*} with \begin{equation*} A = \frac{g^2}{2} \frac{1}{(4 \pi)^{d/2}} \Gamma\left( 2 - \frac{d}{2}\right) \end{equation*} Step 2 gives \begin{align*} \delta_{\phi} &= A\ ( \frac{d}{2} -2) \int_0^1 \!\!\! dx \ \Delta(x,m)^{d/2 - 3} x ( 1 - x) \\ \delta_m &= A \int_0^1 \!\!\! dx \ \Delta(x,m)^{d/2 - 2} + m^2 \delta_{\phi} \end{align*} In step 3 one can then easily calculate the limit \begin{align*} \Pi(p) &= \lim_{d \to 6} \left( A \int_0^1 \!\!\! dx \ \left( \Delta(x,p)^{d/2 - 2} - \Delta(x,m)^{d/2 - 2} \right) +(p^2 - m^2) A\ ( \frac{d}{2} -2) \int_0^1 \!\!\! dx \ \Delta(x,m)^{d/2 - 3} x ( 1 - x) \right)\\ &= \frac{\alpha}{12} (p^2 - m^2) + \frac{\alpha}{2} \int_0^1 \!\!\! dx \ \Delta(x,p) \log \frac{\Delta(x,p)}{\Delta(x,m)} \end{align*} with $\alpha = \dfrac{g^2}{(4 \pi)^3}$. This answer agrees with equation 166 in lecture notes by Einan Gardi. The integral can also be calculated in closed form. The result then agrees with equation 14.44 in Srednicki (but I have not checked the choices of branches).

Remark
In Srednicki's book the integral $\int \dfrac{d^6k}{(2 \pi)^6} \frac{1}{(k^2 - \Delta(x,p) )^2}$ is calculated using spherical coordinates, integrating over the radius, and looking up (or calculating) the volume of a $d-1$ dimensional sphere. I liked the method Einan Gardi uses for this integral: Gardi's method is very quick and the volume of $d-1$ dimensional sphere is not needed.