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I am Studying Electron self-energy using Ryder's textbook, In page 334 we can see

Defining $k'=k-pz$ and avoiding the term linear in $k'$(because it integrates to zero) gives \begin{equation} \Sigma(p)=-ie^2\mu^{4-d}\int_0^1dz\gamma_\mu({\not} p-{\not}p z+m)\gamma^\mu\int\frac{d^dk'}{(2\pi)^d}\frac{1}{[k'^2-m^2z+p^2z(1-z)]^2}.\label{r2.7}\end{equation} [...] This integral is performed with the help of equation (9A.5), giving \begin{equation} \Sigma(p)=\mu^{4-d}e^2\frac{\Gamma(2-\frac{d}{2})}{(4\pi)^{d/2}}\int_0^1dz\gamma_\mu[{\not}p(1-z)+m]\gamma^\nu[-m^2z+p^2z(1-z)]^{d/2-2}. \end{equation}

The equation 9A.5 is \begin{equation} \int\frac{d^dp}{(p^2+2pq-m^2)^{\alpha}}=(-1)^{d/2}\imath\pi^{d/2}\frac{\Gamma\left(\alpha-\frac{d}{2}\right)}{\Gamma(\alpha)}\frac{1}{[-q^2-m^2]^{\alpha-d/2}} .\tag{9A.5} \end{equation} I don't understand how he applied this integral (9A.5) to obtain the result \begin{equation} \Sigma(p)=\mu^{4-d}e^2\frac{\Gamma(2-\frac{d}{2})}{(4\pi)^{d/2}}\int_0^1dz\gamma_\mu[{\not}p(1-z)+m]\gamma^\nu[-m^2z+p^2z(1-z)]^{d/2-2}. \end{equation} please help me to get an idea.

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It's just a matter of applying the result (9A.5) to the integral in $d^d k^\prime$. In fact call $M^2 = m^2z-p^2z(1-z)$ and put $q=0$ in the integral (9A.5) $$ \int\frac{d^dk'}{(2\pi)^d}\frac{1}{[k'^2-m^2z+p^2z(1-z)]^2} = \int\frac{d^dp}{(2\pi)^d}\frac{1}{[p^2-M^2]^2}=\frac{1}{(2\pi)^d}(-1)^{d/2}i\pi^{d/2}\frac{\Gamma\left(2-\dfrac{d}{2}\right)}{\Gamma(2)}\frac{1}{[-M^2]^{2-d/2}}$$

where we just changed the integration variable from $k^\prime$ to $p$ to make it clearer from the result 9A.5. Using the fact that $\Gamma(2) = 1$, using the above definition of $M^2$ and simplifying a bit you get $$\frac{(-1)^{d/2}}{2^d}i\pi^{-d/2}\Gamma\left(2-\frac{d}{2}\right)[-m^2z+p^2z(1-z)]^{d/2-2} = \frac{i(-1)^{d/2}}{(4\pi)^{d/2}}\Gamma\left(2-\dfrac{d}{2}\right)[-m^2z+p^2z(1-z)]^{d/2-2}$$ where we used the fact that $2^d = 4^{d/2}$

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Compare the second integrand in the first equation with ty he in the grand in 9A5. You see that $\alpha \rightarrow 2$, $q \rightarrow 0$, $ -m^2 \rightarrow etc.$ will transform one integrand into the other. Making the same substitutions in the rhs of 9A5 should give you the desired result.

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