Electron's self-energy in QED in arbitrary gauge

Question

Recently I've tried to evaluate electron's self-energy in QED in the second order of perturbation theory by using dimensional regularization. Corresponding 1PI-diagram leads to $$ \Sigma_{1loop} = -ie^{2} \int \frac{d^{4}k}{(2 \pi )^{4}}\frac{\gamma^{\mu}\left( ( p\!\!\!/ -k\!\!\!/ ) + m\right)\gamma^{\nu} \left( g_{\mu \nu} - (1 - \varepsilon ) \frac{k_{\mu} k_{\nu}}{k^{2}}\right)}{((p - k)^{2} - m_{e}^{2} + i\epsilon )(k^{2} - \mu^{2} + i \epsilon)}. $$ Here I've regularized infrared divergences by introducing fictive photon mass $\mu$.

For evaluating of this integral I used following relations for gamma-matrices, $$ \gamma^{\mu}\gamma_{\alpha}\gamma_{\mu} = -2\gamma_{\alpha}, \quad (\gamma_{\mu}k^{\mu})^{2} = k^{2}, $$ and Feynman tricks: $$ \frac{1}{AB} = \int \limits_{0}^{1} \frac{dx}{(A + (B - A)x)^{2}}, \quad k_{\mu} \to k_{\mu} + p_{\mu} (1 - x), \quad A = (p - k)^{2} - m_{e}^{2} + i\epsilon . $$ So the denominator of $(1)$ was reduced to $$ (k^{2} - m_{e}^{2} + p^{2}(1 -x)x + i\epsilon )^{2}. $$ But there was arisen the problem in nominator which forbids the usual evaluation of integral $(1)$ (in the best traditions of simplest dimensional regularization based calculations). After manipulations with gamma-matrices there was left one problematic term, $\gamma_{\mu}k^{\mu} \frac{(p \cdot k)}{k^{2}}(1 - \varepsilon )$. After shifting $k$ it becomes $$ \tag 2 (k_{\mu} + p_{\mu}(1 - x))\gamma^{\mu}\frac{(p \cdot k) + p^{2}(1 - x)}{k^{2} + p^{2} + 2(p \cdot k )(1 - x)}. $$ The denominator leads to impossibility of making the standard calculations.

The question: how to evaluate quantity $$ \tag 3 \int d^{4}k \int \limits_{0}^{1}dx (k_{\mu} + p_{\mu}(1 - x))\gamma^{\mu}\times $$ $$ \times \frac{(p \cdot k) + p^{2}(1 - x)}{k^{2} + p^{2} + 2(p \cdot k )(1 - x)} \frac{1}{(k^{2} - \mu^{2}x- (1-x)m_{e}^{2} + p^{2}x(1 - x) + i\epsilon )^{2}} $$ by using dimensional regularization?

Maybe it would be better before manipulating with shift $k \to k + p(1 - x)$ to use $k_{\alpha} k_{\beta} \to \frac{1}{d}k^{2}g_{\alpha \beta}$? I would be grateful for detailed demonstration of solving of problem.

Answer 1

You don't need to introduce photon's fictive mass: dimensional regularization can be used for IR divergences as well as for UV divergences.

First, use gamma-matrices identities, $$ \gamma_{\mu}\gamma^{\alpha}\gamma^{\mu} = (2 - d)\gamma^{\alpha} ,\quad [\gamma_{\mu}, \gamma_{\nu}]_{+} = 2g_{\mu \nu}, \quad g_{\mu}^{\mu} = d $$ and do simple that you've claimed: you'll get (here I've denoted $\epsilon = 1 - \eta$, where $\eta $ is gauge parameter) $$ -M^{4 - d}q_{e}^{2} \int \frac{d^{d}k}{(2 \pi )^{d}} \frac{(2 - d)(p\!\!\!/ - k\!\!\!/ ) + dm }{(k^{2} + i\varepsilon )((p - k)^{2} - m^{2} + i\varepsilon )} + $$ $$ + \epsilon M^{4 - d}q_{e}^{2} \int \frac{d^{d}k}{(2 \pi )^{d}}\left( \frac{k\!\!\!/ (p^{2} - m^{2}) - (p\!\!\!/ - m )k^{2}}{(k^{2} + i\varepsilon )^{2}((p - k)^{2} - m^{2} + i\varepsilon )} - \frac{k\!\!\!/}{(k^{2} + i\varepsilon )^{2}}\right) . $$ The last summand vanishes as antisymmetric function of $k$ integtated over symmetrical limits. As you can see, gauge terms don't contribute to mass, $\delta m_{\epsilon} = 0$, because mass correction is calculated on mass shell $p\!\!\!/ = m, p^{2} = m^{2}$.

As for the next step you need to introduce Feynman parameters: $$ \frac{1}{(k^{2} + i\varepsilon )^{n}((p-k)^{2} - m^{2} + i\varepsilon )} = $$ $$ =n\int \limits_{0}^{1}\frac{x^{n - 1}dx}{((k - p(1 - x))^{2} -(m^{2}(1 - x) - p^{2}x(1 - x)) + i\varepsilon)^{n + 1}}. $$ After that you need to making a shift $k \to k + p(1 - x)$. Finally, after neglecting linear in k terms, your pre-result should look like $$ -M^{4 - d}q_{e}^{2}\frac{\Gamma \left(2 - \frac{d}{2}\right)}{(4 \pi )^{\frac{d}{2}}} (2 - d)p\!\!\!/\int \limits_{0}^{1}\frac{xdx}{(m^{2}(1 - x) - p^{2}x(1 - x))^{2 - \frac{d}{2}}} $$ $$ -M^{4 - d}q_{e}^{2}dm\int \limits_{0}^{1}\frac{dx}{(m^{2}(1 - x) - p^{2}x(1 - x))^{2 - \frac{d}{2}}} + $$ $$ M^{4 - d}q_{e}^{2}\epsilon (p^{2} - m^{2})\frac{\Gamma \left( 3 - \frac{d}{2} \right)}{(4 \pi )^{\frac{d}{2} }} \int \limits_{0}^{1}\frac{x(1 - x)dx}{(m^{2}(1 - x) - p^{2}x(1 - x))^{3 - \frac{d}{2}}}- $$ $$ - M^{4 - d}q_{e}^{2}\epsilon(p\!\!\!/ - m)\frac{\Gamma \left(2 - \frac{d}{2}\right)}{(4 \pi )^{\frac{d}{2}}} \int \limits_{0}^{1}\frac{dx}{((m^{2}(1 - x) - p^{2}x(1 - x))^{2 - \frac{d}{2}}}. $$ In the third summand you can set $d = 4$, while in the other you may use identity (don't forget to make denominator dimensionless by using $M$) $$ \frac{\Gamma \left( 2 - \frac{d}{2} \right)}{(2 \pi )^{d}}\frac{1}{X^{2 - \frac{d}{2}}} \approx \frac{1}{(4 \pi )^{\frac{d}{2}}}\left( \frac{1}{4 - d} - C + ln(4 \pi ) - ln(X)\right), $$ where $C$ is Euler constant. The result of these manipulations are finite integrals parametrized by $4 - d = \gamma$ and $\mu$.

Answer 2

So in your calculation, you have set $\epsilon$ to be zero. Since the result should be independent of the gauge you choose, why not calculate under $\epsilon=0$, which is equivalent to take the propagator as $-\frac{g^{\mu\nu}}{k^{2}}$? If you still want to calculate with a brute force, I think the problem may arise from the Feynman parameterization, where there should be three factors in the denominator and you ignore $k^{2}$ in the propagator also contributes one factor. So the Feynman tricks should deal with $\frac{1}{ABC}$, then you are allowed to take the shift. In fact, electron-self energy calculation in many textbooks is done with $\epsilon=0$.