QED formulary

2020-04-29

Gianluca Pacchiella

Tags:

$\def\Tr{\hbox{Tr}}$ $\def\slashme#1{\rlap{\backslash}{#1}}$ $\def\pslash{\rlap{\backslash}{p}}$ $\def\partialslash{\rlap{\backslash}{\partial}}$

This post is personal: I found some notes taken more than 10 years ago and I'm worried to lost them, so I write it down here in order to preserve the memory; maybe one day I'll write a post about the physics behind this stuff.

$$ \int d^Nq\,\left(q^2 + \mu^2 -i\epsilon\right)^{-\alpha} = i{\pi^{N/2}\over 2}{\Gamma(N/2)\Gamma(\alpha - N/2)\over\Gamma(\alpha)}\left(\mu^2\right)^{N/2 - \alpha} $$

$$ {1\over\Pi^N_{i=1}a_i} = \Gamma(N)\int^1_0 dx_1\, \int_0^{x_1} dx_2 \dots\int^{x_{N - 2} }_0 dx_{N - 1} \left[a_N x_{N - 1} + \dots + \Bigl(x_{N - 2} - x_{N - 1}\Bigr)a_{N - 1} + \dots + \left(1 - x_1\right)a_1\right]^{-N} $$

$$ \Gamma(\epsilon) = {1\over\epsilon} - \gamma + O(\epsilon) $$

$$ a^\epsilon = 1 + \epsilon\ln a + o(\epsilon) $$

$$ \int d^4k\,\theta(k_0)\delta(k^2 + m^2) \sim \int{d^3k\over2 k_0} $$

Tensorial integrals

For some informations see this slides.

$$ \eqalign{ J_0 &= \int d^Nq\,{1\over\left(q^2 + 2kq+\mu^2\right)^\alpha} = i\pi^{N/2}{\Gamma(\alpha - N/2)\over\Gamma(\alpha)}\left(\mu^2 - k^2\right)^{N/2 - \alpha} \cr J_\mu(k) &= \int d^Nq\,{q_\mu\over\left(q^2 + 2kq+\mu^2\right)^\alpha} = -J_0 k_\mu\cr J_{\mu\nu}(k) &= \int d^Nq{q_\mu q_\nu\over\left(q^2 + 2kq+\mu^2\right)^\alpha} = -J_0\left(k_\mu k_\nu + {1\over2}{\mu^2 - k^2\over\alpha - 1 - N/2}\delta_{\mu\nu}\right)\cr } $$

Gamma matrices

Suppose $D$ is the dimensionality of the space-time

$$ \eqalign{ \Tr(\mathbf{I}_{n}) &= 4\cr \left\{\gamma^\mu,\gamma^\nu\right\} &= 2g^{\mu\nu}\mathbf{I}_{n} \cr \gamma^\mu\gamma_\mu &= g^\mu_\mu = D \cr Tr(\gamma^\mu\gamma^\nu) &= 4g^{\mu\nu} \cr Tr(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma) &= 4\left(g^{\mu\nu}g^{\rho\sigma} - g^{\mu\rho}g^{\nu\sigma} + g^{\mu\sigma}g^{\nu\rho}\right) \cr Tr(\gamma^\mu\gamma^\nu\gamma^\sigma\gamma^\rho\gamma^\xi) &= \epsilon^{\mu\nu\sigma}\quad\hbox{(TO FIX)}\cr } $$

Spinors

$$ \eqalign{ \pslash &= \gamma^\mu p_\mu \cr \left(i\pslash + m\right)u(\vec{p}) &= 0 \cr \left(-i\pslash + m\right)v(\vec{p}) &= 0 \cr \sum_{\hbox{spin}}u(p)\bar{u}(p) = {1\over2 p_0}\left(-i\pslash + m\right) \cr \sum_{\hbox{spin}}v(p)\bar{v}(p) = {1\over2 p_0}\left(i\pslash + m\right) \cr } $$

Lagrangian

$$ L = -{1\over4} F_{\mu\nu}F_{\mu\nu} - {1\over2}\left(\partial_\mu A_\mu\right)^2 - \bar\psi\left(\partialslash + m \right)\psi + ie A_\mu\bar\psi\gamma^\mu\psi $$

1loop photon

$$ \eqalign{ \Pi_{\mu\nu} &= e^2\int d^nq{1\over \left(q^2 + m^2\right)\left(\left(q+p\right)^2 + m^2\right)}\Tr\left\{\gamma^\mu\left(-i\slashme{q} + m\right)\gamma^\nu\left(-i\left(\slashme{p} + \slashme{q}\right) + m\right)\right\} \cr &= -i8\pi^2e^2 \int^1_0dx\,J_0\left(p^2\delta_{\mu\nu} - p_\mu p_\nu\right) x(1 - x)\cr &\sim -i8\pi^2e^2 \int_0^1 dx\,\left(\Delta - \ln \mu^2\right) x\left(1 - x\right)\left(p^2 \delta_\mu\nu - p_\mu p_\nu\right) \cr } $$

QED renormalization

From an analysis using the propagators of photons and fermions and the vertex we can tell that the global degree of divergence of a diagram is given by

$$ D(G) = 4 - {3\over2}E_e - E_\gamma\quad\left\{\eqalign{ &D(G) < 0\quad\hbox{converges} \cr &D(G) \ge0\quad\hbox{diverges} \cr }\right. $$

Weinberg theorem

Given a $G$ such that $D(G) < 0$ and for all its subdiagrams then $G$ converges.