Divergences and Dimensional Regularization of QED

References : 

1. G. t’Hooft and M. Veltman, Nucl. Phys. B44(1972).

2. C.G. Bellini and J.J. Giambiagi, Nuovo Cimento B12(1972).

3. L. H. Ryder, (1996) QFT, 2nd edition.

 

 

Fig 1(L) : self-energy diagram for electron / Fig 2(R) : self-energy diagram for photon

The only particles in QED are photons and electrons. Divergences occur in several types of Feynman diagram, for example the electron and photon self-energy diagrams of Figs 1 and 2. We shall treat these diagrams in the same systematic way, and shall show that there is only a finite number of primitively divergent diagrams, and hence that QED is, in principle, renormalizable. Pauli-Villars regularization doesn't preserve gauge invariance in non-Abelian gauge theories, so it is simplest to use dimensional regularization. The key concept is to generalize our spacetime dimension to $(d-1,1)$ and reduce to $d \rightarrow 4$. In this case, we shall later show it has a pole singularity. The general formula for the superficial degree of divergence $D$ of a Feynman graph in $d$-dimensional spacetime is analogous to

$$D=dL-2P_i-E_i$$

where $L$ is the number of loops, $P_i$ the number of internal photon lines, $E_i$ the number of internal electron lines, and $d$ the dimension of spacetime. And in addition, let $n$ denotes the number of verticies, $P_e$ the number of external photon lines, and $E_e$ the number of external electron lines.

Also, $L$ = the number of independent momenta for integration = number of internal lines - $n$ (because of momentum conservation at each vertex) + 1 (because overall momentum conservation holds) :

$$L=E_i + P_i -n+1$$

Now each vertex gives 2 electron legs. If they are external, they are counted once, and if internal, twice, so

$$2n = E_e + 2E_i$$

The analogous relation for photons is clearly

$$n = P_e + 2P_i$$

Then we have

$$D = (d-1)E_i + (d-2)P_i - d(n-1)$$

which, on substituting for $E_i$ and $P_i$ gives

$$D = d + n \left( \frac{d}{2}-2 \right) - \left( \frac{d-1}{2} \right) E_e - \left( \frac{d-2}{2} \right)P_e$$

When $d=4$, this yields

$$D=4-\frac{3}{2} E_e - P_e$$

It shows $D$ is independent of $n$, the sine qua non for renormalizability.

 

QED Lagrangian in $d$-dimension is the form of (neglecting all unnecessaries)

$$\mathcal{L} = i\bar{\psi} \gamma^\mu \partial_\mu \psi - m \bar{\psi}\psi - e A^\mu \bar{\psi} \gamma_\mu \psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2} (\partial_\mu A^\mu)^2$$

Since the mass dimension of $\mathcal{L}$ is $d$, $[\mathcal{L}]=d$, the fields dimension should be

$$[\psi]=\frac{d-1}{2}, \;\;\;\; [A_\mu]=\frac{d-2}{2}$$

then except the 3rd term, all terms in $\mathcal{L}$ have correct dimension. To get the 3rd one right, we must multiply $e$ by $\mu^{2-d/2}$, where $\mu$ is an arbitrary mass. Hence, 

$$\mathcal{L} = i\bar{\psi} \gamma^\mu \partial_\mu \psi - m \bar{\psi}\psi - e \mu^{2-d/2} A^\mu \bar{\psi} \gamma_\mu \psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2} (\partial_\mu A^\mu)^2$$

The extension to $d$ dimensions is to be made only for internal loops, so external wave functions like $u(p,s)$, for example, may be left in their 4-dimensional form. We now have to consider the $d$-dimensional Clifford algebra. Denoting $g_{\mu\nu}=\mathrm{diag}(+,-,-,\cdots)$, we have

$$\{\gamma_\mu, \gamma_\nu\} = 2g_{\mu\nu}, \;\;\;\; \gamma_\mu \gamma_\nu \gamma^\mu=(2-d)\gamma_\nu$$

The other properties :

$$\mathrm{Tr} \text{(odd gamma's)} = 0, \;\;\;\; \mathrm{Tr} I = f(d), \;\;\;\; \mathrm{Tr} \gamma_\mu \gamma_\nu = f(d) g_{\mu\nu}$$

$$\mathrm{Tr} \gamma_\mu \gamma_\kappa \gamma_\nu \gamma_\lambda = f(d) (g_{\mu\kappa} g_{\nu\lambda} - g_{\mu\nu} g_{\kappa\lambda}+g_{\mu\lambda}g_{\kappa\nu})$$

where $f(d)$ is an arbitrary well-behaved function with $f(4)=4$; its explicit form is irrelavent. What we cannot do in $d$-dimensions is to define the analogue of $\gamma_5$ since the Levi-Civita is specific to $d=4$. This difficulty has bearing on the problem of chiral anomalies. Anyway, let's consider the divergent diagrams, starting from the Fig 1 in $d$-dimensions. (Feynman slash notation applied)

$$\begin{align} \Sigma(p) & =-ie^2 \mu^{4-d} \int \frac{d^d k}{(2\pi)^d} \gamma_\mu \frac{1}{\not\! p - \not\! k - m} \gamma_\nu \frac{g^{\mu\nu}}{k^2} \\ & = -ie^2 \mu^{4-d} \int \frac{d^d k}{(2\pi)^d} \frac{\gamma_\mu (\not\! p - \not\! k +m) \gamma^\mu}{[(p-k)^2-m^2]k^2} \\ & = -ie^2 \mu^{4-d} \int_0^1 dz \int \frac{d^d k}{(2\pi)^d} \frac{\gamma_\mu (\not\! p -\not\! k +m)\gamma^\mu}{[(p-k)^2 z - m^2 z + k^2 (1-z)]^2} \end{align}$$

The last line we used

$$\frac{1}{ab}=\int_0^1 \frac{dz}{[az+b(1-z)]^2}$$

Defining $k'=k-pz$ gives

$$\Sigma(p) = -ie^2 \mu^{4-d} \int_0^1 dz \int \frac{d^d k'}{(2\pi)^d} \frac{\gamma_\mu (\not\! p -\not\! p z -\not\! k'+m)\gamma^\mu}{[{k'}^2 z - m^2 z + p^2 z (1-z)]^2}$$

The term linear in $k'$ integrates to zero, so

$$\Sigma(p) = -ie^2 \mu^{4-d} \int_0^1 dz \gamma_\mu (\not\! p -\not\! p z +m) \gamma^\mu \int \frac{d^d k'}{(2\pi)^d} \frac{1}{[{k'}^2 - m^2 z + p^2 z(1-z)]^2}$$

Note that in 4D, the momentum integral is, as claimed above, logarithmically (not linearly) divergent. The integral is pretty hard to be performed, we first are interested in the type of

$$I_d(q)=\int \frac{d^d p}{(p^2+2pq-m^2)^\alpha}$$

where $p=(p_0, \vec{r})$. Introducing polar coordinates $(p_0, r, \phi, \theta_1, \theta_2, \cdots, \theta_{d-3})$ so that

$$d^d p = dp_0 r^{d-2} dr d\phi \sin \theta_1 d\theta_1 \sin^2 \theta_2 d\theta_2 \cdots \sin^{d-3} \theta_{d-3} d\theta_{d-3}$$

Then

$$I_d (q) = 2\pi \int_{-\infty} ^{\infty} dp_0 \int_0 ^{\infty} r^{d-2} dr \left( \int_0^\pi \right)^{d-3} \frac{\prod_1^{d-3} \sin^k \theta_k d\theta_k}{(p^2+2pq-m^2)^\alpha}$$

We now can use the formula 

$$\int_0^{\pi/2} (\sin \theta)^{2n-1} (\cos \theta)^{2m-1} d\theta = \frac{\Gamma(n)\Gamma(m)}{2 \Gamma(n+m)}$$

and putting $m=\frac{1}{2}$ gives

$$\int_0^\pi (\sin \theta)^k d\theta = \sqrt{\pi} \Gamma\left(\frac{k+1}{2}\right) / \Gamma\left(\frac{k+2}{2}\right)$$

and hence

$$I_d(q) = \frac{(2\pi)^{(d-1)/2}}{\Gamma(\frac{d-1}{2})} \int_{-\infty}^{\infty} dp_0 \int_0^{\infty} \frac{r^{d-2}dr}{(p_0^2-r^2+2pq-m^2)^\alpha}$$

This integral is Lorentz invariant, so we evaluate it in the frame $q_\mu=(\mu,\vec{0})$. Then $2pq=2\mu p_0$. Changing variables to $p'_\mu=p_\mu+q_\mu$, which implies that ${p'}_0^2-q^2=p_0^2 + 2\mu p_0$, we have

$$I_d(q)=\frac{(2\pi)^{(d-1)/2}}{\Gamma(\frac{d-1}{2})} \int_{-\infty}^{\infty} d{p'}_0 \int_0^{\infty} \frac{r^{d-2}dr}{[{p'}_0^2 - r^2 - (q^2+m^2)]^\alpha} \;\;\;\;\;\;\;\;\;\;\;\;\; (1)$$

The Euler beta function is (valid for $\mathrm{Re}(x)>0, \mathrm{Re}(y)>0$)

$$B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} = 2\int_0^\infty dt t^{2x-1} (1+t^2)^{-x-y}$$

so putting

$$x=\frac{1+\beta}{2}, \;\;\;\; y=\alpha - \frac{1+\beta}{2}, \;\;\;\; t=\frac{s}{M}$$

we have

$$\int_0^\infty ds \frac{s^\beta}{(s^2+M^2)^\alpha} = \frac{\Gamma\left( \frac{1+\beta}{2} \right) \Gamma \left( \alpha - \frac{1+\beta}{2} \right)}{2(M^2)^{\alpha-(1+\beta)/2} \Gamma(\alpha)}$$

Substituting this into $(1)$, with $\beta=d-2$, $M^2=-{p'}_0^2 + q^2+m^2$, gives

$$\begin{align} I_d(q) & = (-1)^{-\alpha} \pi^{(d-1)/2} \frac{\Gamma\left( \alpha - \frac{d-1}{2} \right)}{\Gamma(\alpha)} \int_{-\infty}^{\infty} \frac{d{p'}_0}{(q^2+m^2-{p'}_0^2)^{\alpha-(d-1)/2}} \\ & =(-1)^{2\alpha+(d-1)/2} \pi^{(d-1)/2} \frac{\Gamma \left( \alpha - \frac{d-1}{2} \right)}{\Gamma(\alpha)} \int_{-\infty}^{\infty} \frac{d{p'}_0}{[{p'}_0^2-(q^2+m^2)]^{\alpha-(d-1)/2}} \\ & = i \pi^{d/2} \frac{\Gamma\left( \alpha -d/2 \right)}{\Gamma(\alpha)} \frac{1}{[-(q^2+m^2)]^{\alpha-d/2}} \\ & = (-1)^\alpha i \pi^{d/2} \frac{\Gamma\left( \alpha -d/2 \right)}{\Gamma(\alpha)} \frac{1}{(q^2+m^2)^{\alpha-d/2}} \end{align}$$

Hence we get

$$\int \frac{d^d p}{(p^2+2pq-m^2)^\alpha} = (-1)^{d/2} i \pi^{d/2} \frac{\Gamma\left( \alpha -d/2 \right)}{\Gamma(\alpha)}\frac{1}{[-q^2-m^2]^{\alpha-d/2}} \;\;\;\;\;\;\;\;\;\;\;\;\;\; (2)$$

Also differentiating this with respect to $q^\mu$ and putting $\alpha \rightarrow \alpha-1$, we have

$$\int d^d p \frac{p_\mu}{(p^2+2pq-m^2)^\alpha} = (-1)^{1+d/2} i \pi^{d/2} \frac{\Gamma(\alpha-d/2)}{\Gamma(\alpha)} \frac{q_\mu}{(-q^2-m^2)^{\alpha-d/2}} \;\;\;\;\;\;\;\;\;\;\;\;\;\; (3)$$

Differentiating once again with respect to $q^{\nu}$ yields

$$\int d^d p \frac{p^2}{(p^2+2pq-m^2)^\alpha} = (-1)^{d/2} \frac{i \pi^{d/2}}{\Gamma(\alpha) (-q^2-m^2)^{\alpha-d/2}} \left[ q^2 \Gamma\left( \alpha - \frac{d}{2} \right) + \frac{d}{2} (-q^2-m^2) \Gamma\left( \alpha-1-\frac{d}{2} \right) \right] \;\;\;\;\;\; (4)$$

Now we can perform the $\Sigma(p)$ integral with the help of $(2)$, giving

$$\Sigma(p) = \mu^{4-d} e^2 \frac{\Gamma(2-d/2)}{(4\pi)^{d/2}} \int_0^1 dz \gamma_\mu [\not\! p (1-z) + m] \gamma^\mu [-m^2 z + p^2 z(1-z)]^{d/2-2}$$

As $d \rightarrow 4$, $\Gamma(2-d/2)$ develops a pole. Putting $d=4-\epsilon$, and use $d$-dimensional identities for gamma matrices and 

$$\Gamma(2-d/2)=\Gamma(\epsilon/2)=\frac{2}{\epsilon}-\gamma+\mathcal{O}(\epsilon)$$

where $\gamma$ is the Euler-Mascheroni constant :

$$\gamma = \lim_{n\to \infty} \left(1+\frac{1}{2} + \cdots + \frac{1}{n} - \ln n \right) = 0.5772157\cdots$$

Then we get

$$\begin{align} \Sigma(p) & = -\frac{e^2}{16\pi^2} \Gamma(\epsilon/2) \int_0^1 dz \{ 2\not\! p (1-z) -4m -\epsilon[\not \! p (1-z) + m] \} \left( \frac{-m^2 z + p^2 z(1-z)}{4\pi \mu^2} \right)^{-\epsilon/2} \\ & = \frac{e^2}{8\pi^2 \epsilon} (-\not \! p + 4m) + \frac{e^2}{16\pi^2} \left\{ \not\! p (1+\gamma) -2m(1+2\gamma) + 2\int_0^1 dz [\not\! p (1-z) -2m] \ln \left( \frac{-m^2 z + p^2 z(1-z)}{4\pi \mu^2} \right) \right\} \\ & = \frac{e^2}{8\pi^2 \epsilon} (-\not\! p + 4m) + \text{finite} \end{align}$$