6 Quantum electrodynamics

Note that (360) has a gauge symmetry

The gauge-covariant derivative transforms like the field under this symmetry

leaving $\mathcal{L}$ invariant. This means it would have been possible to state $\mathcal{L}$ based on the requirement that the gauge symmetry holds without thinking about minimal coupling at all!

The photon as a consequence of local symmetry

In fact, we can go further and derive the complete Lagrangian, including the photon field, from the fact that

$\displaystyle\mathcal{L}=\bar{\psi}(i\gamma\cdot\partial-m)\psi$

(373)

should be invariant under local gauge transformation

$\displaystyle\psi(x)\to{\rm e}^{i\alpha(x)}\psi(x)\,.$

(374)

The mass term is obviously invariant but what about the derivative? Due to the gauge symmetry, the derivative $\partial_{\mu}$ no longer has any geometric meaning since the phase $\alpha(x)$ could mess things up. Let us therefore define a new derivate $D_{\mu}$ which compares two nearby points along a direction $\hat{n}^{\mu}$

$\displaystyle\hat{n}^{\mu}D_{\mu}\psi=\lim_{\epsilon\to 0}\frac{\psi(x+\epsilon\hat{n})-U(x+\epsilon\hat{n},x)\psi(x)}{\epsilon}\,.$

(375)

Here we had to introduce a new object, $U$, which for the normal derivative $\partial_{\mu}$ is just $U=1$ but accounts for the change $\alpha$. For $\mathcal{L}$ to be invariant, we need this new derivative to transform like the field itself, i.e.

$\displaystyle\hat{n}^{\mu}D_{\mu}\psi\to\lim_{\epsilon\to 0}\frac{{\rm e}^{i\alpha(x+\epsilon\hat{n})}\psi(x+\epsilon\hat{n})-U^{\prime}(x+\epsilon\hat{n},x){\rm e}^{i\alpha(x)}\psi(x)}{\epsilon}\stackrel{{\scriptstyle!}}{{=}}{\rm e}^{i\alpha(x)}\hat{n}^{\mu}D_{\mu}\psi\,.$

(376)

The only way to more this work generally is if $U$ transforms as

$\displaystyle U(y,x)\to U^{\prime}(y,x)={\rm e}^{i\alpha(y)}U(y,x){\rm e}^{-i\alpha(x)}\,.$

(377)

Then we have

$\displaystyle\hat{n}^{\mu}D_{\mu}\psi\to\lim_{\epsilon\to 0}{\rm e}^{i\alpha(x+\epsilon\hat{n})}\frac{\psi(x+\epsilon\hat{n})-U(x+\epsilon\hat{n},x)\psi(x)}{\epsilon}={\rm e}^{i\alpha(x)}\hat{n}^{\mu}D_{\mu}\psi\,.$

(378)

Taylor-expanding $U$ gives us with $U(x,x)=1$

$\displaystyle U(x+\epsilon\hat{n},x)=1-i\epsilon\hat{n}^{\mu}(eA_{\mu}(x))+\mathcal{O}(\epsilon^{2})\,.$

(379)

Here we had to introduce a field $A_{\mu}$ that is the derivative of $U$ as well as an arbitrary constant $e$. It is easy to see that $A_{\mu}$ transforms as required and it is no surprise that it will turn into the photon field.

We can now concatenate four comparison operations into a small square

$\displaystyle\mathcal{U}(x)=U(x,x+\epsilon\hat{n})U(x+\epsilon\hat{n},x+\epsilon\hat{n}+\epsilon\hat{m})U(x+\epsilon\hat{n}+\epsilon\hat{m},x+\epsilon\hat{m})U(x+\epsilon\hat{m},x)\,.$

(380)

It is easy to see that $\mathcal{U}(x)$ is invariant under the transformation. Starting from

$\displaystyle U(x,y)=\exp\Big{(}-ieA\big{(}\frac{x+y}{2}\big{)}\cdot(x-y)+\mathcal{O}\big{(}(x-y)^{3}\big{)}\Big{)}\,,$

(381)

we find

	$\displaystyle\mathcal{U}(x)$	$\displaystyle=\exp\bigg{[}-ie\epsilon\Big{(}-\hat{n}\cdot A\big{(}x+\hat{n}\tfrac{\epsilon}{2}\big{)}-\hat{m}\cdot A\big{(}x+\hat{n}\epsilon+\hat{m}\tfrac{\epsilon}{2}\big{)}+\hat{n}\cdot A\big{(}x+\hat{m}\epsilon+\hat{n}\tfrac{\epsilon}{2}\big{)}+\hat{m}\cdot A\big{(}x+\hat{m}\tfrac{\epsilon}{2}\big{)}\Big{)}\Bigg{]}$
		$\displaystyle=\exp\Big{[}-ie\epsilon\big{(}\partial_{\hat{m}}(\hat{n}\cdot A)-\partial_{\hat{n}}(\hat{m}\cdot A)\big{)}\Big{)}\,.$		(382)

This proofs that $F_{\mu\nu}$ and any functions that depend on $F_{\mu\nu}$ are invariant. However, $A_{\mu}$ itself is not invariant meaning that a mass term like $m_{\gamma}\ A_{\mu}A^{\mu}$ would not be allowed. This is the reason that the photon is massless.

You may now wonder about the $Z$ and $W$ bosons. The same argument still applies and we cannot write down a mass for them. In the Standard Model, their masses are dynamically generated through the Higgs mechanism. Basically, the theory contains a scalar field $\phi$ whose kinetic term includes a covariant deriviative that couples it dynamically to the $W$ and $Z$ bosons. Uniquely among all particles, this field has a non-zero vev meaning that $W$ and $Z$ get a dynamically generated mass.

The following is a rough recipe for evaluating Feynman diagrams. We will shortly discuss each step in an example but it may be convenient to have it all in one place. To calculate an amplitude,

1. 1.
   ​

   draw all the Feynman diagrams.

2. 2.
   ​

   assign momenta to all edges and Lorentz indices to all photon vertices.

3. 3.
   ​

   pick the end of any fermion line and follow the arrow backwards, evaluating as you go.

4. 4.
   ​

   multiply in the polarisation tensors for any external photons and propagators for each internal photon.

5. 5.
   ​

   contract any open index

6. 6.
   ​

   if your diagram has any loops, integrate over the unconstrained momenta. If you have any internal fermion loops, calculate the trace over their gamma matrices.

7. 7.
   ​

   you now have the amplitude $\mathcal{M}$.

To calculate a cross section,

1. 8.
   ​

   square the amplitude as $|\mathcal{M}|^{2}=\mathcal{M}^{\dagger}\times\mathcal{M}$. Make sure to rename any indices to avoid collisions. To calculate $\mathcal{M}^{\dagger}$, you can use the fact that $(A\cdot B)^{\dagger}=B^{\dagger}A^{\dagger}$ and

   $\displaystyle u^{\dagger}=u^{\dagger}\gamma^{0}\gamma^{0}=\bar{u}\gamma^{0}\qquad\text{and}\qquad\bar{u}^{\dagger}=\big{[}u^{\dagger}\gamma^{0}\big{]}^{\dagger}=\big{[}\gamma^{0}\big{]}^{\dagger}u=\gamma^{0}u\,,$

   (383)

   and similarly for $v$. Further,

   $\displaystyle(\gamma^{\mu})^{\dagger}=\gamma^{0}\gamma^{\mu}\gamma^{0}\,.$

   (278)

   As we will see, this just amounts to reversing the spin line.

2. 9.
   ​

   if you calculate unpolarised scattering, sum over final-state and average over initial-state polarisations. For this, the completeness relations (317) and (356) will be helpful

   	$\displaystyle\sum_{r=1,2}u_{r}(p)\bar{u}_{r}(p)=\gamma\cdot p+m\,,$
   	$\displaystyle\sum_{r=1,2}v_{r}(p)\bar{v}_{r}(p)=\gamma\cdot p-m\,,$		(317)
   	$\displaystyle\sum_{\lambda}\epsilon_{\lambda,\mu}(p)\epsilon_{\lambda,\nu}^{*}(p)=-\eta_{\mu\nu}\,.$		(356)

3. 10.
   ​

   for fermions, this will result in a trace of $\gamma$ matrices (see below). Evaluate this trace using trace identities (274)

   	$\displaystyle{\rm tr}(\gamma^{\mu})$	$\displaystyle=0\,,$
   	$\displaystyle{\rm tr}(\underbrace{\gamma^{\mu_{1}}\cdots\gamma^{\mu_{k}}}_{\text{odd}})$	$\displaystyle=0\,,$
   	$\displaystyle{\rm tr}(\gamma^{\mu}\gamma^{\nu})$	$\displaystyle=4\eta^{\mu\nu}\,,$
   	$\displaystyle{\rm tr}\Big{(}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\Big{)}$	$\displaystyle=4\Big{(}\eta^{\mu\nu}\eta^{\rho\sigma}-\eta^{\mu\rho}\eta^{\nu\sigma}+\eta^{\mu\sigma}\eta^{\nu\rho}\Big{)}\,.$		(274)

4. 11.
   ​

   use (235) or (237) to relate this to the cross section or decay rate

   	$\displaystyle{\rm d}\sigma$	$\displaystyle=\frac{1}{(2E_{a})(2E_{b})|\vec{v}|}\underbrace{\Bigg{(}\prod_{n=1}^{f}\frac{{\rm d}^{3}p_{n}}{(2\pi)^{3}2E_{n}}\Bigg{)}(2\pi)^{4}\delta^{(4)}\Bigg{(}p_{a}+p_{b}-\sum_{n=1}^{f}p_{n}\Bigg{)}}_{{\rm d}\Phi_{2\to f}}\ |\mathcal{M}|^{2}\,,$		(235)
   	$\displaystyle{\rm d}\Gamma$	$\displaystyle=\frac{1}{2M}\underbrace{\Bigg{(}\prod_{n=1}^{f}\frac{{\rm d}^{3}p_{n}}{(2\pi)^{3}2E_{n}}\Bigg{)}(2\pi)^{4}\delta^{(4)}\Bigg{(}p_{a}-\sum_{n=1}^{f}p_{n}\Bigg{)}}_{{\rm d}\Phi_{1\to f}}\ |\mathcal{M}|^{2}\,.$		(237)

5. 12.
   ​

   integrate over phase space.

Let us calculate our first real process, $ee\to\mu\mu$. There is just a single $s$-channel diagram contributing so Step 1 and 2 are very simple

We have to apply Step 3 twice, once for the muon and once for the electron. After Step 5, we have

We can now square this object

Let us first work on the first bracket $[\cdots]^{\dagger}$ and use that

This means we now have after re-bracketing things

Let us look at one of those, e.g. the first one where we have $v_{s_{2}}\bar{v}_{s_{2}}$. If we sum over the polarisation states, i.e. calculate $v_{1}\bar{v}_{1}+v_{2}\bar{v}_{2}$, we can write this using the completeness relation as $\gamma\cdot p_{2}-m$.

To make use of this also for $s_{1}$, remember that the bracket is a number under spinor, i.e.

This is sometimes referred to the Casimir trick and it is essential to calculating matrix elements with traces. Therefore,

We now can expand and calculate this trace

Doing the same for the other bracket, we have

and with $s=(p_{1}+p_{2})^{2}$ and $t=(p_{1}-p_{3})^{2}$

where we have used that $\eta^{\mu\nu}\ \eta_{\mu\nu}=4$.

To simplify the discussion of the cross section, let us set $m=0$ (high energy limit) and write $t$ in terms of $\cos\theta$ (cf. (258b))

the velocity of the muon. Now the matrix element squared is

Since we have to average over $2\times 2$ incoming spins, the four cancels. The cross section is

In the high-energy limit, $M=0$ or $\beta=1$, this is

At this point, there are a number of exercises you can do. In order of increasing complexity.

Suggested Exercise

Calculate ${\rm d}\sigma/{\rm d}t$ for $e\mu\to e\mu$ scattering. The amplitude squared is

$\displaystyle\sum|\mathcal{M}|^{2}=\frac{8e^{4}}{t^{2}}\Big{(}2m^{4}+2M^{4}+4m^{2}(M^{2}-s)-4M^{2}s+2s^{2}+2st+t^{2}\Big{)}\,,$

(404)

and the cross section is

$\displaystyle\frac{{\rm d}\sigma}{{\rm d}t}=4\pi\alpha^{2}\frac{(M^{2}+m^{2})^{2}-su+t^{2}/2}{t^{2}\lambda}\,,$

(405)

with $\lambda=m^{4}-2m^{2}M^{2}+M^{4}-2m^{2}s-2M^{2}s+s^{2}$.

Suggested Exercise

Calculate the cross section for $ee\to\gamma\gamma$. The amplitude can be written down as

$\displaystyle\begin{split}\mathcal{M}&=\bar{v}(p_{2})(-ie\gamma^{\mu})\frac{i}{\gamma\cdot(p_{1}-p_{4})-m}(-ie\gamma^{\nu})u(p_{1})\ \epsilon_{\mu}^{*}(p_{3})\epsilon_{\nu}^{*}(p_{4})\\ &+\bar{v}(p_{2})(-ie\gamma^{\nu})\frac{i}{\gamma\cdot(p_{1}-p_{3})-m}(-ie\gamma^{\mu})u(p_{1})\ \epsilon_{\mu}^{*}(p_{3})\epsilon_{\nu}^{*}(p_{4})\,.\end{split}$

(406)

Suggested Exercise

Calculate the cross section ${\rm d}\sigma/{\rm d}t$ for $e^{+}e^{-}\to e^{+}e^{-}$. We find

$\displaystyle\sum|\mathcal{M}|^{2}=8e^{4}\Bigg{(}\frac{8m^{4}-8m^{2}s+2s^{2}+2st+t^{2}}{t^{2}}+\frac{8m^{4}+s^{2}-8m^{2}t+2st+2t^{2}}{s^{2}}+\frac{2(s+t)^{2}-8m^{4}}{st}\Bigg{)}\,.$

(407)

Suggested Exercise

Write down, but do not calculate, the amplitudes at $\mathcal{O}(e^{4})$ for $ee\to\mu\mu$.