3 Quantum Electrodynamics

In this course, we have so far considered spin-0 and spin-1/2 particles. We will postpone a detailed discussion of spin-1 particles until Section 3.2. For the time being, we start from the Maxwell’s equations in the vacuum in relativistic notation (cf. Appendix A for a derivation from first principle)

$J^{\nu}$ is a conserved current, i.e. $\partial_{\nu}J^{\nu}=0$, and $A^{\mu}=(\phi,\vec{A})$ is the four-potential. Maxwell’s equations originate from the following Lagrangian

by evaluating the Euler-Lagrange equations

The Dirac equation also originates from a Lagrangian

We obtain the free part of the QED Lagrangian by summing $\mathcal{L}_{\rm em}$ and $\mathcal{L}_{\rm Dirac}$. However, to get a realistic theory, we need something that couples the photon field $A$ to the spinor $\psi$. The derivation of Appendix A shows that there is only one valid way of doing this which is to keep $\mathcal{L}_{\rm int}$ untouched and to set

This makes a lot of sense. The resulting Lagrangian still follows Maxwell’s equation and, unlike $\bar{\psi}\gamma^{5}\gamma^{\mu}\psi$, $\bar{\psi}\gamma^{\mu}\psi$ is a conserved current

where we have used the Dirac equation. The quantity $-e$ multiplies the vector current so as to be sure that the resulting Coulomb potential arising from the solution of the static Maxwell’s equations is the expected one.

We can now write down the complete QED Lagrangian

Notice how $\mathcal{L}$ is invariant with respect to the gauge transformation

This is called a local $U(1)$ symmetry because we add a different phase ${\rm e}^{i\alpha(x)}$ at every point in spacetime and it is the starting point of the derivation in Appendix A. We can also write $\mathcal{L}$ more compactly as

where we have defined the gauge-covariant derivative

This idea of substituting $p\to p-qA$ is called minimal coupling and is also how one can derive the Lorentz force in classical electrodynamics. The consequences of this idea will be covered in more detail in the Standard Model course.

The gauge invariance (63) means that there are unphysical degrees of freedom in $A^{\mu}$. This is clear from the fact that the massless photon has two physical polarisations but $A^{\mu}$ has four degrees of freedom. In order to eliminate this degeneracy, a gauge fixing condition is required. A possible choice is the Coulomb gauge which requires $\nabla\cdot\vec{A}=0$. While this works, it break Lorentz invariance. Another example is the Lorenz gauge (not to be confused with Lorentz)

In this gauge, free Maxwell equation is $\partial_{\mu}\partial^{\mu}A^{\nu}=0$.

Note how the Lorenz gauge only reduces the number of degrees of freedom to three so that there is still one unphysical mode which we can parametrise by

In the classical case, we would normally remove this remaining degree of freedom by hand. In the quantum case, this does not work because it breaks the covariant canonical commutation relations. The strategy is instead to introduce a gauge-fixing term to the Lagrangian

with the gauge parameter $\xi$. Using this Lagrangian as a starting point, and an extra condition on physical states, only the two physical polarisations propagate on-shell. The Euler-Lagrange equation for $A_{\nu}$ now read

If $\xi$ is left symbolic, this is referred to the $R_{\xi}$ gauge. Specific values of $\xi$ include

$\xi=0$

This recovers the Lorenz gauge.

$\xi=\infty$

This is called the unitary gauge and it has certain advantages when dealing with massive vector bosons.

$\xi=1$

This is called the Feynman gauge and it is by far the most common.

We also briefly need to discuss photons in more detail. The plane-wave solution that we found in (3) and (23) looks like

for photons where $\epsilon^{\mu}$ is the polarisation vector. This has a similar role to the $v$ and $u$ in the case of spinors in (23) and the $A$ coefficient in (3). In the Lorenz gauge of (66) the equation of motion

is automatically satisfied as long as $k^{2}=0$. Further, we have from the gauge condition itself

However, this still does not fully determine the polarisation vector since, if $\epsilon$ is a solution than so is $\epsilon^{\prime}=\epsilon+\lambda k$. This corresponds to the propagation of an extra unphysical longitudinal photon, with a polarisation proportional to $k_{\mu}$. This can be fixed by setting $\epsilon^{0}=0$ such that $\vec{k}\cdot\vec{\epsilon}=0$. The two remaining polarisations are in transverse direction and can be chosen orthonormal.

Feynman diagrams are a very useful tool for organising expressions for scattering amplitudes, even if it has certain downsides. These are constructed from vertices, each corresponding to a term in the interaction Lagrangian $\mathcal{L}_{\rm int}$, and edges between vertices called propagators. To calculate a scattering amplitude $i\mathcal{M}$ we therefore draw all possible diagrams with the correct initial and final states. To turn a diagram into a mathematical equation, we use Feynman rules.

As part of the quantum field theory (QFT) course in this school, you will learn how to derive these for a scalar $\phi^{4}$ theory. It is possible, if unwieldy, to do the same for QED so we will just assume the Feynman rules given and learn how to use them.

A few comments are now necessary

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  Individual pieces of a Feynman diagram are a mixture of matrices, vectors, co-vectors and scalars. They do not commute. The final amplitude is a number and therefore you must follow each fermion line from a spinor (either outgoing particle or incoming anti-particle) through the series of matrices to finish on an anti-spinor (either incoming particle or out-going anti-particle). This corresponds to working backwards along the fermion line. We will see this in the examples which follow. Similarly, all Lorentz indices corresponding to photons have to be contracted.

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  The photon propagator term was given with a free parameter $\xi$. This is due to the gauge freedom we discussed in the previous section. It does not represent a physical degree of freedom and therefore any calculation of a physical observable will be independent of $\xi$. We will most commonly work in Feynman gauge, i.e. $\xi=1$.

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  The propagators come with factors of $i\epsilon$ in the denominator, otherwise they would have poles on the real axis and any integral over $p$ would not be well-defined. The factor of $i\epsilon$ prescribes which direction to travel around the poles. This choice corresponds to the Feynman prescription, which ensures causality.

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  The interaction vertex contains only one flavour of fermion. We know that the emission of a photon does not change an electron to a quark for example. Weak interactions do change the flavour of the quarks, but QED and quantum chromodynamics (QCD) do not.

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  There are additional factors of $(-1)$ in the following scenarios:

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    an anti-fermion line runs continuously from an initial to a final state;

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    there is a closed fermion loop;

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    between diagrams with identical fermions in the final state.

  These arise from the anti-commutation properties of fermionic operators which is beyond the scope of this course. This sign can be important to get the relative phase between diagrams correct, as happens for instance in Bhabha scattering.

As a first process, let us consider the process

If you want, you can think about this in the classical limit where an electron scatters on the potential of a much heavier particle (the muon). This means we are calculating the Coulomb potential using QFT

We start this by drawing the internal and external particles and then need to find all possible ways to connect them. Since we are doing perturbation theory, we assume that $e\ll 1$, i.e. we need to prioritise diagrams with fewer powers of $e$. There is no way to connect a muon directly to an electron, we need to use a photon. There is just a single $t$-channel diagram that does not have more than $e^{2}$ so the amplitude is for now

If we wanted to, we could calculate higher order corrections that have $e^{4}$ or more. This will involve diagrams with loops that will be covered in the phenomenology course. Had we decided to calculate $e^{-}e^{-}\to e^{-}e^{-}$ instead of $e^{-}\mu^{-}\to e^{-}\mu^{-}$ we would have two diagrams (find them!) which we would have to add.

Let us now evaluate this diagram. Following the spin lines of the electron and muon backwards, we find

Here, we have used Feynman gauge $\xi=1$. Just as in quantum mechanics, in order to compute the probability of this process happening, we must calculate $|\mathcal{M}|^{2}$.

Note how we have introduced a new index for the $[\cdots]^{\dagger}$ part to avoid having the same index more than twice. Let us first work on the first bracket $[\cdots]^{\dagger}$ and use that

This means we now have after re-bracketing things

In order to describe an unpolarised physical scattering process, we will average over initial-state spins and sum over final-state spins. Consider for a moment only the $u_{s_{3}}(p_{3})\bar{\bar{u}}_{s_{3}}(p_{3})$ and sum over $s_{3}$

where we have used the completeness relation. To make use of this also for $s_{1}$, remember that the bracket is a number in spinor space. This means we can do the following rearrangement

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

And similarly for the muon line

We now can expand and calculate these traces using the identities from Appendix C

And for the muon line

Therefore, for $|\mathcal{M}|^{2}$

We can re-write this using the Mandelstam variables

These can be solved for the different scalar products appearing in (92). Also note that, due to momentum conservation $p_{1}+p_{2}=p_{3}+p_{4}$, $s$, $t$, and $u$ are not independent. You can easily show that $s+t+u=2m^{2}+2M^{2}$. With these,

Since we need to average over the initial states, rather than sum, we need to divide by 2 for $s_{1}$ and 2 for $s_{2}$, i.e.

The above equation gives the probability that the corresponding process occurs at a given point in phase space. In the next section, we will derive how to calculate a total cross section (or a total decay width) from amplitudes squared.

Consider the Feynman rules (73d) and (73e). These mean that the amplitude for a process with external photons can be written as

$\mathcal{M}$ is a physical quantity and therefore needs to be gauge invariant. We should therefore be able to set $\epsilon+\lambda\,k$ which means

This is called the Ward identity for QED and it is a very strong test of gauge invariance. If we were to square this amplitude and sum over transverse polarisation of the photon (for initial photons for simplicity)

We now are in need for another completeness relation, just like we had for the spinors with $u\bar{u}$ (37). Since $\epsilon_{0}=0$ per assumption, we can write

Note how the indices $i$ and $j$ are only over spatial components. The polarisation-summed amplitude becomes

From the Ward identity, it follows that $\mathcal{M}^{i}k^{i}=\mathcal{M}^{0}k^{0}$ or $\mathcal{M}^{i}\hat{k}^{i}=\mathcal{M}^{0}$. We can therefore write

This suggest a convenient shorthand of setting

Note that this is not an equality without the rest of the matrix element there.