These are notes for a PHY421 tutorial, essentially mirroring the discussion in Thomson’s Modern Particle Physics (middle sections of chapter 6), to complement the derivation (in terms of the explicit spinors and Clifford representations) in Prof. Costanzo’s lecture notes.

Completeness relations

Instead of heading straight to the computation of the relevant Feynman diagram, it is best to lay down now a few expressions that will come in handy later. Most of what follows assumes you’re already familiar with gamma matrices and Dirac spinors, and have some idea of where we’re going with all that (which you should if you have read your notes!).


Particularly important to us will be the object

which is summed over two orthogonal spin states (i.e. and ). From having solved the Dirac equation earlier, we have the Dirac-Pauli representation:

We know how to build from : using the matrix. It should also be pointed out that being built from Pauli matrices, it is of course hermitian. And so:

The spin dependence of and has therefore been isolated to the components. It is easy enough to work out their sum:

We also need:

where we’ve used a useful relation for the Pauli matrices. Finally, we put everything together and write:

If you’re wondering how we read off from the matrix expression, here’s the reasoning: the terms on the diagonal are simply with a sign coming from ; add a (multiplied by the unit matrix) and you get back the correct expressions. The off-diagonal entries are dotted with , which come of course from .

Similarly, we get:

where the minus sign originates from our convention on particles vs antiparticles in the Dirac equation.

Gamma matrices

If you’re not confident with the manipulation of gamma matrices, now is a good time to go back to your notes or read this entry. Otherwise let’s just move on.

Spinors and gamma matrices

An object that pops up all the time when evaluating Feynman diagrams is , where and are spinors, and is some combination of gamma matrices. In our main calculation, we shall deal with e.g. . There are two things we need to say about such objects.

Firstly, taken on their own they are complex numbers. Since the “bar” operation is really a fancy transpose, we’re dealing with a row-vector times a matrix times a column-vector - and that’s just a scalar. However, be warned that they are also understood as vectors (or their combinations as tensors) - where the index is not spatial, but that of the gamma matrix. If you find that confusing, you’re not the only one; but the worked example later on should help relieve some of that confusion (hopefully).

Secondly, we’ll also be interested in taking the hermitian conjugate of that quantity:

where we’ve used several properties of the gamma matrices to help us along, including (which holds for all ’s involved in QED, QCD and EW; i.e. in the case of QED, this is ). You should be happy, this is a nice enough expression (things can get much worse in QFT…).

To the main calculation

Here I’m assuming you’ve read through your notes, know about the Feynman rules for QED and have already written down an expression for the electron-positron annihilation matrix element, waiting with baited breath to see how you’re actually supposed to evaluate that mess. Well, wait no more! This is where the proverbial fun starts

What we’re really after

… is the cross-section corresponding to that particular process. From what we discussed last time, we know that it’s that goes in the expression for the cross-section. As you found out in your notes, it’s actually slightly trickier: we need to average over 4 possible initial helicity configuration (16 combinations, including final state configurations). And so what we want to express is:

Using the Feynman rules for QED and reading off the diagram:

Contracting , we have:

where we’ve dropped the momenta to make clear how we grouped like-terms (which is allowed since each one of them is just a complex number - see above).

Using the properties of

(I did say they show up all the time, didn’t I?)

Using the spin-sum results

You see now why we did everything we did, before jumping into the calculation? Let’s isolate the first part of the last expression (it’s a rank-2 tensor, by the way), and we’ll use symmetry arguments to replicate our results to the second part. It is also important to make explicit the matrix indices, as we shall see in a minute:

where in the last line we’ve just put the various matrix indices back in the right order. Why? because it shows we’re computing the component of some object…

Trace techniques

Just to make it absolutely clear, we’re dealing with:

Using the linearity of the trace:

where the crossed out terms evaluate to zero as they contain an odd number of gamma matrices (remember that !). The two traces we’re left with are two common expressions, see for instance this entry:

Symmetry argument and contractions

As promised, we can obtain a similar contribution from , the second sum we had earlier. Initially, the minus signs are placed differently, but since the cross terms cancel out, the result is the same (up to the lowering of indices and the replacement ). There are three ways in which we will contract indices:

And so:

And thus we’ve completely eliminated gamma matrices and spinors from our original expression. We’re now only left with momenta and masses, which are measurable, physical quantities. But we can still go one step further.


Relativistic limit

After all, this is QFT, so let’s go relativistic! Let’s set:

And rewrite:

Mandelstam variables

By now you should be acquainted with Mandelstam variables:

and so:

We’ve previously worked out an expression for the differential cross section:

Center-of-mass frame

Note that this expression holds in the COM frame, so let’s make the appropriate substitutions. Without loss of generality:

Setting , the fine-structure constant, we get:

The need for a complete electroweak theory

One interesting experimental observation at this stage (that is, considering only QED), is that the above formula we’ve derived predicts a forward-backward symmetry: the escape angle of the muons follows a symmetric distribution. However, early tests of QED in particle accelerators proved this prediction to be wrong!

In fact, to make sense of what is actually a forward-backward asymmetry, one needs to consider contributions from other Feynman diagrams. An immediate hypothesis is that we have to look at the loop-level to find a possible correction to the cross-section; however, it is demonstrable that the tree-level process we’ve computed is accurate to the 99% (you can think about each new interaction vertex being suppressed by a factor of ). No, instead, one must appeal to new physics: a neutral electroweak boson, . (You’ll see later in the course that the same diagram exists in the complete EW theory, with the photon propagator replaced by a propagator)

One last integration

To derive the total cross-section, we simply need to integrate over the solid angle :