# Feynman diagrams for spin 1/2 particles

• Chopin
In summary, this equation says that the total scattering cross section for two particles, each with a spin of 1/2, is proportional to the sum of their individual cross sections.
Chopin
I think I have a pretty good handle on how scalar field scattering works in QFT, so now I'm trying to wrap my head around spin 1/2 particles, and I'm having a bit of trouble with it. For instance, in $N + \phi \rightarrow N + \phi$ scattering, an application of the Feynman rules leads to an expression like

$$(2\pi)^4\delta(p + q - (p' + q'))\bar{u_{p'}}\frac{i}{-\not{p} -\not{q} - m + i\epsilon}u_{p} + crossing$$

So far so good--I see how the Wick expansion plus the definition of the fermion propagator lead to that. What I want to know (and I have a feeling this is going to turn out to be kind of a stupid question) is: how exactly do I go about evaluating this? I know that $\bar{u_p}u_p = 2m$, so that's a start on how to multiply the spinors, but a) that only works if the initial and final momenta are the same, and b) I still have to deal with the slashed momenta in the denominator.

For part a), my first thought is that when the momenta are different, the result might be proportional to the dot product of the momenta--something like $\bar{u_{p'}}u_p \propto p'\cdot p$, but I don't know how to show that. Is that right?

For part b), I know that the u's have been constructed such that $\not{p}u_p = mu_p$, so that would tell me how to evaluate the fraction if the slashed momentum were in the same direction as the spinor. But I'm not sure I understand how to handle the case where they're different, for instance $\not{a}u_p$. Going along the same lines as above, my guess is that it'd work out to something like $\not{a}u_p = (a\cdot p)u_p$ but again I'm not really sure how to prove that. Am I on the right track here, or completely off-base?

Peskin and Schroeder has a section or 2 devoted to this. Basically, you want to get rid of slash operators on the denominator by multiplying top and bottom by the complex conjugate. And then you want to specify a frame in which to evaluate the expression (usually CoM frame), and denote, for example, p=(E,E,0,0) for particle coming in the x-direction. And express the outgoing particle in terms of scattering angles and whatnot.

Thanks Matterwave, that makes sense. I managed to find out a bit more about this, too, which I thought I'd put here for posterity, in case anybody else stumbles across this thread.

Things get simpler when you consider two things: a) usually once you get the S matrix, you're going to square it as part of getting a cross section, and b) usually you don't care about incoming or outgoing spin, so you have to average over incoming spins and sum across outgoing spins. Those two things mean that if your S matrix looks like $\bar{u_{p'}}^{(r)}\not{q}u_p^{(s)}$, what you're really interested is something more like $\frac{1}{2}\sum_{rs}\bar{u_p}^{(s)}\not{q}u_{p'}^{(r)}\bar{u_{p'}}^{(r)}\not{q}u_p^{(s)}$.

That makes things simpler, due to some observations by Feynman. This quantity is a scalar, which is a 1x1 matrix, and 1x1 matrices are equal to their trace. So we really have:

$$\frac{1}{2}\sum_{rs}\bar{u_p}^{(s)}\not{q}u_{p'}^{(r)}\bar{u_{p'}}^{(r)}\not{q}u_p^{(s)} = \frac{1}{2}Tr(\sum_{rs}\bar{u_p}^{(s)}\not{q}u_{p'}^{(r)} \bar{u_{p'}}^{(r)}\not{q}u_p^{(s)})$$

Next, we note that the trace is unaffected by cyclic permutations of its factors. So we can write:

$$\frac{1}{2}Tr(\sum_{rs}\bar{u_p}^{(s)}\not{q}u_{p'}^{(r)}\bar{u_{p'}}^{(r)}\not{q}u_p^{(s)}) = \frac{1}{2}Tr(\sum_{rs}u_p^{(s)}\bar{u_p}^{(s)} \not{q}u_{p'}^{(r)} \bar{u_{p'}}^{(r)}\not{q})$$

Even though the LHS is a 1x1 matrix, and the RHS is a 4x4 matrix, their traces are equal. Now we have p and p' on the same side as each other, which in conjunction with the summations, allows us to use the identity $\sum_{r}u_p^{(r)}\bar{u_p}^{(r)} = \not{p} - m$. So the above quantity reduces to:

$$\frac{1}{2}Tr(\sum_{rs}u_p^{(s)} \bar{u_p}^{(s)}\not{q}u_{p'}^{(r)}\bar{u_{p'}}^{(r)}\not{q}) = \frac{1}{2}Tr((\not{p} - m)\not{q}(\not{p'}-m)\not{q})$$

At this point, you just have to distribute all the factors, and start beating on it with a bunch of gamma matrix trace identities to turn it into a real number, which does in fact involve a bunch of dot products like I was speculating above.

## 1. What is a Feynman diagram for spin 1/2 particles?

A Feynman diagram is a graphical representation of the interactions between elementary particles. It was developed by physicist Richard Feynman and is used in quantum field theory to calculate the probability of different outcomes of particle interactions.

## 2. How do Feynman diagrams represent spin 1/2 particles?

Feynman diagrams use arrows to represent the direction of spin of a particle. For spin 1/2 particles, the arrows point either up or down, corresponding to the two possible spin states of the particle.

## 3. What is the significance of the direction of the arrows in a Feynman diagram for spin 1/2 particles?

The direction of the arrows in a Feynman diagram represents the flow of momentum and energy in the particle interactions. The direction of the arrows can be read as the direction in which the particles are moving in space and time.

## 4. How do Feynman diagrams for spin 1/2 particles differ from those for other spin values?

Feynman diagrams for spin 1/2 particles are unique in that they have two arrows representing the two spin states of the particle. For particles with other spin values, there may be more or less arrows depending on the specific spin value.

## 5. What is the advantage of using Feynman diagrams for spin 1/2 particles?

Feynman diagrams allow physicists to visualize and calculate the probability of complex particle interactions. They also provide a way to easily incorporate the effects of quantum mechanics into calculations, making them a valuable tool in theoretical physics.

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