# Electron positron helicities and gamma-5

1. Jun 2, 2008

### jdstokes

Hi all,

I've read that in the QED process $e^+e^- \to \mu^+\mu^-$, terms in the scattering amplitude coming from electrons and muons of the same helicity do not contribute.

I don't understand why this is the case. Is this not true if the interaction is mediated by the weak force?

2. Jun 3, 2008

### neu

EM propagator is $$\gamma^{\mu}$$

Weak propagator is $$\gamma^{\mu}(1-\gamma^5)$$

(neglecting prefactors) You should find that for a spinor U:

$$U_{left}= \overline{U}(1+\gamma^5)$$
$$U_{right}= (1-\gamma^5)U$$ (CHECK THESE; they may be incorrect)

Using the left and right helicity spinors and their adjoints you should find that the EM interation only involves certain helicity state combinations.

ie input $$U_{left}$$ & $$U_{right}$$ into the interaction term $$\overline{U}\gamma^{\mu}U$$ for the EM amplitude and into $$\overline{U}\gamma^{\mu}(1-\gamma^5)U$$ for the weak amplitude.

I think this is the proof that the weak interaction only involves left chiral particles and right chiral antiparticles where: $$U_{L.chirality} = \sqrt[2]{\frac{1+\beta}{2}}U_{L.Helicity}+\sqrt[2]{\frac{1-\beta}{2}}U_{R.Helicity}$$ and right chiral state is opposite signed

I can't remeber much else off the top of my head, but that is the general idea

Last edited: Jun 3, 2008
3. Jun 4, 2008

### jdstokes

So the em only allows interaction of fermions with specific helicity combinations? Interesting. This was never really apparent to me when taking spin sums of Feynman amplitudes in QED.

Helicity projection operators are defined by $\Pi^{\pm} = \frac{1}{2}(1\pm h_p)$ where $h_p = \hat{p}\cdot\Sigma$ is the helicity operator ($\Sigma^i = (\mathrm{i}/2)\epsilon^{ijk}\gamma_j\gamma_k$), of which the Dirac spinors are chosen to be eigenstates.

Clearly the helicity projection operator has the opposite effect on Dirac spinors corresponding to particles and antiparticles.

In the ultrarelativistic limit $\Pi^\pm = (1/2)(1\pm \gamma_5)$ so it can be moved through $\gamma^\mu$'s changing + to -.

If we're interested in a process involving the a positive helicity electron and a positive helicity positron at the same vertex, then a possible bilinear covariant would be $\bar{v}_1\gamma^\mu u_1$. This is equivalent to $\bar{v} \gamma^\mu \Pi^+ u_1 = v^\dag_1 \gamma^0 \Pi^-\gamma^\mu u_1 = v^\dag_1 \Pi^+\gamma^0 \gamma^\mu u_1$. But this is zero since $\Pi^+ v_1 = 0 \implies v_1^\dag \Pi^+ = 0$.

The bilinear covariants in this case are of the form $\bar{w}\gamma^\mu \Pi^-y$. where w and y can be either u or v. If y is a particle, this will vanish unless y has negative helicity ie it is left-handed. If y is an antiparticle this will vanish unless y has positive helicity. Similarly for w. Do you think this explains why the weak interaction couples only to left-handed particles and right-handed anti-particles?

4. Jun 4, 2008

### neu

Yeah, QED conserves chirality.

i.e:
$$\overline{U}\gamma^{\mu}U = \overline{U_{L}}\gamma^{\mu}U_{L} + \overline{U_{R}}\gamma^{\mu}U_{R}+\overline{U_{R}}\gamma^{\mu}U_{L} + \overline{U_{L}}\gamma^{\mu}U_{R} = \overline{U_{L}}\gamma^{\mu}U_{L} + \overline{U_{R}}\gamma^{\mu}U_{R}$$

as $$\overline{U_{L}}\gamma^{\mu}U_{R}=0$$

So in general expect 50/50 mix of R/L interacting particles.

And yes, for the reason you gave the weak force only involves left chiral particles, and right chiral antiparticles, which is the only rationale I know which accounts for their being no observed right-handed neutrinos.

5. Jun 20, 2008