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Chirality, helicity and the weak interaction

+ 8 like - 0 dislike
987 views

From what I'm understanding about Dirac spinors, using the Weyl basis for the $\gamma$ matrices the first two components behave as a left handed Weyl spinor, while the third and the fourth form a right handed Weyl spinor. By boosting in a direction or in the opposite, I can "asymptotically kill" either the left or right handed part of the (massive) spinor. Since only the left-handed part interacts with the weak force, does that mean that when I see an electron travelling very fast in one direction (same as/opposite to spin) I see/don't see it weakly interacting? This sounds very odd indeed.

I have two hypotheses:

  1. Massive spinors don't have an intrinsic chirality (since they are not eigenstates of chirality operator), the only information I have is about helicity, and the odd thing I described before is actually observed (really odd to me).
  2. Massive particles have an intrinsic chirality, but I don't see how the chirality information gets encoded into the Dirac spinor / how the weak interaction couples to only half of it. To me it seems that only the helicity information is carried by a spinor.
This post imported from StackExchange Physics at 2014-04-13 14:39 (UCT), posted by SE-user kornut
asked Apr 8, 2014 in Theoretical Physics by kornut (40 points) [ no revision ]

1 Answer

+ 3 like - 0 dislike

You are correct that for a massive spinor, helicity is not Lorentz invariant. For a massless spinor, helicity is Lorentz invariant and coincides with chirality. Chirality is always Lorentz invariant.

  • Helicity defined $$ \hat h = \vec\Sigma \cdot \hat p, $$ commutes with the Hamiltonian, $$ [\hat h, H] = 0, $$ but is clearly not Lorentz invariant, because it contains a dot product of a three-momentum.

  • Chirality defined $$ \gamma_5 = i\gamma_0 \ldots \gamma_3, $$ is Lorentz invariant, but does not commute with the Hamiltonian, $$ [\gamma_5, H] \propto m $$ because a mass term mixes chirality, $m\bar\psi_L\psi_R$. If $m=0$, you can show from the massless Dirac equation that $\gamma_5 = \hat h$ when acting on a spinor.

Your second answer is closest to the truth:

The weak interaction couples only with left chiral spinors and is not frame/observer dependent.

A left chiral spinor can be written $$ \psi_L = \frac12 (1+\gamma_5) \psi. $$ If $m=0$, the left and right chiral parts of a spinor are independent. They obey separate Dirac equations.

If $m\neq0$, the mass states $\psi$, $$ m(\bar\psi_R \psi_L + \bar\psi_L \psi_R) = m\bar\psi\psi\\ \psi = \psi_L + \psi_R $$ are not equal to the interaction states, $\psi_L$ and $\psi_R$. There is a single Dirac equation for $\psi$ that is not separable into two equations of motion (one for $\psi_R$ and one for $\psi_L$).

If an electron, say, is propagating freely, it is a mass eigenstate, with both left and right chiral parts propagating.

This post imported from StackExchange Physics at 2014-04-13 14:39 (UCT), posted by SE-user innisfree
answered Apr 8, 2014 by innisfree (285 points) [ no revision ]
If an electron, say, is propagating freely, it is a mass eigenstate, with both left and right chiral parts propagating. Ok, but can't I just send $\psi_L$ to zero using a boost and get a spinor that doesn't interact with the weak force at all? (more precisely: i cannot make $\psi_L = 0$ but i can get as close as i want -> to me it seems that i can make the weakly interacting part arbitrarily small)

This post imported from StackExchange Physics at 2014-04-13 14:39 (UCT), posted by SE-user kornut
No, chirality is boost invariant. The weak force is boost invariant.

This post imported from StackExchange Physics at 2014-04-13 14:39 (UCT), posted by SE-user innisfree
I still don't see how the weak force can couple to a spinor whose left-handed part is driven down to zero! (Thanks for your help)

This post imported from StackExchange Physics at 2014-04-13 14:39 (UCT), posted by SE-user kornut
the left-hand chiral part of $\psi$ cannot be made zero or arbitrarily small by boosts. it is boost invariant.

This post imported from StackExchange Physics at 2014-04-13 14:39 (UCT), posted by SE-user innisfree
In Peskin&Schroeder pag. 46-47 (Sec 3.3), it shows that $\psi=((1,0),(1,0))$ boosts to either $\psi=((1,0),(0,0))$ or $\psi=((0,0),(1,0))$ depending on the direction of the boost. I thought this was just the helicity changing due to my boost, but then i don't see where the chirality information gets encoded in the spinor.

This post imported from StackExchange Physics at 2014-04-13 14:39 (UCT), posted by SE-user kornut
@kornut: The original spinor (unboosted) that you mention is not a chirality eigenstate. The chirality operator, $\gamma_5$, acting on it does not give you an eigenvalue. Try boosting any chirality eigenstate.

This post imported from StackExchange Physics at 2014-04-13 14:39 (UCT), posted by SE-user JeffDror

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