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.

1 Answer 1


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.

  • $\begingroup$ 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) $\endgroup$
    – kornut
    Apr 8, 2014 at 11:32
  • 2
    $\begingroup$ No, chirality is boost invariant. The weak force is boost invariant. $\endgroup$
    – innisfree
    Apr 8, 2014 at 11:33
  • $\begingroup$ 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) $\endgroup$
    – kornut
    Apr 8, 2014 at 11:40
  • $\begingroup$ the left-hand chiral part of $\psi$ cannot be made zero or arbitrarily small by boosts. it is boost invariant. $\endgroup$
    – innisfree
    Apr 8, 2014 at 12:08
  • 1
    $\begingroup$ @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. $\endgroup$
    – JeffDror
    Apr 8, 2014 at 16:11

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