16
$\begingroup$

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.
$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '14 at 11:32
  • 2
    $\begingroup$ No, chirality is boost invariant. The weak force is boost invariant. $\endgroup$ – innisfree Apr 8 '14 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 '14 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 '14 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 '14 at 16:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.