In the Weyl/chiral basis, the four components of the Dirac spinor represent left-chirality spin up, left-chirality spin down, right-chirality spin up, and right-chirality spin down, respectively. When solving the Dirac equation for particles at rest, we find that positive-frequency solutions have polarization $$u_s(p_0) \propto \begin{pmatrix} \xi \\ \xi \end{pmatrix}$$ while negative-frequency solutions, interpreted as positrons, have polarization $$v_s(p_0) \propto \begin{pmatrix} \xi \\ -\xi \end{pmatrix}.$$ This makes sense: a spin up positive-frequency electron, for example, is a superposition of the left-chirality spin up and right-chirality spin up components, because the mass term mixes up the chiralities.

I'm confused about what a state of definite chirality is. A spinor at rest with definite left chirality would look like $$\begin{pmatrix} \xi \\ 0 \end{pmatrix}$$ which is a superposition of the positive-frequency and negative-frequency solutions! This seems to imply that there's no such thing as a right-chiral electron; you have to have a superposition of electron and positron. This has to be wrong, because people talk about left chirality electrons all the time; for example, they are produced in beta decay. Where did I mess up?


1 Answer 1


Another one of these issues where careless terminology is our downfall.

Indeed, on the level of the Dirac equation, such a thing as a "left-handed electron" does not exist. Every pure electron and every pure positron state is an equal mixture of left- and right-handed components. I'll call this electron (positive-frequency solution electron) the "mass basis electron".

However, when one decomposes a Dirac spinor into its irreducible components, the Weyl spinors, one gets something different. The Standard Model is a chiral theory, meaning the Weyl components of the Dirac spinor transform in different representations of the weak force, meaning only e.g. the left-handed component of the mass basis electron actually couples to the W bosons.

On this level, i.e. the level of the SM Lagrangian, it is more natural to associate particles to the individual chiral components of the Dirac spinor field. We get a left-handed "electron-1", a right-handed "electron-2", and their anti-particles, a right-handed "anti-electron-1" and a left-handed "anti-electron-2". The electron-1 and the anti-electron-1 interact with W bosons, the electron-2 and anti-electron-2 do not.

Now, if the Dirac field were massless, these parts would not talk to each other. The electron-1 would stay a electron-1 forever, and there wouldn't be any reason to consider electron-1 and electron-2 as particle and anti-particle of the same Dirac field - they would be just two Weyl fields, each with their own antiparticle of the opposite chirality.

However, the electron Dirac field is not massless, meaning the two chiral components do not decouple in their dynamics - any electron-1 will invariably evolve into an oscillatory mixture of a electron-1 and an electron-2, while the electron-2 will evolve into a mixture of a electron-2 and an electron-1. If we now seek the stationary solutions that do not oscillate, they are exactly the positive and negative frequency solutions of the Dirac equation - the mass basis electron is a mixture of the electron-1 and the electron-2, and likewise for the mass basis positron.

Still, although the mass basis particles are now mixtures of chiral components, this has not changed their interaction with the W boson. It is still only the electron-1 components that interact with W bosons, the electron-2 components do not. Hence, only the left-handed part of a mass basis electron (what we usually call "electron") and the right-handed part of a mass basis positron participate in the weak interaction.

  • $\begingroup$ Thanks for the answer! My remaining confusion is how to relate the chiral and mass bases in the opposite direction. Is it true that the electron-1 is a combination of a mass electron and mass positron? If so, my real confusion is that which particles are 'antimatter' seems to be basis-dependent. $\endgroup$
    – knzhou
    Apr 1, 2016 at 16:39
  • $\begingroup$ @knzhou: Yes, the electron-1 is a linear combination of the mass basis electron and position (and likewise the electron-2). Why does it worry you that "antimatter" is basis dependent? Being "antimatter" is not an observable. $\endgroup$
    – ACuriousMind
    Apr 1, 2016 at 16:45
  • $\begingroup$ Well, pair annihilation is observable! For example, if I have an electron-1 emitted in a weak decay, and it hits a mass electron in my detector, this suggests that there is some amplitude for annihilation to occur. But this doesn't appear to happen. $\endgroup$
    – knzhou
    Apr 1, 2016 at 16:49
  • $\begingroup$ @knzhou: Oh, wait, I said the wrong thing. No, the electron-1 is not a linear combination of electron and positron (it can't be because the positron has the wrong charge). The electron is electron-1+electron-2 and the positron is anti-electron-1+anti-electron-2. You don't get back the electron-i s by combining electron and positron. If you just add electron and positron you get something like electron-1+anti-electron-2, which has definite chirality, but is still a mixture of the two Weyl spinors. $\endgroup$
    – ACuriousMind
    Apr 1, 2016 at 17:14
  • $\begingroup$ I agree with all that, but then, what is the electron-1 in mass basis? By your charge argument, the electron-1 and electron-2 must be built from the two mass-electron states. So it is like, electron-1 = mass-electron-spin-up + mass-electron-spin-down? (That sort of seems to make sense by analogy with how, in light, linear polarizations combine to make circular polarizations.) $\endgroup$
    – knzhou
    Apr 1, 2016 at 17:25

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