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Weyl spinors are massless.

Is the converse also true? Does any massless spin-1/2 fermion have to be a two-component Weyl spinor?

In the Standard model, before symmetry breaking, the electron (for example) is not massless. But we still denote it by a Dirac spinor $\Big($either by its left-handed projection $e_L\equiv\frac{1}{2}(\mathbb{1}-\gamma^5)e$ or right-handed projection $e_R\equiv\frac{1}{2}(\mathbb{1}+\gamma^5)e\Big)$.

Is there a reason for not using two-component Weyl spinors for the electron when it is massless?

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  • $\begingroup$ I'm voting to close this question as off-topic because it shows insufficient prior research. $\endgroup$
    – AccidentalFourierTransform
    Commented May 25, 2018 at 15:40
  • $\begingroup$ Depending on the conventions, Weyl spinors do not have to massless. A term $m\psi\cdot \psi=m\psi^i\varepsilon_{ij}\psi^j$ is invariant. Some like to talk in this context of Majorana spinors, but I find this is confusing. $\endgroup$
    – user178876
    Commented May 25, 2018 at 16:54
  • $\begingroup$ @marmot But Majorana spinors aren't two-component. $\endgroup$
    – SRS
    Commented May 25, 2018 at 17:37
  • $\begingroup$ Well, a Majorana spinor (in the Weyl basis) is $\Psi=(\xi,\bar\xi)$ where $\xi$ is a Weyl spinor. So physically it is precisely a Weyl spinor. (In other dimensions things are slightly different, sometimes you can impose a Majorana condition, sometimes a Weyl condition, sometimes both and sometimes none, that's why it makes sense to distinguish these notions. At any rate, above I wrote down a mass for a Weyl spinor. Of course, the full Lagrangean will contain the Hermitean conjugate, which is precisely $m\overline{\Psi}\Psi^C$. $\endgroup$
    – user178876
    Commented May 25, 2018 at 17:42

1 Answer 1

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A massless spin-1/2 particle can be represented by 2-component Weyl spinors. This can be seen by expressing the Dirac equation with $m=0$ in the Weyl basis. But massless a spin-1/2 particle need not be represented by 2-component Weyl spinors. To see that we can write out the Dirac equation (again for $m=0$) in the Durac-Pauli representation. The solutions are now 4-component Dirac spinors.

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  • $\begingroup$ I think this answer conflates particle and fields, which is rather dangerous. Perhaps a more careful way to phrase this is: what kinds of classical fields, upon quantization, yield 2 massless particles, with helicities $\pm 1/2$? The answer is a Weyl field, or a Dirac field satisfying a reality condition, or many many other possibilities (e.g. 20 copies of the Weyl field stacked on top of each other in a 40-component object). $\endgroup$
    – knzhou
    Commented May 26, 2018 at 13:14
  • $\begingroup$ @knzhou The answer is based on relativistic quantum mechanics. My aim is to convince that a massless spin-1/2 fermion need not necessarily be represented by 2-component objects. $\endgroup$
    – SRS
    Commented May 26, 2018 at 13:17
  • $\begingroup$ That is fair, but I think relativistic QM is inherently dangerous for exactly the reasons I stated. Much of the material of it comes from a time when particle and fields are confused. Even the best sources on relativistic QM I have seen are self-contradictory and only serve to muddle the picture when QFT is introduced. $\endgroup$
    – knzhou
    Commented May 26, 2018 at 13:37
  • $\begingroup$ In any case, if your mathematical question is "can an $x$-component object be represented as a $y$-component object for $y > x$", the answer is clearly yes. A 2-component object, for instance, may be represented by a 200000 component object with the constraints $\psi_1 = \psi_3 = \psi_5 = \ldots$ and $\psi_2 = \psi_4 = \ldots$. $\endgroup$
    – knzhou
    Commented May 26, 2018 at 13:38

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