1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I'm voting to close this question as off-topic because it shows insufficient prior research. $\endgroup$ – AccidentalFourierTransform May 25 '18 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 May 25 '18 at 16:54
  • $\begingroup$ @marmot But Majorana spinors aren't two-component. $\endgroup$ – SRS May 25 '18 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 May 25 '18 at 17:42
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 May 26 '18 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 May 26 '18 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 May 26 '18 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 May 26 '18 at 13:38

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.