9
$\begingroup$

Can't the electron have a mass before electroweak symmetry breaking, which is then just modified by its interaction with the Higgs field, instead of gaining all its mass this way?

Apparently other mechanisms for gaining mass are known or postulated, e.g. for the neutrinos, or the Higgs boson itself, which (as far as I read) has a mass even above the electroweak scale.

Is there a fundamental reason why electrons need to be massless before Higgs?

By the way, I am a layperson, please be gentle with terminology.

$\endgroup$
8
  • $\begingroup$ Yes, there is a very fundamental, but tecnical reason: chiral electroweak symmetry. Highly mathematical,, but covered in all texts. You want a story supplanting the math? $\endgroup$ Commented Mar 9, 2022 at 11:45
  • 4
    $\begingroup$ @Cosmas I am amathematician, but not a physicist. I am happy with math, but have trouble with a lot of physics terminology. I know a bit about the Lagrangian and field theory, so just go ahead :) I try to follow. $\endgroup$
    – M. Winter
    Commented Mar 9, 2022 at 11:48
  • 1
    $\begingroup$ @CosmasZachos I should add, if there is a very fundamental reason, then why is the mass of the gauge bosons considered the main reason for the introduction of the Higgs field. Shouldn't the problem with the lepton masses have caused trouble long before weak forces entered the picture? $\endgroup$
    – M. Winter
    Commented Mar 9, 2022 at 13:26
  • 1
    $\begingroup$ This question is complementary. Ideally, someone should post an answer here showing how Higgs protects fermions' chiral symmetry. $\endgroup$
    – J.G.
    Commented Mar 9, 2022 at 13:51
  • 1
    $\begingroup$ @CosmasZachos Thanks for the clarification. The science reporting wasn't the only problem. I just worked through a complete set of introductory lecture notes on QFT, and masses of fermions where never mentioned as a problem (or I missed that part). But also reading through a lot of threads on SE, it was rarely made explicit that fermion masses are equally troublesome. $\endgroup$
    – M. Winter
    Commented Mar 9, 2022 at 13:51

2 Answers 2

12
$\begingroup$

I have several answers on this site detailing the issue, but it is too hard for me to collect them all. Your source might be this mini-crib-sheet.

In a nutshell, as any half-decent QFT course should hammer in, again, and again, is that kinetic and gauge-boson coupling terms preserve chirality, but mass and Yukawa couplings to scalars do not. This is a generic property of the Lorentz group. Mathematically, kinetic and gauge couplings do not couple $\psi_L$ to $\psi_R$, but mass terms and Yukawa terms do. So mass terms must be there, even though, before 1968, physicists knew quite firmly that the weak interactions they had seen so far only involved $\psi_L$s, and not $\psi_R$s (the Feynman--Gell-Mann theory).

This was a major snag, since the fermion mass terms would prevent a correct theory from being fully (su(2)-) chirally invariant (unchanged under a chiral transformation that leaves $\psi_R$s alone while transforming only $\psi_L$s), and, for technical reasons, would make such a gauge theory inconsistent. Lack of gauge theory would then mean ferocious mixings of energy scales upon renormalization, and complete computational failure of such a bad theory.

  • So, the major existential snag for such theories was the non-invariance of all mass terms, such as $m\overline{\psi_R}\psi_L$. This is your fundamental reason.

In 1968, Weinberg (and Salam) broke the logjam. They utilized the fact that a hypothetical scalar, the Higgs field, which could give gauge bosons an effective mass upon SSB, could also solve the above problem. In more detail, among other complicating factors, if the gauge group contained an su(2) acting only on left fermions and a complex Higgs doublet field, then terms such as $$ g_Y ~\overline{\psi_L}\cdot \phi ~~e_R + \hbox{h.c.} ,\\ \psi_L\equiv \begin{pmatrix} \nu_L\\ e_L \end{pmatrix}, ~~~~~~ \phi \equiv \begin{pmatrix} \phi^+\\ \phi^0 \end{pmatrix} $$ would be invariant under Left su(2), as the leading dot product is an invariant.

Finally, upon SSB, preserving the symmetry but changing its realization, would allow shifting $\phi^0$ by a constant $v\equiv \langle \phi^0\rangle$, and thus inducing a mass term for the electron with $m=g_Y v$.

So mass terms are compatible with chiral Weak Interactions, after all, and save the existence of the gauge theory. They are the linchpin. They have absolutely nothing to do with the Higgs mechanism (only involved in the gauge bosons getting a mass) and are a consequence of just SSB. The Higgs particle's mass also gets its mass in that process (SSB), but in a very different way.

I have been cavalier with normalization factors, weak hypercharge, and the full group, as opposed to the Lie algebra, to steer away from logical distractions. If you want further technical details, I could add a mini appendix.

Actually, the idea was not quite Weinberg's, as it had been triumphantly introduced by Gell-Mann and Levy in 1964 (with an important nudge by Feynman) in their celebrated 1964 σ-model paper, finally explaining nucleon masses in strong-interactions' chiral symmetry breaking. (If your teacher failed to introduce this before the SM, that's the root or your problem right there!) Both Weinberg and Salam were σ-model experts, so this ingredient was not as exotic to them as the then hypothetical Higgs mechanism...

$\endgroup$
1
  • 2
    $\begingroup$ This is a really nice answer. $\endgroup$ Commented Mar 15, 2022 at 20:56
2
$\begingroup$

A mass term is a coupling between left- and right-chiral fields. The fields have to have matching gauge charges to preserve the gauge symmetry.

There are no mass terms for fermions in the Standard Model because there are no matching fields for them to couple. The theory is fundamentally asymmetric. The only exception is the Majorana coupling in the Standard Model extended with a sterile neutrino.

It's unknown why the asymmetry exists, but if any mass terms were allowed, there is no known reason why the masses wouldn't be very large (close to the Planck mass), so a plausible reason why we don't see paired fermion fields is that they all have masses beyond the reach of current experiments. In the case of the Majorana coupling, the large mass is a point in favor of the model because it leads to tiny observed neutrino masses by the seesaw mechanism.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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