# How can an eL become an eR by interacting with the Higgs?

1. Nov 5, 2015

### kimcj

https://www.physicsforums.com/threa...f-fermions-in-interactions-with-higgs.686900/

in the yukawa interaction of the higgs vev (h0), the eL absorbs the higgs boson and becomes a eR which doesnt make sense to me since it doesnt conserve angular momentum(since helicity(real angularmomentum)=chirality in massless limit(since im refering to non physical spinors)).... i need a very slightly technical solution.

+
it would be nice to refer to http://www.quantumdiaries.org/2011/06/19/helicity-chirality-mass-and-the-higgs/
im the above article its somehow hard to understand becauses in some part the ''anti-positron(eR)'' becomes R handed while somewhere its L chiral.

2. Nov 5, 2015

Staff Emeritus
This non-standard terminology "anti-positron" is getting in the way of your understanding. You need to abandon it and use standard terminology if you want to make progress.

3. Nov 5, 2015

### kimcj

okay i should.. then could you explain it with standard terminology??

4. Nov 5, 2015

Staff Emeritus
Since the "anti-positron that is not an electron" is not even a thing in mainstream physics, one cannot explain any of it's properties using mainstream physics. If someone were to ask you how many sides a square circle has, could you answer?

5. Nov 6, 2015

### kimcj

ohh.... then back to the main question how does el become er??

6. Nov 7, 2015

### vanhees71

The Yukawa couplings are $\propto \bar{\psi} \psi \phi$, but $\bar{\psi} \psi=\bar{\psi}_R \psi_L + \bar{\psi}_L \psi_R$, i.e., this term's mixing the left- and right-handed (helicity) components of the fermions. Since the $\phi$ as well as the $\psi_L$'s are wiso-doublets and the $\psi_R$'s are wiso-singulets, this does not break the chiral gauge symmetry but through the Higgs mechanism, which means that $\phi$ has a non-vanishing vacuum expecation value, this makes the leptons and quarks massive. The trick is to have massive particles in a model with strict chiral gauge symmetry. The same mechanism also makes the W and Z boson massive (by "absorbing" the would-be Goldstone modes as the longitudinal component of massive vector fields, without destroying gauge symmetry either).

For a more detailed summary look at my old talks on the subject:

http://fias.uni-frankfurt.de/~hees/publ/ew-talk.pdf
http://fias.uni-frankfurt.de/~hees/publ/neutrinos.pdf

7. Nov 12, 2015

While the terminology that kimcj is referring to is non-standard, it is a terminology that Flip used in his blog post in an attempt to convey they way that hep-theorists actually view massive particles.

If I may, I will say things in a slightly different way that should be complementary to what was written above. Imagine an alternative universe in which the Higgs happens not to couple to the electron at all. In that hypothetical universe, the electron is massless. In fact, there are four distinct particles in that universe that are electron-like. There is (a) a left-chiral particle with charge -1, and (b) its antiparticle which is a right-chiral particle with charge +1. There is also (c) a left-chiral particle with charge +1, and (d) its antiparticle which is right-chiral and charge -1. In Flip's terminology, (a) is called the electron, (b) the anti-electron, (c) the positron, and (d) the anti-positron. Vanadium might be right and this terminology is not helpful, so I won't use it here: I'll just call the particles (a), (b), (c), (d). They are distinct particles because they all have different properties, and one will never turn into the other. E.g. (a) and (b) couple to the W boson while (c) and (d) don't.

In the real world, the Higgs couples to the electron. The consequence of this is that the massless particles (a) and (d) of the alternate universe are seen as one massive particle in our universe, which we call the electron, and the massless particles (b) and (c) of the alternate universe are seen as a single massive positron in our universe. Now in principle, I can still measure the chirality of a particle in our universe (technically, the chirality operator is hermitian and so it corresponds to a potential observable in quantum mechanics). In practice, I can't imagine any apparatus that could do such a thing, but for the sake of argument imagine that I have built a chirality-measuring device (the point is that while I don't know how to build one, there is nothing about it that would violate the rules of quantum mechanics). I put an electron through my machine, and I measure it is left-chiral, say. Some time later I measure the same particle again, and it might turn out to be right-chiral this time. This is possible only because of the particle's interaction with the Higgs, which allows a left-chiral particle to turn into a right-chiral particle. In the alternate universe, particle (a) will always be left-chiral no matter how long I wait.

The key point is that chirality is not helicity. My massive electron that I discussed before did not change it's helicity (which is the direction of spin with respect to the direction of motion) if I did not disturb it. This is the thing that in your initial post were concerned that it should be conserved because of angular momentum. The chirality of a massive particle is something more subtle, and is not conserved. Helicity and chirality are only the same thing for a truly massless particle, like particles (a) to (d) in my alternate universe.

8. Nov 13, 2015

Sorry, I was thinking about this on the walk home, and I realized that the following part of my post was wrong, and I retract it:

9. Nov 13, 2015

### kimcj

1.
so you mean that helicity isnt chirality that im concerned of. but doesnt it leave the problem the same? a leftchiral particle absorbs/emits(intracts with) the higgs field and changes into a right chiral one... (so heres the rweal question does the phrase 'helicity=chirality in massless limit' mean that the (1)(left chiral particle can have only a specific helicity) or (2)(leftchiral is leftchiral in massless limit). )

2.
also in the yukawa interaction term with the vev and fermion, does the barred psi L actually mean some psi with R chirality?
3.
also is the phrase 'w interacts with only Ls' wrong? i know that the barred psi L and psi L are both weak isospin charged.......
4.
my overall thought is that (1) "the L s and R s on the yukawa term actually defined wether its a w interacting one or not. so the particular interaction with the higgs is the only one able in order to conserve weak charge(and hypercharge) . the physical chirality for a daggered L is actually a R which makes the interaction between daggered L and r be chirality conservable." (2)"the interaction isnt really important since they arennt physical. the only thing that matters is whether th massive combination conserves angular momentum(which is helicity that is conserved)"(3)"its too mathmtical that you cannot understand"(4)"no actually..."

tyhanks

10. Nov 13, 2015

### vanhees71

Indeed, for massive particles you have to carefully distinguish chirality from helicity, but that's not the point.

Perhaps it's easier to understand the answer to the question, when thinking a bit more formally. As written above, condensed to a minimum, the Higgs-fermion interaction is a Yukawa coupling $\phi \bar{\psi} \psi=\phi(\bar{\psi}_L \psi_R + \bar{\psi}_R \psi_L)$. Perturbatively you write this in terms of Feynman diagrams. So you have a three-point vertex with a Higgs-boson line, and two fermion lines with one "charge arrow" running in and one running out (i.e., you have either an incoming and an outgoing particle/antiparticle or an incoming or outgoing particle and antiparticle). This is the interpretation of the left-hand side of the above equation. On the other hand you can also look at it in terms of the right-hand side. This shows you that at the vertex you always have left-handed and a right-handed component coupled to the Higgs, i.e., when interacting with a Higgs a left-handed incoming particle can come out as a right-handed particle after the scattering process and vice versa.

Now in the electroweak standard model you have a non-zero vacuum-expectation value of the Higgs field, which means you have a component $\phi_0 \bar{\psi} \psi$ with $\phi_0=\text{const}$. That's a mass term, and looking on it as a perturbation to massless fermions you have a "mixing" between left- and righthanded components of the Dirac fermions.

This is the way one usually looks at neutrinos and neutrino masses. The reason is that the neutrinos are always created as left-handed fermions and in a specific flavor due to the electroweak couplings in the standard model, but on the other hand they have a mass, which leads to a mixing with right-handed components, and the flavor eigenstates are not the same as the mass eigenstates, which leads to the famous "neutrino oscillations", i.e., free neutrinos come as a mixture of flavor eigenstates after going from the point of creation to another distant point, where they are detected. You can again only detect flavor eigenstates, because the detector also responds due to the weak interactions of the neutrinos with the detector material. In this way you can measure the amount of neutrino flavors coming from neutrinos originally created in a single-flavor eigenstate. This is the only true hint we have so far for "physics beyond the standard model".