Changing the chirality of fermions in interactions with Higgs

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1. Apr 20, 2013

Thoros

Am i correct when i say that the fermions get a mass and interaction term with the Higgs from the $SU(2)_{L}\times U(1)_{Y}$ invariant Yukawa interaction
$$-g_{y}\bar\psi_{L}\phi\psi_{R} - g_{y}\bar\psi_{R}\bar\phi\psi_{L}$$
where $\psi$ is the fermion field and $\phi$ the Higgs field.

My questions are:
The Higgs field mixes left and right handed particles into massive Dirac particles?
Does the Higgs particle in interactions change the chirality of particles it interacts with?

If these are correct, how does the Higgs particle, being spin 0, change the chirality of fermions it interacts with?

2. Apr 20, 2013

The_Duck

Yes, the fermions get mass because the left-handed fermions can turn into right handed fermions (and vice versa) by interacting with the Higgs field. A massive Dirac particle is exactly one for which the left-handed component has some amplitude to turn into the right-handed component (and vice versa) as the fermion propagates.

I'm not sure how to take your last question. The structure of the interaction tells you that it turns left-handed fermions and right-handed fermions into each other, and there's not much more to say. Note that this is the only possible interaction you can write down between the SM Higgs and the SM fermions that is both renormalizable and preserves Lorentz symmetry and all the internal symmetries.

3. Apr 21, 2013

kurros

I expect this is a chirality vs helicity confusion. I guess you are thinking that interacting with the Higgs flips the helicity of the fermion because it flips the chirality. This can't happen, as you suspect, because the Higgs can't carry away the required angular momentum. However, helicity and chirality are different, so this isn't quite how the scenario goes.

Unfortunately I have not quite remembered how it does go. I spent a little time playing with my gamma matrices but didn't quite figure it out. Maybe someone else can enlighten us. Part of my confusion has to do with the fact that we should be in the massless limit in this case, so helicity and chirality eigenstates should be the same... unless Yukawa interactions wreck that?

4. Apr 22, 2013

Thoros

Thank you, Kurros, for the excellent reply.

You seem to be correct, the indices L and R, identifying chirality, do not correspond to real handedness. Handedness is given by helicity. In the massless limit the two become identical.

I quess the way to understand this is that in some way, L and R chirality are to weak charge what +/- and neutral are to electric charge. The weak charge seems to interact with L particles and not with R ones.

The point i quess, is that under chiral operators angular momentum is conserved because helicity must be conserved.

I think i'm a little closer to understanding the principles, but it's still a little hard for me to grasp it's fundamental value. So the term $-m\bar\psi_{R}\psi_{L}$, that emerges from the Yukawa interaction with a non-zero vev field, destroys a L particle and creates the correspodning R particle. So the R and L here relate merely to weak charge, even while the helicity could be switched for one of them, that is a L chirality particle being R helicity?

5. Apr 22, 2013

kurros

Err, yes I think that must be the case, but I still haven't gotten around to actually doing the math to check, and I couldn't find anything on the internet where someone had already explained it nicely. This article explains it all heuristically, http://www.quantumdiaries.org/2011/06/19/helicity-chirality-mass-and-the-higgs/, and certainly seems to agree with this proposition (it briefly mentions some states having opposite helicity and chirality at one point), but again I have not proved to myself that yes this is the case. Intuitively it is conflicting with my notion that the helicity and chirality eigenstates are the same in the massless limit, but perhaps this statement is more literally true than I realised, and that the eigenstates are the same but not the eigenvalues. But we will have to calculate them and check. Otherwise there is some other thing wrong with this picture...

Last edited: Apr 22, 2013
6. Apr 23, 2013

kurros

Ahh, ok so I did some math and think I figured out the problem. Basically, I was forgetting that the Hermitian conjugate of a left-hand Weyl spinor is a right-hand Weyl spinor, so in your expression for the Yukawa interaction the 'barred' field actually has the opposite chirality to what it is labelled as (since the label refers to the chirality of the 'unbarred' field), i.e. $\overline\psi_R$ is actually left chiral. So actually the Yukawa interactions flip neither chirality nor helicity. Which is very confusing given that it DOES mix what are the left and right hand components of the corresponding Dirac spinor.

The language and notation makes this whole thing pretty confusing, so let me borrow the story in the quantumdiaries link above and translate it into some math. Also I will use 2-component Weyl notation as in Steve Martin's SUSY primer (arxiv number 9709356, section 2 "Notation").

To summarise the notation:

$\Psi_D = \left(\begin{array}{cc} \xi_\alpha \\ \chi^{\dagger\dot{\alpha}} \end{array} \right)$ - 4-component Dirac spinor in Weyl basis
(this is equal to $\left(\begin{array}{cc} \psi_{L} \\ \psi_{R} \end{array} \right)$ in your notation. In my notation the two component spinors $\xi_\alpha$ and $\chi_\alpha$ are both left-chiral, but their "daggered" (Hermitian conjugated) versions are right-chiral. The indices are also important: undotted indices indicate left-chiral (transforming under left "half" of Lorentz group), dotted indicates right-chiral. Moving indices up and down "transposes" the matrix, if you are imagining the spinors in matrix notation, but we also have to be careful because there are antisymmetric tensors doing the lowering and raising, e.g. $\chi_\alpha=\epsilon_{\alpha\beta}\chi^\beta$, so minus signs can appear when you fiddle with the indices. In general the matrix notation kind of sucks for getting this right, so it is best to stick with the indices. Anyway I say this so you don't get concerned about the $\chi^\dagger$ in the bottom half of $\Psi_D$, it is ok in this notation since the index is 'up'. Note that the hermitian conjugate changes dotted indices to undotted ones and vice versa (which reflects that it changes the chirality of the spinor).

Also note that

$\overline{\Psi}_D = \Psi_D^\dagger \gamma_0 = \left(\begin{array}{cc} \xi^\dagger_\dot{\alpha} & \chi^\alpha \end{array} \right) \left(\begin{array}{cc} 0 & I \\ I & 0 \end{array} \right) = \left(\begin{array}{cc} \chi^\alpha & \xi^\dagger_\dot{\alpha} \end{array} \right)$

Now I spent some time trying to figure out how to prove this claim that the Hermitian conjugate changes the chirality of the spinor (i.e. changes which "half" of the Lorentz group it transforms under) but I didn't succeed. For now we shall just have to believe Martin's claim that this is the case. It is built into the dotted index notation anyhow, so it must be true :p.

So anyway, in this notation our Yukawa interaction looks like this:

$g\phi\overline{\Psi}_D\Psi_D = g\phi\left(\begin{array}{cc} \chi^\alpha & \xi^\dagger_\dot{\alpha} \end{array} \right) \left(\begin{array}{ccc} \xi_\alpha \\ \chi^{\dagger\dot{\alpha}} \end{array} \right) = g\phi\chi^\alpha\xi_\alpha + g\phi\xi^\dagger_\dot{\alpha}\chi^{\dagger\dot{\alpha}}$

So it is more clear in this notation that we get two "mass" terms out of this when the scalar field adopts a VEV; each of them "mixes" the two two-component spinors $\chi$ and $\xi$, which, in the Standard Model, one of which would interact with the weak force while the other would not, but each of the terms doesn't change the "actual" handedness of the fields involved.

The story in the Quantum Diaries blog post I linked above is a nice way of understanding this. Let us map their slightly non-standard names to our two-component fields:

Non-"mustache" fields:
$\xi_\alpha$- electron (left handed, interacts with weak force)
$\xi^{\dagger}_{\dot{\alpha}}$ - anti-electron (right handed, still interacts with weak force)

"mustache" fields
$\chi^\alpha$ - anti-positron (left handed, does NOT interact with weak force)
$\chi^{\dagger \dot{\alpha}}$ - positron (right handed, does NOT interact with weak force)

So what our two mass terms do is 1. "mix" the 'electron' and 'anti-positron' fields, and 2. "mix" the 'positron' and 'anti-electron' field.

But what gives? It looks like chirality and helicity have nothing to do with each other here, contrary to this notion that "they are the 'same' in the massless limit". After all, the spinor indices going over each component of the Weyl fields are, if be choose the basis right, just labelling the + and - spin components of the field, and it looks like we sum over those.

It is a bit more work to prove that LH helicity fields cannot be RH chiral (when +/- signs are defined appropriately) and vice versa in the massless limit. Fortunately I found this webpage where someone does it: http://courses.washington.edu/phys55x/Physics%20557_lec9_App.htm [Broken]. The appropriate section is towards the bottom.

Anyway, so in the massless limit yes, the chirality and helicity operators have the same action on our fields, so no, some fields do not have right chirality but left helicity (in this limit!). *However* due to the way we *label* fields, some of them are *labelled* right chiral when actually they are left chiral. But they are labelled (say) right chiral because their "non-conjugated" field is right chiral, and this is what determines whether they interact with the weak force or not, NOT what their chirality *actually* is. So it is arguably the more important property to be keeping track of, while "physical" chirality/helicity stuff all just works itself out due to the structure of field theory.

P.S. someone please correct me if any of this is wrong, because I want to make sure I understand it once and for all!

Last edited by a moderator: May 6, 2017