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Higgs field mass of fermions

by spookyfish
Tags: fermions, field, higgs, mass
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spookyfish
#1
May3-14, 10:19 PM
P: 52
It is said that the Higgs field is responsible for the mass of all particles. But the Higgs mechanism provides mass to the gauge bosons, and as far as I know the mass of the fermions is put "by hand" into the Lagrangian. Why, then, is the Higgs field responsible for the mass of the fermions?
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Bill_K
#2
May4-14, 05:39 AM
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Quote Quote by spookyfish View Post
It is said that the Higgs field is responsible for the mass of all particles. But the Higgs mechanism provides mass to the gauge bosons, and as far as I know the mass of the fermions is put "by hand" into the Lagrangian. Why, then, is the Higgs field responsible for the mass of the fermions?
Yes, you'll often see the offhanded statement that the Higgs field "gives" mass to the fermions. A better way of putting it is that the Higgs field permits the fermions to have mass. The masses of the gauge bosons are

MW = vg
MZ = v√(g2 + g'2)

where v=246 GeV is the vacuum expectation value of the Higgs field, and g and g' are the weak coupling constants. The masses are proportional to v, but not determined by it. Likewise, the mass of a fermion is conventionally

Mf = v Gf/√2

where Gf is a coupling constant. In all three cases, the Higgs field is responsible for the mass only in the sense that it permits the mass to be nonzero.
ChrisVer
#3
May4-14, 10:59 AM
P: 871
Well let's take everything in an order...
You make a model for the elementary particles (the standard model) which is a yang mills gauge theory, the renowned SU(3)xSU(2)xU(1). In that case you build an invariant under those gauge groups Lagrangian.
In that model, the (12) gauge bosons, existing in the adjoint representation of those groups, are massless. By elementary particles though, you know that the bosons responsible for the weak interactions are massive. It happens that the Higgs mechanism (introducing a new higgs field) can break the symmetry group:
SU(3)xSU(2)xU(1)→ SU(3)xU'(1)
If that's the breaking take place, some of the gauge bosons acquire mass by interacting with that new field's non trivial vacuum expectation value.
That's a brief explanation of what you stated already.

One could add by hand the masses for the fermions... However the result will not be an invariant Lagrangian under those symmetry groups. For example the mass term of a fermion is given by:
[itex]L_{m}= m \bar{\psi_{L}} \psi_{R} [/itex]
terms in the Lagrangian. L and R means Left and Right components of the Dirac spinor (Weyl Spinors). The Left and Right components though, belong to different representations of SU(2), and so such a mass-term would violate the initial symmetry group (it would explicitly break SU(2) invariance). The left is in the 2dimensional representation while the Right is in the 1dimensional, meaning that the right's SU(2) transformation is like a scalar (doesn't transform) while the left is taking a transformation with exp(iajTj). So what would happen to your mass term under an SU(2)?
[itex]L_{m}'= m e^{-iT_{i}a^{i}(x)}\bar{\psi_{L}} \psi_{R} ≠ L_{m}[/itex], [itex]i=1,2,3[/itex]
T denotes the generators of SU(2) -can be pauli matrices.
All in all, Higgs mechanism spontaneously breaks the SU(2) symmetry, while adding a mass by hand breaks it explicitly.

You need to make the distinction of left/right handed fermions, because you know that the neutrinos appear with certain helicity (you don't have both left and right handed neutrinos, something which implies also the parity violation of the weak interactions). In that case, you have:
[itex]L=(v_{eL},e_{L})^{Τ}[/itex] ([itex]T[/itex] is transpose because I don't know how to write columns by Latex hehehe excuse me) and [itex]R=e_{R}[/itex] (e means the fermion spinor, take it as just an electron for now).
These could make that distinction since you only have left-neutrinos.... the subscripts L, and R means again the components of the Dirac spinor: [itex] e_{R/L}=P_{R/L}e_{dirac}[/itex] with [itex]P_{R/L}= \frac{(1\pm \gamma_{5})}{2}[/itex] the projector operators.

But I wrote in bold above something important. You can make an invariant Lagrangian under the Standard Model gauge group.
After inserting the Higgs field, you also have the freedom to add invariant terms which AFTER higgs acquires a vacuum expectation value, they would lead to mass terms !
Those terms are generally called Yukawa terms in the lagrangian, they are invariant since the Higgs field is also in a doublet of SU(2) so it could kill the transformation caused by the Left component. They mix the Left and Right fermions with the Higgs field, and so when the last gets a vev (let's call it [itex]v[/itex]) then you'll get a form resembling the usual mass term you knew:
[itex] G_{e} v \bar{L} R [/itex]
by comparison with my first formula, you get [itex]m=G_{e} v[/itex]
That's also the mass Bill_K supplied you with, with the only difference that in context we get the vev to be [itex]\frac{v}{\sqrt{2}}[/itex] instead of [itex]v[/itex] I took it here to be for convenient typing.

So in fact we could say that the Higgs field is indirectly responsible for the fermion masses... It just allows you to have consistent extra terms in your initial Lagrangian, which would give you the mass terms result...

For more fermions than just electrons, you just need to put some extra indicies denoting them, the constructions are just identical. For quarks things are a bit different, because you have 1 left doublet and 2 right singlets of SU(2) and you have to mix them somehow with the Higgs field , but the idea is almost the same (it's only the calculations which can be more tedious).

If you feel uneasy with anything I stated feel free to ask again

The_Duck
#4
May4-14, 12:04 PM
P: 860
Higgs field mass of fermions

Quote Quote by spookyfish View Post
It is said that the Higgs field is responsible for the mass of all particles. But the Higgs mechanism provides mass to the gauge bosons, and as far as I know the mass of the fermions is put "by hand" into the Lagrangian. Why, then, is the Higgs field responsible for the mass of the fermions?
The fermion mass arises from a coupling between the fermion field and the Higgs field. The mass of an electron, say, is really the energy of its interaction with the Higgs field.
spookyfish
#5
May5-14, 10:12 AM
P: 52
Thanks for the responses. ChrisVer - thank you for the detailed explanation. I want to make sure I understand - Is the idea that the SU(2) symmetry should hold before the Higgs breaks it, so that if we couple the Higgs to the left-handed spinors, then without the Higgs the symmetry holds (since we set the Higgs doublet to zero) and after the symmetry breaking, it gives us the mass term in the form that we anticipated?
ChrisVer
#6
May5-14, 01:35 PM
P: 871
I am not sure I understand your question.

The main thing as I stated above is that you can either break SU(2) explicitly, eg by adding a mass term to your lagrangian, or allow Higgs mechanism to break it spontaneously (after Higgs field acquires a non trivial vev). So yes, you must have the SU(2) in the game to describe weak interactions (after it's broken).
After the Higgs acquires a vev, the symmetry is broken.
[itex] SU(2)_{L} \times U(1)_{Y} \rightarrow U(1)_{Q} [/itex]
By that the 3 massless gauge bosons get a mass (Ws,Z), and you still get one more massless boson that describes the photon.

In order to see how the mass terms for the fermions come after the SSB, you can write down the Yukawa term excplicitly for an electron let's say. In that case you have:
Before the symmetry is broken, you have the Yukawa term for the electron:
[itex] G_{e} \bar{L} H R[/itex]
I already gave the L and R in the previous post. Then let's say that the Higgs field gets a non trivial vev, you can use the unitary gauge to write the Higgs doublet after that as:
[itex] H= (0, \frac{u}{\sqrt{2}})^{T}[/itex]
Insert it in the Yukawa term lagrangian and you get:

[itex] G_{e} (\bar{v_{L}}, \bar{e_{L}}) (0, \frac{u}{\sqrt{2}})^{T} e_{R}[/itex]
doing the multiplication, you get:
[itex] G_{e} \frac{u}{\sqrt{2}} \bar{e_{L}} e_{R} [/itex]
Then compare it with the term you'd have if you instead had used a mass term from the beggining, you immediately see that the mass is the:
[itex] m= G_{e} \frac{u}{\sqrt{2}}[/itex]

Now before Higgs gets a vev, then the lagrangian is SU(2)xU(1) invariant... (I forget SU(3) for now since both the leptons and the higgs belong to SU(3) singlet).
Why is that for the Yukawa term? because each field transforms under SU(2) as:
[itex]\bar{L} \rightarrow \bar{L}'= \bar{L}e^{-iT^{a}b^{a}(x)}[/itex] doublet (but also bar contains the dagger and so the exponential's argument gets the minus)
[itex] R \rightarrow R'=R [/itex] singlet/scalar
[itex] H \rightarrow H'= e^{iT^{a}b^{a}(x)} H [/itex] doublet
So inserting to the yukawa term you can build up, you see that the term remains the same under SU(2)...
the same way you can work with the Hypercharge U(1) transformation, because the Higgs has hypercharge 1, the Left component has hypercharge -1 (the dagger +1) and the right component has hypercharge -2, so 1+1-2=0 so it's a singlet too...

So the yukawa term before higgs gets a vev is SU(2)xU(1) invariant... after the SSB, there's no reason for it to be SU(2) invariant so it can be mass terms...
spookyfish
#7
May5-14, 06:26 PM
P: 52
Excellent, I think I understand. Thank you!


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