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Question re SU(2)xU(1) of the SM

  1. Dec 12, 2014 #1
    I don't quite understand the leptonic SU(2) isospin doublet (electron, neutrino):

    As far as I understand, the masses in any multiplet have to be the same always; assuming the neutrino here is massless, so should the electron in this formalism, right? Does this mean that we take the electron in this doublet to be massless (at the Lagrangian level) and then we say that it acquires mass through spontaneous symmetry breaking? Is this the correct way to look at it (i.e. can the electron sit in the same doublet as the neutrino if it acquires a mass "later")?

    Also, when looking at the quantum numbers, or charges, under SU(2)x[hypercharge U(1)] breaking down to [electromagnetic U(1)], one says that we know the electric charges for (electron, neutrino) to be (-1,0) and the 3-component of the SU(2) charges, coming from the generator [itex]T^{3}=\frac{1}{2}\sigma^{3}[/itex], to be (-1/2,1/2).
    Can someone explain how one works out the eigenvalue of [itex]T^{3}[/itex] for the eigenvectors (electron,neutrino)? Does one have to act with the Pauli matrices onto a Weyl spinor?
     
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  3. Dec 12, 2014 #2

    ShayanJ

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    I don't about your second question but about the first:
    1) Neutrinos are not massless. They are very light but they still have mass. Also we don't know their masses.
    2) The symmetry among elements of a isospin multiplet is only an approximate symmetry, not an exact one.
     
  4. Dec 12, 2014 #3
    Yes, I know that, but surely you can't say that the electron and the neutrino have the same mass (which I guess would be the requirement for them being in the same multiplet). And I doubt that the "approximate" nature of the isospin multiplet would account for the difference in mass. Approximate means that at least they are comparable, which I doubt it's the case..
     
  5. Dec 12, 2014 #4

    Vanadium 50

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    Why would you think particles in the same multiplet have the same mass? It's not true for quarks either.
     
  6. Dec 12, 2014 #5
    Well, because them being in the same multiplet means that you can "rotate" one into each other under the action of that symmetry. How can you have a massive/massless particle becoming massless/massive?
     
  7. Dec 12, 2014 #6

    ChrisVer

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    When you take the symmetry of SU(2)xU(1) your leptons are massless. They get a mass after you break the symmetry indeed. If you write the Lagrangian you see that you don't have any mass term [that would break the symmetry explicity].

    The other question about T3 is almost answered. The matrix is in a diagonal form and thus you get the eigenvalues immediately out of it. You can act with it on YOUR DOUBLET (not Weyl spinors, the doublet contains 2 weyl spinors; i.e. [itex]e_L, \nu_L[/itex]) but that is just a not-needed extra work to be done.
     
    Last edited: Dec 12, 2014
  8. Dec 12, 2014 #7

    ChrisVer

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    That is kinda misleading. One has two factors to take into account. When one works in the flavor space for quarks, he makes the assumption that [itex]m_u \approx m_d \approx 0[/itex] compared to the QCD scale... sometimes one can take also [itex]SU(3)[/itex]-flavor and also take the strange quark in the multiplet, but things stop working in general after SU(4) (with charm).

    On the other hand, in the other case, which brings the CKM matrix, one has to make a distinction between mass-states and flavor-states. If the mass and flavor-transformation matrices commute, then they are simultaneously diagonalizable and thus you can identify the fields in the same flavor multiplet to have the same mass.
    That is not the case for when these matrices don't commute. Then your mass eigenstates will have to be a mixture of your flavor states, and thus the fields in the multiplet don't have to have the same mass. So they don't have the same mass, but you can rotate the one in the other. (The flavor eigenstates are a mixture of mass eigenstates).

    I think that in Standard SM, there is no CKM in the leptonic sector. Of course some other theories support there should be one, because they want to introduce CP-violating phases.
     
    Last edited: Dec 12, 2014
  9. Dec 12, 2014 #8

    Vanadium 50

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    Where in the SM does it say that different components of an SU(2) doublet have the same mass?
     
  10. Dec 12, 2014 #9

    ChrisVer

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    The leptonic sector doublet components are massless.... they appear coupled to the scalar field as in a Yukawa term, that before the SSB has zero vev.
     
  11. Dec 12, 2014 #10

    Vanadium 50

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    You can say the same thing about quarks too. But in any event, what you are saying is that in a theory where they are massless (if the Yukawa were zero) they would have the same mass. That's true, but a) not helpful, b) doesn't describe the universe in which we live, and c) just moves the problem one step back, changing it to "Where in the SM does it say that different components of an SU(2) doublet have the same Yukawa?"
     
  12. Dec 13, 2014 #11

    ChrisVer

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    Well, then can I ask why the quark flavor model cannot be described by [itex]SU(6)[/itex]?

    However I am starting getting your point.
    In fact you are right, they don't have to have the same mass. They exist in the same doublet because gauge interactions can change the one into the other. Now then, if someone wants to take into account mass, they must also have right components. [Un]fortunately in the classical SM there is no right handed helicity neutrinos.
    The equality of masses can only be taken as a statement if the gauge "interactions" commute with the mass matrices.
     
    Last edited: Dec 13, 2014
  13. Dec 14, 2014 #12
    Exactly. They can have different Yukawa couplings because only the left chiralities form a doublet under the SU(2) symmetry. The two independent SU(2) singlet right chiralities allow different Yukawas for the up and down quarks. In a theory where the two right chiralities also form an SU(2) doublet, the up and down yukawas are the same.
     
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