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Is the electric charge of bosons w1 w2 w3 well defined?

  1. Sep 24, 2011 #1

    naima

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    Hi

    In GSW theory we start with two fermions f f' (charge 0 -1) and 3 bosons w1 w2 w3.
    the charge of the bosons in only introduced when one mix w1 and w2 giving w+ and w-.
    Does it mean that before symmetry breaking f and f' were exchanging w+/w- but not
    the basic w1 w2 w3?
     
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  3. Sep 26, 2011 #2

    naima

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    I know gluons have no electric charge.
    In Gut Theories using other groups are there other electric charged bosons?
    Must we then use the same trick (W1 - iw2 style) to define them as electric charged?
     
  4. Sep 26, 2011 #3

    Vanadium 50

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    You're starting from a very bad place if you have fermions in eigenstates of the broken symmetry but are discussing the fields of the unbroken symmetry.
     
  5. Sep 26, 2011 #4

    naima

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    Last edited by a moderator: Apr 26, 2017
  6. Sep 27, 2011 #5
    Disclaimer: I have not thought about this for a while so apologies if I screw it up :)

    Ok, so you are talking about the electroweak lagrangian but just the bit with electron and electron neutrino say, not any other fermions? So, on the wikipedia page you linked to there is the lagrangian [itex]L_f[/itex] in the "before electroweak symmetry breaking" section. The last two terms in this lagrangian describes the interactions between the left handed SU(2) doublet (electron and neutrino) and gauge bosons, and the right handed SU(2) singlet electron and gauge bosons, respectively.

    For the doublet, there are seperate terms for each of the w1 w2 w3 and B fields if you expand this term, because the covariant derivative is just a linear thing with a piece for each field. So it is these basic fields that are being exchanged. For the singlet, the w1 w2 w3 fields are not in there, just the B field, because being an SU(2) singlet it has no SU(2) gauge fields in it's covariant derivative.

    Now, the confusing thing is that during symmetry breaking, it is the w3 and B fields that mix into the Z and A (photon). So the w1 and w2 fields are really the same fields as W+ and W-, which is why they use the same names for them both in the wikipedia article. The only trick is that they pick up a mass term from the higgs kinetic term (first term in [itex]L_h[/itex])


    Well being eigenvectors of T3 and Y means their weak isospin and hypercharge are unchanged when acted on by W3 and B, which since Z and A are linear combinations of W3 and B means Z and A also do not change these things, thus Z and A are neutral current interactions. Or you could say that the electron and neutrino are also eigenstates of the newly created U(1)em, and they are eigenstates of the screwed up SU(2) that remains.

    W1/W+ and W2/W- screw with the SU(2) eigenstate because they are raising and lowering operators really, which is why they turn electrons into neutrinos and vice versa. They move you around in the SU(2) space. But after symmetry breaking the electrons get a mass, so the electrons/neutrino doublet is not a proper SU(2) doublet anymore, since you will notice if you rotate this doublet in SU(2) space. Previously the electron bit looked just the same as the neutrino bit. I think. Not sure what happens with the electric charge, I must think about it.
    Anyway, the W+ and W- will still move you in this approximate SU(2) space just the same, so you can still change electrons into neutrinos with them.

    Ok that last paragraph is a bit of a mess, please someone who is better at group theory come along and explain properly what my mess is trying to get at :).
     
    Last edited by a moderator: Apr 26, 2017
  7. Sep 27, 2011 #6

    Vanadium 50

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    No - you have given the fermions electric charge. Before symmetry breaking, you don't have electric charge. My experience is that you really need to work in one or the other basis unless you really, really know what you are doing.
     
  8. Sep 28, 2011 #7

    naima

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    interesting answers.

    It is the first time i read that the charge appeared with symmetry breaking.
    Could you elaborate or give a link?

    thanks
     
  9. Sep 28, 2011 #8

    Vanadium 50

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    Before symmetry breaking you don't have a photon. The right representation is not the A potential.
     
  10. Sep 28, 2011 #9
    The wiki page you link to alludes to this without explaining it terribly much:

    "In the Standard Model, the W± and Z0 bosons, and the photon, are produced by the spontaneous symmetry breaking of the electroweak symmetry from SU(2) × U(1)Y to U(1)em, caused by the Higgs mechanism (see also Higgs boson).[3][4][5][6] U(1)Y and U(1)em are different copies of U(1); the generator of U(1)em is given by Q = Y/2 + I3, where Y is the generator of U(1)Y (called the weak hypercharge), and I3 is one of the SU(2) generators (a component of weak isospin)."

    I.e. you do not have Q (electric charge) until you have U(1)em, which is only a useful thing to talk about after symmetry breaking happens. Your leptons do however have Y (hypercharge), which tells you things about how they interact with the B gauge bosons, and is basically an exact copy of how electric charge works, but it is in fact different.
     
  11. Sep 28, 2011 #10

    naima

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    I disagree with that (the charge of the leptons before Symmetry Breaking).

    I'll quote not wiki but the handbook "Quantum Field Theory" by Itzykson Zuber p620:

    THE WEIBERG SALAM MODEL
    This is not ambiguous except for the 3 bosons.
    the fermions charge are defined (0 and -1) and a coupling constant is just defined for the bosons.
    The title of the post is linked to this point
     
  12. Sep 28, 2011 #11
    You can define the electric charge before symmetry breaking if you really want to, the U(1)em subgroup still exists in the fundamental SU(2)_L x U(1)_Y group, however there are lots of other U(1) subgroups that you could have picked instead and nothing to distinguish them from each other. So the electric charge is meaningless.

    I refer you again to the wikipedia page. Notice in the "after electroweak symmetry breaking" lagrangian there is a piece called the neutral current, L_N. The electromagnetic current part of that contains the electric charge, and is what defines how the fermions interact with photons, which is the whole point of electric charge. There is no such term in the lagrangian before symmetry breaking, so the electric charge, although you can "define" it just the same, doesn't appear in any term in the lagrangian so you may as well not talk about it.
    There IS a totally analogous piece which defines the hypercharge current, so the interaction between the fermions and the B field, in which the hypercharge appears playing the same role as electric charge.
     
  13. Sep 28, 2011 #12

    naima

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    You do not comment Itzykson handbook.
    He says that the left charged leptons are in a doublet (T3 = -1/2 , 1/2)
    and that a Y = -1/2 hypercharge is assigned to them.
    This give Q = 0 , -1 the "official" charge after symmetry breaking.
    perhaps this is meaningless, perhaps it is not
     
  14. Sep 28, 2011 #13

    naima

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    According to you, what were the conserved "things" (for leptons and their jauge bosons) , when leptons interacted before symmetry breaking?
    In fact this is what i try to understand.
     
  15. Sep 28, 2011 #14

    Vanadium 50

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    Well, OK, but if you're having trouble with something, and that's the root of it, the disbelief is unlikely to be much help.
     
  16. Sep 28, 2011 #15

    naima

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    This is not a question of disbelief.
    I am studying handbooks and some ones introduce charges and mixing angles before spontaneous breaking. that's all.
    Please try to answer the question about the physics before SB.

    thanks a lot.
     
  17. Sep 28, 2011 #16

    naima

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    I found http://en.wikipedia.org/wiki/Electroweak_epoch" [Broken].
    You can see that before Symmetry Breaking, cosmology says there were long range W, Z
    (the Z use the theta mixing angle)
    .
     
    Last edited by a moderator: May 5, 2017
  18. Sep 28, 2011 #17

    arivero

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    the coupling constants of _L and _Y fix the U(1), because you want it to be a non-chiral force. I guess that you could forgot about symbreaking and just define the photon in this way.

    Now, perhaps Vanadium could give more details about how then the complex representations of the SU(2)_LxU(1)_Y group combine to allow to extract a real representation of U(1)_EM. I'd try myself, as really I am interested on the details of this discussion, but I have 8 hours flight in a couple hours, plus jetlag...

    I am particularly worried, once the representation issue is clarified, about the charges of massless winos.
     
  19. Sep 28, 2011 #18
    This is just a semantics issue. He refers to the doublet as the charged lepton doublet because it is the doublet that ends up having a charged lepton in it. It remains the same object, but it's properties change with symmetry breaking.

    Hmm, I can't remember how that works, must go look...

    The leptons and quarks are in the complex representation of U(1)_EM I believe, isn't the real one the 2x2 representation?

    I think this is just the same thing, i.e. they have no electric charge when they are massless because it is before symmetry breaking. In fact they can have a mass and still no electric charge because they have mass terms coming from the soft supersymmetry breaking part of the lagrangian as well as from the higgs sector. I assume you are talking about MSSM winos here.
     
    Last edited: Sep 28, 2011
  20. Sep 30, 2011 #19

    naima

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    This was not semantics
    I sent a mail to the french cowriter of my handbook. He was kind enoug to answer my question.

    I translate the end:

     
  21. Sep 30, 2011 #20

    Vanadium 50

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    Let me try this again.

    You need to decide whether you are working in the broken or the unbroken symmetry basis. (Colloquially, "after" or "before" the break) This is exactly analogous in mechanics to picking your coordinate system. Mixing the basis is exactly as bad as mixing coordinates systems; you might get the right answer, but it would surely be accidental.
     
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