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Is the Higgs mechanism really a spontaneous symmetry breaking?

  1. Jul 27, 2012 #1
    The Higgs mechanism is often explained (both here at PF and in many physics sites including wikipedia) as an example of spontaneous symmetry breaking, but the Nobel winner physicist 't Hooft says in his "for laymen" book about particle physics, "In search of the ultimate building blocks", that the Higgs mechanism is not really an spontaneous symmetry breaking kind of mechanism, and that if it was, the weak interaction bosons would be Goldstone bosons and therefore couldn't be massive as they are, so it wouldn't explain the very thing the Higgs mechanism is there to solve, namely the masses of the particles.

    Could somebody explain if this is so and what is the subtle difference between the true spontaneous symmetry breaking and the Higgs mechanism?
    I guess this is related to this words form wikipedia to explain how the gauge boson acquire mass: "In theories with gauge symmetry, the Goldstone bosons are "eaten" by the gauge bosons."
    I understand this "eating" refers to the existence of the Higgs field with a >0 VEV, but why photons remain massless then?
    Also at high energies all particles are supposed to be massless, why the Higgs field doesn't interact to give particles mass at those enrgies, is it because its VEV is zero there?
     
    Last edited: Jul 27, 2012
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  3. Jul 27, 2012 #2

    Bill_K

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    I think all 't Hooft is saying is that the Higgs mechanism avoids the Goldstone theorem and does not contain massless Goldstone bosons. And for that reason is not an example of spontaneous symmetry breaking in the original sense.
     
    Last edited: Jul 27, 2012
  4. Jul 27, 2012 #3
    Maybe you can post the exact quote so that we could see it in context?
     
  5. Jul 27, 2012 #4
    Yes, and my question was how does the higgs mechanism of spontaneous symmetry breaking exactly differ from the generic spontaneous symmetry breaking?
     
  6. Jul 27, 2012 #5

    Bill_K

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    Well for example here, on Tommaso Dorigo's blog, a simple example is discussed in which a Goldstone boson remains. The difference from the Higgs mechanism is that the symmetry is not locally gauged, and so there is no gauge boson to eat it.
     
  7. Jul 27, 2012 #6

    DrDu

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  8. Jul 27, 2012 #7

    atyy

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    Englert-Brout-Higgs-Guralnik-Hagen-Kibble mechanism

    "Of course, there is an alternative, manifestly covariant way of quantizing the electromagnetic field, namely to use the Lorentz gauge ... Here the Goldstone theorem definitely does apply, and there do exist massless Nambu-Goldstone bosons. But it turns out that they are purely gauge modes, uncoupled to the physical particles of the theory.

    This mechanism is often said to exhibit "spontaneously broken gauge symmetry". That is a convenient shorthand description, but the terminology is potentially somewhat misleading. The process of quantization requires a choice of gauge, i.e., an explicit breaking of the gauge symmetry. However, the resulting theory does retain a global phase symmetry that is broken spontaneously"

    Is electromagnetic gauge invariance spontaneously violated in superconductors?

    "In particular, we emphasize that global U(1) phase rotation symmetry, and not gauge symmetry, is spontaneously violated, and show that the BCS wave function is, contrary to claims in the literature, fully gauge invariant. "
     
  9. Jul 27, 2012 #8

    DrDu

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    I don't consider Greiter's analysis to be correct. Namely he states that an arbitrary N particle ket is invariant under gauge transformations. This is not true. Rather we consider all kets which differ by only a phase as describing the same state.
     
  10. Jul 27, 2012 #9

    atyy

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    Is it Greiter's Eq 13, 14 and the discussion immediately following that you think are wrong? BTW, thanks for the Friedrich reference in the other thread too, I'm slowly reading it.
     
  11. Jul 28, 2012 #10

    DrDu

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    Yes, I think that is the point where he is wrong.
    Let's compare to another example: A fermionic ket is transformed into minus itself under a 360 deg rotation (i.e. a transformation corresponding to identity). This gives rise to the univalence superselection rule that we cannot observe superpositions of bosonic and fermionic particles.
    In the case of a gauge symmetry the change of phase of a ket under a gauge transformation implements the charge superselection rule. This has been nicely worked out long time ago by Roberts, Dopplicher and Haag.
     
  12. Jul 28, 2012 #11
    Ok, but the photon symmetry is locally gauged, why is it not obserbed to be massive like the other three gauge bosons.
    IOW, why don't photons interact with the Higgs field?
     
  13. Jul 28, 2012 #12
    DrDu and atyy, thnks for the references.
    The Friedrich paper seemed directly related to my questions but there seems to be a basic problem with the way the notion of SSB is treated there.
    At times it looks as though the author is talking about explicit symmetry breaking instead pf SSB.
    AFAIK, SSB refers to hidden symmetries rather than truly broken ones, so they would never imply breaking of the gauge symmetry.
    Similarly, fixing of a gauge does not mean the gauge symmetry is broken.
    On the other hand in the Higgs mechanism, everybody agrees that the Goldstone bosons are there as the longitudinal polarization of the gauge bosons, they are just invisible.
    So to me SSB never implies breaking of local gauge or any other symmetry, but merely its hiding and from that pov I don't understand the insistence of the paper(and as I said in the OP of 't Hooft) in saying the Higgs mechanism is not a truly SSB example because it doesn't break gauge symmetry when that is a general feature of SSB.
     
  14. Jul 28, 2012 #13

    Bill_K

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    Because we choose the Higgs field φ to be an isospin doublet with weak hypercharge Y = 1.

    This choice is not forced upon us, and φ could easily be something else. Also we choose the vacuum expectation value φ0 to be (0 v). That is, the only nonzero component is the T3 = -1/2 component. For this component, Q = T3 + Y/2 = 0. Which not only says the Higgs field is neutral, Qφ0 = 0 also means that the U(1)em with generator Q remains unbroken: φ0 → φ'0 = eiα(x)φ0 = φ0 for any value of α(x). Thus the vacuum remains invariant under U(1)em transformations, and the photon remains massless.
     
    Last edited: Jul 28, 2012
  15. Jul 28, 2012 #14
    Ok, I see, it is basically the simplest choice of I3 and and Y for the Higgs to get the desired result for the EM symmetry.
     
  16. Jul 29, 2012 #15

    atyy

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    I found a helpful comment about the Higgs mechanism in Nair's QFT text. It looks like in loose jargon, there are two different objects G and G* that are both called the gauge group.

    "We have already seen that the true gauge group of the theory is the set of all gauge transformations which go to the identity at spatial infinity, namely, G* in the notation of Chapter 10. It should be emphasized that there is no breakdown of this true gauge symmetry here. In fact gauge symmetry is crucial in removing the massless modes. What is broken is the global part of the symmetry, corresponding to G/G*."
     
    Last edited: Jul 30, 2012
  17. Jul 30, 2012 #16

    DrDu

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    This distinction of gauge groups is detailed in
    Struyve, W., Gauge invariant accounts of the Higgs mechanism,
    arXiv.org:1102.0468 (2011),
    at least on a classical level. This has been done already in the early papers by Higgs etc.
    As far as I understood Friederichs paper, it is not easy to implement this splitting in quantized theories.
     
  18. Jul 30, 2012 #17

    DrDu

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    Here some thoughts of mine:
    When a system only has a global gauge symmetry which is broken, then the expectation value of the field <f> is non-vanishing in an irreducible representation of the algebra of local operators. In reducible representations, the expectation value may be 0, nevertheless. Compare for example a magnet. We can still bring it into a superposition of all the possible orientations of his magnetization axis, so that the expectation of the mean field vanishes.
    Enlarging the gauge group from global to local will lead to the irreps to combine into a larger irrep.

    In the case of global symmetry only, the rotation of an (irreducible) ground state |f> into an equivalent one |f'> cannot be implemented in the space of local observables, not even as a limit: Let's assume we rotate f(x) into f'(x) only for x in some finite region V. Then the energy of that surface S will diverge with increasing V, so that the limit cannot be identified with the ground state |f'>. The global gauge symmetry can not be implemented unitarily.
    This is different, when the symmetry is local. Then the energy of the surface S of V can be made to vanish by addapting the vector potential (Goldstone bosons get eaten). The limit converges and all the states |f>, |f'> lie in the same Hilbertspace and the operator of global gauge transformations which interchanges them is unitarily represented.
    Hence averaging over the global gauge group is possible and the expectation value of the non-invariant field in the ground state has to vanish. Hence symmetry is not broken. Nevertheless f^2=f'^2 may be non-zero in the ground state.
     
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