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Understanding the need for the Higgs Field

  1. Hello,

    I am a layman who just read Sean Carroll's The Particle at the End of the Universe and I am trying to understand what I think is the crux of the book, the need for the Higg's boson according to the Standard model. This is my thought process:

    1. The weak interaction only interacts with left-handed fermions
    2. The above implies the weak bosons are massless because only bosons which are massless can travel at light speed and therefore observe only one type of helicity
    3. Experimentally, however the weak bosons are shown to have mass due to the their limited range of interaction.
    4. Therefore, mass can't be inherent to the W and Z bosons but rather acquired through the interaction of another field. (i.e. the Higg's Field)

    Is my thought process even remotely close?

    Thanks

    Ryan
     
  2. jcsd
  3. fzero

    fzero 2,623
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    This is true.

    The fact that the weak interaction only involves left-handed fermions does not itself place a restriction on the mass or helicity of the gauge bosons. Rather, the restriction on the mass comes from something called renormalizability of the gauge theory. Renormalizability, in one sense, refers to the fact that observables of a quantum field theory often involve certain divergences related to momentum integrals. These integrals can diverge because we integrate over all momenta with magnitudes from zero to infinity. The standard methods for making sense of these divergences to extract finite, physical observables comes under the name renormalization.

    However, it can be shown that renormalization of a gauge theory requires that the gauge symmetry be a good symmetry at high energies (which the source of the infinities). A mass term for a gauge boson explicitly breaks the gauge symmetry, so the quantum field theory of a massive gauge boson is not renormalizable. The electroweak theory evades this restriction because the gauge bosons are only massive below the scale where the gauge symmetry is broken. At high energies, the gauge bosons are massless and the full gauge symmetry is restored, allowing for renormalization.

    This is true.

    Again, it is the requirement that the theory be renormalizable that leads to the conclusion that the W and Z bosons can't have an explicit mass, but can have a mass generated at low energy by the Higgs mechanism.
     
  4. Bill_K

    Bill_K 4,159
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    The W boson only interacts with left-handed fermions. The Z boson interacts to a certain extent with right-handed fermions as well.
     
  5. Thanks a lot for your response. I well need to digest your response a little. I appreciate it all the same
     
  6. But, to elaborate, this is because it isn't a purely "weak" object, in that it is a mixture of both SU(2) and U(1)_Y gauge bosons. This is partially just the semantics of what one calls the weak force though.
     
  7. Bill_K

    Bill_K 4,159
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    Of course it is! The Z is a weak boson, and deserves the name every bit as much as the W does. It interacts via the weak force, with weak coupling constants g and g'. And the symmetry group it gauges is weak isospin X weak hypercharge.

    (But feel free to substitute "electroweak" everywhere you see "weak".) :smile:
     
  8. Haha, but if you try to make this argument then the photon is also a "weak" gauge boson. So it seems to me the substitution 'electroweak' is indeed necessary. But anyway, if we want to say things like "the weak force only interacts with left-handed fermions" then it seems like we are talking about SU(2)_L, and to me it seems more intuitive to then explain the Z's interaction with right handed fermions as coming from the "not weak" part of the SM lagrangrian. But it's just words, so whatever. I don't know what is less confusing for people who don't already know what is really meant.
     
    Last edited: Nov 23, 2013
  9. mfb

    Staff: Mentor

    That will fail at "with weak coupling constants g and g'". You can use the photon with "electroweak", but without that unification the coupling constant is the fine-structure constant and the coupled charge is the electric charge.
     
  10. haushofer

    haushofer 975
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    I came to this topic with a question in mind from the Beyond the SM subforum.

    I don't understand why this is valid reasoning, considering my understanding that the SM doesn't need to be renormalizable from an effective field theory-point of view. It just happens to be so.
     
  11. fzero

    fzero 2,623
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    What you say is certainly true. While renormalizability was something demanded by the founders when they were trying to explain the weak interaction and QCD, a decade or two later one would not have been so picky. Today we have various reasons to expect that the SM is not exact and shouldn't be viewed as more than an EFT anyway.

    Now there are certainly philosophical reasons to expect that the gauge bosons don't have a small explicit mass, but from the EFT perspective these would be explained as naturally small by the custodial symmetry. But, I am not sure that the observed bounds on the coefficients of the various higher-dimensional would be natural if the SM were not a renormalizable gauge theory.
     
    haushofer likes this.
  12. haushofer

    haushofer 975
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    So when people state that one reason for the need for the Higgs is keep renormalizablity when introducing mass-terms, this is bad reasoning?
     
  13. fzero

    fzero 2,623
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    Not necessarily. As far as I know, the LHC results are consistent with the SM where the W and Z masses are generated entirely due to the Higgs mechanism, but I'm not aware of an experimental bound on explicit mass terms. An EFT with explicit mass terms would also tend to signal some higher-energy scale where the theory breaks down which might be too small to be consistent with some astrophysical or other data. The SM, all along before the discovery of the Higgs, was extremely predictive based on a relatively small number of input parameters. An EFT is much less predictive in general, since, in general, every term in the Lagrangian is an independent parameter.

    Renormalizability was an educated guess for a principle to guide model building that seems to have worked out quite well.
     
    haushofer likes this.
  14. Vanadium 50

    Vanadium 50 17,637
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    I think renormalizability is taking maybe a bit too prominent a role here. This is of course completely true for fermions - I can introduce a Dirac mass (instead of a Higgs Yukawa) and have a consistent theory, but if I try and renormalize it at higher orders, I make a mess. On the other hand, tthe masslessness of the w1, w2 and w3 is instead a requirement of gauge invariance. The Higgs mechanism finds a way to give the W and Z mass without breaking gauge invariance.

    One could break gauge invariance explicitly. However, this is the thing that keeps the photon massless and electric charge conserved. If you decide to break it because you don't like the Higgs mechanism (a position that became harder to support in 2012) you have to explain why the photon is still massless. This requires two "just so stories". One is that this mechanism leaves the photon alone, and the other is that the electroweak symmetry is broken with exactly the right Weinberg angle to still leave the photon alone.
     
    haushofer likes this.
  15. haushofer

    haushofer 975
    Science Advisor

    Thanks. So it is not bad reasoning, but more subtle than a first glance would expect. Vanadium, the theory would be consistent for the fermions if it wouldn't be for the weak force breaking parity, right? Explicit mass terms mix left and right handed fields, making it impossible to put one of them in an su(2) doublet and the other in a singlet.
     
  16. Vanadium 50

    Vanadium 50 17,637
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    The problem is that right-handed fermions act differently than left-handed fermions, yes. We normally place this in the definition of the fields, which requires them to be massless without a Higgs Yukawa. Alternatively, we can place this in the definition of the interaction, which allows us to have massive fields at tree level, but that messes up renormalizability.
     
    haushofer likes this.
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