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When the gauge symmetry of our Lagrangian breaks spontaneously through the Higgs mechanism, we usually find that $n$ Higgs degrees of freedom become massless through the vacuum expecation value (vev), where $n$ is the number of broken generators. This means if our original gauge group has $N$ generators, the group that leaves the vev invariant has only $N'=N-n$ generators. The $n$ broken generatore become massive through the Higgs vev and in turn we can see that the mass terms for $n$ Higgs degrees of freedom vanish if we put in the vev. Often one says the Goldstone boson get eaten by gauge bosons, which then become massive.

What exactly happens to these massless degrees of freedom in the Higgs potential after symmetry breaking, i.e. after we expand the Higgs fields about the vev? **The mass-terms= quadratic terms vanish as noted above if we put in the vev, but, in general, there are several quartic terms possible. These would describe interactions of the Goldstone bosons with the Higgs bosons and surely there must be a good way to see that these vanish, too or have no influence for some reason?

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    $\begingroup$ Are you familiar with the unitary gauge? If not, look it up? Doesn't it answer your question? $\endgroup$
    – innisfree
    Commented Oct 26, 2015 at 10:07
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    $\begingroup$ This is a small explanatory comment on the above key comment by @innisfree: Think of the SM. In the unitary gauge the 3 goldstons g are simply not in the Higgs potential anymore --they are angular variables that drop out. Where did they go? Look at the covariant derivatives of the Higgs in its kinetic term. They include HHWW terms, and, you guessed it!, these include the HHgg from the belly of the Ws you thought were in the Higgs potential. You may convince yourself there were never any terms linear in g. Can you parse the multi-gauge b amps from their covariant derivatives? $\endgroup$
    – Cosmas Zachos
    Commented Jun 2, 2017 at 23:00

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First, note that, strictly speaking, there is no such thing as spontaneous symmetry breaking in Higgs mechanism. I mean, that below and under the Higgs scale (i.e., at scale, at which non-zero Higgs VEV appears) the lagrangian can be rewritten in a gauge invariant way. How is it possible? The answer is that there are different physical states (i.e., eigenstates of hamiltonian) below and under the Higgs scale. Under the Higgs scale eigenstates of hamiltonian form also representations of the gauge group. But below it eigenstates of hamiltonian don't form representations of the gauge group. These eigenstates are linear combination of transverse and longitudinal degrees of freedom.

So, in some sence, nothing happens with three scalar fields in Higgs doublet. They just don't appear as physical states below the Higgs scale.

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  • $\begingroup$ This is what `spontaneous symmetry breaking' means. The symmetry is still in tact, it is just hidden, because the vacuum transform nontrivially under the symmetry. $\endgroup$
    – flippiefanus
    Commented Oct 2, 2016 at 5:17
  • $\begingroup$ @flippiefanus : since the vacuum has to be gauge invariant, it doesnt. $\endgroup$
    – Name YYY
    Commented Oct 2, 2016 at 10:33
  • $\begingroup$ What does "gauge invariant" mean in this context? Strictly speaking, gauge invariant means it transforms as a singlet under the symmetry transformations, but this is not the case for theories with spontaneous symmetry breaking. The vacuum is genererate, which means it has different possible states all related by gauge transformations. It's not a singlet. $\endgroup$
    – flippiefanus
    Commented Oct 2, 2016 at 10:39
  • $\begingroup$ @flippiefanus : it is the sum over all gauge transformations of the choosen gauge variant state, which is gauge invariant. $\endgroup$
    – Name YYY
    Commented Oct 2, 2016 at 11:51
  • $\begingroup$ Mmm, that makes no sense to me. Usually when you have spontaneous symmetry breaking you get a vacuum expectation value that is not gauge invariant. That can only happen if the original vacuum was not gauge invariant. However, we may be saying the same thing but using the terminology in a different way. $\endgroup$
    – flippiefanus
    Commented Oct 3, 2016 at 11:47
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What exactly happens to these massless degrees of freedom in the Higgs potential after symmetry breaking, i.e. after we expand the Higgs fields about the vev?

You will make a redefinition of the vector fields $V_{\mu}' = V_{\mu} - \partial_{\mu} \eta $, where $\eta$ was your goldstone, Doing that you can see that a term like $$ \nu^{2} (V_{\mu} + \partial_{\mu}\eta)(V^{\mu} - \partial^{\mu}\eta) $$

Will generate a mass term for the vector boson.That's how the goldstones are "eaten" by the vector bosons. Furhtermore, they become a longitudinal degree of freedom of the polarization vector of the gauge bosons.

The mass-terms= quadratic terms vanish as noted above if we put in the vev, but, in general, there are several quartic terms possible. These would describe interactions of the Goldstone bosons with the Higgs bosons and surely there must be a good way to see that these vanish, too or have no influence for some reason?

Well, I guess you as you are making a expansion around the vev you take into account that you want small oscillations of the field.

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