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0 answers
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Unitary Gauge Removing Goldstone Bosons

The Lagrangian in a spontaneously broken gauge theory at low energies looks like $$ \frac{1}{2} m^2 ( \partial_\mu \theta - A_\mu )^2 $$ and the gauge transformations look like $\theta \rightarrow \...
infinity's user avatar
5 votes
2 answers
375 views

Why are there no Goldstone modes in superconductor?

Usually, the absence of Goldstone modes in a superconductor is seen as an example of the Anderson-Higgs mechanism, related to the fact that there is gauge invariance due to the electromagnetic gauge ...
cx1114's user avatar
  • 109
1 vote
0 answers
51 views

Goldstone and longitudinal polarization equivalence between off-shell $W$s

If I understand correctly, the couplings of the longitudinal component of the $W$ and $Z$ bosons should be "equivalent" to the Goldstone bosons they ate after SSB. In practice, this makes ...
avocado_love's user avatar
4 votes
1 answer
492 views

Spontaneous symmetry breaking in the Standard Model. What is "broken"?

I know this question has been asked other times, but I am looking for a confirmation of the following. When we say that the gauge group of the standard model is $G_{SM} = SU(3)_{c} \times SU(2)_{L} \...
Pipe's user avatar
  • 858
4 votes
2 answers
5k views

Spontaneous symmetry breaking of $U(1)$ in superconductivity

In the crystal, infinitesimal translational symmetry breaking makes the phonon, In ferromagnet, time-reversal symmetry breaking makes magnon. I know that in superconductor there is a spontaneous ...
iron fist's user avatar
14 votes
2 answers
4k views

The meaning of Goldstone boson equivalence theorem

The Goldstone boson equivalence theorem tells us that the amplitude for emission/absorption of a longitudinally polarized gauge boson is equal to the amplitude for emission/absorption of the ...
cnzz601's user avatar
  • 141