how many goldstone bosons appear as a spntaneous symmetry breaking in process of higgs mechanism?
The SM SU(3)_{C}xSU(2)_{L}xU(1)_{Y} breaks down to SU(3)_{C}xU(1)_{EM} as a result of spontaneous symmetry breaking. Please explain me, how can we apply Goldstone's theorem in this theory?
SU(3) part remains unbroken. SU(2)L x U(1)Y has 4 generators but U(1)em has 1 generator. In total number of broken generators is 4-1=3. There are three goldstone bosons with charges +1,0,-1. We interpret them as longitudinal polarization of massive W+/- and Z0 boson as if W and Z bosons eat up these goldstone bosons and become heavy.
SU(3) part remains unbroken. SU(2)L x U(1)Y has 4 generators but U(1)em has 1 generator. In total number of broken generators is 4-1=3. There are three goldstone bosons with charges +1,0,-1. We interpret them as longitudinal polarization of massive W+/- and Z0 boson as if W and Z bosons eat up these goldstone bosons and become heavy.
Dear Dr. Adeel,
We can just answer the Quantum Mechanical part qualitatively from our own research into Low Dimensional Physics and that is that you have to consider the velocity of the vector bosons and hence the power of the particle accelerator has to be considered. We have obtained a M Theory for this which is posted on Soumitra K. Mallick and Sandipan Mallick's RG page under Soumitra K. Mallick.
SKM, NH, SM
The Brout-Englert-Higgs mechanism eliminates the Goldstone bosons, that would appear, were the symmetry a global one. And it eliminates as many as the gauge fields that would become massive.
So, in the Standard Model, the gauge group, whose global part undergoes spontaneous symmetry breaking, is SU(2)_L x U(1). There are 4 gauge fields, so one needs, at least, 4 scalars. 3 combinations become massive, thus, three scalars become their longitudinal components. One combination remains massless-the photon-so there's a scalar field that remains, that's the Higgs.
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