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CAF123
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I realize there has been quite a few threads on this topic lately but I wanted to ask some questions that I have here. The following are statements from my lecture notes and I have write my questions after each statement.
In the gauge sector of the electroweak theory, we can write down ##\mathcal L_{\text{gauge}}## and for this to be invariant under local transformations,we introduce a covariant derivative $$D_{\mu} = \partial_{\mu} + ig \frac{1}{2} \mathbf{\sigma^j} \mathbf{W^j_{\mu}} + ig' Y B_{\mu}$$ Is this the analogous covariant derivative from what we define in, say, QCD with ##D_{\mu} = \partial_{\mu} + igA_{\mu}^a t^a## but this time ##j## is a flavour index?
To each broken generator we can find, i.e ##T^a \langle \phi_{\text{VEV}}\rangle \neq 0,## where ##\langle \phi_{\text{VEV}}\rangle## is the VEV of the Higgs field, we have a massless Goldstone boson. Choosing ##\langle \phi_{\text{VEV}}\rangle \propto (0,v)## we find that ##\sigma_1, \sigma_2## and ##\sigma^3-Y## are broken generators and thus give rise to three massless Goldstone bosons by Goldstone's theorem. But can't I say the same about ##\sigma^3## alone and Y = 1/2 alone? Or is it because they are not linearly independent from ##\sigma^3-Y## so should not also be considered in addition?
We find that ##\sigma^3/2+Y = Q## is a broken generator and thus this implies there exists a linear combination of ##W_{\mu}## and ##B_{\mu}## such that $$\exp \left( i \left( \frac{1}{2}\sigma^3 W^3_{\mu} + B_{\mu} \text{Id}\right)\right)\langle \phi_{\text{VEV}}\rangle = 0 $$ Why is this so? It looks like some group element of ##SU(2) \times U(1)## which upon acting on the VEV leaves it invariant. But then if the generic structure of a group element is ##e^{i \lambda_a t^a}## then this seems to imply that ##W_{\mu}## and ##B_{\mu}## are to be treated as group parameters rather than massless gauge fields.
Thanks!
In the gauge sector of the electroweak theory, we can write down ##\mathcal L_{\text{gauge}}## and for this to be invariant under local transformations,we introduce a covariant derivative $$D_{\mu} = \partial_{\mu} + ig \frac{1}{2} \mathbf{\sigma^j} \mathbf{W^j_{\mu}} + ig' Y B_{\mu}$$ Is this the analogous covariant derivative from what we define in, say, QCD with ##D_{\mu} = \partial_{\mu} + igA_{\mu}^a t^a## but this time ##j## is a flavour index?
To each broken generator we can find, i.e ##T^a \langle \phi_{\text{VEV}}\rangle \neq 0,## where ##\langle \phi_{\text{VEV}}\rangle## is the VEV of the Higgs field, we have a massless Goldstone boson. Choosing ##\langle \phi_{\text{VEV}}\rangle \propto (0,v)## we find that ##\sigma_1, \sigma_2## and ##\sigma^3-Y## are broken generators and thus give rise to three massless Goldstone bosons by Goldstone's theorem. But can't I say the same about ##\sigma^3## alone and Y = 1/2 alone? Or is it because they are not linearly independent from ##\sigma^3-Y## so should not also be considered in addition?
We find that ##\sigma^3/2+Y = Q## is a broken generator and thus this implies there exists a linear combination of ##W_{\mu}## and ##B_{\mu}## such that $$\exp \left( i \left( \frac{1}{2}\sigma^3 W^3_{\mu} + B_{\mu} \text{Id}\right)\right)\langle \phi_{\text{VEV}}\rangle = 0 $$ Why is this so? It looks like some group element of ##SU(2) \times U(1)## which upon acting on the VEV leaves it invariant. But then if the generic structure of a group element is ##e^{i \lambda_a t^a}## then this seems to imply that ##W_{\mu}## and ##B_{\mu}## are to be treated as group parameters rather than massless gauge fields.
Thanks!