- #1
michael879
- 698
- 7
I was playing around with a classical field treatment of the SM, and I came across something strange I was hoping somebody could explain to me. Basically, I found that if you plug the EL equations back into the SM lagrangian, you can essentially remove the Higgs field entirely (at least the kinetic and mass terms, which are the most important aspects). I'll briefly show my math below, just in case I screwed up somewhere:
The electroweak lagrangian:
[itex]L=-\dfrac{1}{4}Tr(B_{\mu\nu}B^{\mu\nu} + W_{\mu\nu}W^{\mu\nu}) + (D_\mu\phi)^\dagger D^\mu\phi - \dfrac{\mu^2}{2v^2}(|\phi|^2-\dfrac{1}{2}v^2)^2[/itex]
[itex]D_\mu\phi \equiv (\partial_\mu-ig_1 B_\mu - ig_2 W_\mu)\phi[/itex]
[itex]W_{\mu\nu} \equiv \partial_\mu W_\nu - \partial_\nu W_\mu -ig_2[W_\mu,W_\nu][/itex]
[itex]B_{\mu\nu} \equiv \partial_\mu B_\nu - \partial_\nu B_\mu[/itex]
The "3" EL equations are:
[itex]\partial_\nu B^{\mu\nu} = ig_1\phi^\dagger(D^\mu\phi)-ig_1(D^\mu\phi)^\dagger\phi[/itex]
[itex][D_\nu,W^{\mu\nu}] = ig_2(D^\mu\phi)\phi^\dagger-ig_2\phi(D^\mu\phi)^\dagger[/itex]
[itex]D_\mu D^\mu \phi = -\dfrac{\mu^2}{v^2}(|\phi|^2-\dfrac{1}{2}v^2)\phi[/itex]
From the first equation you can derive the continuity equation:
[itex](D_\mu\phi)^\dagger D^\mu\phi + \phi^\dagger D_\mu D^\mu \phi = 0[/itex]
and then using the third equation you find:
[itex](D_\mu\phi)^\dagger D^\mu\phi = \dfrac{\mu^2}{v^2}(|\phi|^2-\dfrac{1}{2}v^2)\phi[/itex]
if you then substitute this back into L, you find the new Lagrangian:
[itex]L=-\dfrac{1}{4}Tr(B_{\mu\nu}B^{\mu\nu} + W_{\mu\nu}W^{\mu\nu}) + \dfrac{\mu^2}{2v^2}|\phi|^4 - \dfrac{1}{8}\mu^2 v^2[/itex]
so that the new Lagrangian has a completely isolated Higgs field, and no spontaneously broken symmetry! Not only this but the mass term for the Higgs, and the coupling terms for the EW fields have disappeared! Now I am aware that the EL equations are only considered valid on-shell, so this suggests to me that the Higgs mechanism is entirely a quantum effect and would not occur for classical fields... Am I making some mistake here?
The electroweak lagrangian:
[itex]L=-\dfrac{1}{4}Tr(B_{\mu\nu}B^{\mu\nu} + W_{\mu\nu}W^{\mu\nu}) + (D_\mu\phi)^\dagger D^\mu\phi - \dfrac{\mu^2}{2v^2}(|\phi|^2-\dfrac{1}{2}v^2)^2[/itex]
[itex]D_\mu\phi \equiv (\partial_\mu-ig_1 B_\mu - ig_2 W_\mu)\phi[/itex]
[itex]W_{\mu\nu} \equiv \partial_\mu W_\nu - \partial_\nu W_\mu -ig_2[W_\mu,W_\nu][/itex]
[itex]B_{\mu\nu} \equiv \partial_\mu B_\nu - \partial_\nu B_\mu[/itex]
The "3" EL equations are:
[itex]\partial_\nu B^{\mu\nu} = ig_1\phi^\dagger(D^\mu\phi)-ig_1(D^\mu\phi)^\dagger\phi[/itex]
[itex][D_\nu,W^{\mu\nu}] = ig_2(D^\mu\phi)\phi^\dagger-ig_2\phi(D^\mu\phi)^\dagger[/itex]
[itex]D_\mu D^\mu \phi = -\dfrac{\mu^2}{v^2}(|\phi|^2-\dfrac{1}{2}v^2)\phi[/itex]
From the first equation you can derive the continuity equation:
[itex](D_\mu\phi)^\dagger D^\mu\phi + \phi^\dagger D_\mu D^\mu \phi = 0[/itex]
and then using the third equation you find:
[itex](D_\mu\phi)^\dagger D^\mu\phi = \dfrac{\mu^2}{v^2}(|\phi|^2-\dfrac{1}{2}v^2)\phi[/itex]
if you then substitute this back into L, you find the new Lagrangian:
[itex]L=-\dfrac{1}{4}Tr(B_{\mu\nu}B^{\mu\nu} + W_{\mu\nu}W^{\mu\nu}) + \dfrac{\mu^2}{2v^2}|\phi|^4 - \dfrac{1}{8}\mu^2 v^2[/itex]
so that the new Lagrangian has a completely isolated Higgs field, and no spontaneously broken symmetry! Not only this but the mass term for the Higgs, and the coupling terms for the EW fields have disappeared! Now I am aware that the EL equations are only considered valid on-shell, so this suggests to me that the Higgs mechanism is entirely a quantum effect and would not occur for classical fields... Am I making some mistake here?