What are the differences between electroweak interactions before and after unification?

Question

I am very confused by this point, although its mathematical description is not hard. I still cannot see how these two theories are "unified", which term in lagrangian indicates this unification? Can someone give me some simple examples what we can/cannot do before/after unification?

Answer 1

There is no before and after unification. They are unified. This is why Griffiths suggests "hidden symmetry" rather than "broken symmetry."

[the Lagranians before and after symmetry breaking] "describe exactly the same physical system; all we have done is to select a convenient gauge and rewrite the the fields in terms of fluctuations about a particular ground state. We have sacrificed manifest symmetry in favor of a notation that makes the physical content more transparent..."

-Griffiths. Introduction to Elementary Particles. 2nd Edition. Page 380.

Which term in the Lagrangian indicates this unification?

The gauge fields $B^\mu$ and $W^\mu$ in the covariant derivative and $B^{\mu\nu}$ and $W^{\mu\nu}$ (these are also defined in terms of $B^{\mu}$ and $W^{\mu}$).

$L=i \bar{\psi} \gamma_\mu (\partial^\mu -igB^\mu-ig'W^\mu)\psi - \frac{1}{4}Tr(W_{\mu\nu} W^{\mu\nu})-\frac{1}{4}Tr(B_{\mu\nu}B^{\mu\nu}) $

$\psi$ is the Dirac Spinor.

Include the spin-0 doublet $\phi$ in the Lagrangian through

$((\partial^\mu +igB^\mu+ig'W^\mu)\phi^\dagger)((\partial^\mu -igB^\mu-ig'W^\mu)\phi)+\rho^2 \phi^\dagger \phi - \lambda (\phi^\dagger \phi)^2$

where

$V(\phi)=-\rho^2 \phi^\dagger \phi + \lambda (\phi^\dagger \phi)^2$ is the Higgs Potential.

Find where the potential is minimum.

$\frac{\partial V}{\partial \phi} = 0$

Therefore,

$\phi_{\text{min}}=\sqrt{\frac{\rho^2}{2\lambda}} e^{i\theta}$

Break the symmetry at this minimum. This is a very long and tedious calculation. For details, see Physics From Symmetry. 2nd Edition. Page 157.

The end result is a Lagrangian with mass terms

${M^2}_W ({W^+} )_{\mu} ({W^-})^\mu + \frac{1}{2} {M^2 }_Z + 0 {A^2}_{\mu}$

indicating that the photon has zero mass. In this long calculation, $A^\mu$ arises as a linear combination of the gauge fields $W_{\mu}$ and $B_\mu$ .