Imposing SU(2) and U(1) local gauge invariance introduces 4 gauge bosons, two of which correspond to $W^{\pm}$ bosons. The other two gauge fields $W^{\mu}_3$ and $B^{\mu}$ however are said not to correspond to $Z$ and $\gamma$ bosons because of the incorrect chiral coupling of the $Z$ boson if this identification is made.
The "physical" bosons are said to be Weinberg rotated by $\sin^2\theta_W=0.23$, where the angle is obtained from experiment. This gives the observed coupling of $Z$ to both left and right chiralities and also a different mass from the $W$ brothers, again observed experimentally.
Why? Why do the physical $Z$ and $\gamma$ bosons have to be Weinberg rotated? (by Nature that is, not by phenomenologists) If there was no rotation, the three $W$ 's would have the same masses and would couple only to lefties. This doesn't immediately seem particularly disastrous to me.
I don't see how Gell-Mann's Totalitarian Principle can save the day here, because what I'm essentially asking is why our particle detectors observe the Weinberg rotated bosons and not the "natural" ones that come out of the Lagrangian? What is it that makes particular linear combinations of bosons the physical ones? And crucially, why aren't the $W^{\pm}$ bosons rotated in a similar manner? (Or would we not notice if they were?)
Answer
If you break the $$SU(2)\times U(1)$$ symmetry spontaneously
$$\left|-ig\frac{\sigma}{2}W_\mu-ig'B_\mu\phi\right|^2$$
and insert the vacuum expectation value of the scalar field you get terms
$$\frac{1}{2}vgW_\mu^+W^{\mu-}+\frac{1}{8}v^2(W_\mu^3,B_\mu)\left( \begin{array}{cc} g^2 & -gg' \\ -gg' & g'^2 \end{array} \right)\left( \begin{array}{c} W^{3\mu} \\ B^\mu \end{array} \right)$$
if you diagonalize the mass matrix on the right to get the physical fields you get
$$\frac{1}{8}v^2[gW_\mu^3-g'B_\mu]^2+0\cdot[g'W_\mu^3+gB_\mu]^2=\frac{1}{2}m_ZZ_\mu^2+\frac{1}{2}m_AA_\mu^2$$, that is a massive Z boson and a massless photon
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