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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?)

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Let me add another question, all in good Totalitarian principle's spirit; is $\theta_{W}$ allowed to be another field or it must be a universal constant? if its allowed then it must be true right? are there any theoretical reasons to forbid that? – lurscher Apr 23 '11 at 20:39
@lurscher Since $g\mathbf{sin}\theta_W=g'\mathbf{cos}\theta_W$ from electroweak unification, $\mathbf{sin}\theta_W=\frac{g'}{\sqrt{g^2+g'^2}}$ and $\mathbf{cos}\theta_W=\frac{g}{\sqrt{g^2+g'^2}}$. So what you're asking is whether the coupling constants can be treated as dynamical fields. I've heard of that sort of treatment somewhere before, either here on this website or on Lubos Motl's blog. – dbrane Apr 23 '11 at 22:58
@lurscher It was neither -- I read it here . Since I know very little about strings, I will comment no further and wait for someone else to say more about that. – dbrane Apr 23 '11 at 23:01

1 Answer 1

up vote 6 down vote accepted

If you break the $$SU(2)\times U(1)$$ symmetry spontaneously


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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Ah so when you break the symmetry, you get terms whose Feynman diagrams transform one gauge boson to another (clearly fishy), and the combination which diagonalizes the mass matrix is the one that doesn't do that -- hence its physical. Excellent. The universe continues to be safe thanks to you. +1! – dbrane Apr 23 '11 at 22:46
@dbrane: you might also want to read up on Goldstone's theorem, which tells you why you would expect a massless boson to arise out of symmetry breaking. – Jerry Schirmer Apr 24 '11 at 0:16
@Jerry I fixed your link (HTML doesn't work in comments). – David Z Apr 24 '11 at 0:21

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