Why are the masses of $W^{\pm}$ and $Z^0$ different? We know that through the Higgs phenomenon, the weak bosons become massive. In our Lagrangian the $W^\pm$ boson is usually defined as $\frac{1}{\sqrt{2}}(W^1_\mu\mp iW^2_\mu)$ and $Z^o$ is usually defined as $(-B_\mu+W^3_\mu)$ ignoring pre factors and couplings. Because of these definitions the masses of $W^\pm$ and $Z^o$ are different. Is the reason of these definitions purely experimental? Or was there a reason for doing this purely from theoretical grounds?
 A: You could ignore over-all factors, but definitely not couplings. 
The (not quite purely) theoretical mass term in the generic Weinberg-Salam model is, instead, proportional to
$$
(W_\mu^1)^2+ (W_\mu^2)^2+\left (W_\mu^3-\frac{g'}{g} B_\mu\right ) ^2,
$$
(not quite purely, as the form was all but suggested by a skein of experimental facts--a hugely long story involving the neutrality of neutrinos and the chirality of the charged currents, ultimately spelled out by Glashow in 1961). 
But, certainly, the magnitude of the weak mixing, 
$$
\frac{g'}{g} \equiv \tan \theta_W
$$
is a purely experimental fact of life. (Theoretically arbitrary, unless you joined speculative  explanatory schemes in GUTs, etc...)
Theoretically, if nature chose $g'=0$, so $\theta_W=0$, the mass of the Z would be the same as that of the charged Ws, since the argument of the third parenthesis is  $Z_\mu / \cos  \theta_W$. 
But, experimentally, nature chose $\sin^ 2 \theta _W$ = 0.2397 ± 0.0013 instead of 0. "Nobody really knows why"...
