In [itex]SU(2)_{L}\times U(1)_{Y}[/itex] theory, we have two coupling constants [itex]g_{L}[/itex] and [itex]g_{Y}[/itex], and four MASSLESS gauge fields: [itex]W_{\mu}^{a}, \ a = 1,2,3[/itex] and [itex]B_{\mu}[/itex]. We can redefine these fields by introducing two electrically charged fields
[tex]W^{\pm}_{\mu} = \frac{1}{\sqrt{2}}( W^{1}_{\mu} \mp i W^{2}_{\mu}),[/tex]
and two neutral fields
[tex]Z_{\mu} = W^{3}_{\mu}\cos \theta - B_{\mu} \sin \theta[/tex]
[tex]A_{\mu} = W^{3}_{\mu} \sin \theta + B_{\mu} \cos \theta[/tex]
We still have no photon in here, because the gauge group is not [itex]U(1)_{em}[/itex], all fields are still massless and (more important) the two couplings [itex]g_{L}[/itex] and [itex]g_{Y}[/itex] are unrelated.
To break [itex]SU(2)_{L}\times U(1)_{Y}[/itex] down to [itex]U(1)_{em}[/itex], we need to introduce a set of scalar fields [itex]\Phi[/itex] which has [itex]U(1)_{em}[/itex] invariant non-zero vacuum expectation value [itex]< \Phi > = v[/itex], i.e. it vanishes under the action of the [itex]U(1)_{em}[/itex] generator (the electric charge)
[tex]Q_{em}< \Phi > = 0. \ \ \ \ (1)[/tex]
Next, we introduce a small perturbation [itex]H(x)/ \sqrt{2}[/itex] around the VEV of the scalar field [itex]< \Phi >[/itex]. This will provides masses to ALL four gauge fields [itex]W^{\pm}_{\mu}, Z_{\mu}[/itex] and [itex]A_{\mu}[/itex]. So, in order to satisfy eq(1) one of the neutral fields must remain massless, so that it can be identified with the gauge field of the (unbroken) [itex]U(1)_{em}[/itex] group, i.e. the photon. This happens for [itex]A_{\mu}[/itex] provided that we CHOOSE the couplings such that
[tex]g_{Y} = g_{L} \tan \theta .[/tex]
Sam