Not exactly “tomorrow” but “better late than never”. Okay, I hope you will enjoy the story that is if I can get the LaTex to work.
Let \Psi_{1} and \Psi_{2} be two free fermion fields. If m_{1}=m_{2}, we can treat \Psi = \begin{pmatrix}\Psi_{1} \\ \Psi_{2} \end{pmatrix} as a vector in 2-dimensional (internal) complex vector space \mathcal{V}^{(2)}. Since the group SU(2) acts naturally on \mathcal{V}^{(2)}, we can use \Psi and its (Dirac) conjugate \bar{\Psi} to form the following SU(2) invariant Lagrangian
\mathcal{L} = \bar{\Psi}\left(i \gamma^{\mu}\partial_{\mu} + m I_{(2)} \right) \Psi .
Since a mass term does not alter the form of Noether currents, we will set m = 0 and take
<br />
\begin{equation}<br />
\mathcal{L}_{0} = i \bar{\Psi}\gamma^{\mu}\partial_{\mu}\Psi ,<br />
\end{equation}<br />
to be our free Lagrangian. The invariance of \mathcal{L}_{0} under the infinitesimal SU(2),
<br />
\begin{equation*}<br />
\delta \Psi_{i} = -i \epsilon^{a}\left( \frac{\tau^{a}}{2} \right)_{ij} \Psi_{j} , \ \ i = 1,2<br />
\end{equation*}<br />
or,
<br />
\begin{equation}<br />
\delta^{a} \Psi_{i} = -i \left( \frac{\tau^{a}}{2} \right)_{ij} \Psi_{j} , \ \ \ a = 1,2,3<br />
\end{equation}<br />
leads to the following (conserved) Noether currents
<br />
\begin{equation*}<br />
J^{a\mu} = \frac{\partial \mathcal{L}_{0}}{\partial(\partial_{\mu}\Psi_{i})} \delta^{a}\Psi_{i} ,<br />
\end{equation*}<br />
<br />
\begin{equation}<br />
J^{a\mu} = \bar{\Psi}_{i} \gamma^{\mu}\left(\frac{\tau^{a}}{2}\right)_{ij}\Psi_{j} .<br />
\end{equation}<br />
The corresponding time-independent charges are, then, given by
<br />
\begin{equation}<br />
T^{a} = \int d^{3}x J^{a0}(x) = \int d^{3}x \ \Psi^{\dagger}_{i}\left(\frac{\tau^{a}}{2}\right)_{ij} \Psi_{j} .<br />
\end{equation}<br />
Okay, let me pause for a minute and ask you to do the following exercise:
The canonical formalism (for fermions) rest on the following equal-time anti-commutation relations
<br />
\begin{equation}<br />
\big\{ \Psi^{\dagger}_{i}(t,\mathbf{x}) , \Psi_{j}(t,\mathbf{y}) \big\} = \delta_{ij}\delta^{3}(\mathbf{x} - \mathbf{y}),<br />
\end{equation}<br />
<br />
\begin{equation*}<br />
\big\{ \Psi_{i}(t,\mathbf{x}) , \Psi_{j}(t,\mathbf{y}) \big\} = \big\{ \Psi^{\dagger}_{i}(t,\mathbf{x}) , \Psi^{\dagger}_{j}(t,\mathbf{y}) \big\} = 0 .<br />
\end{equation*}<br />
Show that the canonical Noether charges T^{a} generate the correct infinitesimal transformations on the fields
<br />
\begin{equation}<br />
\left[iT^{a} , \Psi_{i}(x)\right] = \delta^{a}\Psi_{i}(x) ,<br />
\end{equation}<br />
and form a representation of the Lie algebra of SU(2)
<br />
\begin{equation}<br />
\left[T^{a} , T^{b}\right] = i \epsilon^{abc} T^{c} .<br />
\end{equation}<br />
And finally, show that the Noether current J^{a\mu} transforms in the adjoint representation of SU(2)
<br />
\begin{equation}<br />
\delta^{a} J^{b}_{\mu}(x) = \left[iT^{a} , J^{b}_{\mu}(x)\right] = - \epsilon^{abc} J^{c}_{\mu}(x) .<br />
\end{equation}<br />
Hint: use the algebra
<br />
\begin{equation}<br />
\left[\frac{\tau^{a}}{2} , \frac{\tau^{b}}{2}\right] = i \epsilon^{abc} \frac{\tau^{c}}{2} ,<br />
\end{equation}<br />
and the identity
<br />
\begin{equation*}<br />
\left[AB , C \right] = A \big\{ B , C \big\} - \big\{ A , C \big\} B .<br />
\end{equation*}<br />
For later use, let’s define the “ladder” currents by
<br />
\begin{equation}<br />
J^{\mu}_{\pm} = J^{1\mu} \pm i J^{2\mu}.<br />
\end{equation}<br />
The corresponding ladder charges are, then, given by
<br />
\begin{equation}<br />
T^{\pm} = T^{1} \pm i T^{2} .<br />
\end{equation}<br />
Indeed, using the anti-commutation relations (5), it is easy to show that T^{\pm} and T^{3} form, as they should, a closed ladder algebra
<br />
\begin{align*}<br />
\left[ T^{+} , T^{-} \right] &= 2 \ T^{3} \\<br />
\left[ T^{\pm} , T^{3} \right] &= \mp T^{\pm} .<br />
\end{align*}<br />
Notice that
<br />
\begin{equation*}<br />
T^{+} = \int d^{3}x \ \Psi_{1}^{\dagger}(x) \Psi_{2}(x) = \left(T^{-}\right)^{\dagger} ,<br />
\end{equation*}<br />
destroys type-2 fermion and creates type-1 fermion.
Of course, any SU(2)-invariant Lagrangian is also invariant under independent global U(1) transformations. Infinitesimally,
<br />
\begin{equation}<br />
\delta \Psi_{i} = i \theta \Psi_{i} ,<br />
\end{equation}<br />
with \theta being an arbitrary constant, leads to a conserved U(1) current
<br />
\begin{equation}<br />
j^{\mu} = \bar{\Psi}_{i} \gamma^{\mu} \Psi_{i} ,<br />
\end{equation}<br />
and time-independent charge
<br />
\begin{equation}<br />
Y = \int d^{3}x j^{0}(x) = \int d^{3}x \ \Psi^{\dagger}_{i}(x)\Psi_{i}(x) .<br />
\end{equation}<br />
Since we don’t yet know the nature of this U(1) symmetry, we will call its charge, Y, the fermion number.
So, the global symmetry group of our theory is the non semi-simple group SU(2) \times U(1). Indeed, using the anti-commutation relations (5), you can easily show that the U(1) charge Y commutes with the SU(2) charges T^{a}
<br />
\begin{equation}<br />
[ Y , T^{a}] = 0 , \ \ \forall a (= 1,2,3) .<br />
\end{equation}<br />
This means that the two particles described by the fields \Psi_{1} and \Psi_{2} must have the same fermion number Y. Using this fact for all possible U(1) groups, one can determine, almost uniquely, the particles content of all possible SU(2) doublets.
Since we are at it, let us gauge our global symmetry group SU(2) \times U(1). The detail of how to gauge the global symmetry of a generic field theory is explained in the PDF below. However, the details are irrelevant for our discussion here. All what we need to know is the fact that gauging SU(2) \times U(1) forces us to introduce four real massless gauge fields: an SU(2) triplet W_{\mu}^{a}, transforming in the adjoint representation, to couple to the matter current J^{a\mu} of SU(2), and a U(1) gauge field B_{\mu} to couple with the U(1) matter current j^{\mu}. That is to say that we need to modify our free Lagrangian \mathcal{L}_{0} by adding the following interaction terms
<br />
\begin{equation}<br />
\mathcal{L}_{1} = gW^{a}_{\mu}J^{a\mu} + \frac{\bar{g}}{2} B_{\mu}j^{\mu} ,<br />
\end{equation}<br />
where g and \bar{g} are the coupling parameters and the factor 1/2 is for convenience.
Of course for a complete dynamical model, we also need to include the kinetic terms for the gauge fields. However, this is, again, an irrelevant issue for us.
Now, from the definition of the ladder currents (10), we have
<br />
\begin{align*}<br />
J^{1\mu} &= \frac{1}{2}(J^{\mu}_{+} + J^{\mu}_{-}) \\<br />
J^{2\mu} &= \frac{1}{2i}(J^{\mu}_{+} - J^{\mu}_{-}) .<br />
\end{align*}<br />
Inserting these currents in (16), we obtain
<br />
\begin{equation}<br />
\begin{split}<br />
\mathcal{L}_{1} =& \frac{g}{\sqrt{2}} J^{\mu}_{+} \left[ \frac{1}{\sqrt{2}}\left( W^{1}_{\mu} - i W^{2}_{\mu} \right)\right] + \frac{g}{\sqrt{2}} J^{\mu}_{-} \left[ \frac{1}{\sqrt{2}}\left( W^{1}_{\mu} + i W^{2}_{\mu} \right)\right] \\<br />
& + g W^{3}_{\mu} J^{3\mu} + \frac{\bar{g}}{2} B_{\mu}j^{\mu} .<br />
\end{split}<br />
\end{equation}<br />
Notice that the first line in (17) contains complex (hence electrically charged) vector fields, and in the second line only real (i.e., electrically neutral) vector fields enter. In other words, the first line represents the charged current interaction Lagrangian \mathcal{L}_{CC}, and the second line is the neutral current interaction Lagrangian \mathcal{L}_{NC}.
Now, we define the complex (i.e., charged) fields
<br />
\begin{equation}<br />
W^{\pm} = \frac{1}{\sqrt{2}}( W^{1}_{\mu} \mp i W^{2}_{\mu} ) ,<br />
\end{equation}<br />
and rewrite \mathcal{L}_{1} as \mathcal{L}_{CC} + \mathcal{L}_{NC}, where
<br />
\begin{equation}<br />
\mathcal{L}_{CC} = \frac{g}{\sqrt{2}} \left( W^{+}_{\mu} J^{\mu}_{+} + W_{\mu}^{-} J^{\mu}_{-} \right) ,<br />
\end{equation}<br />
and
<br />
\begin{equation}<br />
\mathcal{L}_{NC} = g W^{3}_{\mu} J^{3\mu} + \frac{\bar{g}}{2} B_{\mu}j^{\mu} .<br />
\end{equation}<br />
Let us redefine the neutral fields by
<br />
\begin{equation}<br />
\begin{pmatrix} W^{3}_{\mu} \\<br />
B_{\mu} \end{pmatrix}<br />
= \begin{pmatrix}<br />
\cos \theta & \sin \theta \\<br />
- \sin \theta & \cos \theta<br />
\end{pmatrix}<br />
\begin{pmatrix} Z_{\mu} \\<br />
A_{\mu} \end{pmatrix} .<br />
\end{equation}<br />
Since fields redefinition should not increase the number of parameters, \theta must be a function of the couplings g and \bar{g}. Mathematically, \theta (g , \bar{g}) can be any (arbitrary) function of its arguments. However, if our SU(2)\times U(1) model contains, beside the fermion doublet \Psi, a doublet \Phi of complex scalar fields which can develop non-zero vacuum expectation value \langle \Phi \rangle_{0}, remarkable physical consequences follow when we choose
<br />
\begin{equation}<br />
\tan \theta = \frac{\bar{g}}{g} .<br />
\end{equation}<br />
Indeed, with this choice only the A_{\mu} field remains massless. This means that (22) determines the following pattern of symmetry breaking
<br />
\begin{equation}<br />
SU(2) \times U_{Y}(1) \to U_{em}(1) .<br />
\end{equation}<br />
To see this, consider the gauge invariant kinetic part of the scalar field Lagrangian, (D_{\mu} \Phi)^{\dagger}(D^{\mu}\Phi), where the covariant derivative is given by
<br />
\begin{equation*}<br />
D_{\mu}\Phi = (\partial_{\mu} - i \frac{g}{2} W^{a}_{\mu}\tau^{a} - i \frac{\bar{g}}{2} B_{\mu}) \begin{pmatrix} 0 \\ v + h(x)/ \sqrt{2} \end{pmatrix} ,<br />
\end{equation*}<br />
or
<br />
\begin{equation}<br />
D_{\mu}\Phi = \begin{pmatrix} -i \frac{g}{\sqrt{2}} W_{\mu}^{-} \\ \partial_{\mu} + \frac{ig}{2}( W^{3}_{\mu} - \frac{\bar{g}}{g}B_{\mu}) \end{pmatrix} \left(v + \frac{h(x)}{\sqrt{2}}\right) ,<br />
\end{equation}<br />
with \frac{h(x)}{\sqrt{2}} represents a small perturbation (the Higgs field) on the chosen physical vacuum \langle 0|\Phi|0 \rangle \equiv \langle \Phi \rangle_{0} = (0 , v)^{T}.
In terms of the fields (Z_{\mu},A_{\mu}), as given by (21), we find
<br />
\begin{equation}<br />
W^{3}_{\mu} - \frac{\bar{g}}{g}B_{\mu} = Z_{\mu}( \cos \theta + \frac{\bar{g}}{g} \sin \theta ) + A_{\mu}( \sin \theta - \frac{\bar{g}}{g} \cos \theta ) .<br />
\end{equation}<br />
So, choosing \theta as in (22) we find that
<br />
\begin{equation}<br />
W^{3}_{\mu} - \frac{\bar{g}}{g}B_{\mu} = \frac{Z_{\mu}}{\cos \theta} .<br />
\end{equation}<br />
Thus, the field A_{\mu} drops out of the covariant derivative,
<br />
\begin{equation}<br />
D_{\mu}\Phi = \begin{pmatrix} - i \frac{g}{\sqrt{2}}W_{\mu}^{-} \\<br />
\partial_{\mu} + \frac{i g}{2 \cos \theta}Z_{\mu} \end{pmatrix} \left(v + \frac{h(x)}{\sqrt{2}} \right) ,<br />
\end{equation}<br />
and, therefore, it (the A_{\mu}) remains massless. The masses of the other vector bosons, (W^{\pm}_{\mu},Z_{\mu}), are obtained by expanding the kinetic term
<br />
\begin{equation}<br />
\begin{split}<br />
(D_{\mu}\Phi)^{\dagger}(D_{\mu}\Phi) =& \left(\frac{g^{2}v^{2}}{2}\right) \eta^{\mu\nu}W_{\mu}^{-}W^{+}_{\nu} + \frac{1}{2}\left(\frac{g^{2}v^{2}}{2 \cos \theta}\right) Z_{\mu}Z^{\mu} \\<br />
& + \eta^{\mu\nu}\partial_{\mu}H \partial_{\nu}H + \mathcal{L}_{int}(W,H) + \mathcal{L}_{int}(Z,H) .<br />
\end{split}<br />
\end{equation}<br />
From this we can read off the masses of the IVB
<br />
\begin{equation}<br />
M_{W^{\pm}} = \frac{gv}{\sqrt{2}}, \ \ M_{Z} = \frac{M_{W}}{\cos \theta} .<br />
\end{equation}<br />
Okay, now let’s go back to the neutral current interaction lagrangian \mathcal{L}_{NC} of (20). Substituting (21) and the choice (22) in (20), we obtain
<br />
\begin{equation}<br />
\begin{split}<br />
\mathcal{L}_{NC} =& \frac{g}{\cos \theta} Z_{\mu} \left( J^{3\mu}\cos^{2}\theta - \frac{1}{2} j^{\mu} \sin^{2}\theta \right) \\<br />
& + g \sin \theta A_{\mu} \left( J^{3\mu} + \frac{1}{2} j^{\mu} \right) .<br />
\end{split}<br />
\end{equation}<br />
If A_{\mu} is to be the massless photon field, then the second line in (30) must be the pure electromagnetic interaction eA_{\mu}J^{\mu}_{em}. Thus, we are led to the following identifications of the couplings
<br />
\begin{equation}<br />
e = g \sin \theta ,<br />
\end{equation}<br />
and currents
<br />
\begin{equation}<br />
j^{\mu} = 2 (J^{\mu}_{em} - J^{3\mu}) .<br />
\end{equation}<br />
Setting \mu = 0 in (32) and integrating, we obtain an expression for the generator Y of the original U(1) group
<br />
\begin{equation}<br />
Y = 2 (Q_{em} - T^{3}) .<br />
\end{equation}<br />
This relation is exactly the Gell-Mann-Nishijima relation relating the (strong) hypercharge to the electric charge and the third isospin component of hadrons. This is why we call Y the weak hypercharge.
Now, substituting (31) and (32) in (30) leads to
<br />
\begin{equation}<br />
\mathcal{L}_{NC} = e A_{\mu} J_{em} + \frac{g}{\cos \theta} Z_{\mu} J^{\mu}_{NC} ,<br />
\end{equation}<br />
where
<br />
\begin{equation}<br />
J^{\mu}_{NC} \equiv J^{3\mu} - J^{\mu}_{em} \sin^{2} \theta ,<br />
\end{equation}<br />
is the weak neutral current.
The last thing I would like to do is to calculate the size of (vev) the vacuum expectation value v. Using (19), we can write down the charged IVB amplitude
<br />
\begin{equation}<br />
\mathscr{M}^{CC} = \left( \frac{g}{\sqrt{2}} J^{\mu}_{+} \right) \left( \frac{\eta_{\mu\nu} - \frac{q_{\mu}q_{\nu}}{M_{W}^{2}}}{M_{W}^{2} - q^{2}} \right) \left(\frac{g}{\sqrt{2}} J^{\nu}_{-} \right) .<br />
\end{equation}<br />
At very low energies q^{2}_{\mbox{max}} \ll M_{W}^{2}, the amplitude
<br />
\begin{equation}<br />
\mathscr{M}^{CC} \approx (\frac{g}{\sqrt{2}}J^{\mu}_{+}) (\frac{\eta_{\mu\nu}}{M^{2}_{W}}) (\frac{g}{\sqrt{2}}J^{\nu}_{-}) ,<br />
\end{equation}<br />
should be comparable with the charge current weak amplitude of V-A (4 fermions) theory
<br />
\begin{equation}<br />
\mathscr{M}^{CC} = \frac{4 G_{F}}{\sqrt{2}} \eta_{\mu\nu}J^{\mu}_{+}J^{\nu}_{-} ,<br />
\end{equation}<br />
where G_{F} \sim 10^{-5}/m^{2}_{p} is the Fermi coupling constant with m_{p} being the proton mass.
Thus, we can make the identification
<br />
\begin{equation}<br />
\frac{g^{2}}{8M^{2}_{W}} = \frac{G_{F}}{\sqrt{2}} .<br />
\end{equation}<br />
Using (29), this implies that the vacuum expectation value has the size
<br />
\begin{equation}<br />
v \sim \sqrt{\frac{5}{\sqrt{2}}} 10^{2}m_{p} \approx 176 \mbox{GeV} .<br />
\end{equation}<br />