# Why $SU(2) \times U(1)$ for the SM?

1. Dec 2, 2015

### ChrisVer

Just a question that had always been bugging me, yet I didn't have the chance to ask...
How was WGS model of $SU(2) \times U(1)$ come into the game as the theory of leptons? in other words, what led them (gave them hints) in choosing this successful group combination instead of some other?

2. Dec 2, 2015

Staff Emeritus
U(1) looks like electromagnetism, SU(2) looks like isospin. Why not start from what you already know?

3. Dec 2, 2015

### samalkhaiat

For simplicity let us focus just on the electron and its neutrino. It was known that the weak interaction is governed by an IVB theory with $(A-V)$-type charged current
$$J^{\mu} = \bar{\nu}_{e}\gamma^{\mu}(1-\gamma_{5}) e ,$$ with two corresponding weak charges
$$T_{+} = \frac{1}{2} \int d^{3}x \ \nu^{\dagger}_{e}\gamma^{\mu}(1-\gamma_{5}) e = T_{-}^{\dagger} .$$ On the other hand there was the electromagnetic interaction $j^{\mu}A_{\mu}$ of these leptons with the em-current and its charge was given by
$$j^{\mu} = \bar{e}\gamma^{\mu}e , \ \ \ Q = \int d^{3}x \ e^{\dagger}e .$$
As far back as 1957 and based on the vectorial nature of both interactions, Schwinger suggested the idea of weak and electromagnetic unification. Now, the simplest gauge group with 3 generators is $SU(2)$. However, the 3 charges $(T_{\pm},Q)$ do not form a closed algebra. Indeed, it is easy to show that
$$[T_{+},T_{-}] = \frac{1}{2}T_{3} \neq Q . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ The fact that $Q$ cannot be a generator of $SU(2)$ with only two leptons is easy to see: (1) The generators of $SU(2)$ are traceless. So, in order for $Q$ to be a generator the electric charges of an irreducible multiplet must add up to zero. Clearly, the doublet formed out of $\nu_{e}$ and $e$ does not satisfy this condition. (2) $Q$ corresponds to purely vector current, while the weak charges $T_{\pm}$ correspond to $V-A$ type currents.
(i) The unification group is $SU(2)$. This requires adding new fermions to the multiplet and, therefore, modify the currents so that the new charges $t_{\pm}$ and $q$ form a closed $su(2)$ algebra. In the case at hand, a new positively charged lepton $E^{+}$ was needed to form an electrically neutral $SU(2)$-triplet. And another neutral lepton $N$ was also needed to obtain the $V-A$ nature of the weak current at low energies. In fact, Georgi & Glashow (1972) had formulated such model with 2 triplets and a singlet
$$\frac{1}{2}(1-\gamma_{5}) \left( \begin{array}{c} E^{(+)} \\ \nu_{e} \\ e^{(-)} \end{array} \right) , \ \ \frac{1}{2}(1+ \gamma_{5}) \left( \begin{array}{c} E^{(+)} \\ N \\ e^{(-)} \end{array} \right) , \ \ \frac{1}{2}(1+\gamma_{5}) N .$$
This model leads to 2 (electrically) charged weak currents (therefore they couple to charged vector bosons) with the corresponding weak charges $t_{\pm}$ given by
$$t_{+} = \frac{1}{2} \int d^{3}x \ \left( E^{\dagger} (1-\gamma_{5}) \nu_{e} + \nu_{e}^{\dagger}(1-\gamma_{5})e + E^{\dagger}(1+\gamma_{5})N + N^{\dagger}(1+\gamma_{5})e \right) ,$$
and a single neutral current (which could couple to the photon field) with charge given by
$$q = \int d^{3}x \ (E^{\dagger}E - e^{\dagger}e) .$$
It is simple exercise to show that the algebra of these charges is indeed isomorphic to $su(2)$
$$[t_{+} , t_{-}] = 2q .$$
This model was ruled out because it could not account for the weak neutral-current which was observed in 1973. Notice that the only neutral-current in this model is the electromagnetic current.
(ii) The gauge group is $SU(2) \times U(1)$. To achieve this, one only needed to introduce another gauge field to couple to the weak charge $T_{3}$ as given in equation (1). The four generators, $T_{\pm}, T_{3}$ and $Y \propto (Q - T_{3})$ can now generate the Lie algebra of $SU(2) \times U(1)$. Note that the crucial relation here is
$$[T_{i} , Q - T_{3}] = 0 , \ \ i = 1, 2 , 3 . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$
This option was eventually adopted by Weinberg, Salam and Glashow and you should know the rest of the success story of this particular choice for the gauge group, namely anomaly cancellation, unitarity etc.
Okay, that was history and phenomenology, but one can derive the $SU(2) \times U(1)$ model from pure field theoretical arguments. I will do that for you if I find spare time tomorrow.

Last edited: Dec 2, 2015
4. Dec 16, 2015

### samalkhaiat

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
$$\mathcal{L}_{0} = i \bar{\Psi}\gamma^{\mu}\partial_{\mu}\Psi ,$$
to be our free Lagrangian. The invariance of $\mathcal{L}_{0}$ under the infinitesimal $SU(2)$,
$$\begin{equation*} \delta \Psi_{i} = -i \epsilon^{a}\left( \frac{\tau^{a}}{2} \right)_{ij} \Psi_{j} , \ \ i = 1,2 \end{equation*}$$
or,
$$\delta^{a} \Psi_{i} = -i \left( \frac{\tau^{a}}{2} \right)_{ij} \Psi_{j} , \ \ \ a = 1,2,3$$
leads to the following (conserved) Noether currents
$$\begin{equation*} J^{a\mu} = \frac{\partial \mathcal{L}_{0}}{\partial(\partial_{\mu}\Psi_{i})} \delta^{a}\Psi_{i} , \end{equation*}$$
$$J^{a\mu} = \bar{\Psi}_{i} \gamma^{\mu}\left(\frac{\tau^{a}}{2}\right)_{ij}\Psi_{j} .$$
The corresponding time-independent charges are, then, given by
$$T^{a} = \int d^{3}x J^{a0}(x) = \int d^{3}x \ \Psi^{\dagger}_{i}\left(\frac{\tau^{a}}{2}\right)_{ij} \Psi_{j} .$$
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
$$\big\{ \Psi^{\dagger}_{i}(t,\mathbf{x}) , \Psi_{j}(t,\mathbf{y}) \big\} = \delta_{ij}\delta^{3}(\mathbf{x} - \mathbf{y}),$$
$$\begin{equation*} \big\{ \Psi_{i}(t,\mathbf{x}) , \Psi_{j}(t,\mathbf{y}) \big\} = \big\{ \Psi^{\dagger}_{i}(t,\mathbf{x}) , \Psi^{\dagger}_{j}(t,\mathbf{y}) \big\} = 0 . \end{equation*}$$
Show that the canonical Noether charges $T^{a}$ generate the correct infinitesimal transformations on the fields
$$\left[iT^{a} , \Psi_{i}(x)\right] = \delta^{a}\Psi_{i}(x) ,$$
and form a representation of the Lie algebra of $SU(2)$
$$\left[T^{a} , T^{b}\right] = i \epsilon^{abc} T^{c} .$$
And finally, show that the Noether current $J^{a\mu}$ transforms in the adjoint representation of $SU(2)$
$$\delta^{a} J^{b}_{\mu}(x) = \left[iT^{a} , J^{b}_{\mu}(x)\right] = - \epsilon^{abc} J^{c}_{\mu}(x) .$$
Hint: use the algebra
$$\left[\frac{\tau^{a}}{2} , \frac{\tau^{b}}{2}\right] = i \epsilon^{abc} \frac{\tau^{c}}{2} ,$$
and the identity
$$\begin{equation*} \left[AB , C \right] = A \big\{ B , C \big\} - \big\{ A , C \big\} B . \end{equation*}$$
For later use, let’s define the “ladder” currents by
$$J^{\mu}_{\pm} = J^{1\mu} \pm i J^{2\mu}.$$
The corresponding ladder charges are, then, given by
$$T^{\pm} = T^{1} \pm i T^{2} .$$
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
\begin{align*} \left[ T^{+} , T^{-} \right] &= 2 \ T^{3} \\ \left[ T^{\pm} , T^{3} \right] &= \mp T^{\pm} . \end{align*}
Notice that
$$\begin{equation*} T^{+} = \int d^{3}x \ \Psi_{1}^{\dagger}(x) \Psi_{2}(x) = \left(T^{-}\right)^{\dagger} , \end{equation*}$$
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,
$$\delta \Psi_{i} = i \theta \Psi_{i} ,$$
with $\theta$ being an arbitrary constant, leads to a conserved $U(1)$ current
$$j^{\mu} = \bar{\Psi}_{i} \gamma^{\mu} \Psi_{i} ,$$
and time-independent charge
$$Y = \int d^{3}x j^{0}(x) = \int d^{3}x \ \Psi^{\dagger}_{i}(x)\Psi_{i}(x) .$$
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 $T^{a}$
$$[ Y , T^{a}] = 0 , \ \ \forall a (= 1,2,3) .$$
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
$$\mathcal{L}_{1} = gW^{a}_{\mu}J^{a\mu} + \frac{\bar{g}}{2} B_{\mu}j^{\mu} ,$$
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
\begin{align*} J^{1\mu} &= \frac{1}{2}(J^{\mu}_{+} + J^{\mu}_{-}) \\ J^{2\mu} &= \frac{1}{2i}(J^{\mu}_{+} - J^{\mu}_{-}) . \end{align*}
Inserting these currents in (16), we obtain
$$\begin{split} \mathcal{L}_{1} =& \frac{g}{\sqrt{2}} J^{\mu}_{+} \left[ \frac{1}{\sqrt{2}}\left( W^{1}_{\mu} - i W^{2}_{\mu} \right)\right] + \frac{g}{\sqrt{2}} J^{\mu}_{-} \left[ \frac{1}{\sqrt{2}}\left( W^{1}_{\mu} + i W^{2}_{\mu} \right)\right] \\ & + g W^{3}_{\mu} J^{3\mu} + \frac{\bar{g}}{2} B_{\mu}j^{\mu} . \end{split}$$
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
$$W^{\pm} = \frac{1}{\sqrt{2}}( W^{1}_{\mu} \mp i W^{2}_{\mu} ) ,$$
and rewrite $\mathcal{L}_{1}$ as $\mathcal{L}_{CC} + \mathcal{L}_{NC}$, where
$$\mathcal{L}_{CC} = \frac{g}{\sqrt{2}} \left( W^{+}_{\mu} J^{\mu}_{+} + W_{\mu}^{-} J^{\mu}_{-} \right) ,$$
and
$$\mathcal{L}_{NC} = g W^{3}_{\mu} J^{3\mu} + \frac{\bar{g}}{2} B_{\mu}j^{\mu} .$$
Let us redefine the neutral fields by
$$\begin{pmatrix} W^{3}_{\mu} \\ B_{\mu} \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} Z_{\mu} \\ A_{\mu} \end{pmatrix} .$$
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
$$\tan \theta = \frac{\bar{g}}{g} .$$
Indeed, with this choice only the $A_{\mu}$ field remains massless. This means that (22) determines the following pattern of symmetry breaking
$$SU(2) \times U_{Y}(1) \to U_{em}(1) .$$
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
$$\begin{equation*} D_{\mu}\Phi = (\partial_{\mu} - i \frac{g}{2} W^{a}_{\mu}\tau^{a} - i \frac{\bar{g}}{2} B_{\mu}) \begin{pmatrix} 0 \\ v + h(x)/ \sqrt{2} \end{pmatrix} , \end{equation*}$$
or
$$D_{\mu}\Phi = \begin{pmatrix} -i \frac{g}{\sqrt{2}} W_{\mu}^{-} \\ \partial_{\mu} + \frac{ig}{2}( W^{3}_{\mu} - \frac{\bar{g}}{g}B_{\mu}) \end{pmatrix} \left(v + \frac{h(x)}{\sqrt{2}}\right) ,$$
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
$$W^{3}_{\mu} - \frac{\bar{g}}{g}B_{\mu} = Z_{\mu}( \cos \theta + \frac{\bar{g}}{g} \sin \theta ) + A_{\mu}( \sin \theta - \frac{\bar{g}}{g} \cos \theta ) .$$
So, choosing $\theta$ as in (22) we find that
$$W^{3}_{\mu} - \frac{\bar{g}}{g}B_{\mu} = \frac{Z_{\mu}}{\cos \theta} .$$
Thus, the field $A_{\mu}$ drops out of the covariant derivative,
$$D_{\mu}\Phi = \begin{pmatrix} - i \frac{g}{\sqrt{2}}W_{\mu}^{-} \\ \partial_{\mu} + \frac{i g}{2 \cos \theta}Z_{\mu} \end{pmatrix} \left(v + \frac{h(x)}{\sqrt{2}} \right) ,$$
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
$$\begin{split} (D_{\mu}\Phi)^{\dagger}(D_{\mu}\Phi) =& \left(\frac{g^{2}v^{2}}{2}\right) \eta^{\mu\nu}W_{\mu}^{-}W^{+}_{\nu} + \frac{1}{2}\left(\frac{g^{2}v^{2}}{2 \cos \theta}\right) Z_{\mu}Z^{\mu} \\ & + \eta^{\mu\nu}\partial_{\mu}H \partial_{\nu}H + \mathcal{L}_{int}(W,H) + \mathcal{L}_{int}(Z,H) . \end{split}$$
From this we can read off the masses of the IVB
$$M_{W^{\pm}} = \frac{gv}{\sqrt{2}}, \ \ M_{Z} = \frac{M_{W}}{\cos \theta} .$$
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
$$\begin{split} \mathcal{L}_{NC} =& \frac{g}{\cos \theta} Z_{\mu} \left( J^{3\mu}\cos^{2}\theta - \frac{1}{2} j^{\mu} \sin^{2}\theta \right) \\ & + g \sin \theta A_{\mu} \left( J^{3\mu} + \frac{1}{2} j^{\mu} \right) . \end{split}$$
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
$$e = g \sin \theta ,$$
and currents
$$j^{\mu} = 2 (J^{\mu}_{em} - J^{3\mu}) .$$
Setting $\mu = 0$ in (32) and integrating, we obtain an expression for the generator $Y$ of the original $U(1)$ group
$$Y = 2 (Q_{em} - T^{3}) .$$
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
$$\mathcal{L}_{NC} = e A_{\mu} J_{em} + \frac{g}{\cos \theta} Z_{\mu} J^{\mu}_{NC} ,$$
where
$$J^{\mu}_{NC} \equiv J^{3\mu} - J^{\mu}_{em} \sin^{2} \theta ,$$
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
$$\mathscr{M}^{CC} = \left( \frac{g}{\sqrt{2}} J^{\mu}_{+} \right) \left( \frac{\eta_{\mu\nu} - \frac{q_{\mu}q_{\nu}}{M_{W}^{2}}}{M_{W}^{2} - q^{2}} \right) \left(\frac{g}{\sqrt{2}} J^{\nu}_{-} \right) .$$
At very low energies $q^{2}_{\mbox{max}} \ll M_{W}^{2}$, the amplitude
$$\mathscr{M}^{CC} \approx (\frac{g}{\sqrt{2}}J^{\mu}_{+}) (\frac{\eta_{\mu\nu}}{M^{2}_{W}}) (\frac{g}{\sqrt{2}}J^{\nu}_{-}) ,$$
should be comparable with the charge current weak amplitude of $V-A$ (4 fermions) theory
$$\mathscr{M}^{CC} = \frac{4 G_{F}}{\sqrt{2}} \eta_{\mu\nu}J^{\mu}_{+}J^{\nu}_{-} ,$$
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
$$\frac{g^{2}}{8M^{2}_{W}} = \frac{G_{F}}{\sqrt{2}} .$$
Using (29), this implies that the vacuum expectation value has the size
$$v \sim \sqrt{\frac{5}{\sqrt{2}}} 10^{2}m_{p} \approx 176 \mbox{GeV} .$$

File size:
343.6 KB
Views:
91