Electroweak unification

As I said, the standard model is made up of two theories. Why not three, since there are three fundamental interactions at the quantum level (we have seen that gravitation is a whole other story)? We have QCD now (strong interaction), QED (electromagnetism), why not add some theory of weak interactions?

The primary reason is that even today, with their vastly different strength levels, electromagnetism and weak interactions seem thoroughly entangled. Remember that the phenomenon that started investigations in the weak force was beta decay of a neutron into a proton (with the emission of an electron and an anti-neutrino).

\[n \rightarrow p + e^- + \bar{\nu}\]

We now know that since \(n=udd\) and \(p=uud\), what is actually happening is the transformation of a down quark into an up quark.

\[d \rightarrow u + e^- + \bar{\nu}\]

Furthermore, we can interpret antiparticles are particles that experience time in reverse. So the antineutrino being "emitted" can be seen as a neutrino "arriving" to be transformed into an electron.

All in all, the weak force seems to be transforming particles into new particles of different electric charges. That’s new! The strong force could turn a blue \(u\) quark into a red \(u\) quark, but color is the strong charge. All other conserved charges were left alone. In a way, the strong force is minding its own business.

In a beta decay, the overall electric charge is conserved, of course. But the standard model does not handle the mixing of leptons (electrons, neutrinos) with quarks. So we’re really looking at two reactions happening "side-by-side": up-down quarks, and electron-neutrino.

Both of them involve a "charged current", and therefore, an electrically charged boson. Gluons were charged, but only with color, the conserved quantity corresponding to "their" symmetry. It seems that to describe electrically charged weak bosons, we need to incorporate electromagnetism as part of the theory!

We could be quite confident. After all we’ve just learned how to develop an \(SU(N)\) gauge theory before. We know that particles that can transform into each other should be placed in doublets. Perhaps those might do:

\[\begin{pmatrix} e \\ \nu \end{pmatrix} , \begin{pmatrix} d \\ u \end{pmatrix}\]

Alas, things are not that simple. As it turns out we need to consider if those particles are left-handed or right-handed. This can seem like a very eccentric statement, so let me explain.

Every quantum particle has a "spin", which can be presented as a rotation degree of freedom (let’s stick to that rough characterization, lest we’re gone on an even longer tangent). The spin can "point" in the same direction as a particle’s momentum: then that particle’s is right-handed. If it points the other way, the particle is left-handed.

Of course, we are in the quantum world, so everything is about fields in superposed states, rather than well-defined particles. But we can use Dirac’s theory to define operators that "isolate" the left-handed and right-handed parts of the fields:

\[\frac{1}{2}(1-\gamma^5)\psi = L\psi = \psi_L \quad \frac{1}{2}(1+\gamma^5) \psi = R\psi = \psi_R\]

\(\gamma^5\) is deduced from the other Dirac matrices (\(\gamma^5=i\gamma^0\gamma^1\gamma^2\gamma^3\)). With that we can prove that these operators actually do behave like projections:

\[\begin{aligned} L+R &= 1 \\ LR=RL&=0 \\ L^2&=L \\ R^2&=R \end{aligned}\]

From all this we can derive these useful equalities: \[\begin{aligned} \bar{\psi}\psi &= \bar{\psi}_R\psi_L+\bar{\psi}_L\psi_R \\ \bar{\psi}\gamma^\mu \psi &= \bar{\psi}_R \gamma_\mu \psi_R+\bar{\psi}_L \gamma_\mu \psi_L \end{aligned}\]

The first equation denotes a mass term: only massive particles have that one featured in the lagrangian. It shows that the left- and right-handed components are mixed together. The second equation is a gauge current: it shows that conversely, those components are not mixed! When two charged particles interact, the interaction of their left-handed components, and that of their right-handed components are perfectly superposed: there’s no mixing.

So far, so good, but what does this handedness business have to do with the weak force. Well, we know it has a lot to do since to a crucial experiment by Chien Shiung Wu. She proved that weak interaction violate parity, in other words, that it doesn’t apply the same to left- and right-handed particles.

To understand how surprising that is, consider that left-handed and right-handed particles are literally mirror reflections of each other. "Common sense", as well as all our physical observations to that point suggest that it’s an absurd thought. Take any physical phenomenon, and flip everything left to right (momenta, forces, positions...), and everything will still follow the law of physics. If I film Rafael Nadal (a lefty) and flip the video so that the court appears inverted, that is still a physical situation. It doesn’t correspond to our world, because in the "flipped world", Nadal is now a righty, he’s serving on the "wrong" side when the game starts, etc. But no law of physics appears to be violated.

Well, Chien Shung Wu, by observing the decay of Cobalt atoms, showed that beta decay could not be flipped. That is remarkable. If you show a photograph of someone I’ve never met, and then its mirror version, both likely look fine and "physical". But in fact, if I had godlike powers allowing me to "see" the many beta decays happening on those photos (say, of Carbon-14), then I could be able to tell for sure which photo is the original and which one was flipped!

More precisely, what has been shown is that weak interactions only concern the left-handed component of particles. The right way to write our beta decay reaction is then:

\[d_L \rightarrow u_L + e_L + \bar{\nu}_L\]

Therefore, our suggestion to use the doublets \((e,\nu)\) and \((d,u)\) does not work. If it did, the right-handed components of different particles could interact. The real \(N\)-uples are then doublets for the left-handed fermion components, and singlets for the right-handed ones:

\[\begin{pmatrix} e_L \\ \nu_L \end{pmatrix} , \begin{pmatrix} d_L \\ u_L \end{pmatrix} , (e_R),(d_R), (u_R)\]

In what follows, to simplify this brief overview, we will consider only leptons, and define:

\[\mathbf{L} = \begin{pmatrix} \nu_L \\ e_L \end{pmatrix} \quad , \quad \mathbf{R} = (e_R ) \quad , \quad \Psi = \begin{pmatrix} \nu \\ e \end{pmatrix}\]

You may notice that there is no right-handed neutrino. That is the hypothesis the standard model makes: the only neutrinos we’ve observed and identified as such were interacting through the weak force (therefore they were left-handed). Since neutrinos were thought to be massless, and therefore constantly travel at \(c\) (the speed of light), they couldn’t change their helicity and would stay left-handed forever (making it an intrinsic property, unlike for massive fermions).

This suggests that the right symmetry group is not of the form \(SU(N)\). Instead we will use a semi-simple Lie group: \(SU(2) \times U(1)\).

\(SU(2)\) will act on the doublets (left-handed particles) and \(U(1)\) on both the doublets and the singlets.

Just like \(SU(3)\) had \((3^2-1)=8\) generators, \(SU(2)\) has \((2^2-1=3)\) generators, 3 matrices of order 2 called the Pauli matrices.

\[\tau^1=\frac{1}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \quad \tau^2=\frac{1}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \quad \tau_3=\frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\]

The commutation relations are: \[[\tau_1, \tau_2] = i \tau_3 \quad [\tau_2, \tau_3] = i \tau_1 \quad [\tau_3, \tau_1] = i \tau_2\]

The \(U(1)\) generator \(\tilde{Y}\) is required by the properties of the algebra to commute with the \(SU(2)\) generators: \([\tilde{Y},\tau^i]=0\). The only solution is for that generator to be a multiple of the identity matrix: \(\tilde{Y} = Y \text{Id}\).

The factor \(Y\) is arbitrary. Furthermore, \(U(1)\) is acting on both doublet and singlet representations, and in principle it doesn’t need to be the same factor (\(Y_L\neq Y_R\)). Conventionally, we choose \(Y_L=-1\). This is called the weak hypercharge of the doublet.

Now, \(U(1)\) seems very familiar from our QED exploration. Is this the same symmetry that governs electromagnetism? We can see it’s not the case; indeed:

\[\tilde{Y}_L \mathbf{L} = Y_L \mathbf{L} = - \begin{pmatrix} \nu_L \\ e_L \end{pmatrix}\]

This means that the neutrino and electron have the same hypercharge. But we know it’s not the case for the electric charge. so \(U(1)_Y\) is not the symmetry of electromagnetism.

The electric charge operator is supposed to have eigenvalues that correspond to the components’ vrious electric charges. In this case:

\[Q\begin{pmatrix} \nu_L \\ e_L \end{pmatrix} = \begin{pmatrix} 0 \\ -e_L \end{pmatrix}\]

Which suggests:

\[Q= \begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix}\]

Looking at our algebra’s generators, we find that:

\[Q = \tau^3 + \frac{\tilde{Y}_L}{2}\]

This matrix equation is also valid in eigenvalue form. Let’s call \(T_3\) the conserved charge from \(\tau^3\), and we get:

\[q = T_3 + \frac{Y_L}{2}\]

For singlets, there is no action of the Pauli matrices and therefore no \(T_3\), so we can conclude \(Y_R= 2q = -2\) (for the electron singlet).

We continue doing what we did for the non-abelian and abelian cases: for \(SU(2)\) we know there will be 3 force bosons (for the 3 generators), the \(W^i_\mu\), and for \(U(1)\) one boson, \(B_\mu\). Then the covariant derivatives applying to left-handed and right-handed components are.

\[\begin{aligned} (D_L)_\mu &= \partial_\mu + i g \tau_i W^i_\mu + i \frac{g'}{2}Y_L B_\mu = \partial_\mu + i g \tau_i W^i_\mu - i \frac{g'}{2} B_\mu \\ (D_R)_\mu &= \partial_\mu + i \frac{g'}{2} Y_R B_\mu = \partial_\mu - i g' B_\mu \end{aligned}\]

The free electroweak lagrangian (without interaction) will look like the previous one we’ve worked on.

We also define:

\[\tilde{m}= \begin{pmatrix} m_e & 0\\ 0 & m_\nu \end{pmatrix}\]

It will be useful since the two particles involved don’t have the same mass (unlike quarks in a \(SU(3)\) triplet). The free Dirac lagrangian is then:

\[\begin{aligned} \mathcal{L}_{EW} &= \bar{\Psi}(i\gamma^\mu \partial_\mu - \tilde{m}) \Psi\\ &= \overline{e} ( i\gamma^\mu \partial_\mu - m_e) e + \bar{\nu} ( i\gamma^\mu \partial_\mu - m_\nu) \nu \end{aligned}\]

So far, so good: the two particles are unmixed, as we expect in the absence of interactions. However this lagrangian is not invariant under global \(SU(2) \times U(1)\) transformations. That is because the mass term mixes the left- and right-handed components, as we’ve seen. That’s a roadblock right from the start! The only way to make it invariant is to impose \(m_e = m_\nu = 0\). But we know that electrons have mass (and we think neutrinos have)... All we can do is accept it for now. It will be resolved in spectacular fashion later.

The full EW lagrangian (still massless) is then given by:

\[\mathcal{L}_{EW}=\underbrace{ \bar{ \mathbf{L}} i\gamma^\mu (D_L)_\mu \mathbf{L} + \mathbf{R} i\gamma^\mu (D_R)_\mu \mathbf{R} }_\text{Fermion fields} -\underbrace{\tfrac{1}{4} ( W^{\mu \nu}_i W^i_{\mu \nu} + B^{\mu \nu}B_{\mu \nu})}_\text{Kinetic Boson fields}\]

Let’s focus on the fermion fields:

\[\mathcal{L}_{EW}^{fermions}=\underbrace{ \bar{ \Psi} i\gamma^\mu (D_L)_\mu \Psi }_\text{Free fermions} + \underbrace{ g W^\mu_i \bar{\mathbf{L}} \gamma_\mu \tau ^i \mathbf{L} - \frac{g'}{2} B^\mu \bar{\mathbf{L}} \gamma_\mu Y \mathbf{L} - g' B^\mu \bar{\mathbf{R}} \gamma_\mu Y \mathbf{R} }_{\text{Fermions-Bosons interactions}}\]

Let’s consider the interaction terms, starting with the one involving non-diagonal matrices:

\[g\bar{\mathbf{L}} \gamma_\mu (\tau^1 W^\mu_1 + \tau^2 W^\mu_2 ) \mathbf{L} = \frac{g}{2}\bar{\mathbf{L}} \gamma_\mu \begin{pmatrix} 0 & W_\mu^1 -iW_\mu^2 \\ W_\mu^1 + iW_\mu^2 & 0\end{pmatrix} \mathbf{L}\]

It makes sense to rewrite the bosons as \(W_\mu^\pm \hat{=} \frac{1}{\sqrt{2}} W_mu^1 \mp i W^\mu_2\). Then the term above becomes:

\[\begin{aligned} \mathcal{L}_{EW}^{\pm int}& = \frac{g}{2\sqrt{2}} ( \overline{\nu_L} \gamma^\mu e_L W_\mu^+ + \overline{e_L} \gamma^\mu \nu_L W_\mu^- ) \end{aligned}\]

As expected, we have weak charged currents transforming the doublet particles into each other.

Let’s look at the neutral term now:

\[\begin{aligned} \mathcal{L}_{EW}^{0 \, int}& = g \bar{\mathbf{L}} \gamma_\mu \tau^3 W^\mu_3 \mathbf{L} - \frac{g'}{2} B^\mu \bar{\mathbf{L}} \gamma_\mu \mathbf{L} - g' B^\mu \bar{\mathbf{R}} \gamma_\mu \mathbf{R} \\ &= \frac{g}{2} ( \overline{\nu_L} \gamma^\mu \nu_L - \overline{e_L} \gamma^\mu e_L) W_\mu^3 - \frac{g'}{2} ( \overline{\nu_L} \gamma^\mu \nu_L + \overline{e_L} \gamma^\mu e_L + 2\overline{e_R} \gamma^\mu e_R) B_\mu \\ &= g j_3^\mu W^3_\mu + g' j^\mu_Y B_\mu \end{aligned}\]

\(j_3^\mu\) and \(j^\mu_Y\) are the currents corresponding to the two neutral generators, but none of them corresponds to the electromagnetic current, since they involved neutrinos. The EM current looks like this: \[j_{EM}^\mu = \bar{e} \gamma^\mu e = \overline{e_L} \gamma^\mu e_L+ \overline{e_R} \gamma^\mu e_R\]

We can note that the following holds: \[j_{EM}^\mu = - ( j_{3}^\mu+ \frac{j_{Y}^\mu}{2})\]

What if we could re-combine the \(W^3\) and \(B\) bosons to retrieve something like a photon? The idea is to rotate the bosons vector by an angle \(\theta_W\) to get a new coupling involving \(A\) and \(Z\) bosons. We’ll be hopeful, of course, that \(A\) can be identified with the photon.

\[\begin{pmatrix} A_\mu \\ Z_\mu \end{pmatrix} = \begin{pmatrix} \cos{\theta_W} & \sin{\theta_W}\\ -\sin{\theta_W} & \cos{\theta_W} \end{pmatrix} \begin{pmatrix} B_\mu \\ W^3_\mu \end{pmatrix}\]

We can plug this into our lagrangian:

\[\begin{aligned} \mathcal{L}_{EW}^{0 \, int}& = \begin{pmatrix} \frac{g'}{2} j_Y^\mu & g j_3^\mu \end{pmatrix}\begin{pmatrix} B_\mu \\ W^3_\mu \end{pmatrix} \\ &= \begin{pmatrix} \frac{g'}{2} j_Y^\mu & g j_3^\mu \end{pmatrix} \begin{pmatrix} \cos{\theta_W} & -\sin{\theta_W}\\ \sin{\theta_W} & \cos{\theta_W} \end{pmatrix} \begin{pmatrix} A_\mu \\ Z_\mu \end{pmatrix} \\ \end{aligned}\]

This gives us the answer we wanted: to have an electromagnetic-like term, we need to impose:

\[\tan{\theta_W}=\frac{g'}{g} \quad e=g'\cos{\theta_W}=g\sin{\theta_W}\]

And finally:

\[\begin{aligned} \mathcal{L}_{EW}^{0 \, int}& = e j_{EM}^\mu A_\mu + \frac{e}{\sin{\theta_W}\cos{\theta_W}} ( \frac{j_3^\mu}{2} - \sin^2{\theta_W} j_{EM}^\mu ) Z_\mu\\ \end{aligned}\]

The \(Z\) boson then mediates a weak interaction that acts on all particles: left-handed and right-handed electrons, and the neutrino.

We should note that at this stage the switch from \((W^3,B)\) to \((Z,A)\) is just a mathematical "trick". The underlying physics have not changed at all.

Our electroweak theory now describes the following:

• Massless fermions

• 2 massless charged bosons (\(W^\pm_\mu)\)

• 2 massless neutral bosons (\(Z\),\(A\))

This is still not consistent with reality. Obviously, electrons do have mass. And we also know that the weak bosons have mass, because that’s the only way to explain their very short range. If they were massless like the photon, then the interaction would have infinite range and, among other effects, neutron decay would be quasi-instantaneous, making the world a very different place (with no possible physical structure).

Above all else, we know that the bosons’ mass terms would cause the lagrangian to not be invariant anymore under the symmetry group. That means that in the current universe, the symmetry is broken.

The Higgs mechanism, crowning achievement of the Standard Model

This obviously glaring issue was solved by the application of the now-famous Higgs mechanism to the spontaneous symmetry breaking (SSB) of the electroweak theory. This is not the place to come back on the origins of this discovery, so at the risk of making it a deus ex machina, I will simply present the procedure, which I think is still important given its elegance.

We start by introducing a scalar boson doublet, with one positively charged particle, and a neutral one (a scalar boson means it’s a spin-0 particle which is just a complex function of spacetime, and follows the Klein-Gordon equation, unlike the fermions \(\psi\), which are spinors, or the vector bosons like \(A\_mu\) which are 4-vectors).

\[\Phi = \begin{pmatrix} \phi^+ \\ \phi^0 \end{pmatrix}\]

We associate the following potential with it:

\[V (\Phi^\dagger \Phi)= \mu \Phi^\dagger \Phi + \lambda (\Phi^\dagger \Phi)^2\]

It might seem arbitrary, and to an extent it is — this potential was stumbled upon while studying ferromagnetism and then ported to quantum field theory. But it also happens to be the only acceptable potential for a scalar field. Indeed, we know that we need it to be a function of the norm \(\Phi^\dagger \Phi\) (to be gauge invariant). We also know that if the term with highest power is \((\Phi^\dagger \Phi)^n\), \(n\) has to be an even number; otherwise the potential is not physical because it doesn’t have a minimum (and then no stable state). Lastly, we can show that \(n>3\) doesn’t work because it makes the lagrangian non-renormalizable (in other terms: it is impossible to compute any physical quantity).

This is all well and good, but how can this additional field help us explain the broken symmetry?

To understand this, we need to talk briefly about the vacuum expectation value (VEV). QFT is a perturbative theory. It means that all the calculations we make are from "small" fields. This doesn’t mean that the theory is only approximately true. But the (relatively) simple picture we’ve developed of fermions interacting with force-mediating bosons is an approximation that is incredibly accurate for "small" perturbations of the vacuum. Note that "small" here is very relative. Even the energies from the early universe (say, at \(10^{-30}\text{s}\)) would still qualify as small perturbations. Closer to the Big Bang is when the perturbative approach starts to break down, in a way we’ve described previously.

So, the perturbations are small, but compared to what? It seems natural to say they are close to zero. However, that’s not a given because in a QFT, the quantum vacuum can have a non-zero background energy level! And we will see that such a non-zero VEV is precisely what interacts with particles to give them mass.

This explains why we need this scalar field: it is possible to prove that in a QFT with the basic relativistic symmetry, only such terms can potentially generate a VEV other than zero. Simply speaking the VEV is the minimum value of the potential \(V\) (as always in physics, a system tends to "fall" into the potential well).

As it turns out, the value of that minimum depends on the sign of \(\mu\). For \(\mu\) positive, it is simply zero. So there’s no change compared to before.

\[\mu>0 \rightarrow \Phi^\dagger \Phi \rangle_0 = 0 \rightarrow \langle \Phi \rangle_0 = 0\]

But for negative values of \(\mu\), the minimum of the norm is not nil: it takes a value we will call \(v^2\).

\[\mu<0 \rightarrow \langle \Phi^\dagger \Phi \rangle_0 = -\frac{\mu}{2\lambda} \hat{=} v^2\]

This could be verified for any doublet such that \(\phi^+{*}\phi^+ + \phi{^0*}\phi^0=v^2\). Thanks to the \(SU(2)\) symmetry we can pick any value that’s convenient (whatever the exact VEV we can rotate the whole theory to that value).

The key requirement of the symmetry breaking operation is that we preserve the \(U(1)_{EM}\) symmetry, which we know is still preserved. This happens if the doublet is invariant under \(U(1)_{EM}\) transformations. Since those can be written as \(U=\exp{i\theta Q}\), this is equivalent for this doublet to:

\[Q \langle \Phi \rangle_0 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \langle \Phi \rangle_0 = 0\]

We can then choose:

\[\langle \Phi \rangle_0 = \begin{pmatrix} 0 \\ v \end{pmatrix}\]

Remember that QFT is analysing the Lagrangian perturbatively around the VEV. So now instead of looking at \(\Phi\) we can look at the perturbation around \(v\). This perturbation will be called \(H\), the Higgs boson. (Since none of this can break the electromagnetic symmetry, \(H\) needs to be electrically neutral, hence the field is real).

\[\Phi = \begin{pmatrix} 0\\ v + H \end{pmatrix}\]

Now all we need to do is examine our scalar doublet’s lagrangian. This is a regular Klein-Gordon lagrangian, but since we want to enforce \(SU(2)\) gauge invariance, we use the relevant covariant derivative (with \(Y_\Phi=1):\)

\[\begin{aligned} \mathcal{L}_\Phi &= D_\mu \Phi^\dagger D^\mu \Phi - V(\Phi^\dagger \Phi)\\ \end{aligned}\]

Before symmetry breaking, there’s not much to say about it. It’s entirely invariant under \(SU(2)\times U(1)_{Y}\), so it doesn’t bring anything new. If we developed each term, we would find \(\phi^0\) and \(\phi^+\) interacting with each others as well as with the weak bosons.

Things change when we plug in the new value around the post-SSB VEV.

\[\begin{aligned} \mathcal{L}_\Phi &= D_\mu (v+H)^\dagger D^\mu (v+H) - V((v+H)^2) = \dots =\\ =& \underbrace{ \partial^\mu \partial_\mu H}_{\text{Higgs (scalar) boson}} + \underbrace{\frac{g^2}{2} v^2 (W^+_\mu W^{-\mu} + \frac{1}{2\cos^2{\theta_W}}Z_\mu Z^\mu)}_{\text{Weak bosons' mass}} \\ &+ \underbrace{\frac{g^2}{2} (2vH+H^2) (W^+_\mu W^{-\mu} + \frac{1}{2\cos^2{\theta_W}}Z_\mu Z^\mu)}_{\text{Higgs-weak interactions}} \\ & + \underbrace{2\mu H^2}_{\text{Higgs boson mass}} + \underbrace{ \mu^2(\frac{2}{v}H^3 + \frac{1}{2v^2}H^4-\frac{1}{4} ) }_{\text{Cubic and quartic Higgs terms}} \end{aligned}\]

After some calculations, we find mass terms for the weak bosons! Even better, the calculations confirm that the actual fields getting masses are \(W^+, W^-, Z\). That is not a huge surprise, since \(\Phi\) was by construction invariant under \(U(1)_{EM}\), therefore even after symmetry breaking, the interaction with the \(A\mu\) would just cancel out.

Beyond that, we also get the lagrangian for a scalar boson, the Higgs boson, as well as its self-interaction and interactions with the weak bosons.

It might be enlightening to think in terms of degrees of freedom: the original Higgs field was complex scalar doublets: it had 4 real degrees of freedom. The single real scalar \(H\) only has one. But each of the three massive bosons has one extra degree of freedom than its massless predecessor. It’s like the bosons have "absorbed" part of the Higgs field, but we can see that mathematically we are still consistent.

The first role of the Higgs field is then fulfilled. But that’s not all... It will perform a second role! For a scalar field, we can also introduce an interaction term with the fermions, which takes this form:

\[-G_{H} ( \bar{\mathbf{R}} \Phi^\dagger \mathbf{L} +\bar{\mathbf{L}} \Phi \mathbf{R} )\]

Now, remember why the mass term for the fermion wasn’t gauge invariant. We have shown that it would look like this:

\[-m_e \bar{e}e = -m_e (\overline{e_R}e_L + \overline{e_L}e_R )\]

It cannot be \(SU(2) \times U(1)\)-invariant, because it mixes together a singlet and one component only from a doublet. The interaction term we’ve shown above is invariant though, since all the factors involved are invariant in their own right.

After developing we end up with a mass term! There is also a fermion-Higgs interaction which is called a Yukawa term.

\[-G_{H} ( \bar{\mathbf{R}} \Phi^\dagger \mathbf{L} +\bar{\mathbf{L}} \Phi \mathbf{R} ) = \underbrace{- G_H v \bar{e}e}_{\text{Electron mass}} - \underbrace{G_H \bar{e}e H}_{\text{Yukawa interaction}}\]

The theory doesn’t specify fermion masses since the coupling \(G_H\) is arbitrary and we express it in terms of the empirically observed masses. But the Higgs SSB has played its second role in providing gauge-invariant mass terms for the fermions!

To finish about the Higgs field, we need to talk about the process itself. We’ve showed that symmetry was broken \(\mu\) was negative, but we never explain why its sign would change. As it turns out, all the "constants" that appear in QFT, including \(e\), \(g_S\), \(g_W\) change with the energy level, as we’ve explained while discussiong the evolution of forces’ relative strengths. \(\mu\) is no exception, and as the universe cools down, its value will also diminish, down to the crucial zero point when the symmetry breaks...

So, here we are. Symmetry implies dynamics. We have:

• started from Dirac’s QFT lagrangian, and the technical apparatus around it,

• imposed the right symmetry group (for our world): \(SU(3)_C\times SU(2)_L \times U(1)_Y\),

• used a Higgs mechanism to break the symmetry, leaving us with \(SU(3)_C\times U(1)_{EM}\).

Wit that, we get the Standard Model: the crowning achievement, together with general relativity, of modern physics. It is incredibly exhaustive and displays staggering accuracy at the highest energies we can yet produce.

And it still all comes from that simple lagrangian :

\[\mathcal{L}_{SM}=\sum_G{ ( \underbrace{\bar{\Psi_G} (i\gamma^\mu D_\mu - m) \Psi_G}_\text{Covariant Dirac fields} +\underbrace{\tfrac{1}{4g_G^2}[D^{\mu},D^{\nu}] [D_{\mu},D_{\nu}]}_\text{Covariant gauge fields} ) }+ \underbrace{ \mathcal{L}_\Phi}_{\text{Higgs lagrangian}}\]

In the above, we sum over \(G\), representing the three simple components of the Lie group, with \(\Psi_G\) representing the fermion content as multiplets. It is time to present the field content of the standard model in a more algebraic manner.

Each multiplet of fields corresponds to a representation of the symmetry group. We’ve been talking about \(SU(N)\) as a group of \(N\times N\) matrices, but that’s not the case. What defines the group is the algebraic relationship between its elements. The group of size-\(N\) matrices happens to be the definition that we’ve used, so we call it the fundamental representation. Since \(N\times N\) matrices act on \(N\)-uplets, we say those belong to the fundamental representation.

In \(SU(3)_C\times SU(2)_L \times U(1)_Y\), each multiplet is defined by three data points: the representations that it belongs to relative to each of those groups. Conventionally, this is written as \((R_{SU(3)},R_{SU(2)})_Y\), where \(Y\) is the hypercharge, and defines the representation in \(U(1)\).

We end up with this content:

This is a total of 29 particles. We should note that antiparticles also exist (only for the fermions) and that the corresponding representations are written as \(\bar{\mathbf{3}}\), \(\bar{\mathbf{6}}\) etc., unless the representations are their own conjugate.

A few points deserve an explanation:

• Fermions are in the fundamental representation, which for \(SU(N)\) has \(N\) dimensions. Right-handed quarks are \(SU(3)\) triplets but \(SU(2)\) singlets so their representation is \((\mathbf{3},\mathbf{1})\) (the \(Y\) is deduced from our empirical observation about electric charges). Conversely, left-handed leptons form a \(SU(2)\) doublet but a \(SU(3)\) singlet, hence \((\mathbf{1},\mathbf{2})\). Right-handed electrons are singlets under all symmetries. And left-handed quarks are both triplets and doublets, so they live in \((\mathbf{3},\mathbf{2})\).
  The \(i\) indices are here to remind us that there are three generations of fermions (so \(e^3\) would be the \(\tau\) lepton, \(d^2\) the strange quark, etc.).

• Gauge bosons are in the adjoint representation (a group defined by the generators of the Lie algebra, so it has the same number of dimensions as there are generations. For \(SU(N)\) that is \(N^2-1\).

• The scalar Higgs field was defined as a \(SU(2)\) doublet.