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The weak force on its own is responsible for various forms of decay or transmutation from one type of particle to another, most notably in the case of beta decay where for example a proton decays into a neutron within the nucleus. This is a relatively rare event, and the weak force is, after all, “weak”, so in general it perhaps doesn’t get as much attention as it otherwise might.

However, the electroweak unification of the electromagnetic force (i.e., Maxwell’s equations) together with the weak force, along with the central role of the Higgs field in this framework, provides a profoundly different picture of the fundamental nature of an electron and its interaction with the electromagnetic field. This electroweak picture also integrates the neutrinos with their charged lepton partners like the electron, and is intimately connected with the flavors or generations of particles, e.g., muon and tau.

Thus, it is perhaps not an exaggeration to say that the electroweak framework is as revolutionary for understanding the nature of leptons as the quark framework is for understanding the hadrons: it provides an entirely different fundamental basis space, that parsimoniously integrates a wide range of disparate phenomena into a more coherent framework. To understand the most basic nature of the electron and electromagnetic interactions, understanding the electroweak model seems essential.

However, the counter-argument is that the actual physical implications of this revolutionary rearrangement of the quantum furniture are somewhat difficult to detect, which is presumably why this part of the story gets less attention than it otherwise might. The famously successful QED model, which preceded the electroweak model by more than a decade, persists unchanged!

This fact says a lot about how different calculational tools can be used to describe the same phenomena, but if the goal is to understand something about the underlying physical mechanisms, then it seems that there is much to learn from the electroweak framework.

It is strongly recommended to read gauge theory and Higgs before proceeding, as these contain essential background information necessary for understanding what follows. You should understand how gauge theory describes the interaction between particle and force fields, and how this interaction can give rise to a dynamic mass-like factor, which shows up in the conserved current density expression. This is precisely where the mass of the weak bosons will arise in the electroweak framework, for precisely the same reason.

The role of the Higgs field is, first and foremost, to provide the source of a non-zero “fuel” for the current density, through the mechanism of spontaneous symmetry breaking as a result of the specific shape of the Higgs potential. The specific structure of this field, as a scalar complex field (spin 0, i.e., complex KG) with two elements, is also critical for how it interacts with the four different force fields present in the electroweak system, to deal with the otherwise problematic Goldstone boson problem and to allow one effective force field to remain massless (which then represents the Maxwell EM field), while three others take on mass terms, which turn out to be consistent with the measured masses of the weak bosons.

Electroweak

The electroweak framework is based on three fields, which together comprise a total of 4 four-vector potential-like fields, and the Higgs doublet, which has 4 real-valued numbers, organized into two complex values.

• The Higgs doublet \(\Psi\), which has 2 complex field elements (4 real numbers), one of which interacts with electric charge (\(\chi^+\)) and another that is electrically neutral (\(\chi^0\)). As a result of the spontaneous symmetry breaking property of the Higgs potential, the neutral component acquires a stable expected value of \(v_h = 246\) GeV, which is what then provides the mass term for all other wave fields.

• The weak hypercharge four-potential \(B_\mu\), which is essentially just like the EM four-potential \(A_\mu\), with four real-valued numbers, \(B_0\) is the scalar electric potential, and the remaining three terms (\(B_{(x,y,z)}\)) are the vector potential. The source of this potential is the weak hypercharge value \(Y_W\) which acts like the \(Q\) charge for the EM potential. It interacts with the Higgs doublet through the same local gauge invariance mechanism (see gauge theory) that describes minimal coupling with the EM field, exactly as done in the case of the complex KG equation.

• The weak isospin triplet of four-vectors, \(W^a_\mu\) where \(a=1,2,3\), and each such four-vector is mostly like the EM four-potential \(A_\mu\). The first two of these four-vectors, \(W^1_\mu\) and \(W^2_\mu\) together end up as the components of the \(W^\pm\) bosons, while the third one \(W^3_\mu\) contributes to both the final electromagnetic field potential (\(A^\mu\)) and the neutral \(Z^0\) boson.

These fields are mixed together in specific combinations, using a standard unitary rotation matrix defined by an angle \(\theta_w\) known as the Weinberg or weak mixing angle, to obtain the actual observable fields, as follows:

The EM four-potential, which propagates without any mass:

\[ A_\mu = \cos \theta_w B_\mu + \sin \theta_w W^3_\mu \]

The four-potential associated with the weak neutral boson \(Z^0\), which has a (large) mass:

\[ Z_\mu = -\sin \theta_w B_\mu + \cos \theta_w W^3_\mu \]

And the four-potentials associated with the two charged weak bosons \(W^\pm\):

\[ W^+_\mu = \frac{1}{\sqrt2} \left( W^1_\mu - i W^2_\mu \right) \]

\[ W^-_\mu = \frac{1}{\sqrt2} \left( W^1_\mu + i W^2_\mu \right) \]

Furthermore, the electric charge \(Q\) is also a mixture of two quantum numbers that are associated with different types of particles, weak hypercharge (\(Y\)) and weak isospin (third component: \(T^3\)):

\[ Q = Y + T^3 \]

Table 1 shows these quantum numbers for neutrinos and electrons (the leptons). This table, plus the mixing factors shown above, clearly show that the electroweak framework provides an entirely different factorization of the theoretically fundamental lepton particles. There are two fundamentally different versions of the electron, and the left-chiral electron and the neutrino share most of their quantum numbers, so they would seem to be much more similar at the electroweak level than we would otherwise expect given their overall properties.

Thus, it seems that one’s understanding of what an electron really is changes dramatically under this electroweak framework. Furthermore, every one of these electroweak fields, including in addition the particle fields associated with the leptons, is coupling to the very same Higgs doublet field: what are the implications of this mutual interaction through this shared field?

A fundamental question raised by all of this, is why are the observable fields a mixture and not just equivalent to the corresponding weak hypercharge or weak isospin fields that we started with? This is described as being a consequence of the spontaneous symmetry breaking dynamic in the Higgs field, but the story is quite a bit more complicated than that. The symmetry breaking is necessary for these mixture factors to actually matter, because prior to this symmetry breaking, all of the four-potentials are essentially equivalent, acting like the EM massless four-potential. Once the symmetry is broken and the Higgs field has a vacuum expectation value, then all of the differences that were actually built into the system start to matter.

In particular, the gauge theory derivation of how the Higgs field couples to the weak hypercharge and weak isospin fields results in a covariant derivative that cancels out the local gauge factors arising from the gauge fields, of the form:

\[ D_\mu = \partial_\mu - i g' Y B_\mu - i g W^a_\mu T^a \]

Where \(g'\) and \(g\) are arbitrary real-valued coupling parameters for each of the respective fields (\(B_\mu\) and \(W^a_\mu\)), and \(Y\) is the weak hypercharge quantum number value for the Higgs field (a real number), which is set to ¹⁄₂ by convention. The \(T^a\) is a set of 3 different coupling matrices, one for each of the three \(W^a_\mu\) four-vector components, that are defined by the SU(2) unitary rotation group – i.e., a set of basis vectors that perform unitary rotations of a 2-component complex vector, which is what the Higgs doublet is (2 complex values).

These are none other than the Pauli matrices, which are also used in the Dirac equation, and define the property of spin in the quantum world. This is why the \(W\) field is called isospin, because it causes spinning. Each of these is also multiplied by the conventional ¹⁄₂ factor, but it turns out that the third Pauli matrix has a -1 on the bottom-right diagonal, which therefore gives the neutral Higgs doublet component \(\chi^0\) a -¹⁄₂ effective \(T^3\) quantum number.

This negative quantum number means that the \(W^3\) component force field has a negative contribution, while the \(B_\mu\) field has a positive contribution, and any field configuration that has positive values in each of these fields, in the correct ratio as defined below, will end up with no mass, due to the cancellation of these positive and negative contributions.

The covariant derivative is used in the Lagrangian for the Higgs field (see gauge theory for details on how this is all computed, using a simpler single-valued complex KG particle field):

\[ \mathcal{L}_h = |D_\mu \Psi|^2 - V(\Psi) \]

where \(V(\Psi)\) is the Higgs potential that leads to symmetry breaking, as described in detail in Higgs, and the Higgs field \(\Psi\) is the (\(1/\sqrt2\) normalized) complex doublet:

\[ \Psi \equiv \frac{1}{\sqrt2} \binom{\chi^+}{\chi^0} \]

From all of this, we can now see that the reason that electric charge and the electromagnetic field are a mixture of the \(B_\mu\) and \(W^3\) fields is because of the way that these gauge coupling factors work. In particular, the only way for the EM field to remain fully massless (and thus have long-range effects) is for it to have opposite effects on the current density computed by the gauge theory analysis of the electroweak system, so that these effects perfectly cancel out, leaving zero current, and thus, zero mass. We will see this in detail below.

This cancellation is captured qualitatively by the definition of electric charge in terms of the sum of the factors for each field, and quantitatively it is given by the ratio of the coupling parameters \(g'\) and \(g\), which determines the weak mixing angle:

\[ \sin \theta_w = \frac{g'}{\sqrt{g^2 + g'^2}} \]

These coupling parameters are determined experimentally, and are not predicted by the theory itself. Nevertheless, their impacts have far-reaching implications on a range of other parameters and experimental outcomes, all of which have been strongly confirmed. As with everything else in the Standard Model at this point, all of this seemingly crazy complexity has been experimentally validated to high levels of precision, with this electroweak theory making strong predictions for things that had not yet been seen experimentally before.

Furthermore, the gauge theory framework that drives so much of the phenomenology here is fundamentally based on describing the conservation of energy and charge in the equations of motion, derived from the Lagrangian framework. Thus, ultimately, these conservation dynamics are what is driving everything. And the only way to get the math to all work out was to do it with massless gauge fields that work like the EM four-potential, and then have the mass emerge dynamically from within that fundamentally massless context, which the specific form of the Higgs potential accomplishes.

Thus, even though this seems like a very complicated way of going about everything, in the end each step is strongly constrained, and it is difficult to see how a different kind of system could be derived.

Phenomenology of the electroweak system

Go to the electroweak simulation to explore the full electroweak system of coupled fields, with several important demonstrations of its phenomenology.

The EM field is defined by the ratio of local field strengths

In the following analysis, we see precisely how the current density that is computed for the coupled electroweak system will turn out to be zero whenever the local field values for \(B_\mu\) and \(W^3\) are in a specific positive-valued ratio. The result is that any waves that have this particular configuration will propagate over a long range, because they are effectively massless, while all the massive configurations will end up decaying over a relatively short range. Thus, the Higgs field acts much like a filter: the only thing that gets through is light itself (EM radiation over the effectively massless field configuration), while everything else is effectively the massive weak bosonic fields \(W^\pm\) and \(Z^0\).

Everything below concerns only the two neutral gauge fields \(W^3_\mu\) and \(B_\mu\). The charged \(W^{1,2}\) and the Higgs fluctuations play no part. Each gauge field obeys a wave equation driven by a current, and the Higgs doublet is what supplies that current:

\[ \square\, W^3_\mu = j^3_\mu, \qquad \square\, B_\mu = j^Y_\mu \]

\[ j^3_\mu = 2 g\, \mathrm{Im}\!\left[\Phi^\dagger T^3 D_\mu \Phi\right], \qquad j^Y_\mu = 2 g' Y\, \mathrm{Im}\!\left[\Phi^\dagger D_\mu \Phi\right] \]

The key factor in these current equations is the nature of the covariant derivative, which will end up bringing the force fields into the overall current expression, via the seagull term that we saw in the gauge theory derivation.

\[ D_\mu \Phi = \left(\partial_\mu - i G_\mu\right)\Phi, \qquad G_\mu = g W^3_\mu T^3 + g' Y B_\mu \]

With \(T^3 = \mathrm{diag}(+\tfrac12, -\tfrac12)\) and \(Y = \tfrac12\), which is a multiple of the identity, \(G_\mu\) is diagonal:

\[ G_\mu = \frac{1}{2} \begin{pmatrix} g W^3_\mu + g' B_\mu & 0 \\[2pt] 0 & -g W^3_\mu + g' B_\mu \end{pmatrix} \]

\(W^3\) enters with opposite signs in the two entries, because \(T^3\) distinguishes the components. \(B\) enters with the same sign in both, because hypercharge does not. That asymmetry is the origin of everything that follows: positive values of \(W^3\) will end up canceling out positive values of \(B\).

Now use the actual state of the Higgs doublet. It is uniform, so \(\partial_\mu \Phi = 0\) and the covariant derivative is entirely the gauge term. And it is \(\Phi = (0,\ v/\sqrt2)\) – the upper charged component is exactly zero, while the vacuum expectation value is concentrated in the lower neutral component. Note that this specific configuration is essential for all of the mixing logic described here to actually work: the 0 cancels out possible contributions from the other W components. Thus, the upper entry of \(G_\mu\) acts on nothing, due to the 0:

\[ D_\mu \Phi = -i\, G_\mu \Phi = \left(0,\; \tfrac{i}{2}\left(g W^3_\mu - g' B_\mu\right) \tfrac{v}{\sqrt2}\right) \]

Only one scalar survives. Give it a name:

\[ S_\mu \;\equiv\; g W^3_\mu - g' B_\mu \]

Two neutral fields, but the Higgs doublet presents only one non-zero component for them to couple through.

Put that \(D_\mu\Phi\) into the currents. Both reduce to the same scalar \(S_\mu\), with different prefactors:

\[ \square\, W^3_\mu = -\frac{v^2}{4}\, g\, S_\mu, \qquad \square\, B_\mu = +\frac{v^2}{4}\, g'\, S_\mu \]

The resulting current factors act like a mass, and note that the \(S_\mu\) term contains both the \(W^3\) and \(B\) field values, with opposite signs, so this is why the specific field levels in each field determine the resulting effective mass value at each point. There is no sum over \(\mu\) and no magnitude anywhere, which is why it is local to each field component at each point. Each spacetime index \(\mu\) – the scalar potential and the three vector components – has its own \(S_\mu\) and thus its own mass term, computed from the field values at that one lattice site.

Because both equations are driven by the same \(S_\mu\), two particular combinations decouple. Take \(g\) times the first plus \(g'\) times… more simply, form:

\[ \begin{aligned} P_\mu &\equiv g' W^3_\mu + g B_\mu & \square P_\mu &= -\tfrac{v^2}{4} g g' S_\mu + \tfrac{v^2}{4} g g' S_\mu = 0 \\[4pt] Z_\mu &\equiv g W^3_\mu - g' B_\mu & \square Z_\mu &= -\tfrac{v^2}{4}\left(g^2 + g'^2\right) S_\mu \end{aligned} \]

\(P_\mu\) is the photon (EM field), up to normalization: dividing by \(\sqrt{g^2+g'^2}\) turns \((g', g)\) into \((\sin\theta_W, \cos\theta_W)\). \(Z_\mu\) is the \(Z\), likewise up to normalization, and it obeys a massive wave equation with \(M_Z^2 = \tfrac{v^2}{4}(g^2+g'^2)\).

In summary, when you work through the final bit of math, there are two specific combinations of wave state magnitudes, one that results in zero effective mass (the photon) and one that has mass (the Z boson).

\[ \begin{aligned} \text{photon:}\quad (W^3, B) &\propto (g',\, g) & S &= g g' - g' g = 0 \\[2pt] \text{Z:}\quad (W^3, B) &\propto (g,\, -g') & S &= g^2 + g'^2 \neq 0 \end{aligned} \]

By virtue of the filtering argument given above, nothing specifically requires that precisely the right ratio of field values must be present, because anything that is not well-aligned will be filtered out! For example, a pure \(W^3\) pulse produces both photon and Z fields, moving at different speeds.

Does the electron make only EM field waves?

It does not — and it is worth being clear that this is the gauge coupling’s business, not the Yukawa’s. The Yukawa term gives the electron a mass; what it couples to is fixed entirely by \(T^3\) and \(Y\).

Rewrite the fermion’s gauge matrix in the photon and \(Z\) directions, using \(W^3 = \sin\theta_W A + \cos\theta_W Z\) and \(B = \cos\theta_W A - \sin\theta_W Z\):

\[ g W^3 T^3 + g' Y B \;=\; \underbrace{e\,Q}_{\text{to } A}\, A \;+\; \underbrace{\sqrt{g^2+g'^2}\left(T^3 - \sin^2\!\theta_W\, Q\right)}_{\text{to } Z}\, Z \]

with \(Q = T^3 + Y\) and \(e = g g'/\sqrt{g^2+g'^2}\).

The photon coupling collapses to \(e Q\) — that is what \(Q = T^3 + Y\) buys, and it is zero for the neutrino. The \(Z\) coupling is a different combination, and it vanishes for nothing here:

So an electron sources both fields at once. Putting a lump of each in an empty box and reading off the fields it creates:

The neutrino makes no photon field and a large \(Z\) field; the electron makes both. Their \(Z\) fields stand in the ratio \(1.805\), against the \(0.500 / 0.277 = 1.805\) the couplings predict.

Note also that \(e_L\) and \(e_R\) have \(Z\) couplings of opposite sign while their photon couplings are identical. That difference is parity violation, and it is why the weak interaction distinguishes handedness while electromagnetism does not.

Then why does only electromagnetism reach us?

Because of the mass, not the coupling. A massless field around a static source falls off as \(1/r\); a massive one is screened,

\[ \frac{1}{r} \;\longrightarrow\; \frac{e^{-M_Z r}}{r} \]

so the \(Z\) part of an electron’s field is gone beyond \(r \sim 1/M_Z\), about \(2\times10^{-3}\) fm. Every electron is radiating \(Z\) field as vigorously as photon field, but it remains highly localized.

A random excitation of \(W^3\) and \(B\) splits into both modes, and so does the field around an electron. What makes electromagnetism look like the only long-range neutral force is not that anything is emitted selectively, but that one of the two modes has a mass and the other does not, so the massive mode is screened (filtered) while the EM field radiates.

The right-handed chiral electron is a singlet, untouched by W fields

Due to the way that the coupling works in the SU(2) structure of the electroweak symmetry structure, the W fields only interact with the left-chiral fermions, electron and neutrino. The right chiral electron (there is no corresponding neutrino) is a singlet that only couples with Z and B fields, not the W’s. This is consistent with extant data. This requires more investigation here to understand the implications.

The weak force W bosons are hard to activate

An electron generates a current that serves as a source for \(B\) and \(W^3\), but, critically, not the \(W^1\) and \(W^2\) fields, which are the ones that underlie the \(W^\pm\) bosons.

Here’s Claude’s explanation that remains to be fully digested:

The weak isospin currents that drive the gauge fields is given by:

\[ J^{a\mu} = \bar\psi_L\gamma^\mu T^a\psi_L \]

This goes into the Lagrangian like so:

\[ \mathcal{L}_{CC} = \frac{g}{\sqrt2}\left[\bar\nu_L\gamma^\mu e_L\,W^+_\mu + \text{h.c.}\right] \]

Look at what that current is: it turns an electron into a neutrino. It is off-diagonal in particle species.

This means that there is no charged W current for a single electron: $\( \langle e |\, J^{+\mu} \,| e \rangle = 0 \)$

because the operator maps the electron to a neutrino and \(\langle e|\nu\rangle = 0\). There is no charged-current density for an electron to carry, so there’s no static \(W^\pm\) field around it. W fields appear only in processes that actually change particle identity — beta decay, muon decay, and so on.

At loop level (i.e., wrt virtual particles generated by high energy transients in the vacuum) the W absolutely is present around an electron. The electron self-energy includes a \(W\)–\(\nu\) loop, and the electron is dressed in a virtual cloud of them. This contributes measurably to the anomalous magnetic moment — the electroweak contribution to the muon’s \(g-2\) is around \(154\times10^{-11}\), small but well above experimental precision and essential to the \(g-2\) analysis.

So: no classical W field, but a real quantum one. The difference between “tree-level source” and “loop-level dressing” is the difference between a field configuration you could put on a lattice and a correction to a propagator.

If you add fermions, the source term for the SU(2) gauge field is:

\[ J^a_\mu = g\,\bar\psi\gamma_\mu T^a\psi \]

and the off-diagonal pieces \(J^1, J^2\) are proportional to products like \(\bar\nu_L\gamma_\mu e_L\). Classically, \(W^\pm\) is sourced by the interference between the electron and neutrino fields. Excite only an electron field and \(J^1 = J^2 = 0\) identically — you get a \(Z\) field and nothing else. Excite both coherently and the \(W\) fields light up. That’s the classical field-theory shadow of “the charged current changes particle identity.”

Worth noting the electron also sources a Higgs field, via \(\nabla^2 h - m_h^2 h = -y_e\bar\psi\psi\). But \(y_e = \sqrt2\,m_e/v \approx 2.9\times10^{-6}\), so it’s utterly negligible. For a top quark, \(y_t \approx 1\), and the Higgs cloud is as significant as anything else around it.

\(J^{+\mu} = \bar\nu_L\gamma^\mu e_L\) is a bilinear in two different fields. Write them schematically as \(\psi_e = A\,e^{i\varphi_e}\) and \(\psi_\nu = B\,e^{i\varphi_\nu}\), and

\[ J^+ \sim AB\,e^{i(\varphi_e - \varphi_\nu)} \]

It depends on the relative phase. Both fields nonzero at the same point isn’t enough — they must be phase-locked. If the phases are uncorrelated, any average over space or time returns zero.

In quantum language: \(\langle\Psi|\,J^{+\mu}\,|\Psi\rangle \ne 0\) requires \(|\Psi\rangle = \alpha|e\rangle + \beta|\nu\rangle\) with both coefficients nonzero. A state in which the particle’s identity is genuinely indefinite.

Here’s the crux, and it’s not about neutrino scarcity at all. \(|e\rangle\) has electric charge \(-1\) and \(|\nu\rangle\) has charge \(0\). A coherent superposition of those is a superposition of different total electric charge, which charge superselection forbids.

The physical mechanism is worth stating, because it makes the rule feel less like an axiom: the long-range Coulomb field acts as a permanent measuring apparatus. The electric field out at infinity already “knows” whether you have an electron or a neutrino, so any such superposition decoheres immediately and completely. You cannot prepare one, anywhere, ever.

So there is no classical \(W^\pm\) field, not because neutrinos are rare but because the required coherence is prohibited in principle.

They are, and in impressive numbers — the cosmic neutrino background sits at roughly \(336/\text{cm}^3\), and the solar flux through your thumbnail is about \(6.5\times10^{10}\) per second. But every one of them is in a definite particle-number eigenstate, uncorrelated in phase with any electron.

What that abundance does produce is a large \(\langle\bar\nu\gamma^\mu\nu\rangle\), a diagonal current. Neutrinos carry \(T^3 = +1/2\), so the neutrino background genuinely sources a \(Z\) field. What it produces is exactly zero of \(\langle\bar\nu\gamma^\mu e\rangle\).

So what actually drives \(W^\pm\)? Two things, and neither is a static field:

Virtual, internal to an amplitude. The W appears as an off-shell internal line connecting two currents — say \(\bar\nu_L\gamma e_L\) at one vertex and \(\bar u_L\gamma d_L\) at the other, which is beta decay. Its job is to ferry electric charge between them. It never exists as a field configuration you could photograph; it’s a propagator inside a process.

Real and on-shell, above threshold. Collide something with \(\sqrt{s} > 80\) GeV and you make actual W bosons — LEP2, the Tevatron, the LHC, and the occasional cosmic-ray shower. They then decay in about \(3\times10^{-25}\) seconds, travelling a distance far smaller than a nucleus.

What happens to the other 3 components of the Higgs field?

Derivation of gauge coupling

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

\[ T^a = \frac{\sigma^a}{2} \]

\[ Y_\Phi = +\tfrac12 \]

\[ \mathcal{L} = (\partial_\mu\Phi)^\dagger(\partial^\mu\Phi) - V(\Phi^\dagger\Phi) \]

Take a constant transformation \(\Phi \to U\Phi\) with

\[ U = e^{i\alpha^a T^a}\,e^{i\beta Y} \in SU(2)\times U(1) \]

The potential is fine because \(U\) is unitary: \(\Phi^\dagger U^\dagger U \Phi = \Phi^\dagger\Phi\). The kinetic term is fine because \(U\) passes through the derivative: \(\partial_\mu(U\Phi) = U\partial_\mu\Phi\).

Now let \(\alpha^a(x)\), \(\beta(x)\) vary. The potential still survives — unitarity holds pointwise. But

\[ \partial_\mu\left(U(x)\Phi\right) = U\,\partial_\mu\Phi + \underbrace{(\partial_\mu U)\,\Phi}_{\text{the problem}} \]

That second term wrecks the kinetic term. Everything that follows exists to cancel it.

Introduce a derivative \(D_\mu\) and demand that \(D_\mu\Phi\) transform exactly like \(\Phi\) itself:

\[ \boxed{\;(D_\mu\Phi)' = U\,(D_\mu\Phi)\;} \]

If that holds, the kinetic term is automatically invariant, since \((D_\mu\Phi)'^\dagger(D^\mu\Phi)' = (D_\mu\Phi)^\dagger U^\dagger U (D^\mu\Phi)\). Write the ansatz

\[ D_\mu = \partial_\mu - i\,\mathcal{G}_\mu \]

\[ \mathcal{G}_\mu \equiv g\,W^a_\mu T^a + g'\,Y B_\mu \]

and impose the box. Left side:

\[ \partial_\mu(U\Phi) - i\mathcal{G}'_\mu U\Phi = U\partial_\mu\Phi + (\partial_\mu U)\Phi - i\mathcal{G}'_\mu U\Phi \]

Right side:

\[ U\partial_\mu\Phi - iU\mathcal{G}_\mu\Phi \]

Equate, cancel \(U\partial_\mu\Phi\), and strip \(\Phi\) (it’s arbitrary):

\[ (\partial_\mu U) - i\,\mathcal{G}'_\mu U = -i\,U\mathcal{G}_\mu \]

Right-multiply by \(U^\dagger\):

\[ \mathcal{G}'_\mu = U\,\mathcal{G}_\mu\,U^\dagger - i\,(\partial_\mu U)U^\dagger \]

That single equation is the entire gauge-field transformation law, and it was forced on us — we chose nothing.

Split into the SU(2) and U(1) pieces

Write \(U = U_2 U_1\) with \(U_2 = e^{i\alpha^aT^a}\) and \(U_1 = e^{i\beta Y}\). Since \(U_1\) is a phase times the identity it commutes with everything, and

\[ (\partial_\mu U)U^\dagger = (\partial_\mu U_2)U_2^\dagger + i(\partial_\mu\beta)\,Y \]

Matching the traceless-matrix part against the identity part:

\[ g\,W'^a_\mu T^a = U_2\left(g W^a_\mu T^a\right)U_2^\dagger - i(\partial_\mu U_2)U_2^\dagger \]

\[ B'_\mu = B_\mu + \frac{1}{g'}\partial_\mu\beta \]

Put \(U_2 \approx 1 + i\alpha^aT^a\). For the homogeneous piece, using \([T^a,T^b] = i\varepsilon^{abc}T^c\):

\[ U_2\left(W^bT^b\right)U_2^\dagger \approx W^bT^b + i\alpha^a W^b\left(i\varepsilon^{abc}T^c\right) = W^bT^b - \varepsilon^{abc}\alpha^a W^b T^c \]

For the inhomogeneous piece, \(-i(\partial_\mu U_2)U_2^\dagger \approx (\partial_\mu\alpha^a)T^a\). Collecting:

\[ \delta W^a_\mu = \frac{1}{g}\partial_\mu\alpha^a + \varepsilon^{abc}W^b_\mu\,\alpha^c, \qquad \delta B_\mu = \frac{1}{g'}\partial_\mu\beta, \qquad \delta\Phi = i\left(\alpha^aT^a + Y\beta\right)\Phi \]

Note the structural difference. \(B\) gets only the inhomogeneous shift. \(W\) gets that plus a homogeneous rotation \(\varepsilon^{abc}W^b\alpha^c\) — because \(W\) lives in the adjoint representation, i.e. it is charged under its own group. That extra term is the seed of every W self-interaction.

Also note the gauge field is not a tensor. The \(\partial_\mu\alpha/g\) term is inhomogeneous, which is precisely what makes it a connection and what lets it absorb the offending \((\partial_\mu U)\Phi\).

Apply \(\delta\) to \(D_\mu\Phi = \partial_\mu\Phi - ig W^aT^a\Phi - ig'YB\Phi\), writing \(\Lambda \equiv \alpha^aT^a + Y\beta\):

\[ \delta(D_\mu\Phi) = \underbrace{i(\partial_\mu\Lambda)\Phi + i\Lambda\partial_\mu\Phi}_{\text{from }\delta\Phi} \underbrace{- i(\partial_\mu\alpha^a)T^a\Phi - ig\,\varepsilon^{abc}W^b\alpha^c T^a\Phi}_{\text{from }\delta W} \underbrace{+\, gW^aT^a\Lambda\Phi}_{\text{from }\delta\Phi\text{ in the }W\text{ term}} \underbrace{- iY(\partial_\mu\beta)\Phi}_{\text{from }\delta B} + \underbrace{g'YB\Lambda\Phi}_{\text{from }\delta\Phi\text{ in the }B\text{ term}} \]

The inhomogeneous terms cancel. Since \(\partial_\mu\Lambda = (\partial_\mu\alpha^a)T^a + Y\partial_\mu\beta\), the \(i(\partial_\mu\Lambda)\Phi\) piece exactly kills the \(-i(\partial_\mu\alpha^a)T^a\Phi\) and \(-iY(\partial_\mu\beta)\Phi\) terms. That’s the whole job of the \(\partial\alpha/g\) in \(\delta W\).

The commutator terms cancel too. What’s left must equal \(i\Lambda(D_\mu\Phi)\), which requires

\[ -ig\,\varepsilon^{abc}W^b\alpha^c T^a \;=\; gW^a\left[\Lambda, T^a\right] \]

Evaluate the right side: only the \(SU(2)\) part of \(\Lambda\) contributes, giving \(gW^a\alpha^b\left(i\varepsilon^{bac}T^c\right)\). Relabelling both sides so the generator index is \(c\), the left gives \(-\varepsilon^{cab}W^a\alpha^b\) and the right gives \(+\varepsilon^{bac}W^a\alpha^b\). Since \(\varepsilon^{cab} = \varepsilon^{abc}\) and \(\varepsilon^{bac} = -\varepsilon^{abc}\), both equal \(-\varepsilon^{abc}\). They match.

\[ \Longrightarrow \quad \delta(D_\mu\Phi) = i\Lambda\,(D_\mu\Phi) \quad\checkmark \]

\(D_\mu\Phi\) transforms exactly like \(\Phi\). The kinetic term \((D_\mu\Phi)^\dagger(D^\mu\Phi)\) is now locally invariant.

\[ \mathcal{G}_\mu = gW^a_\mu T^a + g'Y B_\mu = \frac12\begin{pmatrix} gW^3_\mu + g'B_\mu & g\left(W^1_\mu - iW^2_\mu\right)\\[4pt] g\left(W^1_\mu + iW^2_\mu\right) & -gW^3_\mu + g'B_\mu\end{pmatrix} \]

Everything from here is reading off entries. Note the \(\pm g W^3\) on the diagonal — that sign flip is \(T^3 = \pm\tfrac12\), and it is why the two components feel opposite \(W^3\) and identical \(B\).

Higgs with the vacuum expectation

Put \(\Phi \to \langle\Phi\rangle = (0,\, v/\sqrt2)^T\). Then \(\partial_\mu\langle\Phi\rangle = 0\), so only the last term of

\[ |D_\mu\Phi|^2 = (\partial_\mu\Phi)^\dagger(\partial^\mu\Phi) - i(\partial_\mu\Phi)^\dagger\mathcal{G}^\mu\Phi + i\Phi^\dagger\mathcal{G}_\mu(\partial^\mu\Phi) + \Phi^\dagger\mathcal{G}_\mu\mathcal{G}^\mu\Phi \]

survives. Acting on the VEV picks out the second column:

\[ \mathcal{G}_\mu\langle\Phi\rangle = \frac{v}{2\sqrt2}\begin{pmatrix} g\left(W^1_\mu - iW^2_\mu\right) \\[3pt] -gW^3_\mu + g'B_\mu \end{pmatrix} \]

\[ \left|\mathcal{G}_\mu\langle\Phi\rangle\right|^2 = \frac{v^2}{8}\left[g^2\left(W^1_\mu W^{1\mu} + W^2_\mu W^{2\mu}\right) + \left(gW^3_\mu - g'B_\mu\right)^2\right] \]

With \(W^\pm_\mu = (W^1_\mu \mp iW^2_\mu)/\sqrt2\), so that \(W^1W^1 + W^2W^2 = 2W^+_\mu W^{-\mu}\):

\[ \Longrightarrow\quad \frac{g^2v^2}{4}\,W^+_\mu W^{-\mu} \;+\; \frac{v^2}{8}\left(gW^3_\mu - g'B_\mu\right)^2 \]

\[ m_W = \frac{gv}{2}, \qquad m_Z = \frac{v\sqrt{g^2+g'^2}}{2}, \qquad m_\gamma = 0 \]

The photon combination is simply absent from that expression — there is no term to give it a mass.

What the full expansion contains

Restoring \(\Phi = \left(G^+,\ (v+h+iG^0)/\sqrt2\right)^T\), the same \(|D_\mu\Phi|^2\) generates, in one stroke:

• • kinetic terms for \(h\) and the Goldstones
• • the mass terms above, from \(v^2\)
• • \(hVV\) and \(hhVV\) couplings, from the \(2vh\) and \(h^2\) pieces of \((v+h)^2\)
• • the bilinear mixing \(m_W W^\mu\partial_\mu G^\mp + m_Z Z^\mu\partial_\mu G^0\) — the term \(R_\xi\) gauge exists to cancel
• • assorted \(VGG\) vertices

All of it from one term, with no free parameters beyond \(g\), \(g'\), \(v\). That rigidity is the point.