How does the Higgs mechanism generate mass for the $W$ and $Z$ gauge bosons?

Question

I came across this discussion point about how the Higgs mechanism generates mass for the $W$ and $Z$ gauge bosons (see attached problem below). Regarding the Higgs field factor $$\Phi^2 = \frac{1}{2}(v+h)^2$$ I think it is quite straight-forward, but I was unsure how to handle the $D_{\mu}$ expansion. Is it correct to use $$D_{\mu} = \partial_{\mu} + i \frac{g}{2}\tau W_{\mu} + i \frac{g'}{2}B_{\mu}$$ for the covariant derivative? I saw some calculated mass terms in this link (on page 9), but no explicit calculations.

Consider the kinetic term of the Higgs field $$\Phi\mathcal{L}=|D_{\mu}\Phi|^2=(D_{\mu}\Phi)^*(D^{\mu}\Phi)$$ and expand it along the minimum of the Higgs potential $$\Phi=\frac{1}{\sqrt{2}}\begin{pmatrix}0\\v+h\end{pmatrix}$$ where $v$ is the vacuum expectation value (VEV) and $h$ is the Higgs boson.

1. Derive the coefficients of the operators representing the gauge boson masses $m_W$, $m_Z$ and $m_A$ in terms of the gauge couplings and $v$ (you can use the expressions of $A_{\mu}$ and $Z_{\mu}$ in terms of $B_{\mu}$ and $W_{\mu}^3$ without deriving them explicitly).

2. Derive the coefficients of the trilinear and quadrilinear interactions between the gauge bosons $W$ and $Z$ and the $h$ boson in terms of the gauge boson masses and of $v$.

Answer 1

The covariant derivative is $$ D_\mu\phi=\left(\partial_\mu+i\frac g2\tau^i W^i_\mu+i\frac {g'}2B_\mu\right)\phi $$ up to normalization of the generators. Then when the $\mu^2$ term in the Higgs potential becomes positive, the Higgs field develops a constant VEV at the bottom of the potential which can be taken as $$ \langle\phi\rangle=\frac1{\sqrt2}\begin{pmatrix}0\\v\end{pmatrix} $$ Then fluctuations are parameterized by the Higgs boson $h$: $$ \phi=\frac1{\sqrt2}\begin{pmatrix}0\\v+h\end{pmatrix} $$

Finally, via a routine computation, we substitute the definition of the covariant derivative and the broken Higgs into the $|D^\mu\phi|^2$ term in the Lagrangian. If we wish to only determine the gauge boson masses, then we can safely ignore the dynamical $h$ in the Higgs interaction, since it will generate $h$-interactions rather than mass contributions at tree-level.

$$ |D^\mu\phi|^2=\left|\left(\partial_\mu+i\frac g2\tau^i W^i_\mu+i\frac {g'}2B_\mu\right)\frac1{\sqrt2}\begin{pmatrix}0\\v+h\end{pmatrix}\right|^2 $$ $$ \cong\frac {v^2}8 \left|\begin{pmatrix}g W_\mu^1-igW_\mu^2\\-gW_\mu^3+g'B_\mu\end{pmatrix}\right|^2\tag{modulo $h$-interactions} $$ $$ =\frac{v^2g^2}8\left((W_\mu^1)^2+(W_\mu^2)^2\right)+\frac{v^2}8(gW_\mu^3-g'B_\mu)^2 $$

The field redefinitions $$ W_\mu^\pm=\frac1{\sqrt2}(W_\mu^1\mp iW_\mu^2) \\Z_\mu=\frac1{\sqrt{g^2+g'^2}}(gW_\mu^3-g'B_\mu) \\A_\mu=\frac1{\sqrt{g^2+g'^2}}(gW_\mu^3+g'B_\mu) $$

diagonalize the mass matrix, and we can immediately read the mass terms of the new fields:

$$ \frac12\left(\frac{gv}{2}\right)^2 \left(W_\mu^+\right)^2+\frac12\left(\frac{gv}{2}\right)^2 \left(W_\mu^-\right)^2+\frac12\left(\frac{v\sqrt{g^2+g'^2}}{2}\right)^2Z_\mu^2+0\cdot A_\mu^2 $$

Field	Mass
$W_\mu^+$	$gv/2$
$W_\mu^-$	$gv/2$
$Z_\mu$	$v\sqrt{g^2+g'^2}/2$
$A_\mu$	$0$

Note that a different choice of Higgs VEV would lead to exactly the same field content, albeit with a different diagonalization required to get there.

Answer 2

The question is actually more generic than you realize and more so than the replies would suggest. In general, a scalar field when coupled to a Maxwell or massless gauge field and make it a Proca field or the gauge field version of one, endowing it with the appearance of mass. This has nothing per se to do with quantum theory, as it's a process that's generic to field theory, itself. What you see arise, in the context of quantum field theory, is the quantum version of that process, rather than something that's quantum in origin.

In schematic form, the equations for a non-Abelian gauge field can be written as $F = dA + A^2$, where $F$ denotes the field strength and $A$ the potential. It is Abelian if the $A^2$ is absent, and that includes the Maxwell field as a case in point. The equations for a scalar field coupled to the gauge field can be written schematically as $v = dq + Aq$, where $q$ is the scalar field potentials and $v$ its field strength. The operator $D = d + A$ is the "gauge-covariant" derivative for the scalar field.

The dynamics for the fields contain the second set of equations. When they are generated from an action principle, given by an action integral $\int L$ and a Lagrangian $L$ (represented as a 4-form $L = d^4x$, where $d^4x = dt∧dx∧dy∧dz$ and $$ is the "Lagrangian density"), then the variational of $L$ gives rise to the "response fields", $$δL = δA·J - δF·G + δq·f + δv·p.$$

Normally $L$ is assumed to be a function only of $F$, $q$ and $v$, only, not $A$, and is expressible in separable form as $L = L_F(F) + L_v(v) + L_q(q)$, with the first two components being linear in $F^2$ and $v^2$. That's the case for Maxwell and Yang-Mills fields. For Proca fields and their gauge-field analogue, there would also be a dependence on $A$, and a fourth component $L = ⋯ + L_A(A)$ would arise, that's also linear in $A^2$.

Correspondingly, the response fields are linear and can be written schematically as $G = k F$, $p = e v$, where $k$ and $e$ are sets of constant structure coefficients. Additionally, if $L_A(A)$ is present, $J = m A$, for some additional set $m$ of constant structure coefficients.

The remaining term $L_q(q) = -V(q) d^4 x$ is the "potential" for the scalar field. When symmetry breaking arises, $V(q)$ is not convex and does not have minima at $q = 0$, but at any of a range of $q = q_0$. The choice of $q_0$ will break the symmetry of the gauge field, since $q_0$, itself, need not have the gauge symmetry. So, the scalar field is stuck in an on-position.

The Euler-Lagrange equations, in terms of the response field will have the form $$dG + AG - GA = J + qp,\quad dp + pA = f.$$

Now, we can trace through the chain of dependence:

• Since $p$ is linear in $v$, we have $qp = q e v = q e (dq + Aq)$. In its quiescent mode, $q = q_0$, and $qp$ reduces to $q_0 e A q_0$.
• This adds a term linear in $A$ to the right-hand side of the field equation for $dG + AG - GA = J + qp$. For Proca fields, $J$ would actually contain a term proportional to $A$, while for Maxwell and Yang-Mills fields, $J = 0$.
• The extra contribution arising from $q_0 e A q_0$ yields the effect of a Proca field with $q_0 e q_0$ playing the role of the above-mentioned $m$ structure coefficients. It alters the $m$ coefficients of $J$, if they're non-zero, or else creates the appearance of $m$ coefficients anew.

Given time, I'll fill in with more detail in a later edit, perhaps also linking it to the Yang-Mills-Higgs fields, as cases in points and illustrative examples.