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

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 $$v$$.

• Related: physics.stackexchange.com/q/17944/2451 and links therein. May 27, 2021 at 10:29
• @Jonas - You have made a great edit to this question. Worth appreciating.
– SG8
May 27, 2021 at 21:40

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.

• Perfect, thank you! However, should the masses correspond to $W_{\mu}^+$, $W_{\mu}^-$ and $Z_{\mu}$? May 27, 2021 at 11:09
• @sailew Sorry, I had a typo in the table. Does my edit fix what you were asking, or are you asking if e.g. $gv/2$ corresponds to the experimentally-measured value of $W_\mu^+$? May 27, 2021 at 11:16
• Yes, your edit fixed it now! Thanks again! May 27, 2021 at 11:40
• Why does the derivative $\partial_{\mu}$ disappear after the first line for $|D_{\mu} \phi|^2$? May 27, 2021 at 15:17
• @sailew $v$ is just a constant, so its partial derivative is $0$. To be clear: this $\partial_\mu$ acts only on the Higgs boson part, but these don't contribute to the uncorrected (tree-level) masses of the gauge bosons May 27, 2021 at 15:40