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 $v$.

  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/17944/2451 and links therein. $\endgroup$
    – Qmechanic
    May 27, 2021 at 10:29
  • $\begingroup$ @Jonas - You have made a great edit to this question. Worth appreciating. $\endgroup$
    – SG8
    May 27, 2021 at 21:40

1 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.

  • $\begingroup$ Perfect, thank you! However, should the masses correspond to $W_{\mu}^+$, $W_{\mu}^-$ and $Z_{\mu}$? $\endgroup$
    – sailew
    May 27, 2021 at 11:09
  • $\begingroup$ @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 $W_\mu^+$? $\endgroup$ May 27, 2021 at 11:16
  • $\begingroup$ Yes, your edit fixed it now! Thanks again! $\endgroup$
    – sailew
    May 27, 2021 at 11:40
  • $\begingroup$ Why does the derivative $\partial_{\mu}$ disappear after the first line for $|D_{\mu} \phi|^2$? $\endgroup$
    – sailew
    May 27, 2021 at 15:17
  • $\begingroup$ @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 $\endgroup$ May 27, 2021 at 15:40

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