1
$\begingroup$

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 $v$.

$\endgroup$
2
1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Perfect, thank you! However, should the masses correspond to $W_{\mu}^+$, $W_{\mu}^-$ and $Z_{\mu}$? $\endgroup$ – sailew May 27 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 $W_\mu^+$? $\endgroup$ – Nihar Karve May 27 at 11:16
  • $\begingroup$ Yes, your edit fixed it now! Thanks again! $\endgroup$ – sailew May 27 at 11:40
  • $\begingroup$ Why does the derivative $\partial_{\mu}$ disappear after the first line for $|D_{\mu} \phi|^2$? $\endgroup$ – sailew May 27 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 $\endgroup$ – Nihar Karve May 27 at 15:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.