0
$\begingroup$

Let's consider the Higgs Lagrangian \begin{equation} \mathcal{L}= -(\partial^\mu\phi+B^\mu\phi)^\dagger(\partial_\mu\phi+B_\mu\phi)-V(\phi^\dagger\phi) \end{equation} where $B^\mu$ is the $\mu$ component of the gauge field (a Lie-algebra-valued field, i.e. an anti-Hermitian matrix), and the potential $V$ is the usual "Mexican hat" function. (I simplified the notation a bit, I hope it's clear. The gauge field $B$ refers to the whole $\mathrm{SU}(2)\otimes\mathrm{U}(1)$ group, i.e. it's the sum of the three weak bosons and the hypercharge boson.) I am given a minimum \begin{equation} \phi_0= \frac{v}{\sqrt{2}} \begin{pmatrix} 0\\ 1 \end{pmatrix} \end{equation} of the potential $V$ about which I should expand the Lagrangian density, writing $\phi$ as $\phi_0+\phi'$. The part with the covariant derivative becomes \begin{equation} -(\partial^\mu\phi'+B^\mu\phi'+B^\mu\phi_0)^\dagger(\partial_\mu\phi'+B_\mu\phi'+B_\mu\phi_0) \end{equation} which generates the kinetic term for $\phi'$, the mass quadratic terms for the gauge bosons, and a whole lot of other cross-terms. Now, some of the books on which I studied this straight up ignore those cross-terms:

  • Weinberg's The Quantum Theory of Fields vol. 2 (eq. 21.1.5) suggests to "expand the Lagrangian to second order in $\phi'$ and $B$";
  • Peskin and Schroeder's Introduction to QFT (eq. 20.61) says that "the relevant terms" are those that generate the bosons' masses and says nothing about the others;
  • Ellis (et al.)'s QCD and Collider Physics (eq. 8.26) just writes the Lagrangian without all those cross-terms.

What should I do with those cross-terms? Do they cancel out? In the unitarity gauge I was able to show that some of them are zero, but others, like $(\partial^\mu\phi')^\dagger B_\mu\phi'$, are not. I thought that they may become interaction terms between the Higgs field and the gauge bosons, but judging from the Feynman rules for boson interactions in Ellis' book (figure 8.2, page 277) they are not there.

I'm starting to think that the line of reasoning here is that I should just discard non-quadratic terms, and be done with it. But I can't find any deeper explanation for it than "do it because everyone else does it". Why can they be ignored?

$\endgroup$
0
$\begingroup$

I believe the problem here is that the Lagrangian in the question is not gauge invariant, owing to the way that the gauge covariant derivative is written. Rather, defining the Lagrangian

\begin{equation} \begin{split} \mathcal{L} &= -(D_\mu \phi)^\dagger(D^\mu \phi) + V(\phi^\dagger,\phi) \\ & = (\partial_\mu\phi-iB_\mu \phi)^\dagger(\partial^\mu\phi-iB^\mu \phi), \end{split} \end{equation}

and expanding in the unitary gauge you should pick up a term of the form

\begin{equation} \mathcal{L} \subset i\big\{\phi'(\partial_\mu \phi'^\dagger) - \phi'^\dagger(\partial_\mu \phi')\big\}B^\mu. \\ \end{equation}

The last step is to recall that the Higgs is a real valued scalar field, such that the term in curly brackets should vanish.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Your Lagrangian is pretty much the same as mine, after the substitution $B_\mu\to -iB_\mu$ (you probably write $B$ as a Hermitian matrix, whereas mine is anti-Hermitian... I'll make this clear in the question body). Now that I think about it... is $B$ in your answer a matrix at all? I don't get how it can be moved at the right end of your last expression. $\endgroup$ – yellon Apr 30 at 19:47
  • $\begingroup$ No, apologies. I had assumed that we're working with $U(1)_Y$ gauge symmetry such that $B_\mu$ is the weak hypercharge gauge field. $\endgroup$ – nuLab Apr 30 at 21:36
  • $\begingroup$ Yeah, I guess my notation can be a bit misleading. I'll have to explain it more thoroughly. $\endgroup$ – yellon Apr 30 at 21:41

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.