Imagine that you have some model with an enlarged scalar potential, such that there is, for instance, a quartic coupling $\kappa$ between the Higgs charged component and three other scalars, which do not get a vev.

After electroweak symmetry breaking in the Feynman-t’ Hooft gauge ($\xi=1$), one should have then a similar quartic coupling $\kappa$ between the three scalars and the Goldstone boson associated to $W$ with mass $m_W$.

However, in the unitary gauge ($\xi \rightarrow \infty$), the Goldstone boson gets “eaten” by the $W$ and the coupling $\kappa$ disappears, as the mass of the Goldstone is infinity in this gauge.

So my question is, where does the information of the coupling $\kappa$ go in the unitary gauge? Is it some how hidden in the momenta structure that appears in the $W$ propagator in the unitary gauge?

For clarification I give an example: Consider adding an $SU(2)_L$ doublet with hypercharge $3/2$, $\eta \equiv (\eta^{++},\eta^+)$. Along with an $SU(2)_L$ singlet with hypercharge $-1$, $S \equiv S^-$.

For this particle content the full scalar potential is, \begin{eqnarray} \mathcal{V} = \mathcal{V}_{SM} &+& m_{\eta} \eta^\dagger \eta + m_S S^* S \\ &+& ( \, \mu_2 \, \eta^\dagger H S + \kappa \, H \eta S S + \text{h.c.} \, ) \\ &+& \lambda_{\eta} \, (\eta^\dagger\eta)^2 + \lambda_{\eta S} \,(S^* S)^2 \\ &+& \lambda_{H\eta,1} \, (H^\dagger H)(\eta^\dagger\eta) + \lambda_{H\eta,3} \, (H^\dagger\eta^\dagger)(H\eta) + \lambda_{HS} \, (H^\dagger H)(S^* S) + \lambda_{\eta S} \, (\eta^\dagger\eta)(S^* S) \, . \end{eqnarray}

Regarding the couling $\kappa \, H \eta S S$,

\begin{equation} H \eta S S \supset H^+ \eta^+ S^- S^- \, , \end{equation} which is the coupling I am talking about in the main text of the question.

  • $\begingroup$ It would make a significant difference in your answers if you actually wrote down the extra pieces of the potential you are talking about... you are imposing that the extra scalars do not couple to the surviving Higgs, but how do you arrange EW invariance here? $\endgroup$ Apr 26, 2020 at 21:22
  • $\begingroup$ I think is not importat but... let's say that there is an $SU(2)_L$ doublet $\eta$ with hypercharge $3/2$ and a singly charged $SU(2)_L$ singlet $S \equiv S^+$, so that the coupling $H \eta S^* S^*$ is invariant under the Standard Model. For this particle content, the only extra pieces in the potential are this couling, a trilinear coupling $\eta^\dagger H S$, and the usual quartic couplings $( \eta^\dagger \eta)^2$, $H^\dagger H \eta^\dagger \eta$, etc.This helps you? Also, what do you mean by "you are imposing that the extra scalars do not couple to the surviving Higgs"? $\endgroup$
    – ceperic
    Apr 27, 2020 at 10:39
  • $\begingroup$ Not a hypercharge singlet. Not electrically neutral. $\endgroup$ Apr 27, 2020 at 10:42
  • $\begingroup$ Of course not, it is charged. I meant an $SU(2)_L$ singlet. I edit it. $\endgroup$
    – ceperic
    Apr 27, 2020 at 10:46
  • $\begingroup$ I mean the potential term! Write down your potential explicitly and precisely. In the question. Right now it is contradictory. Might express H in unitary gauge Before gauging... $\endgroup$ Apr 27, 2020 at 10:49

1 Answer 1


After some forensic examination of your inconsistent potential and squaring off your $\mu_2$ versus $\kappa$ terms, I have concluded you have meant to write something like $$ \kappa {\tilde H}^\dagger \eta SS + \hbox{h.c.}, $$ where $$ H= \begin{pmatrix} H^+ \\ H^0 \end{pmatrix} \qquad \implies {\tilde H}= \begin{pmatrix} H^{0 ~*} \\ -H^- \end{pmatrix} , $$ based on the conjugate representation, that is, in your nonstandard half-scale hypercharge notation, $$ Y(\eta)=3/2 , Y(S)=-1,Y(H)=1/2,\implies Y(\tilde H)=-1/2, \implies Y({\tilde H}^\dagger) =1/2. $$ This way, your quartic can conserve weak isospin, weak hypercharge, and hence charge.

In component language, this amounts to something like $$ \kappa S^- S^- \begin{pmatrix} H^{0} \\ -H^+ \end{pmatrix} \cdot \begin{pmatrix} \eta^{++} \\ \eta^+ \end{pmatrix} , $$ where I've been cavalier with factors and signs. But you must recognize this coupling as the exact group-theoretic analog of the Yukawa term for up-type quarks (and, today, Dirac neutrinos).

Now the crucial step is to appreciate that, in the "polar" representation, just as the goldstons can assemble exclusively in $$ H= \exp (i\vec \xi\cdot \vec \tau /2v)~~ \begin{pmatrix} 0 \\ v+h \end{pmatrix} , $$ a similar exponential matrix U performs the same function in the conjugate representation, $$ \tilde H ^\dagger U^\dagger = \begin{pmatrix} v+h \\ 0 \end{pmatrix}, $$ but I won't work it out for you in the conjugate representation: should be something like $\tau^2 \exp (i\vec \xi\cdot \vec \tau /2v) \tau^2$.

What you are meant to appreciate is that, now, in the unitary gauge,

  • the actual states of the η doublet are the $U\eta$, that is, the physical states $\eta^+, \eta^{++}$ already contain the absorbed goldstons $\vec \xi$ in their definition.

That is to say, even though now the $\eta^+$ vanishes from quartic couplings, its doublet brother survives with a vengeance, $$ \kappa (v+h) \eta^{++}S^-S^-, $$ containing all the relevant information of κ.

Unitarity works in mysterious ways, never scanting information.


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