Where does the Goldstone couplings go in the unitary gauge?

Question

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.

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.