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I just started having a look at 2 Higgs Doublet Models (2HDM) and came across this:

The general idea seems pretty simple. Introduce a second Higgs doublet that acquires a vev.

The general Yukawa interaction term (ignoring leptons) for both doublets $\Phi^1$ and $\Phi^2$ then looks like: $$ \mathcal{L}_Y = Y_{ij}^{U1}\overline{Q}_{L,i}\tilde{\Phi}^1U_{R,j} + Y_{ij}^{D1}\overline{Q}_{L,i}\Phi^1D_{R,j} + Y_{ij}^{U2}\overline{Q}_{L,i}\tilde{\Phi}^2U_{R,j} + Y_{ij}^{D2}\overline{Q}_{L,i}\Phi^2D_{R,j} $$ Now there are several types of 2HDMs two of which go by the simple names Type I and Type II:

If I got it right the short summary of both is the following:

Type I: One doublet is just like the standard model Higgs i.e. it is responsible for both fermion and vector boson masses while the other doublet only interacts with the gauge bosons and does not have any Yukawa couplings
i.e. we set $Y^{U1}$ and $Y^{D1}$ $\rightarrow0$.

Type II: Again both doublets are responsible for the gauge boson masses while in the fermionic sector they split up such that one doublet is responsible for the up-type masses while the other one takes care of the down-type masses (including charged leptons).
Or in short we set $Y^{U1}$ and $Y^{D2}$ $\rightarrow0$.


My question is now:
Which mechanism drives the corresponding yukawa couplings to zero such that the 2 doublets interact in the type I or II way.

For the $\rho$-Parameter to be 1 at least at tree level both doublets need to have the Standard Models Higgs hypercharge of 1. Therefore naturally both doublets should act like the SM Higgs i.e. have Yukawa interactions with up- and down-type fermions.

Is this just some ad hoc Ansatz assuming some more fundamental model will explain it or did I miss something?

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Type I and Type II obey different global phase symmetries.

Type I respects global symmetry:

  • $Q_{L,j} \rightarrow Q_{L,j}$
  • $\Phi_1 \rightarrow e^{-i\theta}\Phi_1$
  • $\Phi_2 \rightarrow e^{i\theta}\Phi_2$
  • $U_{R,j} \rightarrow e^{i\theta}U_{R,j}$
  • $D_{R,j} \rightarrow e^{-i\theta}D_{R,j}$

Whereas Type II respects global symmetry:

  • $Q_{L,j} \rightarrow Q_{L,j}$
  • $\Phi_1 \rightarrow e^{-i\theta}\Phi_1$
  • $\Phi_2 \rightarrow e^{i\theta}\Phi_2$
  • $U_{R,j} \rightarrow e^{i\theta}U_{R,j}$
  • $D_{R,j} \rightarrow e^{i\theta}D_{R,j}$

Either of the two types upholds one symmetry and spoils the other symmetry. The technical naturalness argument is therefore: both symmetries will be respected by a hitherto unknown underlying theory, which experiences some symmetry breaking mechanism, so that the presumed "zeros" are actually non-zero but small quantities protected by the weakly broken symmetries.

The choice of type I or type II depends on the symmetry breaking pattern.

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  • $\begingroup$ Thanks for the answer. It should be exactly the other way around right (at least if I did not mix up type I and II in my question above) ? To summarize your answer one imposes zero couplings (which is equivalent to imposing above global symmetries) without any further justification (which is of course completely fine from a phenomenological point of view) except maybe a naturalness like argument and one expects that those symmetries will be explained by e.g. the gauge structure of a more fundamental theory. $\endgroup$ – Katermickie Nov 18 '19 at 19:30
  • $\begingroup$ @Katermickie, either of the two types upholds one symmetry and spoils the other symmetry. The naturalness argument would assume that the said symmetries will be respected by a hitherto unknown underlying theory, which experiences some symmetry breaking mechanism, so that the presumed "zeros" are actually small but non-zero quantities protected by the weakly broken symmetries. $\endgroup$ – MadMax Nov 18 '19 at 19:36
  • $\begingroup$ That's why I said naturalness like :-D $\endgroup$ – Katermickie Nov 18 '19 at 19:40

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