# When does a VEV induce another VEV?

If we have a model with more than one Higgs-doublet, when do the VEV of a scalar field of one of those doublets must induce a nonzero VEV on a scalar field of another of those doublets?

• In general, whenever a VEV of a field is allowed by symmetry, it will be induced, either already at tree level due to the choice of potential, or by radiative corrections. So if the VEV of one of the Higgs fields spontaneously breaks the symmetry that protects the other Higgs field, then the latter will develop a VEV as well. In short, the answer is that the VEV of one field can induce the VEV of another field if both fields transform nontrivially under the same symmetry. It doesn't matter whether or not the fields are directly coupled in the Lagrangian. – Tomáš Brauner Jul 13 at 18:50
• @TomášBrauner can you answer the question about a concrete example I made on the commentary to MadMax's answer? I still haven't fully understood this. – jmaguire Jul 16 at 17:01
• Let me reply here. With a nonzero VEV of $\phi_1$, the interaction term $\phi_1^\dagger\phi_1\phi_2^\dagger\phi_2$ will contribute to the mass term of $\phi_2$. Depending on the sign and magnitude of the coupling, it may force $\phi_2$ to condense as well. This is a good example, it shows that my above comment gives a natural mechanism for a VEV of one field to induce a VEV of another field, but it's not the only possibility. With a biquadratic coupling of the type $\phi_1^2\phi_2^2$, the VEV of $\phi_1$ may induce a VEV of $\phi_2$ regardless of how symmetries act on the fields. – Tomáš Brauner Jul 16 at 21:27
• @TomášBrauner Then what would be the conditions on the potential so that a vacuum structure in which only $\phi_1$ has a VEV on the second component is possible? – jmaguire Jul 17 at 9:25
• I'm not sure one can answer such a detailed question without having a specific potential with a finite-dimensional parameter space. If you do have a concrete potential though, then all you have to do is to ensure that your favorite vacuum is a (local) minimum of the potential, right? Checking this is a mere exercise on the eigenvalues of the Hessian matrix of the potential. – Tomáš Brauner Jul 17 at 11:12

It depends.

In the two Higgs doublets models, there are two cases:

1. The two doublets are independent
2. There is a cross potential term between these doublets a la: $$\phi_1^2\phi_2^2$$

For case 1, the two VEVs are independent.

For case 2, the two VEVs are are coupled to each other. But it is hard to tell which is inducing which.

• Let me then give an example, imagine a 3-HDM with a $\mathbb{Z}_3$ symmetry such that $\phi_1 \rightarrow \phi_1$, $\phi_2 \rightarrow \exp{(2\pi i/3)} \phi_2$, $\phi_3\rightarrow \exp{(4\pi i/3)} \phi_3$. If the second component of $\phi_1$ acquires a VEV $v$, will that induce a VEV on the other fields? – jmaguire Jul 13 at 20:21
• It depends on the specific couplings. Is the $\mathbb{Z}_3$ somehow related to the Tribimaximal mixing? – MadMax Jul 13 at 20:25
• No. It is just a simple model with no more restrictions other than being renormalizable and being symmetric under the transformation I wrote above. – jmaguire Jul 13 at 20:32