Why can you set the charged field VEV's to zero when minimizing the potential?

Question

When minimizing a scalar potential how come we are allowed to set the VEV of charged components to zero?

I stumbled upon this when studying the type-II seesaw model in this paper (though the question should be self-contained). The potential couples a Higgs scalar, $ \Phi = \left( \phi ^+ , \phi ^0 \right) $ with the new isospin triplet, $ \xi = \left( \xi ^{ + + } , \xi ^+ , \xi ^0 \right) $. The charges on the scalars indicate their $ T _3 + Y $ values and hence their charge after EW symmetry breaking. The most general potential is,
\begin{align} 
V &= m ^2 \Phi ^\dagger \Phi + M ^2 \xi ^\dagger \xi + \frac{1}{2} \lambda _1 ( \Phi ^\dagger \Phi ) ^2 + \frac{1}{2} \lambda _2 \left( \xi ^\dagger \xi \right) ^2 + \lambda _3 \left( \Phi  ^\dagger \Phi \right) \left( \xi ^\dagger \xi \right) \\ 
& \qquad + \mu \left( \bar{\xi} ^0 \phi ^0 \phi ^0 + \sqrt{2} \xi ^- \phi ^+ \phi ^0 + \xi ^{ - - } \phi ^+ \phi ^+ h.c. \right) 
\end{align} 
Then instead of doing a 10 (4 degrees of freedom in $ \Phi $ and 6 in $ \xi $) field minimization, the authors write $ \left\langle  \phi ^0 \right\rangle = v $ and $ \left\langle \xi ^0 \right\rangle = u $. They completely ignore the possibility of the charged components getting a VEV. Now of course, these should not get a VEV to break $ SU(2) \times U(1) _Y \rightarrow U(1) _{ EM } $, but shouldn't this be a output of the model instead of something that is forced upon it? Otherwise I think the resulting potential could be unstable since it may not be symmetric around the charged components. 

Note that this situation is distinct from the Higgs field where we are able to choose where the VEV's go since the Higgs is the only scalar and can be transformed using the SU(2)xU(1) gauge to get a VEV in whichever component we please.

Edit: I think they are also setting the VEV of the imaginary component of each uncharged field to zero. Is there a reason why this would be justified as well?

Answer 1

Charged scalar fields $\Phi(x)$ usually appear in the action as a function of $\Phi(x)^*\Phi(x)$, which implies that there is a stationary point at $\Phi(x)=0$. More generally, the gradient of any term containing two components of $\Phi$ vanishes, hence makes $\Phi=0$ stationary. Thus it is a shortcut to assume this from the start. 

If one wants to be sure that no other stationary points exist, one cannot take the shortcut. In general, there might be other stationary points with nonzero $\Phi(x)$. This may happen if there are linear or terms of higher order. But these are not candidates for a vacuum state, as the vacuum is, by definition, uncharged. 

Comments

That makes perfect sense, thanks! So the idea is unless you have a term that's only linear in the field, there will exist a minima at the origin and we are choosing this solution to expand around? 

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Sorry one more question. In the above example the $\xi^-$ field appears linearly in the term, $\xi^- \phi^+\phi^0$. So in this case is it necessary to be more careful when minimzing the potential?

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''unless you have a term that's only linear in the field, there will exist a minima at the origin'' - at least a stationary point. It may be a saddle if the Hessian is indefinite; then the zero state will be broken.

''In the above example the $ξ^−$ field appears linearly in the term, $ξ^−ϕ^+ϕ^0$. So in this case is it necessary to be more careful'' - Only if one allows a charged vacuum, which is a strange situation.

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Ah, I now see. The uncharged vacuum state is still one of the minima in the potential and this will always be the case if SU(2)xU(1) is breaking to a U(1). Thanks for your help.

Answer 2

I look at the convention for broken symmetry the other way around. In the Glashow-Weinberg-Salam model: whatever the vev of the doublet, there's always some generator (in $SU(2) \times U(1) = U(2)$) which annihilates it. That's the generator of an unbroken long-ranged force (which I choose to call EM). This is a worthy model, because we observe such a force in nature. Wlog, that doublet can be rotated to a basis where the vev takes the conventional form that you've decribed in the question.

As to the comment of whether any potential without a linear term will have a minimum at the origin: Yes, but that isn't the complete story. If you had a cubic in the field, then you could get a secondary minimum at some other location (which might lead to tunnelling and a first order phase transition, etc.)

Comments

Thanks for your response. In this viewpoint it doesn't seem that we are minimizing the potential at all. How do we know that the spontaneously broken state is a minima? By setting a VEV and rotating to a different basis, aren't you apriori assuming that such a VEV exists instead of finding it?

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The "radial" value of your vev is still a free parameter which you need to determine by minimizing the potential. I've just exploited the "rotational" (gauge) symmetry in the doublet to get rid of it's "angular" part.

If you find that the minima is at $\nu = 0$ then that's the origin and symmetry is unbroken... and for nonzero values we have a broken symmetry. Gauge symmetry/redundancy implies that all possible angular directions are equivalent. Only $\langle \phi^\dagger \phi \rangle$ can be physically relevant.

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I see why this must hold for the typical case. But what about in the case above where we have a term in the Lagrangian,

$$\Psi^T \xi \Psi = \bar{\xi} ^0 \phi ^0 \phi ^0 + \sqrt{2} \xi ^- \phi ^+ \phi ^0 + \xi ^{ - - } \phi ^+ \phi ^+ h.c. $$

Here we don't only find $\Phi^\dagger\Phi,\xi^\dagger\xi  $ terms in the Lagrangian so aren't there other physical contributions? When we have more then one scalar with more than one VEV, don't you need to be more careful?

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$\phi^\dagger \phi $ was just an example of a gauge-invariant combination of the scalars. In your example, $\Phi^T \xi \Phi$ seems to be precisely such a combination -- so that combination can pick up a vev. Maybe I don't quite understand your confusion.

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Ah okay, I think I understand now. Thanks again!