# Alternative potentials in the context of spontaneous symmetry breaking

Consider a one-dimensional real scalar field $$\phi$$. Usually when spontaneous symmetry breaking (SSB) with such field is discussed, the following potential is assumed:

$$V=-\mu^2\phi^2+\lambda \phi^4$$

which is a Mexican hat potential for $$\mu^2, \lambda > 0$$.

I see, that this is probably the most simple potential to discuss spontaneous symmetry breaking.

So my question is: if we assume a classical (not-quantized) field $$\phi$$: what are the restrictions we must impose on the potential? What about terms like $$\sim\phi$$ or $$\sim\cos(\phi)$$?

Probably we must have an adequate minimum, to get the SSB running? Also, I think there was something about ginsburg-landau-theory not allowing for odd powers of $$\phi$$, which is related to the symmetry $$\phi \rightarrow-\phi$$ of $$V$$?

Let $$V(\phi)$$ be the potential for a scalar field $$\phi(x)$$. The scalar can be real or complex, it can have an arbitrary number of components, the important thing is just that it is a scalar under the Poincare group. We will think of this as the scalar field being a map $$\phi: \mathcal{M} \to \mathcal{F}\,,$$ where $$\mathcal{M}$$ is the spacetime manifold you defined your theory on, and $$\mathcal{F}$$ parameterizes the possible values your field can take, often called "field space".

A concrete example of this rather abstract definition is to take $$\mathcal{M} =\mathbb{R}^{1,3}$$, i.e. Minkowski space, and $$\mathcal{F} =\mathbb{R}$$, i.e. the field is a real scalar. Other common examples include $$\mathcal{F} =\mathbb{C}$$ or $$\mathcal{F} =\mathbb{R}^N$$. I'm keeping it general to show you just how few assumptions we need to understand spontaneous symmetry breaking.

Let $$\mathcal{F}_0 \subset \mathcal{F}$$ be the subspace of field space which minimizes $$V(\phi)$$. In your example of $$V(\phi) = - \mu^2 \phi^2 + \lambda \phi^4\,,$$ this subspace is just $$\mathcal{F}_0 = \begin{cases}\left\{-\sqrt{\frac{\mu^2}{2\lambda}}, \sqrt{\frac{\mu^2}{2\lambda}} \right\} & \mu^2 >0 \,,\\ \{0\} & \mu^2 <0 \,.\end{cases}$$

$$\mathcal{F}_0$$ is often called the vacuum manifold, and as you can see in the above examples, it can change if you change what potential you have. This process of "changing the vacuum manifold" can happen dynamically in nature and can be a kind of phase transition, depending on how the geometry of $$\mathcal{F}_0$$ changes. In particular, in the example you gave, there can be a transition from a unique vacuum state when $$\mu^2<0$$ to two equivalent vacuum states when $$\mu^2>0$$.

Spontaneous symmetry breaking occurs whenever $$\mathcal{F}_0$$ is not a point. It occurs because any point of $$\mathcal{F}_0$$ has just as much right as any other to be called the vacuum state, and so we must choose one to perturb around. This choice breaks the symmetry of $$\mathcal{F}_0$$ (unless it was just a point to begin with), hence the name.

The only restrictions on the potential for SSB to occur is that $$\mathcal{F}_0$$ exists and is not a point. That requires that $$V(\phi)$$ is bounded below, and has at least two distinct global minima.

• Thx for the answer! I thought the answer would be much more restrictive than this, but I like it a lot :-). I have two questions: I thought, the potential should somehow have a symmetry, for SSB to occur, shouldn't it? So in that real field case if I add a term proportional to $\phi$ I would lose the symmetry $\phi\rightarrow -\phi$ of the Lagrangian, or doesn't this matter? The other is: if you say "at least two distinct global minima" this somehow implies that the minima must be on the same level, or could one global and another local minimum do it as well? Commented Jun 22 at 20:45
• It's true that you could find tune parameters to have multiple vacuua without a symmetry, but generically the only way to get a non-trivial vacuum manifold is through a symmetry. In your example, when you add a term linear in $\phi$, the degeneracy between the two vacuua is resolved. Try plotting the potential on Desmos with a slider for the coefficient of the linear term to see what I mean. Commented Jun 22 at 21:16
• As for your second question, when we talk about vacuua we usually want that state to be stable under long time evolution. That's only possible if it's a global minima of the potential. If the state you're considering is only a local minima, it is called a meta-stable state. A meta stable vacuum can tunnel into a different vacuum with lower energy. Commented Jun 22 at 21:18
• Thx for the clarification. Just for me to understand one term: "trivial vacuum" means that there is one global minimum. And "non-trivial vacuum" then means there is at least two of them, right? Commented Jun 23 at 9:08