More on Spontaneous Symmetry Breaking

Last week I talked about SSB and focused on its physical interpretation and effects, looking at fields and potentials while avoiding (as much as possible) to mention group theory. This is why I’ve decided to make the following into a separate post, where I’ll briefly re-emphasize a point I’ve already made in the last post (how we count massless modes in the broken theory) from a group-theoretic perspective.

A few definitions

Proper subgroup

$H\subset G$ if $H$ is a subset of $G$ and follows the same group axioms. It’s a group within a group.

Left coset

. It’s the set of all possible left multiplications of $H$ with $G$.

Right coset

. It’s the set of all possible right multiplications of $H$ with $G$. Duh.

Normal subgroup

if $gH=Hg$ $\forall g\in G$. Happens when the left and right cosets of $H$ coincide.

Vacuum manifold

. The set of all possible vacua (has cardinality one if the symmetry is unbroken).

Invariant/stability group

. $H$ leaves the vacuum manifold… invariant…

Putting them together

Let’s consider a Lagrangian with symmetry group $G$, spontaneously broken to $H\subset G$, that is with $\lvert\Phi_0\rvert>1$. As we’ve seen in the previous post, $H$ is not any subgroup of $G$, it’s the corresponding stability group. Or rather, having picked some $\phi_0\in\Phi_0$, we define $H$ to be the corresponding invariant subgroup; this is with the idea in mind that different vacua might have different invariant subgroups. In fact, because we can still find some $g\in G$ to transform a given $\phi_0$ to another $\phi_0’$, these subgroups have to be isomorphic. Consider for example $(H’,\phi_0’)$ and $(H,\phi_0)$; then $\phi_0’=g\phi_0$ if and only if $H’\cong gHg^{-1}$ :

Furthermore, it shouldn’t make any difference whether we act invariantly (with $H$) or not at all on a vacuum before we transform it into another one (with $G$), and so we are led to define equivalence classes : $g_1\sim g_2$ if $\exists h\in H$ such that $g_1=g_2h$. Eventually, all $\phi_0\in\Phi_0$ are assigned an equivalence class so that there is only one way to transform a vacuum into another, up to invariant operations. We sum all of this up by writing

(Note that because $\Phi_0$ is a manifold, $G/H$ has to be a group and not a mere set; this is only achieved when $H$ is a normal subgroup). Now, acting infinitesimally on $\phi$ allows us to write down the usual linear perturbation

where $a$ ranges from $1$ to $\dim G$ and $t^a$ are the generators of the Lie algebra of $G$. In terms of the potential :

where $r,s=1,\ldots,N$ are components of $\phi$. Around a minimum $\phi_0$ :

Compare this to the usual $\frac{1}{2}m^2\phi^2$ and we can interpret the matrix of second derivatives as a mass matrix $M_{rs}^2$. Now taking the derivative of the previous equation (evaluated at the minimum) :

If the theory is unbroken and the vacuum is unique, the first bracket vanishes (since $g\phi_0=\phi_0$ $\forall g\in G$). On the other hand, if the symmetry is broken, the bracket becomes an eigenvector of the mass matrix with eigenvalue (i.e. mass) zero (for some $g\in G$).

How many such eigenvectors are there? Let’s call $T^i$ the generators of $G$ for which $T^i\phi_0=0$. There are exactly $\dim H$ of them, since by definition they generate a subgroup $H\subset G$. All the Lie groups involved in the Standard Model are compact and semi-simple, so it is always possible to define a scalar product and the notion of orthogonality. We can then write the remaining generators of $G$ as $\tau^j$ and make them orthogonal to the $T^i$ : $\text{Tr}\left(T^i\tau^j\right)=0$. The full basis of the Lie algebra of $G$ is therefore $t^a=\left(T^i,\tau^j\right)$. The $\tau$’s are of interest, because they form the eigenvectors we’re after; it is now easy enough to count them : there are exactly $\dim G-\dim H$ of them !

The massless modes corresponding to these eigenvectors are precisely the Goldstone bosons we described in the previous post; finally, since $M^2$ is an $N\times N$ matrix, there are at most $N-\left(\dim G-\dim H\right)$ massive modes.

Note : I might write something about the full quantum proof of Goldstone’s theorem in the future, but this is already a good starter. The essential difference lies in the axioms the proof is built upon – in particular, the requirement of states with positive definite norms, which is broken by gauge theories (ghosts). This is why we end up needing the Higgs mechanism for the Standard Model.

comments powered by Disqus