In what sense do Goldstone bosons live in the coset?

Goldstone's theorem says that if a group, $G$, is broken into its subgroup, $H$, then massless particles will appear. The number of massless particles are given by the dimension of the coset, $G/H$. It is then often said that the Goldstone boson's live in the coset. In what sense is this statement true? The Lagrangian is not invariant under transformations of the coset so what does this "living" explicitly mean?

To be explicit we can consider the linear sigma model: $${\cal L} = \frac{1}{2} \partial _\mu \phi ^i \partial^\mu \phi ^i - \frac{m ^2 }{2} \phi ^i \phi ^i - \frac{ \lambda }{ 4} ( \phi ^i \phi ^i ) ^2$$

We define, \begin{align} & \phi _i \equiv \pi _i \quad \forall i \neq N\\ & \phi _N \equiv \sigma \end{align} and give $\sigma$ a VEV.

The spontaneously broken Lagrangian is, $${\cal L} = \frac{1}{2} \partial _\mu \pi _i \partial ^\mu \pi _i + \frac{1}{2} ( \partial _\mu \sigma ) ^2 - \frac{1}{2} ( 2 \mu ^2 ) \sigma ^2 - \lambda v \sigma ^3 - \frac{ \lambda }{ 4} \sigma ^4 - \frac{ \lambda }{ 2} \pi _i \pi _i \sigma ^2 - \lambda v \pi _i \pi _i \sigma - \frac{ \lambda }{ 4} ( \pi _i \pi _i ) ^2$$ The Goldstone bosons, $\pi_i$, exibit a $O(N-1)$ symmetry, but this is not the coset group symmetry. So where in the Lagrangian do we see this symmetry?

• I don't feel like I understand this topic well enough myself to give a proper answer, but there is a discussion of this in Weinberg's book, vol II, Chapter 19. In particular, section 19.6, and cosets are introduced on page 214. – Robin Ekman Apr 17 '14 at 2:42
• @RobinEkman: Thanks for letting me know. Unfortunately, I don't have the book. I'll take a look when I can get my hands of a copy. – JeffDror Apr 17 '14 at 12:30

I understand the statement in the following way:

Pions, which are pseudo-goldstone bosons of chiral symmetry breaking, are described by the introduction of a unitary matrix $U(x)$, defined as

$$U(x)=\text{exp}\left(2i\pi^a(x)T^af_\pi^{-1}\right),$$

where $\pi^a$ is the pion field, $f_\pi$ is the pion decay constant and $T^a$ are the generators of the broken symmetry, i.e. the coset space. The pion Lagrangian can be written down in terms of $U(x)$

$$\mathcal{L}=-\frac14 f_\pi^2\text{Tr}\partial^\mu U^\dagger\partial_\mu U,$$

which by expanding the exponential form results in

$$\mathcal{L}=-\frac12\partial^\mu \pi^a\partial_\mu \pi^a+\dots,$$

where dots denote higher order terms. Thus, the statement that goldstone bosons live in the coset space can be related to the fact that the fields themselves are linked to the generators of the coset.

This can be understood in terms of Goldstone's theorem: if the original Lagrangian exhibits a continuous symmetry, the number of goldstone bosons is equal to the number of generators of the broken symmetry. Take for example the linear sigma model: if your original theory is $O(N)$-symmetric, it has $N(N-1)/2$ symmetries. If the symmetry is broken spontaneously, you end up with $O(N-1)$, leaving you with $(N-1)(N-2)/2$ symmetries. The amount of broken symmetries is the difference, i.e. $N-1$. But this is precisely the number of pions you have in your theory. We can conclude that the pions are linked directly to the broken symmetries, i.e. the coset space.

• Thank you very much for your response. It makes things a clearer. I have seen the matrix $U(x)$ but I didn't make the connection. The way I thought of $U(x)$ was as a convenient redefinition of the pion field that leads to a nice power counting in the momentum of the pions. Is there a more to it then that? – JeffDror Apr 17 '14 at 14:24
• The point is that the pion field is defined by the generators of what you call the coset, i.e. $\pi=\pi^a T^a$. I just explained it in terms of U(x) because that is a common way of describing them. – Frederic Brünner Apr 17 '14 at 14:56
• What exactly makes the pion field defined by the generators of the coset? Is it just that they have the same dimension so we can contract their indices as $\pi=\pi^aT^a$ which leads to a convenient field redefinition in our Lagrangian? – JeffDror Apr 17 '14 at 15:03
• I have edited my answer. – Frederic Brünner Apr 17 '14 at 15:25
• Thanks, I think its starting to make sense. So essentially the statement that the Goldstones live in the coset arise from the fact that the number of Goldstones will be equal to the dimension of the coset and we can use this fact to make a convenient field redefinition in terms of the coset generators. – JeffDror Apr 17 '14 at 16:07

It is in fact a very simple matter if you use a different parametrization of the fields. Since we care about the Goldstone bosons only, just send $\lambda\rightarrow \infty$ so that the Higgslike state decouples. Moving to the following parametrization
$$\phi_i(x)=U(x)\langle \phi_i\rangle \,,\qquad U(x)=e^{i \hat{T}^a \pi^a(x)}\,,\qquad \langle\phi_i\rangle=\left(0,0,0\ldots,v\right)^T$$ (where $\hat{T}^a$ are the broken generators) you immediately see that there is a gauge redundancy in the definitions of the pion fields $\pi^a(x)$ since we are allowed rotate them with an $x-$dependent transformation $h(x)$ of the unbroken group $H$, namely $$\phi_i(x)=U(x)\langle \phi_i\rangle=U(x)h(x)\langle \phi_i\rangle\,.$$ In other words, the pion field is defined only up to this equivalence $U(x)\sim U(x)h(x)$, which is the statement that they live on the coset space $G/H$.

• Ah, very interesting! How do we know that we are able to apply a gauge transformation on the vacuum using only the broken generators and get back $\phi_i$? Or is this just the definition of the parametrization of the pion fields, $\phi_i$? – JeffDror Apr 18 '14 at 14:13
• Maybe I am misunderstanding what you are asking. $h(x)$ is a transformation of the unbroken group built with the unbroken generators, not the broken ones. This is why we say that they live I the right coset $G/H$. – TwoBs Apr 18 '14 at 16:28
• Sorry, I don't think I was clear. I meant how do we know that we can write, $\phi_i (x) = U (x)\langle \phi_i\rangle$ or is that just the definition of the reparametrization? – JeffDror Apr 18 '14 at 16:29
• In the limit $\lambda\rightarrow\infty$ you can move only around the minimum orbits, which is what the action of $U$ does for you. For finite $\lambda$ you can also move away from the minimum. This exhaust all possibilities. – TwoBs Apr 18 '14 at 17:44