# Residual symmetry group of a scalar field theory

Given a Lagrangian

$$\frac{1}{2} (\partial_\mu \phi)^2 - \frac{\lambda}{4!}(\phi^2 - v^2)^2$$

for a real scalar field theory with $$\vec{\phi} = (\phi_1,\phi_2,...,\phi_n)^T$$ and $$O(n)$$ symmetry. Why is the residual symmetry group (or little group) given by $$O(n\!-\!1)$$ when spontaneous symmetry is broken?

• Little group? Where is your vector? What leaves it invariant? How? Why? Show your work. Nobody is going to throw a pretty story at you which makes sense, unless you specify your language and what troubles you. This is routine stuff explained in most good texts. Commented Apr 23, 2021 at 19:08
• @CosmasZachos actually that's part of my confusion as well (I found this statements in my lecture notes). The vector in this case would be the vector with compoenents $\phi_a$ where $a = 1,2,3,...,N$. As far I understand, the vacuum state here of the theory would coincide with the minimum of the potential so when $\phi_a \cdot \phi_a = v^2$. Commented Apr 23, 2021 at 19:46
• So the rotations would belong to $O(n)$ the n-dimensional group of rotations. What I'm after is finding is to show that vacuum states given by the field $\phi$ (which must obey the condition in my previous comment) has $O(n-1)$ as the largest group that when acted upon leaves the state vector invariant. Commented Apr 23, 2021 at 20:01
• My understanding was that we have a continuum of vacuum states so we can choose one arbitrarily. Then I would find the largest subgroup of O(n) that leaves this state invariant and this should hold for the other vectors (as essentially all vacuum states are the "same"). Looking around I found that a similar statement is stated in "Gauge Theories of the Strong, Weak,and Electromagnetic Interactions" on page 90 (problem 5.6). I'm not sure what more details you think I should provide. Commented Apr 23, 2021 at 21:45
• $\uparrow$ Is $N=n$? Commented Apr 24, 2021 at 3:25

O(n) means you may rotate any n-vector to any other of the same length, or a suitably normalized combination of others. So you make a choice to rotate your reference vector to say, $$\phi_1=v(1,0,0,..,0)^T$$.
Its little group rotating the n-1 components indexed by 2,3,...,n among themselves is thus O(n-1), and it has the obvious $$(n-1)(n-2)/2$$ generators of that group acting linearly on your fields. The ones you "lost" (not really, the symmetry generators are still there, transforming $$\phi_1$$ to the other components, in a nonlinear manner) are the $$n-1$$ ones realized nonlinearly, corresponding to massless Goldstone bosons (show this!). Your Goldstone bosons are $$\phi_a$$ with a=2,3,...,n-1, while $$\phi_1$$ is massive, the σ or Higgs.
Specifically, $$\Delta_{ij}\phi_k= -\Delta_{ji}\phi_k= \theta_{ij}(\delta_{ik}\phi_i - \delta_{jk}\phi_j),$$ So $$\Delta_{ij}\phi_1= 0$$ For the O(n-1) Δs involving only indices 2,3,...n. Further, $$\Delta_{1j}\phi_1= \theta_{1j}\phi_j,$$ for only one index, j in that set: these do not leave your reference vector invariant. There are n-1 of them and shift the $$\phi_j$$ s by a constant when you redefine $$\phi'_1$$ to have a vanishing vacuum value.