# 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. 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 $\phi_a \cdot \phi_a = v^2$. 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. 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. Apr 23, 2021 at 21:45
• $\uparrow$ Is $N=n$? 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.