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?

  • $\begingroup$ 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. $\endgroup$ Commented Apr 23, 2021 at 19:08
  • $\begingroup$ @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 $\phi_a \cdot \phi_a = v^2$. $\endgroup$
    – mathripper
    Commented Apr 23, 2021 at 19:46
  • $\begingroup$ 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. $\endgroup$
    – mathripper
    Commented Apr 23, 2021 at 20:01
  • $\begingroup$ 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. $\endgroup$
    – mathripper
    Commented Apr 23, 2021 at 21:45
  • 1
    $\begingroup$ $\uparrow$ Is $N=n$? $\endgroup$
    – Qmechanic
    Commented Apr 24, 2021 at 3:25

1 Answer 1


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.


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