# How does the choice of a particular vacuum in a field theory problem decide the number of Goldstone bosons?

How does the field expansion method (by this I mean expanding your fields about a chosen VEV and plugging into a given potential so that the masses of the fields are given by the coefficients in front) corrolate with the vacuum you choose. In SU(3), choosing v = (0,0,v) gives 5 NGB, while v = (v,0,0) gives 6 NGB (using the Gell-Mann matrices as a basis). However, when you do a field expansion and plug it into the potential, for either of the two vacua above, you get 5 massless fields and 1 massive field. How does this match up and where have I gone wrong?

• Could you maybe describe in more detail what you did? Jun 14, 2014 at 17:20
• So the potential $V = m^2 \phi^{\dagger} \phi + \frac{\lambda}{4} (\phi^{\dagger} \phi)^2$. Plug in $\phi = (h_1 + i h_2, h_3 + i h_4, v_0 + h_5 + i h_6)$, where the h_i are small fluctuations about the vacuum. Looking at the h_i^2 terms in the Lagrangian gives the masses. h_5 is the only massive excitation with mass = $\sqrt{2} m$. Clearly, the same results occur for VEV = (v,0,0) and (0,v,0). But the breaking pattern isn't the same for the different vacua. (I calculate the breaking pattern by acting on the VEV with each of the generators and the broken gens are given by $t_i v \neq 0$). Jun 14, 2014 at 19:52

## 1 Answer

Where you have gone wrong is in accounting for the dimensionality of your surviving subalgebra. In the first case, as you correctly see, the surviving SU(2) is, indeed, manifestly spanned by $\lambda_{1,2,3}$, so the remaining 5 generators are broken.

In the 2nd case, however, you should have noticed the less manifest SU(2) spanned by $\lambda_{6,7}$ and $(\sqrt{3} \lambda_8-\lambda_3)/2$, in the 2-3 subspace, now leaving the 1st component alone, and breaking the rest, including, of course, $(\sqrt{3} \lambda_3+\lambda_8)/2$.

Lie Algebras are linear structures.