Goldstone theorem states that when a continuous symmetry is broken there is a massless mode for each broken generator. To exemplify the theorem, many references consider the complex scalar theory with an $U(N)$ symmetry. The potential for the theory is

$$-m^2H^\dagger H+\frac{\lambda^2}{2}(H^\dagger H)^2$$

when $m>0$ the vacuum expectation value of the theory $v^2\equiv\langle H^\dagger H\rangle=\sqrt{m/\lambda}$ will be non-zero and will spontaneously break $U(N)$. If we expand the field $H$ around its vacuum as

$$H=\begin{pmatrix} v+\chi_1+i\eta_1 \\ \chi_2+i\eta_2 \\ \vdots \\ \chi_n+i\eta_n \end{pmatrix}$$

we will find that only $\chi_1$ have a mass term in the expanded potential, and that $\chi_2,...,\chi_n,\eta_1,...,\eta_{n}$ remain massless. This agrees with the general statement of the Goldstone theorem since we are breaking $U(N)$ to $U(N-1)$ and therefore we should have $N^2-(N-1)^2=2N-1$ massless modes. However, if instead, I expand the field as

$$H=\begin{pmatrix} v/\sqrt{2}+\chi_1+i\eta_1 \\ v/\sqrt{2}+\chi_2+i\eta_2 \\ \chi_3+i\eta_3\\ \vdots \\ \chi_n+i\eta_n \end{pmatrix}$$

then both $\chi_1$ and $\chi_2$ will have mass terms in the expanded potential. More generally, as I split $v$ among the different components of $H$ the respective component will get a mass term.

So I want to understand better what is going on. I am changing the symmetry breaking pattern as I split $v$ among the different components of $H$? If so, what are the symmetry breaking patterns? Alternatively, maybe this is not the right way to see if I have massless modes in the theory. If so, what is the right way and what is the relation with this way?


1 Answer 1


If you work out the mass term using your second $H$, you will see that it is of the form $$m^2(\chi_1+\chi_2)^2$$

Only the combination $\chi_1+\chi_2$ has a mass, and in particular the combination $\chi_1-\chi_2$ is massless.

  • $\begingroup$ Thank you very much! So the more general procedure to find the massless modes is to diagonalize the matrice associated with quadratic terms and see if there are zero eigenvalues. Is this correct? $\endgroup$ Commented Oct 20, 2020 at 13:01
  • $\begingroup$ Yes, that would be a way to definitively determine which modes have mass. $\endgroup$
    – fewfew4
    Commented Oct 20, 2020 at 14:49
  • $\begingroup$ I can conclude from your answer that it is not possible to break more the $U(N)$ symmetry using just a scalar in the fundamental representation, right? I would need to have other fields to have a different breaking pattern from $U(N)$ to $U(N-2)$ for example? $\endgroup$ Commented Oct 20, 2020 at 22:55

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