 Solution

Let $V(\varphi )=m^2{\varphi }^2/2+\lambda ({\varphi }^2{)}^2/4$, where ${\varphi }^2={\sum }_i{\varphi }_i^2$. Its stationary-point equation is $(m^2+\lambda {\varphi }^2){\varphi }_i=0$.

For $m^2>0$, the unique minimum is $\varphi =0$. The quadratic mass term is $m^2{\delta }_{ij}$, so there are $n$ real scalar fields, all of mass $\sqrt{m^2}$, and the global special orthogonal group $SO(n)$ remains unbroken.

For $m^2<0$, write $v^2=-m^2/\lambda$. The vacuum manifold is the sphere ${\varphi }^2=v^2$. Choose $⟨\varphi ⟩=v{e}_n$. The transformations preserving this vacuum expectation value rotate the first $n-1$ components, giving $SO(n)\to SO(n-1)$. At this vacuum the potential's Hessian matrix isWriting $\varphi =({\pi }_1,\dots ,{\pi }_{n-1},v+\eta )$, the radial mode $\eta$ therefore hasThese are one massive radial scalar and $n-1$ physical Goldstone bosons. The Goldstone theorem applies to the broken global generators; the flat angular directions of the vacuum manifold explain their zero masses. The spontaneous symmetry breaking is a choice of vacuum, not a change in the invariant Lagrangian. For $n=1$ there are no continuous angular directions: only the radial scalar remains, and the potential's discrete $\varphi \mapsto -\varphi$ symmetry is broken.