Does the number of broken generators in SSB depend on the choice of VEV? I take the Lagrangian,
$$\mathcal{L}=\frac{1}{2}\partial_\mu \phi^T\,\partial^\mu\phi\,-\, \frac{1}{2}\mu^2\phi^T\phi-\frac{\lambda}{4}(\phi^T\phi)^2~,$$
where $\phi=(\phi_1,\,\phi_2,\,\phi_3)$ (real scalar field). $\mathcal{L}$ satisfies SO(3) symmetry. The potential is minimized for $\phi_0^T\phi_0=\mu^2/\lambda\equiv v^2$. When I take $\phi_0=(v,0,0)$, and then replace $\phi(x)=(v+\sigma(x),\pi_1(x),\pi_2(x))$ in the Lagrangian, I get $\pi_1$ and $\pi_2$ as massless and $\sigma$ as massive. Also I can see that for this choice of $\phi_0$ one of the three generators of $SO(3)$ gives $L_1\,\phi_0=0$, where $L_1$ is the generator with all element in the first column as zero.
Now if I choose $\phi_0 = (v/\sqrt{3},\,v/\sqrt{3},\,v/\sqrt{3})$ and then replace $\phi(x)=(v/\sqrt{3}+\sigma(x),v/\sqrt{3}+\pi_1(x),v/\sqrt{3}+\pi_2(x))$ into $\mathcal{L}$ then $\pi's$ are also becoming massive along with $\sigma$. Also none of the enerators giving $L_i\,\phi_0=0$.
But this seems not to be right. Where am I going wrong?
[I am reading from Peskin - chapter 11]
 A: No, the number of broken generators does not depend on the choice of vev.
When we list the generators of a group, we are just writing the basis of its Lie algebra. Writing these three elements does allow us to deduce all of the remaining ones, but it is not a comprehensive list.
In other words, the generators of $SO(3)$ are not just $L_1$, $L_2$, and $L_3$. In fact, all of the linears combinations $\alpha L_1 + \beta L_2 + \gamma L_3$ are generators. Hence, what you want to find is not an $i$ such that $L_i \phi_0 = 0$, but rather three numbers $\alpha$, $\beta$, and $\gamma$ such that $(\alpha L_1 + \beta L_2 + \gamma L_3)\phi_0 = 0$.
At the Lagrangian level, the issue is similar. One you figure out which is the broken generator, you'll be able to find the appropriate linear combination of $\sigma$, $\pi_1$, and $\pi_2$ that does become a massless particle. I'm also guessing, since it's been some time since I last saw this calculation, but I believe there is a term looking like $\propto \sigma \pi_1$ somewhere in your expression. This in an indication of the fact that the mass terms are not diagonalized. Collecting these terms of the Lagrangian in the form
$$\begin{pmatrix}\sigma & \pi_1 & \pi_2\end{pmatrix}\begin{pmatrix}m_{\sigma \sigma} & m_{\sigma \pi_1} & m_{\sigma \pi_2} \\ m_{\pi_1 \sigma} & m_{\pi_1 \pi_1} & m_{\pi_1 \pi_2} \\ m_{\pi_2 \sigma} & m_{\pi_2 \pi_1} & m_{\pi_2 \pi_2}\end{pmatrix}\begin{pmatrix}\sigma \\ \pi_1 \\ \pi_2\end{pmatrix}$$
and diagonalizing the matrix is another way of finding the appropriate combinations with definite masses.
A: The $\pi$ fields have to be chosen so that they lie along the directions of symmetry. In your second example, none of the fields lie along a direction of symmetry. For $SO(N)$ symmetries, the direction of symmetry is always perpendicular to $\phi_0$, and neither $(0,1,0)$ nor $(0,0,1)$ are perpendicular to $\frac{v}{\sqrt{3}}(1,1,1)$. As a result, the spontaneous symmetry breaking is still hidden, just like it is in the original Lagrangian. In fact, the $SO(3)$ symmetry isn't even obvious anymore.
To properly define the $\pi_i$ fields, you need to choose vectors of the form $u_i=L_i\phi_0$, where $L_i$ are the generators. This guarantees that the vectors lie along a direction symmetry around $\phi_0$. Note that when you do this, not all of the $u_i$ will be linearly independent. Depending of your choice of generators and $\phi_0$, some $u_i$ may either be zero or linear combinations of the other $u_i$. After eliminating linearly dependent $u_i$, you can define $\phi=\sum_j\sigma_jv_j+\sum_iu_i\pi_i$, where the $v_j$ vectors form a basis for everything not covered by the $\{u_i\}$ basis.
