Let's say we are considering the case of Abelian symmetry, with the Lagrange density given by:
$$\mathcal{L}=\frac{1}{2}(\partial_\lambda \sigma)^2 + \frac{1}{2} (\partial_\lambda \pi)^2-V(\sigma^2 +\pi^2),$$
where $$V(\sigma^2+\pi^2)=-\frac{\mu^2}{2}(\sigma^2 +\pi^2)+\frac{\lambda}{4}(\sigma^2 +\pi^2 )^2. $$
The Lagrangian has $U(1)\cong O(2)$ symmetry. By minimizing the potential we get the minimum of the potential given by (for $\mu^2>0)$:
$$\sigma^2+\pi^2 =v^2,\qquad v=\sqrt{\frac{\mu^2}{\lambda}}.$$
In my literature it then says:
The minima consist of points on a circle with radius $v$ in the ($\sigma$,$\pi$) plane. They are related to each other through $O(2)$ rotations. Hence they are all equivalent and there are an infinite number of degenerate vacua. Any point on this circle may be chosen as the true vacuum. We can pick, for example, $$\langle 0|\sigma|0\rangle =v,\qquad\langle 0|\pi|0\rangle =0.$$
Then small oscillations around the true minimum are considered and a shifted field is defined: $\sigma'=\sigma-v$, which then leads to massless particles, those are the result of vacuum symmetry-non-invariance.
My questions:
1. Why do we get only one Goldstone particle? Why can't we take some other two points on the mentioned circle, for example,
$$ \langle 0|\sigma|0\rangle =\alpha, \qquad\langle 0|\pi|0\rangle =\beta$$
so that $\alpha^2+\beta^2=v^2$. Then we could define: $\sigma'=\sigma-\alpha$ and $\pi'=\pi-\beta$ and get two different types of Goldstone, massless particles? Or more in general, we know that: #broken symmetry operators (charges) = #Goldstone bosons, why is that so?
2. How are oscillations around the minimum connected to particle creation?
Remark: I know that the result of massless particles can be obtained more formally using the Goldstone theorem which leads to the same result, i.e. we only get one massless particle.