Suppose we have relativistic system with spontaneously broken symmetry. For simplicity, let choose broken $U(1)$ symmetry:

$$

L = \frac{1}{2}(\partial_{\mu}\varphi )^{2} - V(|\varphi |) , V(|\varphi | = \varphi_{0}) = 0 \qquad (1)

$$

Let us parametrize Goldstone degree of freedom, $\varphi = \varphi_{0}e^{i\theta}$. Then instead of $(1)$ we will get

$$

L = \frac{\varphi_{0}^{2}}{2}(\partial_{\mu}\theta )^{2} \qquad (2)

$$

In fact, this leaves us with undetermined VEV of Goldstone degree of freedom. This VEV corresponds to the coherent state of $\theta$-particles with zero momentum.

Suppose also that instead of $(1)$ we have the lagrangian with small explicit symmetry breaking term, which after substitution of anzats $\varphi = \varphi_{0}e^{i\theta}$ leads to appearance $-\frac{\varphi_{0}^{2}m^{2}}{2}\theta^2$ extra term in $(2)$:

$$

L' = \frac{\varphi_{0}^{2}}{2}(\partial_{\mu}\theta )^{2} - \frac{\varphi_{0}^{2}m^{2} \theta^{2}}{2} \qquad (2')

$$

The first question: in case of absense of explicit symmetry breaking, can we fix this VEV by some properties of underlying theory? Or it is completely undertermined and may take arbitrary values from $0$ to $2 \pi$?

The second question: in case of presence of explicit symmetry breaking term, can we immediately state that VEV of $\theta$ field is zero due to presence of mass term in $(2')$ which makes zero contribution in classical energy of $\theta$ field only if $\rangle \theta \langle = 0$?

The third extra question. Suppose that we know only effective theory with $\theta$ field and its interactions with other fields, and the lagrangian takes the form

$$

L{''} = \frac{1}{2}(\partial_{\mu}\theta )^{2} - \sum_{i}\partial_{\mu}\theta J^{\mu}_{i}c_{i}, \qquad (3)

$$

where $J_{\mu}^{i}$ denotes some vector or pseudovector currents, in dependence of what $\theta$ is. May we strictly state that due to scale invariant form of the lagrangian $(3)$ $\theta$ field arises in corresponding effective theory in a way nothing but as goldstone degree of freedom?