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Goldstone phase initial value

+ 3 like - 0 dislike
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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?

asked Jan 5, 2016 in Phenomenology by NAME_XXX (1,010 points) [ revision history ]
edited Jan 5, 2016 by NAME_XXX

Or it is completely undertermined and may take arbitrary values from 00 to 2π2π?

It is completely arbitrary.

n case of presence of explicit symmetry breaking term, can we immediately state that VEV of θθ field is zero due to presence of mass term in (2′)(2′) 

Seems so. This sometimes is called vacuum alignment.

But why it is completely arbitrary? $U(1)$ group is compact group.

But why it is completely arbitrary? U(1)U(1) group is compact group.

Why not? All vacuua are completely congruent to each other if you look at the potential. And what does it have to do with compactness here?

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