Broken-Symmetric Phase:

In this case, if there is no external source, then instead of zero we have for $v_{0}$ the non-trivial solution that we will denote here by $v_{0,{\rm SSB}}$, which according to Equation (7) is given by

$$\lambda v_{0,{\rm SSB}}^{2} = - \alpha - \lambda \le... ...-1) \sigma_{0}^{2} + 3 \sigma_{\mathfrak{N}}^{2} \right],$$

which is a positive quantity in this phase. Substituting this for the term $\lambda v_{0}^{2}$ in Equation (5) we get for the transversal renormalized mass parameter

$$\alpha_{0} = 2\lambda \left( \sigma_{0}^{2} - \sigma_{\mathfrak{N}}^{2} \right).$$ (D.11)

Since $\sigma_{0}^{2}$ and $\sigma_{\mathfrak{N}}^{2}$ become identical in the continuum limit, this seems to indicate that $\alpha_{0}$ tends to zero in the limit and thus corresponds to zero mass $m_{0}$ in that limit. However, $\alpha_{0}$ always goes to zero in the continuum limit, and the fact that it does so is not enough to guarantee that $m_{0}$ is zero in the limit. Therefore, further analysis of the limit in necessary, which we will do later.

Going back to the case in which there is an external source $j_{0}$, we see that $v_{0}$ will be somewhat larger that the solution $v_{0,{\rm SSB}}$. In this case we may add and subtract $\lambda v_{0,{\rm SSB}}^{2}$ in Equation (5) and therefore write $\alpha_{0}$ as

$$\alpha_{0} = \lambda \left( v_{0}^{2} - v_{0,{\rm SSB... ...\left( \sigma_{0}^{2} - \sigma_{\mathfrak{N}}^{2} \right),$$ (D.12)

showing once more that the mass increases with the variation of $v_{0}$ beyond its spontaneous symmetry-breaking value $v_{0,{\rm SSB}}$, and hence that it increases with the introduction of the external source. This represents the variation of $\alpha_{0}$ as a consequence of a variation of $v_{0}$ beyond its spontaneous symmetry breaking value. In terms of the mass $m_{0}$ this variation is not linear, but quadratic in nature.