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 $v_{0,{\rm SSB}}$, 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 (6) we get for the longitudinal renormalized mass parameter

$$\alpha_{\mathfrak{N}} = -2 \left\{ \alpha + \lambda ... ...a_{0}^{2} + 3 \sigma_{\mathfrak{N}}^{2} \right] \right\}.$$ (D.13)

This is a positive quantity in this phase, and in general corresponds to a non-zero mass $m_{\mathfrak{N}}$. If, however, there is an external source, then $v_{0}$ will be somewhat larger that the solution $v_{0,{\rm SSB}}$. In this case we may add and subtract $3\lambda v_{0,{\rm SSB}}^{2}$ to Equation (6) and therefore write $\alpha_{\mathfrak{N}}$ as

$$\alpha_{\mathfrak{N}} = 3\lambda \left( v_{0}^{2} - v... ...a_{0}^{2} + 3 \sigma_{\mathfrak{N}}^{2} \right] \right\},$$

showing once more that the mass increases with the variation of $v_{0}$ beyond its spontaneous symmetry-breaking value, and hence that it increases with the introduction of the external source. If we denote the value of $\alpha_{\mathfrak{N}}$ without the presence of the source by $\alpha_{\mathfrak{N},{\rm SSB}}$, we may write for the variation of $\alpha_{\mathfrak{N}}$ due to the external source

$$\alpha_{\mathfrak{N}} - \alpha_{\mathfrak{N},{\rm SSB}} = 3\lambda \left( v_{0}^{2} - v_{0,{\rm SSB}}^{2} \right).$$ (D.14)

Observe that once again this variation is three times larger than the corresponding variation of $\alpha_{0}$.