non-trivially and all the others have been set to constants. We write ${\bf P}={\bf
P}_{(\t ,a)}$ and $\z^2=4\sum_n \z_{(\t ,n,a)}$ for simplicity of notation. The effective potential consists of only one term and takes the form $$V=\z^2 e^{{\bf P}\cdot\bph}\; .$$ This situation corresponds to an $SU(2)$ Toda model so the general solution to Eq.  can be found. It is given by $$\bph = {\bf p}^{(i)}\ln (x) +({\bf p}^{(f)}-{\bf p}^{(i)})
        \ln (1+x^\d )^{1/\d } + {\bf k} \label{singles}$$ subject to the constraints $$\begin{aligned}
 |{\bf p}^{(i)}|^2 &=& \frac{4}{3} \label{C1}\\
 \d &=& {\bf P}\cdot{\bf p}^{(i)} \label{C2}\\
 {\bf p}^{(f)} &=& {\bf p}^{(i)}-\frac{2{\bf P}\cdot{\bf p}^{(i)}}
                   {|{\bf P}|^2}{\bf P} \label{C3}\\
 \exp ({\bf P}\cdot{\bf k}) &=& \frac{3E\d^2}{|{\bf P}|^2\z^2}\; .
 \label{C4}\end{aligned}$$ Further, $$x=\frac{|t-t_0|}{T}$$ is the rescaled proper time and $t_0$ and $T$ are constants. Note that Eqs.  and imply that $$|{\bf p}^{(f)}|^2= \frac{4}{3}\; .$$ This solution is invariant under the exchange of ${\bf p}^{(i)}$ and ${\bf p}^{(f)}$ and we remove this ambiguity by requiring $\d\geq 0$ in the positive-time branch and $\d\leq 0$ in the negative-time branch. The interpretation of these solutions is well-known [@Lukas:1996ee; @Lukas:1996iq]. They interpolate between two, generally different, “free" solutions , one with ${\bf q}={\bf p}^{(i)}$ at early time and one with ${\bf q}={\bf 