A. Zeros of the potential and critical value of the parameter

The zeros of the potential $\bar{V} (x)$ can be obtained in closed form, since they are the zeroes of the numerator, i.e. the solutions of the equation

$$\bar{V} (x) = 0 \quad \Rightarrow \quad x ^{2} (x ^{4} - 4 a x + 4 a ^{2}) = 0$$ (31)

with $a > 0$ and $- x ^{3} + 2 x - 2 a \neq 0$. We are interested in the non-negative solutions, which, apart from the $x = 0$ one, can be determined exactly as solutions of the fourth order equation

$$x ^{4} - 4 a x + 4 a ^{2} = 0 .$$ (32)

By setting

$${\mathcal{B}} = \left\{ 9 a ^{2} + \left( 81 a ^{4} - 192 a ^{6} \right) ^{1/2} \right\} ^{1/3}$$

we have that for $0 < a \leq (3/4) ^{3/2}$

$$x _{\stackrel{\mathrm{\scriptstyle{}min}}{\mathrm{\scriptsty... ...{1/3} a ^{2}}{{\mathcal{B}}} + {\mathcal{B}}}} } \right\} .$$ (33)

These expressions are not very enlightening: the one for $x _{\mathrm{min}}$ has been used to exactly evaluate the upper integration limit in the numerical evaluation of the integral that gives the classical action.

To see when the quartic part of the potential has two positive roots we can use the expression above, but also a smarter procedure, as follows. Clearly the limiting case is the one in which the potential is tangent to the positive $x$ axis, i.e. the two solutions coincide. In this case the quartic part must be of the form

$$(x - \alpha) ^{2} (x ^{2} + \beta x + \gamma) = x ^{4} + ... ...^{2} - 2 \alpha \gamma) x + \alpha ^{2} \gamma \nonumber ,$$

which by comparison with $x ^{4} - 4 a x - 4 a ^{2}$ gives the set of equations

$$\left \{ \matrix{ \beta - 2 \alpha = 0 \hfill \cr \gam... ...ll \cr \alpha ^{2} \gamma = 4 a ^{2} \hfill } \right . .$$

Then $\alpha = \beta / 2$ and $\gamma = 3 \beta ^{2} / 4$ from the first two equations. This gives $\alpha = a ^{1/3}$ from the third and $\alpha = (4 a ^{2} / 3) ^{1/4}$ from the fourth. These last two relations are compatible for non-vanishing $a$, if and only if $a = (3/4) ^{3/2}$, which is thus the critical value of $a$.