Calculation of $\xi$

In order to calculate $\xi$ , we solve Eq. (79) Notice that $\cosh 2\xi >1$ since $\mu_{ev}>\nu$ . Using the definition of the $\cosh$ function, we can write

$$e^{2\xi}+e^{-2\xi}=2\frac{\mu_{ev}}{\sqrt{\mu_{ev}^2-\nu^2}},$$

and denoting $t=e^{2\xi}$ , one obtains a quadratic equation for $t$ with two solutions

$$t=\frac{\mu_{ev}\pm\vert\nu\vert}{\sqrt{\mu_{ev}^2-\nu^2}}.$$

After we take into account Eq. (80), we obtain the following expression for $\xi$

$$\xi = \frac{1}{4}\ln{\frac{\mu_{ev}-\nu}{\mu_{ev}+\nu}}.$$ (91)