It turns out that the imaginary part is always negative. Therefore we define

$$E=E_R-iE_{Im},
\label{87}$$

with $E_{Im}>0$. As already mentioned in the introduction the imaginary part is connected to the instability of the states. This instability corresponds the analogue fact in the framework of particle mechanics that a particle which has passed the horizon cannot get back to the area outside the horizon and disappears in the singularity at the centre of the black hole. In the case of quantum mechanics this means there is an effective inward current. By recalling now Eqs. $(\ref{91})$ and $(\ref{92})$ and using the eigenvalue equation we obtain

$$| \psi (t_2) \rangle={\rm exp}\left[-iE(t_2-t_1)\right] | \psi (t_1) \rangle.
\label{93}$$

With the help of Eq. $(\ref{87})$ we can transform Eq. $(\ref{93})$ to

$$| \psi (t_2) \rangle=\left\{{\rm exp}\left[-iE_R(t_2-t_1)\right] | \psi (t_1) \rangle\right\} {\rm exp}\left[-E_{Im}(t_2-t_1)\right].
\label{94}$$

So it can be seen directly that a negative imaginary energy part results in an exponential decrease of the probability density. This is only another way to express the fact that probability density vanishes with time. Such a phenomena can only be interpreted as the effect of a current sink at the singularity of the centre. If probability density extends to the area 