So, it is not a leap? Is the electron wavefunction defined during this transition?
Let $\vert \psi_+\rangle$ and $\vert\psi_-\rangle$ the wavefunctions of the electron before, and after the quantum transfer. You therefore have a superposition of the wavefunctions
$\vert \Psi \rangle = a(t) \vert \psi_+\rangle + b(t)\vert\psi_-\rangle$ and we know that the norm of that superposition must be 1. We also can assume
$\vert \Psi_{t=0} \rangle = \vert \psi_+ \rangle $
and
$\vert \Psi_{t=T} \rangle = \vert \psi_- \rangle $
Due to the constraint of $\Vert \vert \Psi \rangle \Vert = 1$ we therefore know that $a(t)^2 + b(t)^2 = 1 \Rightarrow b(t)^2 = 1 - a(t)^2$
In the case of "leap" the function a, b are step functions. But instead of step any function goes for which $a(t)^2 + b(t)^2 = 1$. That could be sin and cos for example, but there a few others as well.
There you have it: The wave function of the electron in transistion state is the superposition of the wavefunctions before and after the transition, weighted by a transfer function. To determine the exact nature of the transfer function one has to dive into QED.