Lets say we have a potential step with regions 1 with zero potential $W_p\!=\!0$ (this is a free particle) and region 2 with potential $W_p$. Wave functions in this case are:
\begin{align} \psi_1&=Ae^{i\mathcal L x} + B e^{-i\mathcal L x} & \mathcal L &\equiv \sqrt{\frac{2mW}{\hbar^2}}\\ \psi_2&=De^{-i\mathcal K x} & \mathcal K &\equiv \sqrt{\frac{2m(W_p-W)}{\hbar^2}} \end{align}\begin{align} \psi_1&=Ae^{i\mathcal L x} + B e^{-i\mathcal L x} & \mathcal L &\equiv \sqrt{\frac{2mW}{\hbar^2}}\\ \psi_2&=De^{-\mathcal K x} & \mathcal K &\equiv \sqrt{\frac{2m(W_p-W)}{\hbar^2}} \end{align}
Where $A$ is an amplitude of an incomming wave, $B$ is an amplitude of an reflected wave and $D$ is an amplitude of an transmitted wave. I have sucessfuly derived a relations between amplitudes in potential step:
\begin{align} \frac{A}{D} &= \frac{i\mathcal L-\mathcal K}{2i\mathcal L} & \frac{A}{B}&=-\frac{i \mathcal L - \mathcal K}{i \mathcal L + \mathcal K} \end{align}
I know that if i want to calculate transmittivity coefficient $T$ or reflexifity coefficient $R$ i will have to use these two relations that i know from wave physics:
\begin{align} T &= \frac{j_{trans.}}{j_{incom.}} & R &= \frac{j_{trans.}}{j_{incom.}} \end{align}
Question 1: I know that $j = \frac{dm}{dt} = \frac{d}{dt}\rho V \propto \rho v \propto \rho k$ But what is a density $\rho$ equal to?
Question 2: I noticed that $\mathcal L$ and $\mathcal K$ are somehow (i dont know how) connected to the wavevector $k$ from the equation in 1st question but how? How can i make it obvious?
