Probability Flux for the Potential Step *

The probability flux is given by

\begin{displaymath}\bgroup\color{black} j(x,t)={\hbar\over 2mi}
...\partial x}-{\partial\psi^*\over\partial x}\psi\right] .\egroup\end{displaymath}

We can save some effort by noticing that this contains an expression minus its complex conjugate. (This assures that term in brackets is imaginary and the flux is then real.)

\begin{displaymath}\bgroup\color{black} j={\hbar\over 2im}\left[u^*{du\over dx}-...
...\right]={\hbar\over 2im}\left[u^*{du\over dx}-CC\right] \egroup\end{displaymath}

For \bgroup\color{black}$x<0$\egroup

j&=&{\hbar\over 2im}[(e^{-ikx}+R^*e^{ikx})(ike^{ikx}-ikRe^{-ik...
...2ikx}-R^*R] + CC \\
j&=&[1-\vert R\vert^2]{\hbar k\over m} .\\

The probability to be reflected is the reflected flux divided by the incident flux. In this case its easy to see that its \bgroup\color{black}$\vert R\vert^2$\egroup as we said. For \bgroup\color{black}$x>0$\egroup

\begin{displaymath}\bgroup\color{black} j=\vert T\vert^2{\hbar k'\over m} .\egroup\end{displaymath}

The probability to be transmitted is the transmitted flux divided by the incident flux.

\begin{displaymath}\bgroup\color{black} \vert T\vert^2{\hbar k'\over m}{m\over\hbar k}={4k^2\over(k+k')^2}{k'\over k}={4kk'\over(k+k')^2} \egroup\end{displaymath}

again as we had calculated earlier.

Jim Branson 2013-04-22