= Step potential
{title2=$V(x)=V_0\Theta(x)$}
{wiki}
= Potential step
{synonym}
For constant <mass> $m$, a finite jump from <potential energy> zero at $x<0$ to $V_0>0$ at $x>0$ requires continuity of the <wavefunction> and its first <derivative>. With incidence only from the left and $E>V_0$, put $k=\sqrt{2mE}/\hbar$ and $q=\sqrt{2m(E-V_0)}/\hbar$. Matching $A+B=C$ and $k(A-B)=qC$ gives reflected <probability amplitude> $B/A=(k-q)/(k+q)$ and <transmission amplitude> $C/A=2k/(k+q)$. The <transmission probability> is $(q/k)|C/A|^2$, conserving <probability current>.
For $0<E<V_0$, replace $q$ by $i\kappa$, where $\kappa=\sqrt{2m(V_0-E)}/\hbar$ is the <tunnelling decay constant>. The right-hand <wavefunction> is $Ce^{-\kappa x}$, with zero <probability current>. Its <probability density> decays on length $1/(2\kappa)$, while its amplitude decays on length $1/\kappa$. The semi-infinite raised region gives total reflection, unlike a finite <potential barrier> allowing <quantum tunnelling> into a second allowed region.
Back to article page
