# Finite drop potential

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.35.

The problem of the finite step potential can be inverted to give a finite drop potential by replacing ${V_{0}}$ by ${-V_{0}}$, so the potential is given by

$\displaystyle V(x)=\begin{cases} 0 & x\le0\\ -V_{0} & x>0 \end{cases} \ \ \ \ \ (1)$

If we assume particles coming in from the left, then we must have ${E>0}$ (otherwise the wave function would decay exponentially inside the barrier and we couldn’t have particles coming in from infinity on the left). In this case the reflection coefficient is

$\displaystyle R=\left[\frac{E-\sqrt{E\left(E+V_{0}\right)}}{E+\sqrt{E\left(E+V_{0}\right)}}\right]^{2} \ \ \ \ \ (2)$

and the transmission coefficient is

$\displaystyle T=\frac{4E^{3/2}\sqrt{E+V_{0}}}{\left(E+\sqrt{E\left(E+V_{0}\right)}\right)^{2}} \ \ \ \ \ (3)$

For ${V_{0}=0}$ the problem reduces to that of the free particle, and ${R=0}$, ${T=1}$ as we’d expect. As ${V_{0}}$ gets very large, ${R\rightarrow1}$, ${T\rightarrow0}$.

If we take ${E=V_{0}/3}$, then ${R=1/9}$.

Although the graph of the potential looks like a cliff, it doesn’t represent the behaviour of an object, such as a car, falling over a cliff. Classically, the energy of a car is kinetic + potential, which in the absence of other forces, remains a constant. If a car had a speed ${v_{1}}$ in a region where ${V=0}$, then its total energy is kinetic: ${E=K_{1}=\frac{1}{2}mv_{1}^{2}}$. If it suddenly encounters a region where ${V=-V_{0}}$, then we’d have ${E=K_{2}-V_{0}}$, so ${K_{2}}$ is larger than ${K_{1}}$, meaning that the car would instantaneously increase its speed, which of course doesn’t happen. In reality, a car driving off a cliff encounters a potential energy of ${-mgy}$ where ${y}$ is the distance it has fallen, so its kinetic energy increases gradually. Besides, a car falling off a cliff is essentially a two-dimensional problem, so trying to analyze it in one dimension won’t work.

A slightly more realistic case is that of a neutron which is fired at an atomic nucleus. The neutron experiences a sudden drop in potential from ${V=0}$ outside the nucleus to ${V=-V_{0}=-12}$ MeV inside. If we give the neutron an initial kinetic energy of ${E=4}$ MeV, then the probability of transmission into the nucleus is

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{4\times4^{3/2}\sqrt{4+12}}{\left(4+\sqrt{4(4+12)}\right)^{2}}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{128}{144}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{8}{9} \ \ \ \ \ (6)$