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)

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