Finite step potential – scattering

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.34.

A variant of the finite square well is the finite step, which has the potential

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

where ${V_{0}}$ is a positive constant energy.

There are two distinct cases here:

1. Energy below the barrier: ${0\le E\le V_{0}}$
2. Energy greater than the barrier: ${E>V_{0}}$

We’ll consider first the case where ${0\le E\le V_{0}}$.

In this case, the Schrödinger equation for ${x>0}$ is

 $\displaystyle -\frac{\hbar^{2}}{2m}\psi''+V_{0}\psi$ $\displaystyle =$ $\displaystyle E\psi\ \ \ \ \ (2)$ $\displaystyle \psi''$ $\displaystyle =$ $\displaystyle \mu^{2}\psi \ \ \ \ \ (3)$

where

$\displaystyle \mu=\frac{\sqrt{2m(V_{0}-E)}}{\hbar} \ \ \ \ \ (4)$

This has solution

$\displaystyle \psi(x)=Ce^{\mu x}+De^{-\mu x} \ \ \ \ \ (5)$

To keep the solution finite as ${x\rightarrow\infty}$ we must have ${C=0}$ so the solution is an exponentially decaying wave function:

$\displaystyle \psi(x)=De^{-\mu x} \ \ \ \ \ (6)$

To the left of the barrier, the Schrödinger equation is

$\displaystyle \psi''=-\frac{2mE}{\hbar^{2}}\psi \ \ \ \ \ (7)$

Assuming particles coming in from the left, we have

$\displaystyle \psi(x)=Ae^{ikx}+Be^{-ikx} \ \ \ \ \ (8)$

where

$\displaystyle k=\frac{\sqrt{2mE}}{\hbar} \ \ \ \ \ (9)$

Since the potential is finite everywhere, both ${\psi}$ and ${\psi'}$ are continuous everywhere, which gives us two boundary conditions at ${x=0}$.

 $\displaystyle A+B$ $\displaystyle =$ $\displaystyle D\ \ \ \ \ (10)$ $\displaystyle ik\left(A-B\right)$ $\displaystyle =$ $\displaystyle -\mu D \ \ \ \ \ (11)$

This has solution

 $\displaystyle B$ $\displaystyle =$ $\displaystyle \frac{ik+\mu}{ik-\mu}A\ \ \ \ \ (12)$ $\displaystyle D$ $\displaystyle =$ $\displaystyle \frac{2ik}{ik-\mu}A \ \ \ \ \ (13)$

The reflection coefficient is

 $\displaystyle R$ $\displaystyle =$ $\displaystyle \frac{\left|B\right|^{2}}{\left|A\right|^{2}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (15)$

That is, the probability of an incoming particle being reflected is 1. This is because the wave function for ${x>0}$ is exponentially decaying, so the probability of a particle reaching infinity is zero, thus no particles can be transmitted.

For ${E>V_{0}}$ the Schrödinger equation for ${x>0}$ is

 $\displaystyle -\frac{\hbar^{2}}{2m}\psi''+V_{0}\psi$ $\displaystyle =$ $\displaystyle E\psi\ \ \ \ \ (16)$ $\displaystyle \psi''$ $\displaystyle =$ $\displaystyle -\kappa^{2}\psi \ \ \ \ \ (17)$

where

$\displaystyle \kappa=\frac{\sqrt{2m(E-V_{0})}}{\hbar} \ \ \ \ \ (18)$

We now get travelling wave solutions instead of exponentially decaying ones:

$\displaystyle \psi(x)=Ce^{i\kappa x}+De^{-i\kappa x} \ \ \ \ \ (19)$

Assuming incoming particles arrive only from the left, we can set ${D=0}$. Applying the boundary conditions, we get

 $\displaystyle A+B$ $\displaystyle =$ $\displaystyle C\ \ \ \ \ (20)$ $\displaystyle ik\left(A-B\right)$ $\displaystyle =$ $\displaystyle i\kappa C \ \ \ \ \ (21)$

with solutions

 $\displaystyle B$ $\displaystyle =$ $\displaystyle \frac{k-\kappa}{k+\kappa}A\ \ \ \ \ (22)$ $\displaystyle C$ $\displaystyle =$ $\displaystyle \frac{2k}{k+\kappa}A \ \ \ \ \ (23)$

In this case, the reflection coefficient is

 $\displaystyle R$ $\displaystyle =$ $\displaystyle \frac{\left|B\right|^{2}}{\left|A\right|^{2}}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{k-\kappa}{k+\kappa}\right)^{2} \ \ \ \ \ (25)$

Substituting the expressions for ${k}$ and ${\kappa}$ we get

$\displaystyle R=\left[\frac{E-\sqrt{E\left(E-V_{0}\right)}}{E+\sqrt{E\left(E-V_{0}\right)}}\right]^{2} \ \ \ \ \ (26)$

From this we can get the transmission coefficient

 $\displaystyle T$ $\displaystyle =$ $\displaystyle 1-R\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4E^{3/2}\sqrt{E-V_{0}}}{\left(E+\sqrt{E\left(E-V_{0}\right)}\right)^{2}} \ \ \ \ \ (28)$

Note that this is not equal to ${\left|C\right|^{2}/\left|A\right|^{2}=4E^{2}/\left(E+\sqrt{E\left(E-V_{0}\right)}\right)^{2}}$. Have we done something wrong?

The answer lies in the fact that the wave for ${x>0}$ is not the same as the wave for ${x<0}$, since the net energy on the right is ${E-V_{0}}$ while on the left it is just ${E}$. One way of looking at it is in terms of the probability current for the free particle. The probability current must be conserved; this is just a way of saying that particles cannot vanish into, nor arise from, thin air. Since the probability current for a free particle with stationary state

$\displaystyle \Psi(x,t)=Ae^{ikx}e^{-i\hbar k^{2}t/2m} \ \ \ \ \ (29)$

is

$\displaystyle J=\frac{\hbar k}{m}\left|A\right|^{2} \ \ \ \ \ (30)$

the conservation law implies, for the case of the step potential

$\displaystyle \frac{\hbar k}{m}\left[\left|A\right|^{2}-\left|B\right|^{2}\right]=\frac{\hbar\kappa}{m}\left|C\right|^{2} \ \ \ \ \ (31)$

That is, the influx of particles from the left minus the reflected beam must equal the transmitted beam. Dividing through by ${\frac{\hbar k}{m}\left|A\right|^{2}}$ we get

$\displaystyle \frac{\left|B\right|^{2}}{\left|A\right|^{2}}+\frac{\kappa}{k}\frac{\left|C\right|^{2}}{\left|A\right|^{2}}=1 \ \ \ \ \ (32)$

The first term is the reflection coefficient we calculated in 26. The second term is the transmission coefficient, which works out to

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E-V_{0}}{E}}\frac{\left|C\right|^{2}}{\left|A\right|^{2}}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E-V_{0}}{E}}\frac{4E^{2}}{\left(E+\sqrt{E\left(E-V_{0}\right)}\right)^{2}}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4E^{3/2}\sqrt{E-V_{0}}}{\left(E+\sqrt{E\left(E-V_{0}\right)}\right)^{2}} \ \ \ \ \ (35)$

which is what we got earlier.

10 thoughts on “Finite step potential – scattering”

1. growescience

No, since in the first part, we’re looking at ${E, so for ${\mu}$ to be real we need ${\mu=\frac{\sqrt{2m(V_{0}-E)}}{\hbar}}$.

1. gwrowe Post author

Which equation are you referring to and what is ${\alpha}$? (I don’t use ${\alpha}$ anywhere.)

1. Lise

How to understand the change in velocity of wave in physics？How V0 influence the wave function in physics instead of in math？Thanks！