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. Pingback: Finite drop potential « Physics tutorials

  2. 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!

    Reply

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