Solution of the Schrödinger Equation for a Step Potential
Overview
Let’s examine how a particle moves when the potential is a step function as shown in the figure above. The potential $U$ is
$$ U(x) = \begin{cases} 0 & x<0 \\ U_{0} & x>0 \end{cases} $$
The time-independent Schrödinger equation for this potential $U(x)$ is
$$ \dfrac{d^2 u(x)}{dx^2}+\frac{2m}{\hbar ^2} \Big[ E-U(x) \Big]u(x)=0 $$
Solution1
$E<0$
No solution exists if the energy is less than the potential so it need not be considered.
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$0 < E < U_{0}$
Part 2-1. $x<0$
The Schrödinger equation in this region is given by
$$ \dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}Eu=0 $$
Since $\frac{2m}{\hbar^2}E$ is positive, substituting into $k^2$,
$$ \dfrac{d^2 u}{dx^2}+k^2u=0 $$
This is a very simple second-order ordinary differential equation. Solving the differential equation, the solution is as follows.
$$ u_{I}(x)=A_{+}e^{ikx} + A_{-}e^{-ikx} $$
Here, $A_{+}$, $A_{-}$ are constants. In this context, $u=A_{+}e^{ikx}$ refers to the incident wave, and $A_{-}e^{-ikx}$ refers to the reflected wave.
Part 2-2. $x>0$
The Schrödinger equation in this region is given by
$$ \dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}(E-U_{0})u=0 $$
Since $E-U_{0}<0$, substituting into $\frac{2m}{\hbar^2}(E-U_{0})=-\kappa ^2$,
$$ \dfrac{d^2 u}{dx^2}-\kappa^2 u=0 $$
Here, $\kappa$ represents the Greek letter “kappa”. It is a different character from $k$ (kappa). The solution of this differential equation is as follows.
$$ u_{II}(x) = B_{+}e^{\kappa x}+B_{-}e^{-\kappa x} $$
The wave function must be square integrable so it should be $\lim \limits_{x \rightarrow \infty}u_{II}(x)=0$. Since $\lim \limits_{x \rightarrow \infty} B_{+}e^{\kappa x}=\infty $, it must be $B_{+}=0$. Therefore, the transmitted wave is $u_{trans}=B_{-}e^{-\kappa x} $
Part 2-3. Boundary Conditions
Assuming that the wave function is smooth, it is continuous at $x=0$ and the derivative of the wave function (slope) is also continuous at $x=0$. Hence, the following conditions are obtained.
$$ \begin{align*} u_{I}(0)=u_{II}(0) \quad \implies& \quad A_{+}+A_{-} = B_{-} \\ u_{I}^{\prime}(0)=u_{II}^{\prime}(0) \quad \implies& \quad ikA_{+}-ikA_{-} = -\kappa B_{-} \end{align*} $$
Combining the two equations, the following is obtained:
$$ \frac{A_{-}} {A_{+}} = \frac{ ik+\kappa}{ik -\kappa } $$
Part 2-4. Reflection coefficient and Transmission coefficient
To calculate the reflection and transmission coefficients, let’s find the probability current of the incident, reflected, and transmitted waves
$$ j_{inc}=\frac{\hbar k}{m}|A_{+}|^2,\quad j_{ref}=\frac{\hbar k}{m}|A_{-}|^2,\quad j_{trnas}=0 $$
Thus, the reflection and transmission coefficients are
$$ R=\left| \frac{j_{ref}}{j_{inc}}\right|=\left| \frac{A_{-}} {A_{+}} \right|^2=\frac{\kappa^2+k^2}{\kappa^2 + k^2}=1 $$
$$ T=\left| \frac{ j_{trnas} }{j_{inc}}\right|=\frac{ 0}{\frac{\hbar k}{m}|A_{+}|^2}=0 $$
According to the above results, reflection occurs completely and transmission does not occur. However, since $B_{-} \ne 0$, there is a probability of finding the particle in $x>0$. In macroscopic terms, it can be seen that the incident wave is entirely reflected, but microscopically, part of the incident wave has tunneled into a classically forbidden region. This is known as tunneling or quantum tunnel effect.
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$ U_{0} < E$
Part 3-1. $x<0$
The time-independent Schrödinger equation in this region is
$$ \dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}Eu=0 $$
Since $\frac{2m}{\hbar^2}E$ is positive, substituting into $k^2$,
$$ \dfrac{d^2 u}{dx^2}+k^2u=0 $$
The solution has already been obtained in Part 2-1.:
$$ u_{I}(x)=A_{+}e^{ikx} + A_{-}e^{-ikx} $$
The incident wave is $u_{inc}=A_{+}e^{ikx}$ and the reflected wave is $u_{ref}=A_{-}e^{-ikx}$.
Part 3-2. $x>0$ The time-independent Schrödinger equation in this region is
$$ \dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}(E-U_{0})u=0 $$
Since $E-U_{0}>0$, substituting into $\frac{2m}{\hbar^2}(E-U_{0})=\kappa ^2$,
$$ \dfrac{d^2 u}{dx^2}+\kappa ^2u=0 $$
Since it is similar to Part 3-1., the solution is
$$ u_{II}(x)=B_{+}e^{i \kappa x} + B_{-}e^{-i \kappa x} $$
Here, there is no reflected wave in the $x>0$ region, so it is $B_{-}=0$. Therefore,
$$ u_{II}(x)=B_{+}e^{i \kappa x} $$
which is the transmitted wave.
Part 3-3. Boundary Conditions
Assuming that the wave function is smooth, it is continuous at $x=0$ and the derivative of the wave function (slope) is also continuous at $x=0$. Hence,
$$ u_{I}(0)=u_{II}(0) \quad \implies \quad A_{+}+A_{-} = B_{+} $$
$$ u_{I}^{\prime}(0)=u_{II}^{\prime}(0) \quad \implies \quad ikA_{+}-ikA_{-} = i\kappa B_{+} $$
Combining the two equations, the following is obtained.
$$ \frac{A_{-}} {A_{+}} = \frac{ k-\kappa}{k +\kappa } $$
$$ \frac{B_{+}} {A_{+}} = \frac{ 2k}{k +\kappa } $$
Part 3-4. Reflection coefficient and Transmission coefficient
The probability currents of the incident, reflected, and transmitted waves are
$$ j_{inc}=\frac{\hbar k}{m}|A_{+}|^2,\quad j_{ref}=\frac{\hbar k}{m}|A_{-}|^2,\quad j_{trnas}=\frac{\hbar \kappa}{m}|B_{+}|^2 $$
Thus, the reflection and transmission coefficients are
$$ R=\left| \frac{j_{ref}}{j_{inc}}\right|=\left| \frac{A_{-}} {A_{+}} \right|^2=\left( \frac{k-\kappa}{ k + \kappa }\right) ^2= \frac{k^2 -k\kappa +\kappa^2}{ (k + \kappa )^2 } $$
$$ T=\left| \frac{j_{trans}}{j_{inc}}\right|=\frac{\kappa}{k}\left| \frac{B_{+}} {A_{+}} \right|^2=\frac{\kappa}{k} \left( \frac{2k}{ k + \kappa }\right) ^2= \frac{2k \kappa }{ (k + \kappa )^2 } $$
It is easy to confirm that $R+T=1$ holds.
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Stephen Gasiorowicz, Quantum Physics (Korean translation by the Physics Department at Sogang University) (3rd Edition, 2005), p79-83. ↩︎