📂Quantum Mechanics

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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