📂Quantum Mechanics

Solution of the Schrödinger Equation for a Potential Barrier

Overview

Let’s explore how a particle behaves when the potential is in the form of a wall as shown in the figure above. The potential $U$ is

$$U(x) = \begin{cases} 0 & x<-a \\ U_{0} & -a < x <a \\ 0 &a<x \end{cases}$$

The time-independent Schrödinger equation when the potential is $U(x)$ is

$$\dfrac{d^2 u(x)}{dx^2}+\frac{2m}{\hbar ^2} \Big[ E-U(x) \Big]u(x)=0$$

Solution¹

$E<0$

Since there is no solution if the energy is less than the potential, it does not need to be considered.

■

$0 < E < U_{0}$

Part 2-1. $x<-a$ In this region, the time-independent Schrödinger equation is

$$\dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}Eu=0$$

As $\frac{2m}{\hbar^2}E$ is positive, by substituting $k^2$

$$\dfrac{d^2 u}{dx^2}+k^2u=0$$

Solving the equation yields the solution

$$u_{1}(x)=A_{+}e^{ikx} + A_{-}e^{-ikx}$$

Here, $A_{+}$ and $A_{-}$ are constants.

• Part 2-2. $-a<x<a$

  In this region, the time-independent Schrödinger equation is

  $$\dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}(E-U_{0})u=0$$

  Since $E-U_{0}<0$, substituting $\frac{2m}{\hbar^2}(E-U_{0})=-\kappa ^2$

  $$\dfrac{d^2 u}{dx^2}-\kappa^2 u=0$$

  Here, $\kappa$ is the Greek letter ‘kappa’ and is different from $k$ (k). Solving the equation yields

  $$u_{2}(x) = B_{+}e^{\kappa x}+B_{-}e^{-\kappa x}$$

  Here, $B_{+}$ and $B_{-}$ are constants.

• Part 2-3. $a<x$

  In this region, the time-independent Schrödinger equation is

  $$\dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}Eu=0$$

  Since $\frac{2m}{\hbar^2}E=k^2$ was substituted in Part 2-1.

  $$\dfrac{d^2 u}{dx^2}+k^2u=0$$

  Solving the above differential equation yields

  $$u_{3}(x)=C_{+}e^{ikx} + C_{-}e^{-ikx}$$

  As there are no reflective waves in this region, $C_{-}=0$.

• Part 2-4. Boundary Conditions

  Assuming wave functions are smooth, they are continuous at $x=-a$ and $x=a$ and the slope (derivative) of the wave functions is also continuous at $x=-a$ and $x=a$. Therefore,

  $$\begin{cases}u_{1}(-a)=u_{2}(-a) \\ u_{2}(a)=u_{3}(a) \end{cases} \quad \implies \begin{cases} A_{+}e^{-ika}+A_{-}e^{ika} = B_{+}e^{-\kappa a}+B_{-}e^{\kappa a} \quad \cdots (1) \\ B_{+}e^{\kappa a}+B_{-}e^{-\kappa a} = C_{+}e^{ika}+0 \cdot e^{-ika} \ \quad \cdots (2) \end{cases}$$

  $$\begin{cases}u_{1}^{\prime}(-a)=u_{2}^{\prime}(-a) \\ u_{2}^{\prime}(a)=u_{3}^{\prime}(a) \end{cases} \quad \implies \begin{cases} ikA_{+}e^{-ika}-ikA_{-}e^{ika} = \kappa B_{+}e^{-\kappa a}-\kappa B_{-}e^{\kappa a} \quad \cdots (3) \\ \kappa B_{+}e^{\kappa a}-\kappa B_{-}e^{-\kappa a} = ik C_{+}e^{ika}+0 \cdot ik e^{-ika} \quad \cdots (4) \end{cases}$$

  Representing $(1)$ and $(3)$ as matrices,

  $$\begin{pmatrix} e^{-ika} & e^{ika} \\ ike^{-ika} & -ike^{ika} \end{pmatrix} \begin{pmatrix} A_{+} \\ A_{-} \end{pmatrix} = \begin{pmatrix} e^{-\kappa a} & e^{\kappa a} \\ \kappa e^{-\kappa a} & -\kappa e^{\kappa a} \end{pmatrix} \begin{pmatrix} B_{+} \\ B_{-} \end{pmatrix}\quad \cdots (5)$$

  Representing $(2)$ and $(4)$ as matrices,

  $$\begin{pmatrix} e^{\kappa a} & e^{-\kappa a} \\ \kappa e^{\kappa a} & -\kappa e^{-\kappa a} \end{pmatrix} \begin{pmatrix} B_{+} \\ B_{-} \end{pmatrix} = \begin{pmatrix} e^{ik a} & e^{-ik a} \\ ik e^{ik a} & -ik e^{-ika } \end{pmatrix} \begin{pmatrix} C_{+} \\ 0 \end{pmatrix}\quad \cdots (6)$$

  We are interested in the waves reflected and transmitted by the potential barrier and thus need to find $A_{+}$, $A_{-}$, and $C_{+}$.

  By eliminating $\begin{pmatrix} B_{+}\\ B_{-}\end{pmatrix}$ to join $(5)$ and $(6)$, the leftmost matrix of $(5)$ is called $\mathbb{A}$ and the leftmost matrix of $(6)$ is called $\mathbb{B}$. Then the two equations can be simplified as follows.

  $$\begin{pmatrix} A_{+} \\ A_{-} \end{pmatrix} =\mathbb{A}^{-1} \begin{pmatrix} e^{-\kappa a} & e^{\kappa a} \\ \kappa e^{-\kappa a} & -\kappa e^{\kappa a} \end{pmatrix} \begin{pmatrix} B_{+} \\ B_{-} \end{pmatrix} \quad \cdots (7)$$

  $$\begin{pmatrix} B_{+} \\ B_{-} \end{pmatrix} =\mathbb{B}^{-1} \begin{pmatrix} e^{ik a} & e^{-ik a} \\ ik e^{ik a} & -ik e^{-ika } \end{pmatrix} \begin{pmatrix} C_{+} \\ 0 \end{pmatrix}\quad \cdots (8)$$

  By substituting $(8)$ into $(7)$,

  $$\begin{pmatrix} A_{+} \\ A_{-} \end{pmatrix} =\mathbb{A}^{-1} \begin{pmatrix} e^{-\kappa a} & e^{\kappa a} \\ \kappa e^{-\kappa a} & -\kappa e^{\kappa a} \end{pmatrix} \mathbb{B}^{-1} \begin{pmatrix} e^{ik a} & e^{-ik a} \\ ik e^{ik a} & -ik e^{-ika } \end{pmatrix} \begin{pmatrix} C_{+} \\ 0 \end{pmatrix} \quad \cdots (9)$$

  Now computing $\mathbb{A}^{-1}$ and $\mathbb{B}^{-1}$ and the product of the matrices, we can obtain the coefficients. The computation process is detailed in Appendix Q-1. Through thorough calculations, we get the following results:

  $$\dfrac{A_{-}}{A_{+}} = \frac{-i \xi \sinh (2\kappa a)e^{-i2ka} } {\cosh (2\kappa a) + i\eta \sinh (2\kappa a)}$$

  $$\dfrac{C_{+}}{A_{+}}=\frac{e^{-i2ka} }{\cosh (2\kappa a) +i \eta \sinh (2\kappa a)}$$

  Here, $\frac{\kappa}{k} -\frac{k}{\kappa}=2\eta$ and $\frac{\kappa}{k} + \frac{k}{\kappa}=2\xi$.

• Part 2-5. Reflection Coefficient and Transmission Coefficient

  Incident wave, reflected wave, and transmitted wave are respectively

  $$u_{inc}=A_{+}e^{ikx},\quad u_{ref}=A_{-}e^{-ikx},\quad u_{trans}=C_{+}e^{ikx}$$

  To calculate the reflection and transmission coefficients, we find the probability current density for each:

  $$j_{inc}=\frac{\hbar k}{m}|A_{+}|^2,\quad j_{ref}=\frac{\hbar k}{m}|A_{-}|^2,\quad j_{trnas}=\frac{\hbar k}{m}|C_{+}|^2$$

  Therefore, the reflection and transmission coefficients are

  $$R=\left| \frac{j_{ref}}{j_{inc}}\right|=\left| \frac{A_{-}} {A_{+}} \right|^2= \frac{ \xi^2 \sinh ^2 (2\kappa a) } {\cosh^2 (2\kappa a) + \eta^2 \sinh ^2 (2\kappa a) }$$

  $$T=\left| \frac{ j_{trnas} }{j_{inc}}\right|=\left| \frac{ C_{+} }{A_{+}} \right|^2=\frac{1}{\cosh^2 (2\kappa a) + \eta^2 \sinh^2 (2\kappa a)}$$

$U_{0} < E$

When looking at the diagram, it can be seen that except for Part 2-2., the rest of the results are the same as in 2. $0 < E < U_{0}$. Hence, using the previously obtained results,

• Part 3-1. $x<-a$

  $$u_{1}(x)=A_{+}e^{ikx} + A_{-}e^{-ikx}$$

• Part 3-2. $-a< x< a$

  $$\dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}(E-U_{0})u=0$$

  Here, substituting $E-U_{0} >0$ into $\frac{2m}{\hbar^2}(E-U_{0})=\kappa ^2$,

  $$2 \sqrt{ \frac{2m}{\hbar ^2}(E-u_{0})} a=n\pi$$

  $$\implies \frac{8m}{\hbar ^2} (E-U_{0}) a^2 =n^2\pi^2$$

  $$\implies E=\frac{n^2 \pi^2 \hbar ^2}{8ma^{2}_{}}+U_{0}$$