Rectangular Potential Barrier

Rectangular potential barriers, also called square potential barriers, are formed by energy potentials which create wall-like barricades for particles. Essentially, a potential barrier is a potential step except the energy potential returns to zero at some finite positive $x$-position, $a$, and remains zero beyond that point. Here, we'll derive the wave function of a particle facing a rectangular potential barrier, then find the transmission and reflection coefficients of the particle upon encountering the barrier.

Schrodinger Equations

Key to solving for the wave function of a particle hitting a potential barrier is finding the Schrodinger equations which describe the system. First, define the energy potential, $V(x)$, of the system as this:

\begin{equation}
V(x) = \begin{cases} 0, & x < 0 \\ V_0, & 0 < x < a \\ 0, & x > a \end{cases}
\end{equation}

Writing the wave function of the particle as $\psi_1(x)$ for $x < 0$, $\psi_2(x)$ for $0 < x < a$, and $\psi_3(x)$ for $x > a$, the Schrodinger equations for $x < 0$, $0 < x < a$, and $x > a$ are respectively:

\begin{align}
E\psi_1(x) & = -\frac{\hbar^2}{2m} \frac{d^2}{{d x}^2} \psi_1(x) \\
E\psi_2(x) & = -\frac{\hbar^2}{2m} \frac{d^2}{{d x}^2} \psi_2(x) + V_0\psi_2(x) \\
E\psi_3(x) & = -\frac{\hbar^2}{2m} \frac{d^2}{{d x}^2} \psi_3(x)
\end{align}

This can be simplified, considering the wavenumbers, $k_1$ and $k_2$, of the wave function for inside and outside the barrier respectively. Since ${k_1}^2 = 2mE/\hbar^2$ and ${k_2}^2 = 2m( E - V_0) /\hbar^2$, this can be said of the wave function of a particle with $E \geq V_0$.

\begin{align}
0 & = \frac{d^2}{{d x}^2} \psi_1(x) + {k_1}^2 \psi_1(x) \\
0 & = \frac{d^2}{{d x}^2} \psi_2(x) + {k_2}^2 \psi_2(x) \\
0 & = \frac{d^2}{{d x}^2} \psi_3(x) + {k_1}^2 \psi_3(x)
\end{align}

Notice, however that if $E < V_0$, $k_2$ is imaginary and thus no longer an observable. By convention therefore, $\kappa$, defined by ${\kappa}^2 = 2m(V_0-E) /\hbar^2$, is used instead for $E < V_0$. The differential equations defining the wave function of a particle with insufficient energy are thus:

\begin{align}
0 & = \frac{d^2}{{d x}^2} \psi_1(x) + {k_1}^2 \psi_1(x) \\
0 & = \frac{d^2}{{d x}^2} \psi_2(x) - {\kappa}^2 \psi_2(x) \\
0 & = \frac{d^2}{{d x}^2} \psi_3(x) + {k_1}^2 \psi_3(x)
\end{align}

If There Is Sufficient Energy

For $E\geq V_0$, to find the wave function of the particle, equations $(5)$, $(6)$, and $(7)$ must be solved. These are homogeneous second-order linear differential equations and have the following general solutions:

\begin{align}
\psi_1(x) & = Ae^{r_Ax} + Be^{r_Bx} \\
\psi_2(x) & = Ce^{r_Cx} + De^{r_Dx} \\
\psi_3(x) & = Fe^{r_Fx} + Ge^{r_Gx}
\end{align}

where $A$, $B$, $C$, $D$, $F$, and $G$ are constants and $r_{A}=r_{F}$ and $r_{B}=r_{G}$ are the two solutions to the equation $r^2+{k_1}^2=0$ while $r_C$ and $r_D$ are the two solutions to the equation $r^2+{k_2}^2=0$.

\begin{align}
\psi_1(x) & = Ae^{ik_1x} + Be^{-ik_1x} \\
\psi_2(x) & = Ce^{ik_2x} + De^{-ik_2x} \\
\psi_3(x) & = Fe^{ik_1x} + Ge^{-ik_1x}
\end{align}

Notice, considering Euler's formula, that $Ae^{ik_1x}$, $Ce^{ik_2x}$, and $Fe^{ik_1x}$ represent waves travelling in the positive direction while $Be^{−ik_1x}$, $De^{−ik_2x}$, and $Ge^{-ik_1x}$ represent waves travelling in the negative direction. Since reflection by the barrier is conceivable, it is possible to have wave components travelling in the negative direction for $x < a$ , but there is no reason to have waves doing so for $x > a$. Thus, $G=0$.

\begin{align}
\psi_1(x) & = Ae^{ik_1x} + Be^{-ik_1x} \\
\psi_2(x) & = Ce^{ik_2x} + De^{-ik_2x} \\
\psi_3(x) & = Fe^{ik_1x}
\end{align}

To solve for $B$ and $F$ in relation to $A$, impose these four boundary conditions to ensure that the wave function is a smooth curve as $x\to 0$ and as $x\to a$:

\begin{align*}
\lim_{x\to0^-} \psi_1(x) &= \lim_{x\to0^+} \psi_2(x) \\
\lim_{x\to0^-} \frac{d}{d x}\psi_1(x) &= \lim_{x\to0^+} \frac{d}{d x}\psi_2(x) \\
\lim_{x\to a^-} \psi_2(x) &= \lim_{x\to a^+} \psi_3(x) \\
\lim_{x\to a^-} \frac{d}{d x}\psi_2(x) &= \lim_{x\to a^+} \frac{d}{d x}\psi_3(x)
\end{align*}
\begin{align}
A + B &= C + D \\
ik_1A - ik_1B &= ik_2C - ik_2D \\
Ce^{ik_2a} + De^{-ik_2a} &= Fe^{ik_1a} \\
ik_2Ce^{ik_2a} - ik_2De^{-ik_2a} &= ik_1Fe^{ik_1a}
\end{align}
\begin{align}
k_1A + k_1B &= k_1C + k_1D \\
k_1A - k_1B &= k_2C - k_2D \\
k_2Ce^{ik_2a} + k_2De^{-ik_2a} &= k_2Fe^{ik_1a} \\
k_2Ce^{ik_2a} - k_2De^{-ik_2a} &= k_1Fe^{ik_1a}
\end{align}
\begin{align}
2k_1A &= (k_1+k_2)C + (k_1-k_2)D \\
2k_1B &= (k_1-k_2)C + (k_1+k_2)D \\
2k_2Ce^{ik_2a} &= (k_1+k_2)Fe^{ik_1a} \\
2k_2De^{-ik_2a} &= (k_2-k_1)Fe^{ik_1a}
\end{align}

To solve for $F$ in relation to $A$, consider equations $(28)$, $(30)$, and $(31)$.

\begin{equation}
2k_1A = \frac{(k_1+k_2)^2}{2k_2}Fe^{i(k_1-k_2)a} - \frac{(k_1-k_2)^2}{2k_2}Fe^{i(k_1+k_2)a}
\end{equation}
\begin{equation}
4k_1k_2e^{-ik_1a}A = (k_1+k_2)^2Fe^{-ik_2a} - (k_1-k_2)^2Fe^{ik_2a}
\end{equation}

Using Euler's formula to expand $e^{-ik_2a}$ and $e^{ik_2a}$, the following can be derived:

\begin{equation}
4k_1k_2e^{-ik_1a}A = \left(-2i{k_1}^2\sin{k_2a} + 4k_1k_2\cos{k_2a} - 2i{k_2}^2\sin{k_2a}\right)F
\end{equation}
\begin{equation}
F = \frac{2k_1k_2e^{-ik_1a} A}{2k_1k_2\cos{k_2a} - i\left({k_1}^2 + {k_2}^2\right)\sin{k_2a}}
\end{equation}

To solve for $B$ in relation to $A$, consider equations $(29)$, $(30)$, and $(31)$.

\begin{equation}
2k_1B = \frac{{k_1}^2 - {k_2}^2}{2k_2}Fe^{i(k_1-k_2)a} - \frac{{k_1}^2-{k_2}^2}{2k_2}Fe^{i(k_1+k_2)a}
\end{equation}
\begin{equation}
\frac{4k_1k_2e^{-ik_1a}B}{{k_1}^2 - {k_2}^2} = \left(e^{-ik_2a} - e^{ik_2a}\right)F
\end{equation}

Using Euler's formula to expand $e^{-ik_2a}$ and $e^{ik_2a}$, the following can be derived:

\begin{equation}
\frac{2k_1k_2 e^{-ik_1a} B}{-i\left({k_1}^2 - {k_2}^2\right)\sin{k_2a}} = F
\end{equation}

Comparing equations $(35)$ and $(38)$, $B$ is solved for in relation to $A$.

\begin{equation}
B = \frac{-i\left({k_1}^2 - {k_2}^2\right)\sin(k_2a) A}{2k_1k_2\cos{k_2a} - i\left({k_1}^2 + {k_2}^2\right)\sin{k_2a}}
\end{equation}

Considering equations $(30)$ and $(31)$ alongside equation $(35)$, $C$ and $D$ can also be solved for in relation to $A$, but since only $A$, $B$, and $F$ are needed to calculate the reflection and transmission coefficients, the derivations of $C$ and $D$ are omitted here. In order to find the reflection and transmission coefficients, the wave function must be first written in terms of its incident, reflected, and transmitted components, $\psi_i(x)$, $\psi_r(x)$, and $\psi_t(x)$ respectively.

\begin{align}
\psi_i(x) & = Ae^{ik_1x} \\
\psi_r(x) & = Be^{-ik_1x} \\
\psi_t(x) & = Fe^{ik_1x}
\end{align}

The reflection and transmission coefficients, $R$ and $T$ respectively, are defined as follows:

\begin{align}
R & = -\frac{j_r}{j_i} \\
T & = \frac{j_t}{j_i}
\end{align}

where $j_i$, $j_r$, and $j_t$ are the incident, reflected, and transmitted probability currents respectively.

\begin{align}
R & = -\frac{\psi_r(x) \frac{d}{d x} {\psi_r}^*(x) - {\psi_r}^*(x) \frac{d}{d x} \psi_r(x)}{\psi_i(x) \frac{d}{d x} {\psi_i}^*(x) - {\psi_i}^*(x) \frac{d}{d x} \psi_i(x)} \\
T & = \frac{\psi_t(x) \frac{d}{d x} {\psi_t}^*(x) - {\psi_t}^*(x) \frac{d}{d x} \psi_t(x)}{\psi_i(x) \frac{d}{d x} {\psi_i}^*(x) - {\psi_i}^*(x) \frac{d}{d x} \psi_i(x)}
\end{align}
\begin{align}
R & = \frac{|B|^2}{|A|^2} \\
T & = \frac{|F|^2}{|A|^2}
\end{align}

Applying the solutions for $B$ and $F$ found in equations $(39)$ and $(35)$ respectively gives:

\begin{align}
R & = \frac{\left({k_1}^2 - {k_2}^2\right)^2\sin^2{k_2a}}{4{k_1}^2{k_2}^2\cos^2{k_2a} + \left({k_1}^2 + {k_2}^2\right)^2\sin^2{k_2a}} \\
T & = \frac{4 {k_1}^2 {k_2}^2 }{4{k_1}^2 {k_2}^2 \cos^2{k_2a} + \left({k_1}^2 + {k_2}^2\right)^2 \sin^2{k_2a}}
\end{align}
\begin{align}
R & = \frac{\left({k_1}^2 - {k_2}^2\right)^2\sin^2{k_2a}}{4{k_1}^2{k_2}^2 + \left({k_1}^2 - {k_2}^2\right)^2\sin^2{k_2a}} \\
T & = \frac{4 {k_1}^2 {k_2}^2 }{4{k_1}^2 {k_2}^2 + \left({k_1}^2 - {k_2}^2\right)^2 \sin^2{k_2a}}
\end{align}
\begin{align}
R & = \left[ \frac{4{k_1}^2{k_2}^2}{\left({k_1}^2 - {k_2}^2\right)^2\sin^2{k_2a}} + 1\right]^{-1} \\
T & = \left[ \frac{\left({k_1}^2 - {k_2}^2\right)^2 \sin^2{k_2a}}{4{k_1}^2 {k_2}^2} + 1 \right]^{-1}
\end{align}

Interestingly contrary to classical mechanics, quantum mechanics suggests that the particle may actually be reflected by the potential barrier, despite having a total energy of equal or greater value than $V_0$.

If There Is Insufficient Energy

For $E < V_0$, equations $(8)$, $(9)$, and $(10)$ must be solved to find $\psi_1(x)$, $\psi_2(x)$, and $\psi_3(x)$. To do this, follow the methodology employed in the previous section, "If There Is Sufficient Energy". The solutions of equations $(8)$, $(9)$, and $(10)$ are identical to those of $(5)$, $(6)$, and $(7)$ respectively save for the use of $i\kappa$ in the place of $k_2$.

\begin{align}
\psi_1(x) & = Ae^{ik_1x} + Be^{-ik_1x} \\
\psi_2(x) & = Ce^{-\kappa x} + De^{\kappa x} \\
\psi_3(x) & = Fe^{ik_1x}
\end{align}

Applying the same boundary conditions as in the previous section and manipulating algebra in the same manner, it can also be found that:

\begin{align}
2ik_1A &= (ik_1 - \kappa)C + (ik_1 + \kappa)D \\
2ik_1B &= (ik_1 + \kappa)C + (ik_1 - \kappa)D \\
2\kappa Ce^{-\kappa a} &= (\kappa - ik_1)Fe^{ik_1a} \\
2\kappa De^{\kappa a} &= (ik_1 + \kappa )Fe^{ik_1a}
\end{align}

To solve for $F$ in relation to $A$, consider equations $(58)$, $(60)$, and $(61)$.

\begin{equation}
2ik_1A = - \frac{(ik_1 - \kappa)^2}{2\kappa}Fe^{(ik_1 + \kappa)a} + \frac{(ik_1 + \kappa)^2}{2\kappa}Fe^{(ik_1 - \kappa)a}
\end{equation}
\begin{equation}
4ik_1\kappa e^{-ik_1a} A = - (ik_1 - \kappa)^2Fe^{\kappa a} + (ik_1 + \kappa)^2Fe^{ -\kappa a}
\end{equation}
\begin{equation}
4ik_1\kappa e^{-ik_1a} A = \left[ \left( {k_1}^2 - \kappa^2 \right)\left(e^{\kappa a} - e^{ -\kappa a}\right) + 2ik_1\kappa\left(e^{\kappa a} + e^{ -\kappa a}\right) \right] F
\end{equation}
\begin{equation}
F = \frac{ 2ik_1\kappa e^{-ik_1a} A }{ \left( {k_1}^2 - \kappa^2 \right)\sinh{\kappa a} + 2ik_1\kappa\cosh{\kappa a} }
\end{equation}

(In case you are unfamiliar with hyperbolic functions, $\sinh{u} = (e^{u} - e^{-u})/2$ is the hyperbolic sine function and $\cosh{u} = (e^{u} + e^{-u})/2$ is the hyperbolic cosine function.) To solve for $B$ in relation to $A$, consider equations $(59)$, $(60)$, and $(61)$.

\begin{equation}
2ik_1B = \frac{{k_1}^2 + \kappa^2}{2\kappa}Fe^{(ik_1 + \kappa)a} - \frac{{k_1}^2 + {\kappa}^2}{2\kappa} Fe^{(ik_1 - \kappa) a}
\end{equation}
\begin{equation}
\frac{4ik_1\kappa e^{-ik_1a} B}{ {k_1}^2 + \kappa^2 } = Fe^{\kappa a} - Fe^{-\kappa a}
\end{equation}
\begin{equation}
\frac{2ik_1\kappa e^{-ik_1a} B}{ \left({k_1}^2 + \kappa^2\right) \sinh{\kappa a}} = F
\end{equation}

Comparing equations $(65)$ and $(68)$, $B$ is solved for in relation to $A$.

\begin{equation}
B = \frac{ \left({k_1}^2 + \kappa^2\right) \sinh(\kappa a) A }{ \left( {k_1}^2 - \kappa^2 \right)\sinh{\kappa a} + 2ik_1\kappa\cosh{\kappa a} }
\end{equation}

As in the previous section, the wave function written in terms of its incident, reflected, and transmitted components is:

\begin{align}
\psi_i(x) & = Ae^{ik_1x} \\
\psi_r(x) & = Be^{-ik_1x} \\
\psi_t(x) & = Fe^{ik_1x}
\end{align}

Furthermore, the reflection and transmission coefficients, derivable using the same method as in the previous section, are again given by:

\begin{align}
R & = \frac{|B|^2}{|A|^2} \\
T & = \frac{|F|^2}{|A|^2}
\end{align}
\begin{align}
R & = \frac{ \left({k_1}^2 + \kappa^2\right)^2 \sinh^2{\kappa a} }{ \left( {k_1}^2 - \kappa^2 \right)^2\sinh^2{\kappa a} + 4{k_1}^2\kappa^2\cosh^2{\kappa a} } \\
T & = \frac{ 4{k_1}^2\kappa^2 }{ \left( {k_1}^2 - \kappa^2 \right)^2\sinh^2{\kappa a} + 4{k_1}^2\kappa^2\cosh^2{\kappa a} }
\end{align}
\begin{align}
R & = \frac{ \left({k_1}^2 + \kappa^2\right)^2 \sinh^2{\kappa a} }{ \left( {k_1}^2 + \kappa^2 \right)^2\sinh^2{\kappa a} + 4{k_1}^2\kappa^2 } \\
T & = \frac{ 4{k_1}^2\kappa^2 }{ \left( {k_1}^2 + \kappa^2 \right)^2\sinh^2{\kappa a} + 4{k_1}^2\kappa^2 }
\end{align}
\begin{align}
R & = \left[ \frac{4{k_1}^2\kappa^2}{\left( {k_1}^2 + \kappa^2 \right)^2\sinh^2{\kappa a}} + 1\right]^{-1} \\
T & = \left[ \frac{\left( {k_1}^2 + \kappa^2 \right)^2\sinh^2{\kappa a}}{4{k_1}^2\kappa^2} + 1 \right]^{-1}
\end{align}

Contrary to classical expectations which would suggest that the particle has zero probability of travelling beyond $x=0$, quantum mechanics asserts that the particle has a non-zero probability of tunneling through the rectangular potential barrier, despite having a total energy less than $V_0$. This phenomenon marks a major difference between quantum and classical mechanics.