Potential Step

Potential steps are created by energy potentials which form step-like barricades for particles. Before the potential step, the energy potential is uniformly zero, but at the step, the energy potential rises instantaneously to a finite value and remains constant at that value for all positions beyond the step. Here, we'll derive the wave function of a particle facing a potential step, then find the transmission and reflection coefficients of the particle upon encountering the step.

Schrodinger Equations

To find the wave function of a particle hitting a potential step, we need to first write our Schrodinger equations - one for the positions before the step and another for after. Let's define our energy potential, $V(x)$, as this:

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

where $V_0$ is the constant energy potential for positions beyond $x=0$. Writing our wave function as $\psi_1(x)$ for $x < 0$ and $\psi_2(x)$ for $x>0$, our Schrodinger equations look like this:

\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)
\end{align}
\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)
\end{align}

In equation $(4)$, $k_1$ is the wavenumber of the particle's wave function before the potential step, defined by ${k_1}^2 = 2mE/ \hbar^2$. Note that $k_1$ is always real. Meanwhile, in equation $(5)$, $k_2$ is the wavenumber of the particle's wave function beyond $x=0$, defined by ${k_2}^2 = 2m(E - V_0)/ \hbar^2$. Note that $k_2$ is real if $E \geq V_0$, but imaginary otherwise. Because of this, if $E < V_0$, $\kappa^2 = 2m(V_0 - E)/ \hbar^2$ is used instead since the wavenumber $k_2$ would no longer be an observable. So if $E < V_0$,

\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) - {\kappa}^2 \psi_2(x)
\end{align}

If There Is Sufficient Energy

For $E \geq V_0$, to find $\psi_1(x)$ and $\psi_2(x)$, equations $(4)$ and $(5)$ must be solved. Equations $(4)$ and $(5)$ are homogeneous second-order linear differential equations and respectively 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}
\end{align}

where $A$, $B$, $C$, and $D$ are constants and $r_A$ and $r_B$ 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}
\end{align}

Notice, considering Euler's formula, that $Ae^{ik_1x}$ and $Ce^{ik_2x}$ represent waves travelling in the positive direction while $Be^{-ik_1x}$ and $De^{-ik_2x}$ represent waves travelling in the negative direction. While it is possible to have wave components travelling in the negative direction for $x < 0$ since it is conceivable that part of the incident wave component is reflected by the step, there is no reason to have waves doing so for $x > 0$. Thus, $D=0$.

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

To solve for $B$ and $C$ in relation to $A$, impose these two boundary conditions to ensure that the wave function is a smooth curve as $x\to0$:

\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)
\end{align*}
\begin{align}
A+B &= C \\
ik_1A - ik_1B &= ik_2C
\end{align}
\begin{align}
B &= \frac{k_1 - k_2}{k_1 + k_2}A \\
C &= \frac{2k_1}{k_1 + k_2}A
\end{align}
\begin{align}
\psi_1(x) & = Ae^{ik_1x} + \frac{k_1 - k_2}{k_1 + k_2}Ae^{-ik_1x} \\
\psi_2(x) & = \frac{2k_1}{k_1 + k_2}Ae^{ik_2x}
\end{align}

There is not much else that can be done to further solve for $\psi_1(x)$ and $\psi_2(x)$ since $A$ can not be found by normalizing the wave function. Yet, it is still possible to calculate the reflection and transmission coefficients, $R$ and $T$ respectively, with what is there. $R$ and $T$ are defined as:

\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. At this point, it is helpful to rewrite our wave function as $\psi_i(x)$, $\psi_r(x)$, and $\psi_t(x)$ - its incident, reflected, and transmitted components respectively.

\begin{align}
\psi_i(x) & = Ae^{ik_1x} \\
\psi_r(x) & = \frac{k_1 - k_2}{k_1 + k_2}Ae^{-ik_1x} \\
\psi_t(x) & = \frac{2k_1}{k_1 + k_2}Ae^{ik_2x}
\end{align}
\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 & = \left(\frac{k_1 - k_2}{k_1 + k_2}\right)^2 \\
T & = \frac{4k_1k_2}{(k_1 + k_2)^2}
\end{align}

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

If There Is Insufficient Energy

For $E < V_0$, equations $(6)$ and $(7)$ must be solved to find $\psi_1(x)$ and $\psi_2(x)$. To do this, follow the methodology employed in the previous section, "If There Is Sufficient Energy". The solutions of equations $(6)$ and $(7)$ are identical to those of $(4)$ and $(5)$ 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}
\end{align}

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

\begin{align}
B &= \frac{ik_1 + \kappa}{ik_1 - \kappa}A \\
C &= \frac{2ik_1}{ik_1 - \kappa}A
\end{align}
\begin{align}
\psi_1(x) & = Ae^{ik_1x} + \frac{ik_1 + \kappa}{ik_1 - \kappa}Ae^{-ik_1x} \\
\psi_2(x) & = \frac{2ik_1}{ik_1 - \kappa}Ae^{-\kappa x}
\end{align}

The wave function, written in terms of its incident, reflected, and transmitted components, is then:

\begin{align}
\psi_i(x) & = Ae^{ik_1x} \\
\psi_r(x) & = \frac{ik_1 + \kappa}{ik_1 - \kappa}Ae^{-ik_1x} \\
\psi_t(x) & = \frac{2ik_1}{ik_1 - \kappa}Ae^{-\kappa x}
\end{align}

Using the same method as in the previous section, the reflection and transmission coefficients can be derived to be:

\begin{align}
R & = 1 \\
T & = 0
\end{align}

This result is consistent with classical predictions, but does it guarantee that there is zero probability the particle will penetrate the potential step? After all, $\psi_t(x)$, though not a wave, does exist for $x > 0$. To see whether or not it is possible that the particle may have gone beyond $x=0$, the transmitted probability density, $\rho_t(x)$, of the particle must be found.

\begin{equation}
\rho_t(x) = |\psi_t(x)|^2
\end{equation}
\begin{equation}
\rho_t(x) = \frac{4{k_1}^2}{{k_1}^2 + \kappa^2}|A|^2e^{-2\kappa x}
\end{equation}

Although the probability density of the particle for $x > 0$ decays rapidly after $x=0$, being nevertheless non-zero means that the particle has a non-zero probability of tunneling into the potential step, despite having a total energy less than $V_0$. This phenomenon of tunneling marks a major difference between quantum and classical mechanics.