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When a classical particle with energy $E$ reaches a potential step of height $V_1$, it will go over the potential step if $E > V_1$ and it will be reflected if $E < V_1$. In quantum mechanics, for $E > V_1$, the wave function is partially transmitted and partially reflected at the potential step. For $E < V_1$, the wave function decays exponentially in the region where the energy is smaller than the potential. The plot below shows the wave function of a quantum particle at a potential step.

$E$ [eV] | $\psi$ | |

$x$ Å |

10 Å^{-1} 10 Å^{-1}

If the wave is incident from the left, we can assume that the transmitted wave has the form $\psi=A_t\exp(ik_1x)$. Other forms for the transmitted wave are possible such as $\psi=A_t\cos(k_1x)$ but $\psi=A_t\exp(ik_1x)$ has the nice feature that the probability of find the particle is the same everywhere, $\psi^*\psi =1$. With this form for the transmitted wave, the wave function to the left of the potential step, $(x < 0)$, is given by the matrix equation,

$$\begin{bmatrix} \psi(x)\\ \frac{d\psi(x)}{dx} \end{bmatrix} = \begin{bmatrix} \cos(k_0x) & \frac{\sin(k_0x)}{k_0}\\ -k_n\sin(k_0x) & \cos(k_0x) \end{bmatrix} \begin{bmatrix} A_t\\ ik_1A_t \end{bmatrix}.$$ $$\psi = A_t\cos(k_0x) + ik_1A_t\sin(k_0x)/k_0.$$Using the identities $\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$ and $\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$, this can be written in terms of a right-moving incident wave $A_i\exp(ik_0x)$ and a left-moving reflected wave $A_r\exp(-ik_0x)$.

$$\psi = \frac{A_t}{2}\left(1+\frac{k_1}{k_0}\right)e^{ik_0x} + \frac{A_t}{2}\left(1-\frac{k_1}{k_0}\right)e^{-ik_0x}.$$Here we identify,

$$A_i = \frac{A_t}{2}\left(1+\frac{k_1}{k_0}\right),\qquad\text{and}\qquad A_r = \frac{A_t}{2}\left(1-\frac{k_1}{k_0}\right).$$The transmission coefficient is the probability density of the transmitted wave $A_t^*A_t$ times the velocity of the transmitted wave, $v_1=\frac{\hbar k_1}{m}$, divided by the probability density of the incoming wave $A_i^*A_i$ times the velocity of the incoming waves, $v_0=\frac{\hbar k_0}{m}$. $$T=\frac{|A_t|^2v_1}{|A_i|^2v_0}=\frac{4 k_0k_1}{\left(k_0+k_1\right)^2}$$

Similarly, the reflection coefficient is,

$$R=\frac{|A_r|^2v_0}{|A_i|^2v_0}=\frac{\left(k_0-k_1\right)^2}{\left(k_0+k_1\right)^2}.$$and $R+T=1$.

$T$ | |

$E$ [eV] |

The transmission is perfect if $V_0 = V_1$ and there is no potential step. For $E > > V_0, V_1$, the transmission approaches 1 but there is always some reflection even when the potential step is down $(V_1 < V_0)$. Curiously, the reflection gets larger if there is a larger step down.