Finite Potential Step

Classical physics says that a particle with enough energy clears a step and a particle without it bounces back. Quantum mechanics says something richer: even when the particle has more energy than the step, there is a non-zero probability of reflection. Setting up the problem carefully is the first step toward understanding this, and toward the far stranger phenomenon of tunneling through a barrier.

The potential

Place a step potential along the $x$-axis:

where $V_0 \gt 0$ is the height of the step. A particle of mass $m$ and total energy $E$ travels from left to right. We focus on the over-barrier case $E \gt V_0$ first.

The stationary-state approach

Because the potential is piecewise constant, the time-independent Schrödinger equation

reduces to a simple second-order ODE with constant coefficients in each region. Define two wave-numbers:

Both $k_1$ and $k_2$ are real and positive when $E \gt V_0$. In region I the general solution is a superposition of a right-moving incident wave and a left-moving reflected wave:

In region II, which extends to $x \to +\infty$, there is no surface to reflect from, so only a transmitted wave propagates to the right:

The amplitudes $A$, $B$, and $C$ are complex numbers. By convention $A$ is treated as known (the incoming beam is prepared with amplitude $A$), and we solve for $B$ and $C$.

Matching conditions at the boundary

The wave function and its first derivative must both be continuous everywhere, including at the step at $x = 0$. Continuity of $\psi$:

Continuity of $d\psi/dx$:

These two equations in two unknowns ($B$ and $C$) are the entire algebra of the problem. Solving them gives

Why reflection occurs at all

Classically, a particle with $E \gt V_0$ always passes the step; reflection is impossible. Quantum mechanically, $B \neq 0$ whenever $k_1 \neq k_2$, i.e., whenever $V_0 \neq 0$. The physical reason is that the wave-number changes abruptly at $x = 0$: the wave is propagating at wavelength $\lambda_1 = 2\pi/k_1$ on the left and $\lambda_2 = 2\pi/k_2$ on the right. Whenever two media with different propagation speeds meet at a sharp boundary, partial reflection is inevitable — this is the same physics that partially reflects light at a glass surface, expressed in wave mechanics.

Sub-barrier case and the approach to tunneling

When $E \lt V_0$ the quantity $E - V_0$ is negative, so $k_2$ would be imaginary. Write $\kappa = \sqrt{2m(V_0 - E)}/\hbar$ (real, positive). The solution in region II becomes an evanescent (exponentially decaying) wave:

The matching conditions still hold, and the algebra shows $|B| = |A|$: the step reflects the particle with probability 1, matching the classical prediction. However, $\psi_\text{II}$ is not zero — the wave function penetrates into the classically forbidden region with a characteristic length $1/\kappa$.

This penetration depth is the seed of tunneling: if region II has finite width rather than extending to infinity, the wave function can reach the far side before decaying to zero, and transmission becomes possible. That scenario — the rectangular barrier — is the subject of the next lesson.

Summary of the setup

The finite potential step is solved in three moves:

1. Write down $k_1$ and $k_2$ (or $\kappa$ in the sub-barrier case) from the kinetic energy in each region.
2. Write the most general wave function in each region consistent with the boundary conditions at infinity (no incoming wave from the right).
3. Match $\psi$ and $\psi'$ at $x = 0$ to find the reflection and transmission amplitudes.

The reflection coefficient $R = |B/A|^2$ and the transmission coefficient $T = (k_2/k_1)|C/A|^2$ will be derived in the next lesson; they satisfy $R + T = 1$ as required by probability conservation.

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