Reflection and Transmission

When a quantum particle travels from a region of constant potential into a region with a different potential, the wavefunction does not simply continue undisturbed. Even when the particle has enough energy to enter the new region classically, quantum mechanics predicts partial reflection — a wholly non-classical effect with no analogue in Newtonian mechanics.

Setting up the problem

Consider a particle of mass $m$ and total energy $E$ moving in one dimension. For $x < 0$ the potential is zero; for $x \geq 0$ the potential is $V_0$ with $0 < V_0 < E$ . The particle approaches from the left.

In each region the time-independent Schrödinger equation reduces to a free-particle equation:

-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = (E - V)\psi.

Defining the wave numbers

k_1 = \frac{\sqrt{2mE}}{\hbar}, \qquad k_2 = \frac{\sqrt{2m(E - V_0)}}{\hbar},

the general solutions are:

\psi_1(x) = A e^{ik_1 x} + B e^{-ik_1 x} \quad (x < 0),

\psi_2(x) = C e^{ik_2 x} \quad (x \geq 0).

The term $A e^{ik_1 x}$ is the incident wave travelling right, $B e^{-ik_1 x}$ is the reflected wave travelling left, and $C e^{ik_2 x}$ is the transmitted wave. We drop the left-going term in region 2 because there is nothing to reflect from on the right.

Boundary conditions

Continuity of $\psi$ and its derivative at $x = 0$ gives two equations:

A + B = C,

k_1(A - B) = k_2 C.

Solving for the amplitude ratios:

\frac{B}{A} = \frac{k_1 - k_2}{k_1 + k_2}, \qquad \frac{C}{A} = \frac{2k_1}{k_1 + k_2}.

Reflection and transmission coefficients

The physically meaningful quantities are the probability currents, not the amplitudes alone. The probability current for a plane wave $e^{ikx}$ is proportional to $k$ , so:

R = \left|\frac{B}{A}\right|^2 = \left(\frac{k_1 - k_2}{k_1 + k_2}\right)^2,

T = \frac{k_2}{k_1}\left|\frac{C}{A}\right|^2 = \frac{4k_1 k_2}{(k_1 + k_2)^2}.

Probability is conserved: $R + T = 1$ . We can verify this algebraically:

R + T = \frac{(k_1-k_2)^2 + 4k_1 k_2}{(k_1+k_2)^2} = \frac{(k_1+k_2)^2}{(k_1+k_2)^2} = 1. \checkmark

A worked example

Let $E = 9\,\text{eV}$ and $V_0 = 5\,\text{eV}$ . Then

k_1 \propto \sqrt{9} = 3, \qquad k_2 \propto \sqrt{9-5} = 2.

R = \left(\frac{3-2}{3+2}\right)^2 = \left(\frac{1}{5}\right)^2 = 0.04,

T = \frac{4 \cdot 3 \cdot 2}{(3+2)^2} = \frac{24}{25} = 0.96.

Four percent of the probability is reflected even though the particle classically had more than enough energy to cross the step. This is a direct consequence of the wave nature of matter.

Try it

This is a numerical exercise — return a number. A particle with energy $E = 5\,\text{eV}$ hits a potential step of height $V_0 = 1\,\text{eV}$ . Compute the reflection coefficient $R$ . (Hint: $k_1 \propto \sqrt{E}$ and $k_2 \propto \sqrt{E - V_0}$ ; the common proportionality constant cancels in the formula for $R$ .)

Run your code to see the quantum state.