Checkpoint: Tunneling Problems

This checkpoint consolidates everything from the Quantum Tunneling and Barriers module. You will derive the transmission coefficient for a particle tunneling through a rectangular potential barrier from first principles and then evaluate it numerically.

The rectangular barrier

Consider a particle of mass $m$ and energy $E$ moving in one dimension. A rectangular barrier of height $V_0 > E$ occupies the region $0 \leq x \leq a$ :

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

There are three regions. In regions I ( $x < 0$ ) and III ( $x > a$ ) the potential is zero and the wave function oscillates freely:

\psi_\text{I}(x) = A e^{ik_1 x} + B e^{-ik_1 x}, \qquad k_1 = \frac{\sqrt{2mE}}{\hbar}.

\psi_\text{III}(x) = F e^{ik_1 x},

where we set the amplitude of any incoming wave from the right to zero (a particle incident only from the left). In region II ( $0 \leq x \leq a$ ) the kinetic energy is negative, so the wave number would be imaginary. We define the real decay constant

\kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar},

and write

\psi_\text{II}(x) = C e^{-\kappa x} + D e^{\kappa x}.

This is an evanescent wave: it does not propagate but decays (or grows) exponentially inside the barrier. The particle can reach the far side only because the barrier has finite width $a$ .

Matching at both boundaries

Four boundary conditions — continuity of $\psi$ and $d\psi/dx$ at both $x = 0$ and $x = a$ — relate the five amplitudes $A$ , $B$ , $C$ , $D$ , $F$ . After the algebra, the transmission amplitude is

\frac{F}{A} = \frac{e^{-ik_1 a}}{\cosh(\kappa a) + \tfrac{i}{2}\left(\tfrac{\kappa}{k_1} - \tfrac{k_1}{\kappa}\right)\sinh(\kappa a)}.

The transmission coefficient is $T = |F/A|^2$ . Taking the modulus squared of the denominator gives

T = \frac{1}{1 + \dfrac{V_0^2 \sinh^2(\kappa a)}{4E(V_0 - E)}}.

This is the central result for tunneling through a rectangular barrier. The Griffiths derivation (§2.6) and the MIT OCW lectures 13-14 work through both derivations in full.

A worked numerical check

Let $E = 4\,\text{eV}$ , $V_0 = 5\,\text{eV}$ , $a = 0.5\,\text{Å}$ , and use $\hbar^2/(2m) = 3.81\,\text{eV}\cdot\text{Å}^2$ for a free electron. Then

\kappa = \sqrt{\frac{V_0 - E}{3.81}} = \sqrt{\frac{1}{3.81}} \approx 0.512\,\text{Å}^{-1}.

\kappa a \approx 0.256, \qquad \sinh(0.256) \approx 0.259.

T = \frac{1}{1 + \dfrac{25 \times 0.0671}{4 \times 4 \times 1}} = \frac{1}{1 + 0.1044} \approx 0.906.

The transmission probability is about 91%, much higher than for a thicker or taller barrier. This illustrates how strongly $T$ depends on $\kappa a$ .

Try it

This is a numerical exercise — return a number. A free electron with energy $E = 1.0\,\text{eV}$ encounters a rectangular barrier of height $V_0 = 3.0\,\text{eV}$ and width $a = 1.0\,\text{Å}$ . Compute the transmission coefficient $T$ using the exact formula above with $\hbar^2/(2m) = 3.81\,\text{eV}\cdot\text{Å}^2$ .

Run your code to see the quantum state.