Exponential Decay in the Barrier

The classically-forbidden region

Imagine sending a classical particle toward a hill whose height in energy units exceeds the particle's total energy. The particle cannot cross — it bounces back. Quantum mechanics tells a different story. The time-independent Schrödinger equation,

$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi,$

must still hold inside the barrier. Rearranging for the second derivative gives

$\frac{d^2\psi}{dx^2} = \frac{2m(V_0 - E)}{\hbar^2}\,\psi.$

When $E < V_0$ the coefficient on the right is positive. A positive coefficient means the solution is not oscillatory (as it is outside the barrier) but exponential. The two independent solutions are $e^{+\kappa x}$ and $e^{-\kappa x}$, where we define the decay constant

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

For a particle coming from the left, the growing exponential $e^{+\kappa x}$ must be discarded because it would diverge — only the decaying solution $e^{-\kappa x}$ survives inside a thick barrier. The wavefunction therefore falls off as

$\psi_{\text{barrier}}(x) \propto e^{-\kappa x},$

and the probability density decays as $e^{-2\kappa x}$. The word "exponential" in the lesson title refers exactly to this spatial decay.

What $\kappa$ tells you

The decay constant $\kappa$ has units of inverse length. Its reciprocal $1/\kappa$ is the decay length — the distance over which the amplitude falls by a factor of $e \approx 2.718$.

For an electron with $V_0 - E = 1.0\,\text{eV}$, inserting $m_e = 9.109 \times 10^{-31}\,\text{kg}$ and $\hbar = 1.055 \times 10^{-34}\,\text{J}\cdot\text{s}$ yields

$\kappa = \frac{\sqrt{2 \times (9.109 \times 10^{-31}\,\text{kg}) \times (1.602 \times 10^{-19}\,\text{J})}}{1.055 \times 10^{-34}\,\text{J}\cdot\text{s}} \approx 5.12 \times 10^9\,\text{m}^{-1},$

so the decay length is $1/\kappa \approx 0.20\,\text{nm}$ — roughly the diameter of two hydrogen atoms. Tunneling currents therefore vanish on the scale of a few atoms, which is precisely why scanning tunneling microscopes achieve atomic resolution.

Connecting $\kappa$ to the transmission probability

For a rectangular barrier of width $L$ and when $\kappa L \gg 1$, the transmission probability simplifies to the WKB approximation

$T \approx e^{-2\kappa L}.$

This factor of $2\kappa$ in the exponent explains why the tunneling probability is extraordinarily sensitive to barrier parameters — doubling $L$ squares $T$. The exact expression from matching boundary conditions is more involved, but in the thick-barrier limit the exponential dominates and $\kappa$ is the one number that governs tunneling.

Try it

This is a numerical exercise — your code should return a number, not a circuit. Compute $\kappa$ in $\text{m}^{-1}$ for an electron facing a barrier with $V_0 - E = 1.0\,\text{eV}$.

Sign in on the full site to ask questions and join the discussion.