Tunneling in WKB

In a classically forbidden region the WKB wavefunction decays exponentially rather than oscillating. For a particle hitting a barrier it is too low to surmount, the wavefunction does not vanish — it leaks through. WKB turns this leakage into a simple, exponentially sensitive estimate of the tunneling probability.

The decaying solution under a barrier

Where $E < V(x)$ the local momentum is imaginary, $p(x) = i|p(x)|$ with $|p(x)| = \sqrt{2m[V(x) - E]}$ . The WKB wavefunction there is real and decaying,

\psi(x) \approx \frac{C}{\sqrt{|p(x)|}}\, \exp\!\left(-\frac{1}{\hbar}\int_{x_1}^{x}|p(x')|\,dx'\right),

so the amplitude drops by a factor set by the integral of $|p|$ across the barrier.

The tunneling exponent

Define the dimensionless tunneling exponent (sometimes called the Gamow factor exponent)

\gamma = \frac{1}{\hbar}\int_{x_1}^{x_2} |p(x)|\,dx = \frac{1}{\hbar}\int_{x_1}^{x_2} \sqrt{2m\,[V(x) - E]}\;dx,

where $x_1, x_2$ are the entry and exit points of the forbidden region. The probability amplitude is suppressed by $e^{-\gamma}$ , so the transmitted probability is suppressed by the square:

T \approx e^{-2\gamma}.

This formula is exponentially sensitive: doubling the barrier width, or raising it so $|p|$ grows, slashes $T$ by orders of magnitude. That sensitivity is exactly why tunneling rates (alpha decay, scanning tunneling microscopy, tunnel diodes) span such enormous dynamic ranges.

A rectangular barrier

For a barrier of constant height $V_0$ and width $L$ with $E < V_0$ , the integrand is constant inside, so

\gamma = \frac{L}{\hbar}\sqrt{2m\,(V_0 - E)}, \qquad T \approx \exp\!\left(-\frac{2L}{\hbar}\sqrt{2m\,(V_0 - E)}\right).

With $\hbar = m = 1$ , $V_0 = 2$ , $E = 1$ , $L = 1$ :

\gamma = 1\cdot\sqrt{2(2-1)} = \sqrt{2} \approx 1.414, \qquad T \approx e^{-2\sqrt{2}} \approx 0.0591.

So roughly $6\%$ of the incident probability tunnels through — a fully quantum effect with no classical counterpart, since classically a particle with $E < V_0$ always reflects.

Try it

Compute the WKB transmission probability for the barrier above ( $V_0 = 2$ , $E = 1$ , $L = 1$ , units $\hbar = m = 1$ ) and return it as a number between 0 and 1. The answer is about $0.0591$ .

Run your code to see the quantum state.