WKB Barrier Tunneling Worked Example

This worked example estimates tunneling through a one-dimensional rectangular barrier using the WKB exponent, then compares the result with the exact opaque-barrier limit.

Use

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

with

$$0<E<V_0.$$

The Rectangular Barrier as an Exponent Test

The exact rectangular barrier is not smooth at the edges, so it is not the ideal setting for WKB connection formulas. It is still a useful benchmark because the under-barrier exponent is simple and the exact result is known.

The WKB estimate should reproduce the leading exponential dependence when the barrier is opaque:

$$\alpha a\gg1,$$

where

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

is the decay constant inside the barrier.

WKB Calculation

The forbidden region is $0<x<a$. The WKB barrier action in momentum units is

$$K(E) = \int_0^a \sqrt{2m(V_0-E)} \,dx.$$

Since the integrand is constant,

$$K(E) = a\sqrt{2m(V_0-E)} = \hbar\alpha a.$$

The leading WKB transmission probability is

$$T_{\mathrm{WKB}} \approx \exp \left[ - \frac{2K(E)}{\hbar} \right],$$

so

$$T_{\mathrm{WKB}} \approx e^{-2\alpha a}.$$

Exact Comparison

The left and right asymptotic potentials are equal, so their velocities are equal and $T=|t|^2$. Matching the wavefunction and its derivative at each interface gives the exact rectangular-barrier result:

$$T_{\mathrm{exact}} = \left[ 1+ \frac{V_0^2\sinh^2(\alpha a)} {4E(V_0-E)} \right]^{-1}.$$

In the opaque limit,

$$\sinh^2(\alpha a) \approx \frac14e^{2\alpha a}.$$

Therefore

$$T_{\mathrm{exact}} \approx \frac{16E(V_0-E)}{V_0^2} e^{-2\alpha a}.$$

The WKB estimate captures the leading exponential:

$$T_{\mathrm{exact}} \propto T_{\mathrm{WKB}}$$

with an energy-dependent prefactor. At fixed $E/V_0$, making $a$ larger does not make the bare-exponent estimate relatively exact: $T_{\mathrm{exact}}/T_{\mathrm{WKB}}\to16E(V_0-E)/V_0^2$. What becomes exact is the coefficient of $a$ in $\log T$. This distinction matters when estimating a measured probability rather than its exponential scale.

Numerical Illustration

Take units where

$$\hbar=1, \qquad 2m=1, \qquad V_0=1, \qquad E=\frac14, \qquad a=6.$$

Then

$$\alpha = \sqrt{V_0-E} = \frac{\sqrt3}{2},$$

and

$$2\alpha a = 6\sqrt3 \approx 10.39.$$

The WKB exponent gives

$$T_{\mathrm{WKB}} \approx e^{-10.39} \approx 3.1\times10^{-5}.$$

The opaque-limit exact prefactor is

$$\frac{16E(V_0-E)}{V_0^2} = 3.$$

Thus

$$T_{\mathrm{exact}} \approx 9.2\times10^{-5}$$

in the opaque approximation. The two estimates differ by a prefactor but agree on the exponential scale.

Validity Discussion

The leading WKB exponent is useful when

$$\frac{K}{\hbar}\gg1.$$

For the rectangular barrier, edge matching is abrupt, so the prefactor is not expected to be captured by the simplest smooth-barrier WKB expression. The comparison is still valuable because it isolates the most robust part of tunneling: exponential suppression by the forbidden-region action.

At fixed width, the approximation fails as $E\to V_0^-$ because $\alpha a\to0$. The exact limit is finite,

$$T(E\to V_0^-)=\left(1+\frac{mV_0a^2}{2\hbar^2}\right)^{-1},$$

whereas the bare exponential approaches one. The thick-barrier and barrier-top limits are different limits.

Cross-Checks

The exponent $2\alpha a$ is dimensionless.

Increasing $a$ suppresses $T$ exponentially.

Increasing the mass $m$ suppresses $T$ because $\alpha$ grows like $\sqrt m$.

Increasing $E$ toward $V_0$ reduces $\alpha$ and increases $T$.

Exercises

1. In the same units as the numerical example, double the barrier width from $a=6$ to $a=12$. By what factor does the WKB transmission change?

Solution

The WKB probability scales as

$$T_{\mathrm{WKB}} \propto e^{-2\alpha a}.$$

Doubling $a$ multiplies the transmission by

$$e^{-2\alpha a} = e^{-6\sqrt3} \approx 3.1\times10^{-5}$$

relative to the original value.

2. Explain why the exact opaque-limit prefactor does not invalidate the WKB exponent.

Solution

When $K/\hbar\gg1$, the exponential factor can change by many orders of magnitude as parameters vary. A prefactor of order one, or even a modest numerical factor, is usually much less important than the exponent. The exact prefactor matters for precision, but the WKB exponent captures the leading asymptotic dependence.

References

• C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer, 1999.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.