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

$$0<E<V_0.$$

Method Choice

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 exact transmission coefficient for the rectangular barrier is

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

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.

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.

The approximation becomes poor near the top of the barrier, $E\approx V_0$, because $\alpha$ becomes small and the barrier is no longer opaque.

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 $T$.

Common Mistakes

• Using $e^{-\alpha a}$ for the probability instead of $e^{-2\alpha a}$.
• Comparing the bare WKB exponent to the exact result and expecting the prefactor to match.
• Calling the rectangular barrier a smooth WKB problem.
• Forgetting that the left and right velocities matter if the asymptotic potentials differ.
• Trusting the opaque approximation when $\alpha a$ is not large.

Cross-Links

• Worked Problems and Model Calculations
• Barrier Penetration and Tunneling
• Turning Points and Connection Formulas
• Rectangular Barrier Tunneling
• Small Parameters and Error Estimates
• Double-Well Tunneling

References

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

Further 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.