WKB Approximation

Semiclassical approximation for quantum waves in slowly varying potentials.

Conceptual context

The WKB approximation is used when a quantum wave moves through a $V$ landscape that changes slowly with $x$. Instead of solving the full Schrödinger equation exactly, it treats the wave locally: where $E$ exceeds $V$, the phase oscillates; where $V$ exceeds $E$, the amplitude fades. The central question is how that fading controls $T$ when $L$ changes. WKB turns a barrier into an action cost. Each forbidden slice adds cost, and the transmitted wave pays that cost exponentially.

🟢 Essential level

The essential idea is that a quantum particle does not stop sharply at a barrier. If the region where $V$ exceeds $E$ has a small $L$, the wave keeps a noticeable tail on the far side.

If $L$ grows, that tail becomes smaller much faster than a linear rule would suggest. WKB estimates this reduction by accumulating a local difficulty, the $\kappa$, along $x$ inside the barrier. That accumulated difficulty becomes the $\gamma$, and from it comes the $T$. Therefore the usual graph is not just any decreasing curve: it is an exponential fall. Doubling the barrier width usually does not halve transmission; it can suppress it by many orders of magnitude. Width acts like a magnifier of quantum cost.

George Gamow used this same idea in 1928 to explain alpha decay: a heavy nucleus holds its alpha particles behind a potential barrier, and it is precisely WKB tunneling that predicts, with remarkable accuracy, why some nuclei decay in microseconds while others take billions of years.

🔵 Formal level

Basic formulas

In the allowed region, where $E$ is greater than $V$, the local oscillation of the wavefunction is summarized by $k_local$.

Local wavenumber in allowed region

In the forbidden region, where $V$ is greater than $E$, the central quantity becomes $\kappa$, the attenuation per unit $x$.

Decay coefficient in forbidden region

Real student questions

Why is it called tunneling if the particle does not drill a hole? Because the wavefunction has an evanescent amplitude inside the forbidden region and can reappear on the far side.

Does the particle lose $E$ inside the barrier? Not in the stationary model. The spatial form of the wave changes, not the energy of the state.

Why does $\hbar$ matter so much? Because it converts action into phase and makes the forbidden cost dimensionless.

Does WKB always improve for a wide barrier? Not necessarily. Slow variation helps, but a wide barrier with poorly treated turning points needs connection formulas.

Final synthesis

WKB translates a quantum barrier into accumulated semiclassical cost. As $L$ grows, $\gamma$ grows and $T$ falls exponentially. The right answer is not only a calculation; it is a reading of scales, validity, and physical meaning.