WKB Quantization

A bound state must fit smoothly between its two classical turning points. WKB turns this "fitting" requirement into a precise condition on the accumulated phase — the Bohr–Sommerfeld quantization rule — which produces approximate (and sometimes exact) energy levels.

The phase-fitting condition

Between two turning points $x_1$ and $x_2$ the WKB wavefunction oscillates with the local momentum $p(x) = \sqrt{2m[E - V(x)]}$. For the wavefunction to match the decaying exponentials in the forbidden regions on both sides, the connection formulas (covered later) require the total phase swept between the turning points to equal a half-integer number of half-cycles. The result is

The integer $n$ counts the nodes; the $+\tfrac{1}{2}$ is the Maslov correction, contributing $\pi/4$ of phase from each soft turning point. (For a hard wall the correction at that wall changes.)

Equivalently, in terms of the closed classical orbit in phase space,

which is the historical Bohr–Sommerfeld statement: the enclosed phase-space area is quantized in units of Planck's constant.

Applying it to the harmonic oscillator

Take $V(x) = \tfrac{1}{2}m\omega^2 x^2$, in natural units $\hbar = m = \omega = 1$, so $V(x) = x^2/2$. A particle of energy $E$ turns around where $V = E$, i.e. at $x = \pm a$ with $a = \sqrt{2E}$. The momentum is $p(x) = \sqrt{2E - x^2}$, and the action integral is a standard one:

Setting this equal to $(n + \tfrac{1}{2})\pi$:

Restoring units, $E_n = (n + \tfrac{1}{2})\hbar\omega$ — the exact harmonic-oscillator spectrum.

Try it

Solve the quantization rule $\int_{-a}^{a}\sqrt{2E - x^2}\,dx = (n + \tfrac12)\pi$ for the $n = 2$ level and return the energy. You can integrate numerically and bisect on $E$, or use the closed form $\pi E = (n+\tfrac12)\pi$. Either way the answer is $2.5$.

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