Questions: WKB Quantization and Bohr-Sommerfeld Rule

Question 1

Using the Bohr-Sommerfeld rule ∮ p dx = (n + ½)πℏ, what is the ground-state energy (n=0) of a harmonic oscillator with angular frequency ω?

• A. 0 — no energy at the lowest quantum state
• B. ℏω/2 — a non-zero zero-point energy
• C. ℏω — one full quantum of energy
• D. The rule cannot be applied to a harmonic oscillator

Answer: (B)

With n=0, the rule gives ∮ p dx = ½πℏ = ½h, yielding E₀ = ℏω/2. This non-zero zero-point energy is a direct consequence of the ½ in the quantization condition. The original Bohr rule (∮ p dx = nh) with n=0 gives zero, which is wrong — it predicts the oscillator can be completely at rest, violating the uncertainty principle.

Question 2

The ½ in the Bohr-Sommerfeld rule ∮ p dx = (n + ½)πℏ originates from which physical effect?

• A. The kinetic energy averaging over a half-cycle of the classical orbit
• B. Phase shifts of π/2 at each of the two classical turning points
• C. A relativistic correction to the non-relativistic momentum
• D. Spin degeneracy of the electron states

Answer: (B)

At each turning point, the WKB wavefunction must be matched across the classically forbidden region using connection formulas. Each turning point contributes a phase shift of π/2 (a quarter-wavelength). With two turning points per orbit, the total extra phase is π, which translates into the ½ in the quantization rule. This is called the Maslov index contribution.

Question 3

Bohr's original semiclassical quantization rule ∮ p dx = nh correctly predicts the harmonic oscillator energy spectrum.

• T. True
• F. False

Answer: False

Bohr's original rule predicts energies Eₙ = nℏω (with n = 1, 2, 3, ...) and incorrectly predicts zero ground-state energy for n = 0. The modern Bohr-Sommerfeld rule ∮ p dx = (n + ½)πℏ gives Eₙ = (n + ½)ℏω, which matches the exact quantum mechanical result. The ½ correction comes from phase shifts at the turning points — a quantum effect that the original Bohr model neglected.

Question 4

The WKB approximation breaks down near classical turning points because the semiclassical condition λ|dp/dx| ≪ p² is violated there.

• T. True
• F. False

Answer: True

At a turning point, the classical momentum p(x) → 0 while dp/dx remains finite. This makes the ratio λ|dp/dx|/p² diverge, violating the condition that the de Broglie wavelength changes slowly compared to p itself. This is why special 'connection formulas' (Airy functions) are needed to stitch the WKB solutions across turning points — the WKB form alone is not valid there.

Question 5

Explain in physical terms why the Bohr-Sommerfeld quantization rule contains the term (n + ½) rather than simply n.

Model answer: The ½ accounts for the quantum-mechanical phase shifts that occur at the two classical turning points. Each turning point contributes a phase advance of π/2 to the wavefunction as it penetrates into the classically forbidden region and is reflected. Two turning points give a total extra phase of π, equivalent to half a quantum of action, which adds the ½ to the quantization condition.

This Maslov index correction distinguishes the semiclassical WKB theory from Bohr's cruder model. Its practical consequence is the zero-point energy ℏω/2 of the harmonic oscillator — the quantum system cannot sit at the bottom of the potential well because the wavefunction must have minimum curvature consistent with the boundary conditions. The ½ is not a correction to the orbit — it is a statement that the quantum wavefunction 'feels' the turning points as reflecting walls with phase delay.