3.3.3 Bohr-Sommerfeld Quantization#
Prompts
How does the requirement of single-valuedness for the WKB wavefunction over a closed classical orbit lead to Bohr-Sommerfeld quantization \(\oint p\,\mathrm{d}x = 2\pi\hbar(n + 1/2)\)?
Where does the \(1/2\) (Maslov correction) come from, and why does it give zero-point energy?
Apply Bohr-Sommerfeld to the harmonic oscillator. Why is the result exact, and what spectrum does the same rule give for hard walls?
How is Bohr-Sommerfeld the path-integral statement of “constructive interference of periodic orbits”?
Lecture Notes#
Overview#
Apply the WKB connection formulas of §3.3.2 to a bound particle. The wave traverses the classical orbit between two turning points and must return to itself: this single-valuedness condition picks out discrete energies. The result is the Bohr-Sommerfeld rule \(\oint p\,\mathrm{d}x = 2\pi\hbar(n + 1/2)\), where the \(1/2\) — the Maslov correction from connection-formula phase shifts — gives the zero-point energy. We close the chapter by showing that this rule is precisely the path-integral statement that periodic orbits interfere constructively with themselves.
The Quantization Rule#
A particle of energy \(E\) trapped between two soft turning points \(x_a < x_b\) oscillates classically in \(x_a\le x\le x_b\). Quantum mechanically, the WKB wavefunction must return to itself after one complete classical orbit — a single-valuedness condition that picks out discrete energies \(E_n\).
From §3.3.2, each soft turn contributes an effective extra phase \(\pi/2\). A closed orbit has two turns (one at each end), so the round trip carries a total extra phase of \(\pi\) beyond the bare WKB action \(\oint p\,\mathrm{d}x/\hbar\). Single-valuedness of the wavefunction requires the total phase per orbit to be a multiple of \(2\pi\):
which rearranges into the Bohr-Sommerfeld rule.
Bohr-Sommerfeld quantization
For a 1D bound state with two soft turning points:
The integral is around the full classical orbit at energy \(E_n\), and \(p(x) = \sqrt{2m(E_n - V(x))}\).
The Maslov correction \(\tfrac{1}{2}\) is bookkeeping for the connection phases accumulated on one full classical orbit. Each soft turning point contributes \(\pi/2\) from the Airy matching (§3.3.2); with two such ends the round trip picks up an extra \(\pi\) beyond the semiclassical phase \(\oint p\,\mathrm{d}x/\hbar\). Requiring the total orbit phase to change in units of \(2\pi\) is the same as absorbing that extra \(\pi\) as a shift \(\tfrac{1}{2}\) in the quantum number, which is (114).
If the two boundaries are not the same type, package the end contributions with Maslov indices \(\mu_a\) and \(\mu_b\) (our convention: \(\mu=\tfrac{1}{2}\) at a soft turning point, \(\mu=1\) at a hard wall where \(\psi=0\)). Then
The table below lists phase shifts and the Maslov indices in common 1D geometries.
Phase-shift counting at a glance
Process |
Phase shift |
Maslov index |
|---|---|---|
Soft matching |
\(\pi/4\) |
— |
Soft turn |
\(\pi/2\) |
\(\mu = \tfrac{1}{2}\) |
Hard turn |
\(\pi\) |
\(\mu = 1\) |
Closed orbit (two soft turns) |
\(\pi\) |
\(\mu_a=\mu_b=\tfrac{1}{2}\) |
Closed orbit (two hard walls) |
\(2\pi\equiv 0\) |
\(\mu_a=\mu_b=1\) |
Closed orbit (ring, no turn) |
\(0\) |
\(0\) |
Bohr-Sommerfeld follows from setting (orbit phase) \(+\) (bare action \(\oint p\,\mathrm{d}x/\hbar\)) \(= 2\pi n\). Two soft turns give the \((n+\tfrac12)\) Maslov correction; two hard walls or a ring give pure integer \(n\) (because \(2\pi\equiv 0\)).
Quantization = constructive interference
Discrete energy levels arise because only certain energies make the classical periodic orbit interfere constructively with itself under repeated traversal. All other energies dephase and are forbidden as quantum eigenstates.
Discussion: WKB exactness and surprises
WKB is an approximation, yet it gives the exact energy spectrum of the harmonic oscillator (next section) and of hydrogen. For which potentials is WKB exact, and what hidden structure makes that possible? When does WKB fail dramatically?
Application — Harmonic Oscillator#
For \(V(x) = \tfrac{1}{2}m\omega^{2}x^{2}\), the turning points at energy \(E\) are \(x = \pm a\) with \(a = \sqrt{2E/(m\omega^{2})}\).
Example: harmonic oscillator
The momentum is \(p(x) = m\omega\sqrt{a^{2} - x^{2}}\). The orbit integral is
Applying (114):
This matches the exact quantum spectrum, including the zero-point energy. The harmonic oscillator is special because its quadratic potential makes WKB exact at all orders.
The same rule predicts the exact hydrogen spectrum \(E_{n} = -13.6\,\text{eV}/n^{2}\) via a special cancellation in the Coulomb potential (HW 3.3.3.6).
This closes the semiclassical arc of Chapter 3: the path integral (§3.2) sums over all paths; stationary phase (§3.3.1) selects the classical path; WKB (§3.3.2) builds the corresponding wavefunction; Bohr-Sommerfeld extracts discrete spectra from constructive interference of periodic orbits.
Discussion: when does Bohr-Sommerfeld fail?
The rule works exactly for the harmonic oscillator and (via a separate cancellation) for hydrogen. It fails for helium and for chaotic systems where no closed action variable exists. Why is integrability — the existence of as many conserved quantities as degrees of freedom — required for a clean Bohr-Sommerfeld-type rule? What happens to the periodic-orbit sum when classical orbits proliferate exponentially?
Poll: physical content of Bohr-Sommerfeld
The Bohr-Sommerfeld rule (114) quantizes bound-state energies via \(\oint p\,\mathrm{d}x = 2\pi\hbar(n + 1/2)\). Which statement most precisely captures why this is a quantization rule?
(A) Only certain energies make the WKB standing wave single-valued after one full classical orbit.
(B) The classical action is an adiabatic invariant, so quantizing it makes the spectrum discrete by construction.
(C) The integral \(\oint p\,\mathrm{d}x\) counts the number of de Broglie wavelengths that fit around the orbit — quantization picks energies for which a half-integer number fit.
(D) Both (A) and (C).
Summary#
Bohr-Sommerfeld rule: \(\oint p\,\mathrm{d}x = 2\pi\hbar(n + 1/2)\) for two soft turning points; the integral is over the full classical orbit at energy \(E_{n}\).
Maslov correction \(1/2\): each soft turn contributes \(\pi/2\); two turns sum to \(\pi\); divided by the \(2\pi\) single-valuedness unit gives \(1/2\). This is the zero-point energy.
Other boundaries: the phase-counting table covers hard walls (\(\pi\) per wall) and rings (no turn); soft+hard mixes follow from (115).
Exact for harmonic oscillator (HO is special because \(V\) is quadratic) and for hydrogen (Coulomb cancellation).
Path-integral meaning: discrete energies are precisely the energies at which a classical periodic orbit interferes constructively with itself under repeated traversal.
See Also
3.3.2 WKB Approximation: Turning points, connection formulas, and quantization integrals that lead to the Bohr–Sommerfeld rule used here.
3.2.1 Path Integral Formulation: Semiclassical constructive interference on periodic orbits—the path-integral reading stated in this lesson’s Summary.
3.2.3 Free Particle Propagator: Exact quadratic path integral as a worked oscillatory integral before turning-point matching.
Homework#
1. Harmonic oscillator. For the harmonic oscillator \(V(x) = \frac{1}{2}m\omega^{2}x^{2}\), the classical turning points at energy \(E\) are \(x = \pm a\) with \(a = \sqrt{2E/(m\omega^{2})}\).
(a) Compute the orbit integral \(\oint p\,\mathrm{d}x = 2\int_{-a}^{a}\sqrt{2m(E - \tfrac{1}{2}m\omega^{2}x^{2})}\,\mathrm{d}x\) and show that it equals \(2\pi E/\omega\).
(b) Apply the Bohr-Sommerfeld rule (114) to derive \(E_{n} = \hbar\omega(n + 1/2)\). Note that this is the exact quantum result.
2. Particle in a box. A particle of mass \(m\) is confined between hard walls at \(x = 0\) and \(x = a\). Inside the box \(V = 0\), so \(p = \sqrt{2mE}\).
(a) For hard walls, the Maslov index is \(\mu = 1\) per wall (versus \(\mu = 1/2\) at a soft turning point), so use the generalized rule (115) with \(\mu_{a} = \mu_{b} = 1\).
(b) Show that this gives \(E_{n} = (n+1)^{2}\pi^{2}\hbar^{2}/(2ma^{2})\) for \(n = 0, 1, 2,\ldots\), which is the exact spectrum (with the conventional relabeling \(n\to n-1\)).
(c) Explain physically why the hard-wall Maslov index differs from the soft case: what does the wave do at a rigid boundary versus at a smooth turning point?
3. Linear potential. A particle in a half-line gravitational-like potential \(V(x) = Fx\) for \(x \geq 0\), with a hard wall at \(x = 0\).
(a) At energy \(E\), find the soft turning point \(x_{a} = E/F\) and compute \(\int_{0}^{x_{a}}\sqrt{2m(E - Fx)}\,\mathrm{d}x = \tfrac{2}{3F}(2mE)^{3/2}\).
(b) Write the Bohr-Sommerfeld rule with the correct Maslov indices (\(\mu = 1\) at the hard wall, \(\mu = 1/2\) at the soft turning point) and solve for \(E_{n}\).
(c) Show that the energy spacing \(\Delta E_{n}\) decreases with \(n\), unlike the harmonic oscillator. Why does a linear potential produce non-uniform level spacing?
4. Anharmonic oscillator. Consider \(V(x) = \tfrac{1}{2}m\omega^{2}x^{2} + \lambda x^{4}\) with \(\lambda > 0\) small.
(a) Explain qualitatively why the quartic term steepens the potential at large \(\vert x\vert\), shrinking the classical turning points compared to the pure harmonic case at the same energy.
(b) Argue that the orbit integral \(\oint p\,\mathrm{d}x\) is therefore smaller than for the harmonic oscillator at the same energy, so Bohr-Sommerfeld predicts higher energy levels. Are the energy spacings larger or smaller than \(\hbar\omega\)?
(c) For which regime (low \(n\) or high \(n\)) is the quartic correction most important? Use the ratio \(\lambda x^{4}/(\tfrac{1}{2}m\omega^{2}x^{2})\) at the turning point.
5. Two-dimensional anisotropic oscillator. A particle moves in \(V(x,y) = \tfrac{1}{2}m\omega_{x}^{2}x^{2} + \tfrac{1}{2}m\omega_{y}^{2}y^{2}\) with \(\omega_{x} \neq \omega_{y}\).
(a) The motion separates: \(x\) and \(y\) each oscillate independently. Apply Bohr-Sommerfeld to each direction to derive \(E_{n_{x},n_{y}} = \hbar\omega_{x}(n_{x} + \tfrac{1}{2}) + \hbar\omega_{y}(n_{y} + \tfrac{1}{2})\).
(b) For \(\omega_{x} = 2\omega_{y} = 2\omega\), list the first five distinct energy levels in units of \(\hbar\omega\) and identify any degeneracies.
(c) The same construction fails for non-separable (in particular chaotic) potentials. In one sentence, explain why the rule “quantize each closed action variable independently” requires the system to be integrable (as many conserved quantities as degrees of freedom).
6. Hydrogen atom (circular orbits). For the Coulomb potential \(V(r) = -q_{e}^{2}/(4\pi\epsilon_{0}r)\), consider circular orbits with angular momentum \(L = n_{\varphi}\hbar\).
(a) From the force balance \(mv^{2}/r = q_{e}^{2}/(4\pi\epsilon_{0}r^{2})\) and \(L = mvr = n_{\varphi}\hbar\), show that \(r_{n_{\varphi}} = n_{\varphi}^{2}a_{0}\) where \(a_{0} = 4\pi\epsilon_{0}\hbar^{2}/(m q_{e}^{2})\) is the Bohr radius.
(b) Use the virial theorem (\(E = -T = V/2\) for the Coulomb potential) to derive \(E_{n_{\varphi}} = -13.6\,\text{eV}/n_{\varphi}^{2}\).
(c) Adding the radial Bohr-Sommerfeld quantization with a radial quantum number \(n_{r}\) promotes \(n_{\varphi}\) to \(n = n_{r} + n_{\varphi}\). Explain why this reproduces the same formula \(E_{n} = -13.6\,\text{eV}/n^{2}\) with additional degeneracy.
7. Correspondence principle. In the classical limit (\(n \gg 1\)), the energy spacing \(\Delta E = E_{n+1} - E_{n}\) should equal \(\hbar\omega_\text{cl}\), where \(\omega_\text{cl}\) is the classical orbital frequency.
(a) For the harmonic oscillator, verify \(\Delta E = \hbar\omega\) for all \(n\).
(b) For a power-law potential \(V(x) = C\vert x\vert^{\alpha}\), Bohr-Sommerfeld gives \(E_{n} \propto n^{2\alpha/(\alpha+2)}\). Show that \(\Delta E/E_{n}\to 0\) as \(n\to\infty\), so the spectrum becomes quasi-continuous in the classical limit.
(c) For hydrogen (\(E_{n}\propto -1/n^{2}\)), compute the classical orbital frequency \(\omega_\text{cl} = \Delta E/\hbar\) at large \(n\) and show it scales as \(n^{-3}\). This is the radiation frequency in transitions between adjacent high-\(n\) Rydberg levels.