3.3.3 Bohr-Sommerfeld Quantization#

Prompts

How does the requirement of a single-valued wavefunction lead to Bohr-Sommerfeld quantization \(\oint p\,\mathrm{d}x = 2\pi\hbar(n+\tfrac{1}{2})\)?
Where does the \(\tfrac{1}{2}\) (Maslov index) come from, and why does it give zero-point energy?
Apply Bohr-Sommerfeld to the harmonic oscillator — why is the result exact?
How does Bohr-Sommerfeld connect to the path integral picture: quantization as constructive interference of periodic orbits?

Lecture Notes#

Overview#

Bohr-Sommerfeld quantization applies the WKB approximation to bound states. Single-valuedness of the wavefunction after a complete classical orbit requires \(\oint p\,\mathrm{d}x = 2\pi\hbar(n + \tfrac{1}{2})\). The \(\tfrac{1}{2}\) arises from \(\pi/4\) phase shifts at each turning point — the Maslov index — and explains zero-point energy. For the harmonic oscillator, this gives the exact spectrum \(E_n = (n+\tfrac{1}{2})\hbar\omega\). Geometrically, quantization is constructive interference of periodic orbits.

Derivation from WKB#

For a bound state oscillating between turning points \(x_a\) and \(x_b\) (where \(E = V(x)\)), the WKB wavefunction must return to itself after one complete orbit.

The phase accumulated going from \(x_a\) to \(x_b\) is \(\frac{1}{\hbar}\int_{x_a}^{x_b} p(x)\,\mathrm{d}x\). Each turning point contributes an additional \(\pi/4\) phase shift from the connection formulas (§3.3.2). The total round-trip phase is

\[ \Phi_\mathrm{total} = \frac{2}{\hbar}\int_{x_a}^{x_b} p(x)\,\mathrm{d}x + 2 \times \frac{\pi}{4} + 2 \times \frac{\pi}{4} = \frac{1}{\hbar}\oint p\,\mathrm{d}x + \pi \]

Single-valuedness requires \(\Phi_\mathrm{total} = 2\pi n\), giving:

Bohr-Sommerfeld Quantization Rule

\[ \oint p(x)\,\mathrm{d}x = 2\pi\hbar\left(n + \frac{1}{2}\right), \quad n = 0, 1, 2, \ldots \]

where \(p(x) = \sqrt{2m(E - V(x))}\) and the integral is over the full classical orbit. The \(\tfrac{1}{2}\) is the Maslov correction from turning-point phase shifts.

Discussion: Semiclassical Quantization

The Bohr-Sommerfeld condition looks almost classical — an integral over a closed orbit. Yet it predicts quantum energy levels. This raises foundational questions:

The condition works for hydrogen but fails for helium. What changes when there are two interacting electrons?
The \(\tfrac{1}{2}\) gives zero-point energy. Is this purely quantum, or does it have a wave-optics analogue?
For chaotic (non-integrable) systems, there are no closed action variables. How does quantum mechanics handle the spectrum when semiclassical intuition breaks down?

Application: Harmonic Oscillator#

For \(V(x) = \frac{1}{2}m\omega^2 x^2\), the turning points are \(x = \pm a\) with \(a = \sqrt{2E/(m\omega^2)}\).

Example: Harmonic Oscillator Spectrum

The momentum is \(p(x) = m\omega\sqrt{a^2 - x^2}\). The orbit integral:

\[ \oint p\,\mathrm{d}x = 2\int_{-a}^{a} m\omega\sqrt{a^2 - x^2}\,\mathrm{d}x = \pi m\omega a^2 = \frac{2\pi E}{\omega} \]

Applying Bohr-Sommerfeld:

\[ \frac{2\pi E_n}{\omega} = 2\pi\hbar\left(n + \frac{1}{2}\right) \implies E_n = \hbar\omega\left(n + \frac{1}{2}\right) \]

This is exactly the quantum result. The harmonic oscillator is special because its quadratic potential makes WKB exact — all higher-order corrections vanish identically.

Application: Hydrogen Atom#

For the Coulomb potential \(V(r) = -e^2/(4\pi\epsilon_0 r)\), combining angular momentum quantization (\(L = n_\phi\hbar\)) with radial Bohr-Sommerfeld quantization reproduces the Bohr energy levels:

\[ E_n = -\frac{13.6\;\mathrm{eV}}{n^2}, \quad n = 1, 2, 3, \ldots \]

The result is exact due to a special cancellation of WKB errors in the Coulomb potential. For general potentials, Bohr-Sommerfeld is an excellent approximation that improves with increasing quantum number \(n\).

EBK Quantization

For systems with \(N\) degrees of freedom that are integrable (possessing \(N\) independent constants of motion), the Einstein-Brillouin-Keller (EBK) generalization quantizes each action variable separately:

\[ J_i = \oint p_i\,\mathrm{d}q_i = 2\pi\hbar(n_i + \alpha_i) \]

where \(\alpha_i\) is the Maslov index for each degree of freedom. For the 2D isotropic oscillator, this gives \(E = \hbar\omega(n_x + n_y + 1)\), exactly matching the quantum result. EBK fails for non-integrable (chaotic) systems where action variables do not exist.

Connection to the Path Integral#

Bohr-Sommerfeld quantization is a direct consequence of the path integral. The propagator for a bound state receives contributions from periodic orbits — paths that close on themselves after time \(T\). The condition for constructive interference among these periodic orbits is precisely \(\oint p\,\mathrm{d}x = 2\pi\hbar(n + \tfrac{1}{2})\).

Quantization = Constructive Interference

Discrete energy levels arise because only certain energies produce periodic orbits whose accumulated phase is a multiple of \(2\pi\). All other energies lead to destructive interference and are forbidden.

This completes 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 semiclassical wavefunctions; Bohr-Sommerfeld extracts discrete spectra from the constructive interference condition.

Summary#

Bohr-Sommerfeld rule: \(\oint p\,\mathrm{d}x = 2\pi\hbar(n+\tfrac{1}{2})\) quantizes bound-state energies from single-valuedness of the wavefunction
Maslov index: the \(\tfrac{1}{2}\) comes from \(\pi/4\) phase shifts at each turning point, giving zero-point energy
Exact for harmonic oscillator: \(E_n = \hbar\omega(n+\tfrac{1}{2})\); also exact for hydrogen by special cancellation
Path integral origin: quantization = constructive interference of periodic orbits, closing the semiclassical arc of Chapter 3

Homework#

1. (Bohr-Sommerfeld for Particle in a Box) A particle of mass \(m\) is confined in a 1D box of width \(a\) with infinite potential walls at \(x=0\) and \(x=a\).

(a) Inside the box, the potential is \(V(x) = 0\) and \(p(x) = \sqrt{2mE}\). Show that the Bohr-Sommerfeld quantization condition

\[\oint p(x) dx = 2\pi\hbar\left(n + \frac{1}{2}\right)\]

gives \(E_n = \frac{\pi^2\hbar^2 n^2}{2ma^2}\) (note the factor of \(n\), not \(n+1/2\)). Why does this differ from the exact quantum result by a factor in \(n\)?

(b) Explain why the WKB phase shifts at the hard walls are different from those at soft turning points. What is the correct Maslov index for this system?

2. (Linear Potential) Consider a particle in a linear potential \(V(x) = F x\) for \(x \geq 0\) and \(V(x) = \infty\) for \(x < 0\) (Airy potential).

(a) At energy \(E\), the turning point is at \(x_a = E/F\). Show that

\[\int_0^{x_a} p(x) dx = \int_0^{E/F} \sqrt{2m(E - Fx)} dx = \frac{2}{3F}(2mE)^{3/2}\]

(b) Apply the Bohr-Sommerfeld rule to find the energy levels \(E_n\). How do the spacings \(\Delta E_n = E_{n+1} - E_n\) scale with \(n\)?

3. (Beyond Bohr-Sommerfeld — WKB Corrections) The next-order correction to Bohr-Sommerfeld arises from higher-order WKB terms. For a potential with two turning points \(x_a\) and \(x_b\), the leading correction is approximately

\[\frac{1}{\hbar}\int_{x_a}^{x_b} p(x) dx \approx \pi\left(n + \frac{1}{2}\right) + \frac{1}{24}\left(\frac{1}{p(x_a)} \frac{\mathrm{d}V}{dx}\bigg|_{x_a} - \frac{1}{p(x_b)} \frac{\mathrm{d}V}{dx}\bigg|_{x_b}\right)\]

For the harmonic oscillator \(V(x) = \frac{1}{2}m\omega^2 x^2\):

(a) Compute \(\frac{\mathrm{d}V}{dx}\) at the turning points \(x = \pm a = \pm\sqrt{2E/(m\omega^2)}\) and \(p(\pm a) = 0\). What goes wrong with the correction formula at the turning points?

(b) Explain why Bohr-Sommerfeld is exact for the harmonic oscillator and doesn’t require WKB corrections. (Hint: think about which WKB orders contribute.)

4. (Anharmonic Oscillator — Qualitative Analysis) Consider a potential \(V(x) = \frac{1}{2}m\omega^2 x^2 + \lambda x^4\) with \(\lambda > 0\) (quartic anharmonicity).

(a) For small \(\lambda\), the turning points are approximately at \(x_{\pm} \approx \pm\sqrt{2E/(m\omega^2)}\) (same as harmonic oscillator). Explain why the Bohr-Sommerfeld integral will differ from the harmonic result.

(b) Will the energy levels of the anharmonic oscillator be more or less spaced than the harmonic oscillator? Use a physical argument (does the fourth-order term make the well steeper or shallower?).

(c) For which energy range (low \(n\), high \(n\), or both) will Bohr-Sommerfeld for the anharmonic oscillator be less accurate than for the harmonic oscillator? Why?

5. (EBK Quantization — 2D Anisotropic Oscillator) A particle moves in a 2D anisotropic harmonic potential

\[V(x, y) = \frac{1}{2}m\omega_x^2 x^2 + \frac{1}{2}m\omega_y^2 y^2\]

with \(\omega_x \neq \omega_y\). The system is integrable with action variables \(J_x\) and \(J_y\).

(a) For the \(x\)-direction alone, \(\oint p_x dx = 2\pi\hbar(n_x + 1/2)\). Similarly for \(y\). Find the total energy \(E_{n_x, n_y}\) in terms of quantum numbers \(n_x, n_y\) and the frequencies.

(b) Suppose \(\omega_x = 2\omega_y = 2\omega\). Find the first five energy levels (in units of \(\hbar\omega\)) and identify any degeneracies.

(c) Now suppose \(\omega_x/\omega_y = \sqrt{2}\) (irrational). Explain why EBK quantization can still be applied but the system may exhibit complicated dynamics if perturbed. Does irrationality of frequency ratio imply chaos?

6. (Maslov Index and Hard Boundaries) The Maslov index \(\mu\) corrects the phase shift at turning points. For a soft turning point (where \(V(x) = E\)), \(\mu = 1/2\). For a hard wall (infinite potential), \(\mu = 1\).

Consider a particle in potential \(V(x) = V_0\) for \(0 < x < a\) and \(V(x) = \infty\) elsewhere, with total energy \(E > V_0\).

(a) Inside the potential well, \(p(x) = \sqrt{2m(E - V_0)}\). Write down the Bohr-Sommerfeld condition with correct Maslov indices for both hard walls.

(b) Show that the quantization condition simplifies to

\[k a = n\pi, \quad k = \sqrt{2m(E - V_0)}/\hbar\]

and thus \(E_n = V_0 + \frac{n^2\pi^2\hbar^2}{2ma^2}\). Compare to Problem 1.

7. (Hydrogen via Bohr-Sommerfeld (Circular Orbits)) In the Bohr model, a circular orbit with radius \(r\) and angular momentum \(L = m v r = n_\phi\hbar\) is stable. For the Coulomb potential \(V(r) = -e^2/(4\pi\epsilon_0 r)\), the centripetal force condition is

\[\frac{m v^2}{r} = \frac{e^2}{4\pi\epsilon_0 r^2}\]

(a) Show that the circular orbit energy is \(E = -\frac{e^2}{8\pi\epsilon_0 r}\) (half the potential energy, by the virial theorem).

(b) For an azimuthal quantization with \(n_\phi\hbar\), use the centripetal condition to eliminate \(v\) and show that

\[r_{n_\phi} = \frac{4\pi\epsilon_0 n_\phi^2 \hbar^2}{m e^2} = n_\phi^2 a_0\]

where \(a_0\) is the Bohr radius.

(c) Find the energy \(E_{n_\phi}\) for each azimuthal quantum number. Why is the hydrogen atom result often written as \(E_n = -13.6 \text{ eV}/n^2\) with a principal quantum number \(n\) that sums all degrees of freedom?

8. (Classical Limit and Correspondence Principle) In the classical limit (\(n \gg 1\)), the energy spacing \(\Delta E = E_{n+1} - E_n\) should match the classical frequency \(\omega_\text{cl}\) via \(\Delta E = \hbar\omega_\text{cl}\) (Bohr correspondence).

(a) For the harmonic oscillator, \(E_n = \hbar\omega(n + 1/2)\), so \(\Delta E = \hbar\omega\). Show that this agrees with \(\hbar\omega_\text{cl}\) where \(\omega_\text{cl}\) is the classical frequency of oscillation.

(b) For a power-law potential \(V(x) = C|x|^\alpha\) (e.g., \(\alpha = 2\) for harmonic, \(\alpha = 4\) for quartic), the Bohr-Sommerfeld result is \(E_n \propto n^{2\alpha/(\alpha+2)}\). Show that \(\Delta E / E_n \to 0\) as \(n \to \infty\), implying the energy levels become quasi-continuous—the classical limit.

(c) Estimate the quantum number \(n^*\) above which Bohr-Sommerfeld should be accurate (say, within \(10\%\)) for a typical atomic potential. Use the condition \(\hbar\omega / E \lesssim 0.1\) where \(\omega\) is a characteristic frequency scale.