Questions: Hamilton-Jacobi Theory and Action-Angle Variables

Question 1

For a harmonic oscillator, the action variable is J = ∮p dq = 2πE/ω. What frequency ν does ∂H/∂J give?

• A. ν = ω
• B. ν = ω/2π
• C. ν = 2πω
• D. ν = E/J

Answer: (B)

Solving for the energy gives H = Jω/2π. Then ν = ∂H/∂J = ω/2π, the oscillation frequency in cycles per unit time.

Question 2

Evaluated along the actual path of a system, Hamilton's principal function S changes at the rate dS/dt = L, so S equals the action integral of the Lagrangian.

• T. True
• F. False

Answer: True

dS/dt = Σ(∂S/∂q)q̇ + ∂S/∂t = Σpq̇ − H = L along the true motion. Up to a constant, S is the action.

Question 3

The string of a swinging pendulum is shortened very slowly. Which quantity stays approximately constant?

• A. The energy E
• B. The frequency ν
• C. The ratio E/ν, which equals the action J
• D. The amplitude of the swing

Answer: (C)

For slow changes the action variable is an adiabatic invariant. For small oscillations J = E/ν, so E rises in step with ν as the string shortens.

Question 4

Apply the old quantum rule ∮p dq = nh to the harmonic oscillator. What energies does it give, and how do they differ from the quantum result?

Model answer: E = nhν = nħω. The full quantum result is (n + ½)ħω, so the old rule misses the zero-point energy ½ħω.

Setting J = 2πE/ω equal to nh gives E = nhω/2π = nħω. The WKB method corrects this by adding a half-integer term from the turning points.