Questions: Canonical Transformations and Generating Functions
4 questions to test your understanding
Score: 0 / 4
Question 1 Multiple Choice
For one degree of freedom, which transformation is canonical?
AQ = p, P = q
BQ = p, P = −q
CQ = 2q, P = 2p
DQ = q², P = p
Test {Q, P} = ∂Q/∂q ∂P/∂p − ∂Q/∂p ∂P/∂q. For Q = p, P = −q this is 0·0 − (1)(−1) = 1. The others give −1, 4, and 2q.
Question 2 True / False
The evolution of a Hamiltonian system from time t to time t + Δt is itself a canonical transformation of phase space.
TTrue
FFalse
Answer: True
Hamilton's equations generate a flow that preserves Poisson brackets. The Hamiltonian acts as the generator of this transformation, which is why phase-space volume is conserved (Liouville's theorem).
Question 3 Short Answer
Use the type-2 generating function F₂(q, P) = qP, with p = ∂F₂/∂q and Q = ∂F₂/∂P. What transformation results?
Think about your answer, then reveal below.
Model answer: The identity: Q = q and P = p.
∂F₂/∂q = P, so p = P. ∂F₂/∂P = q, so Q = q. This is why qP is the starting point for building transformations close to the identity.
Question 4 Multiple Choice
For the oscillator H = p²/2m + mω²q²/2, the transformation q = √(2P/mω) sin Q, p = √(2mωP) cos Q gives K = ωP. Why does this make the problem easy?
AK no longer depends on Q, so P is constant and Q grows linearly in time
BK no longer depends on P, so Q is constant
CK is zero, so nothing moves
DThe transformation removes the mass from the problem
Q is cyclic, so dP/dt = −∂K/∂Q = 0 and P = E/ω is constant. Then dQ/dt = ∂K/∂P = ω, so Q = ωt + φ. Substituting back gives the familiar sinusoidal motion.