GATE PH 2025 • Q53 MSQ • 2 Marks

Section 2: Classical Mechanics

Lagrangian & Hamiltonian Dynamics

Subtopic: Poisson Brackets & Canonical Transformations

Question Statement:

In a one -dimensional Hamiltonian system with position q and momentum p, consider the canonical transformation (q , p) → (Q = 1 p , P = q p2), where Q and P are the new position and momentum , respectively. Which of the following option(s) regarding the generating function F is/are correct?

Options:

  • (A) F = F1(q, Q) = q Q
  • (B) F = F2(q, P) = √Pq
  • (C) F = F3(p, Q) = 2 p Q
  • (D) F = F4(p, P) = P p

Official Answer Key:

A;D

Step-by-Step Solution & Explanation

### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.53)** **1. Canonical Transformation & Generating Functions**: - Transformation: $(q, p) \to (Q = 1/p, P = q p^2)$. **2. Analysis of Generating Functions (MSQ)**: - **Option (A) $F_1(q, Q) = \frac{q}{Q}$ [TRUE]**: - Canonical equations for Type-1: $p = \frac{\partial F_1}{\partial q} = \frac{\partial}{\partial q}\left(\frac{q}{Q}\right) = \frac{1}{Q} \implies Q = \frac{1}{p} \quad (\text{Matches!})$ $P = -\frac{\partial F_1}{\partial Q} = -\left(-\frac{q}{Q^2}\right) = \frac{q}{Q^2} = q\left(\frac{1}{p}\right)^{-2} = q p^2 \quad (\text{Matches!})$ - **Option (D) $F_4(p, P) = \frac{P}{p}$ [TRUE]**: - Canonical equations for Type-4: $Q = \frac{\partial F_4}{\partial P} = \frac{\partial}{\partial P}\left(\frac{P}{p}\right) = \frac{1}{p} \quad (\text{Matches!})$ $q = -\frac{\partial F_4}{\partial p} = -\left(-\frac{P}{p^2}\right) = \frac{P}{p^2} \implies P = q p^2 \quad (\text{Matches!})$ **3. Official Verification**: The verified official IIT answer is **Options (A) and (D)** (`['A', 'D']`).

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