Question

In a one-dimensional Hamiltonian system with position \( q \) and momentum \( p \), consider the canonical transformation \( (q, p) \to (Q = \frac{1}{p}, P = q p^2) \), 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?

• A. \( F = F_1(q, Q) = \frac{q}{Q} \)
• B. \( F = F_2(q, P) = \sqrt{Pq} \)
• C. \( F = F_3(p, Q) = \frac{2}{pQ} \)
• D. \( F = F_4(p, P) = \frac{P}{p} \)

Hint:

The generating function in a canonical transformation helps define the new variables in terms of the old ones. It satisfies the partial derivative relations with respect to the original coordinates and momenta.

Answer 1

The Correct Option is A, D

Step 1: The canonical transformation is given by the relations: \[ Q = \frac{1}{p}, \quad P = q p^2. \] The generating function \( F \) satisfies the following relations: \[ \frac{\partial F}{\partial q} = P, \quad \frac{\partial F}{\partial p} = -Q. \] Step 2: For \( F_1(q, Q) \), we choose the generating function such that: \[ \frac{\partial F_1}{\partial q} = P = \frac{q}{Q}. \] This satisfies the relation, and thus: \[ F_1(q, Q) = \frac{q}{Q}. \] Step 3: For \( F_4(p, P) \), we need to satisfy: \[ \frac{\partial F_4}{\partial p} = -Q \quad \Rightarrow \quad F_4(p, P) = \frac{P}{p}. \]