GATE 2025 PH – Question 53

Correct answer: (A) $F_1(q,Q)=q/Q$; (D) $F_4(p,P)=P/p$

Explanation

For a canonical transformation $(q,p)\to(Q,P)=(1/p,\ qp^2)$, test each generating function with its defining relations.

**$F_1(q,Q)$:** $p=\dfrac{\partial F_1}{\partial q}$, $P=-\dfrac{\partial F_1}{\partial Q}$.
- For $F_1=q/Q$: $p=\dfrac1Q$, so $Q=1/p$ ✓. $P=-\dfrac{\partial}{\partial Q}\left(\dfrac qQ\right)=\dfrac{q}{Q^2}=qp^2$ ✓. **Correct.**

**$F_4(p,P)$:** $q=-\dfrac{\partial F_4}{\partial p}$, $Q=\dfrac{\partial F_4}{\partial P}$.
- For $F_4=P/p$: $q=\dfrac{P}{p^2}$, so $P=qp^2$ ✓. $Q=\dfrac1p$ ✓. **Correct.**

**$F_2(q,P)$:** requires $p=\dfrac{\partial F_2}{\partial q}$ and $Q=\dfrac{\partial F_2}{\partial P}$. For $F_2=\sqrt{Pq}$ these give $p=\tfrac12\sqrt{P/q}$, which does not reproduce $P=qp^2$ (it gives $P=4qp^2$). ✗

**$F_3(p,Q)$:** for $F_3=2p/Q$ the relations $q=-\partial F_3/\partial p=-2/Q$ and $P=-\partial F_3/\partial Q=2p/Q^2$ do not match the transformation. ✗

Answer **A and D**.