# Find canonical transformation for $P$ given canonical transformation for $Q$

I have just started to study canonical transformations and I am trying to solve this exercise. Given the following canonical transformation for $$Q$$, $$Q = q^{-2}$$ I have to find the corresponding canonical transformation for $$P$$. The determinant of the matrix $$M = \begin{pmatrix} -\frac{2}{q^3} & 0 \\ \frac{\partial P}{\partial q} & \frac{\partial P}{\partial p} \end{pmatrix}$$ has to be $$1$$, from which it follows that $$\frac{\partial P}{\partial p} = -\frac{q^3}{2} \iff P = -\frac{q^3p}{2}+k(q)$$ My question is: can I just take $$k(q)=0$$ to solve the problem or is there a unique canonical transformation given that $$Q$$? Are there any more conditions that $$P$$ needs to verify in order to be the requested canonical transformation?

• Have you tried the method of generating functions for canonical transformations? It is not so easy to get the correct behaviour and so I really only trust answers going via that route because it is simple. May 10, 2023 at 9:56
• What's holding you back? What's spooking you? you preserved the phase-space volume, didn't you? May 10, 2023 at 12:28

1. If we assume that a canonical transformation here means a symplectomorphism, then indeed OP's method is correct: the determinant of the Jacobian is 1, and it is easy to check that OP's solution is a symplectomorphism. We should keep the "integration constant" $$k(q)$$, which may be determined by possible further conditions.
2. If $$q\to Q=q^{-2}$$ is viewed as a point transformation of the base manifold $$M$$ of a cotangent bundle $$T^{\ast}M$$, then the momentum transforms as components of a covector $$p~=~\frac{\partial Q}{\partial q}P~=~-\frac{2}{q^3}P,$$ i.e. $$k(q)=0$$.