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?
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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. – naturallyInconsistent May 10 '23 at 09:56
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What's holding you back? What's spooking you? you preserved the phase-space volume, didn't you? – Cosmas Zachos May 10 '23 at 12:28
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TL;DR: The answer depends on the context.
Examples:
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.
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. $k(q)=0$.
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