Change in Hamiltonian under Non-Canonical transformation

[Saptarshi Sarkar]

Homework Statement

Check whether the following transformation $[(q,p)]\rightarrow (Q,P)]$ is canonical for not.

$$Q=\frac{1}{q}, P=pq^2$$

If the Hamiltonian of the system is given by

$$H=\frac{p^2}{2m}+\frac12kq^2$$

(where m is the mass of the particle, k is a constant), determine the form of the Hamiltonian in the transformed coordinates (Q,P).

Relevant Equations

$$Q=\frac{1}{q}, P=pq^2$$

I know that if the transformation was canonical, the form of Hamilton's equation would remain invariant. If the generating function for the transformation was time independent, then the Hamiltonian would be invariant and we could directly replace q and p with the transformation equations to get the new Hamiltonian.

But here, I found that the transformation is not canonical. I was not sure how to get the new Hamiltonian. I knew that the Lagrangian must be invariant, so I tried to find the new Hamiltonian by going through the Lagrangian.

I do not know if what I did is correct and if there is another simpler way to solve the problem.

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My attempt is shared below

The given transformation equations are:
$$Q=\frac{1}{q}, P=pq^2$$

If the transformation is canonical, the Poisson bracket must be invariant and should be equal to 1.
The Poisson bracket wrt q,p for the given transformation is:

$$[Q,P]=\frac{-q^2}{q^2} = -1$$

So, the transformation is not canonical.

The given Hamiltonian is:

$$H=\frac{p^2}{2m}+\frac12kq^2$$

As the transformation is not a restricted canonical transformation, H is not form invariant. So, I tried to find the new Hamiltonian by going through the Lagrangian.

The corresponding Lagrangian should be:

$$L=\frac{p^2}{2m}-\frac12kq^2$$

The inverse transformations are:

$$q=\frac1Q,p=PQ^2$$

As the Lagrangian is invariant, we get the new Lagrangian by plugging in the above equations.

$$L'=\frac{P^2Q^4}{2m}-\frac12\frac{k}{Q^2}$$

We now get the new Hamiltonian through a Legendre transformation of the Lagrangian

$$H'=P\dot Q-\frac{P^2Q^4}{2m}+\frac12\frac{k}{Q^2}$$

Now, $$\dot Q=\frac{-\dot q}{q^2}=\frac{-p}{mq^2}=\frac{-PQ^4}{m}$$

using this, we get the new Hamiltonian

$$H'=-\frac{3P^2Q^4}{2m}+\frac12\frac{k}{Q^2}$$

 

[mark726]

Thus, the new Hamiltonian is not equal to the original Hamiltonian and the transformation is not canonical. The method used to find the new Hamiltonian is correct, but it is not a simpler way to solve the problem. A simpler way to find the new Hamiltonian in this case would be to directly use the transformation equations to express q and p in terms of Q and P, and then substitute them into the original Hamiltonian to get the new Hamiltonian. However, this method may not work for more complex transformations and in such cases, going through the Lagrangian would be a more reliable approach.