# Question about canonical transformation

1. Dec 4, 2013

### mjordan2nd

I was going through my professor's notes about Canonical transformations. He states that a canonical transformation from (q, p) to (Q, P) is one that if which the original coordinates obey Hamilton's canonical equations than so do the transformed coordinates, albeit for a different Hamiltonian. He then considers, as an example the Hamiltonian

$$H=\frac{1}{2}p^2,$$

with a transformation

$$Q = q,$$
$$P = \sqrt{p} - \sqrt{q}.$$

The notes state that "this transformation is locally canonoid with respect to H," and that in the transformed coordinates the new Hamiltonian is

$$K = \frac{1}{3} \left( P + \sqrt{Q} \right)^3.$$

I don't understand how we know that this is locally canonical, or what it really even means to be locally canonical. Also, where do we get K from. Since the inverse transformation would be

$$q=Q$$
$$p=\left( P + \sqrt{Q} \right)^2$$

why isn't the new Hamiltonian

$$K= \frac{1}{2} \left(P + \sqrt{Q} \right)^4,$$

where all I've done is plug the inverted transformation into the original Hamiltonian. I'm a bit confused by all this. Would appreciate any help.

Thanks.