homework and exercises - Question about canonical transformation

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 then 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 canonical 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? Considering that the inverse transformation would be:

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

Why isn't the new Hamiltonian this:

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

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.

Answer

The original coordinates satisfy the equations of motion when the integral of $p\, \dot{q} - H(p,q)$ is minimized, and the new coordinates satisfy the equations of motion when the integral of $P\, \dot{Q} - K(P,Q)$ is minimized. There is no requirement that $H$ and $K$ be numerically equal.

The transformation is canonical if the Poisson bracket remains invariant.

The EOMs are

$\dot{p} = 0$

$\dot{q} = p$

and from the new Hamiltonian, we get

$\dot{P} = -(P+\sqrt{Q})^2 \frac{1}{2\sqrt{Q}} = - \frac{p}{2\sqrt{q}} = \frac{d}{dt} \left(\sqrt{p} - \sqrt{q} \right)$

$\dot{Q} = (P+\sqrt{Q})^2 = \dot{q}$

thus the equations of motion are numerically equal.