Question about canonical transformation

[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

[tex]H=\frac{1}{2}p^2,[/tex]

with a transformation

[tex]Q = q,[/tex]
[tex]P = \sqrt{p} - \sqrt{q}.[/tex]

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

[tex] K = \frac{1}{3} \left( P + \sqrt{Q} \right)^3.[/tex]

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

[tex]q=Q[/tex]
[tex]p=\left( P + \sqrt{Q} \right)^2[/tex]

why isn't the new Hamiltonian

[tex]K= \frac{1}{2} \left(P + \sqrt{Q} \right)^4,[/tex]

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.

 

[Valerie Park]

To say that the transformation is "locally canonical" with respect to H means that it preserves the canonical structure of the system, i.e. the Poisson brackets between the original coordinates q and p are preserved in the transformed coordinates Q and P. This property ensures that the equations of motion in the new coordinates are still Hamilton's equations, and therefore that the new Hamiltonian K is a valid one. The reason why the new Hamiltonian takes the form you have shown is that this is the most general form of the Hamiltonian compatible with the transformation given. We can determine this by noting that the transformation can be written asP = f(q,p)where f is some function of q and p. The new Hamiltonian K must then be a function of just Q and P, i.e.K = g(Q,P)Substituting in the expression for P, we getK = g(Q,f(q,p))Expanding this out, we getK = g(Q,f(Q,P + \sqrt{Q}))Since f is some arbitrary function, the most general form of K compatible with this transformation isK = g(Q,P + \sqrt{Q})Plugging in the expression for the original Hamiltonian H, we getK = \frac{1}{2} \left(P + \sqrt{Q} \right)^2Finally, since the transformation is canonical, we know that K must be a homogeneous function of degree two, i.e.K = \frac{1}{3} \left(P + \sqrt{Q} \right)^3Hope this helps.