Poisson brackets for simple harmonic oscillator

[Siberion]

Homework Statement

Considering the Hamiltonian for a harmonic oscillator:

[itex]H=\frac{p^2}{2m}+\frac{mw^2}{2}q^2[/itex]

We have seen that the equations of motion are significantly simplified using the canonical transformation defined by [itex]F_1(q,Q)=\frac{m}{2}wq^2cot(Q)[/itex]

Show explicitly that between both coordinates, the following identities are satisfied

1) [itex][q,p]_{qp}=[Q,P]_{QP}[/itex]
2) [itex][q,p]_{QP}=[q,p]_{QP}[/itex]
3) [itex][Q,P]_{qp}=[Q,P]_{QP}[/itex]

where [ ] denote Poisson Brackets

Homework Equations

This is a well known Hamiltonian frequently used to introduce canonical transformations in many books (e.g. Goldstein 378). Given the generating function presented above, we find the following expresions for the new coordinates:

[itex]p=\sqrt{2Pmw}cos(Q)[/itex]
[itex]q=\sqrt{\frac{2P}{mw}}sin(Q)[/itex]

Also, from the definition of Poisson Brackets,
[itex][u,v]_{qp}=\frac{\partial{u}}{\partial{q}}\frac{\partial{v}}{\partial{p}}-\frac{\partial{u}}{\partial{p}}\frac{\partial{v}}{\partial{q}}[/itex]

The Attempt at a Solution

Well, by definition,

[itex][q,p]_{qp}=1=[Q,P]_{QP}[/itex]Also, using the expressions for p and q presented above, we have

[itex][q,p]_{QP}=cos^2(Q)+sin^2(Q)=1=[q,p]_{qp}[/itex]

Am I missing some fundamental formalism in here? I'm far from confident about this topic, so I'm not sure I followed the right procedure.

Also, for 3) I was thinking about obtaining expressions for Q and P straight from the presented expressions for p and q, and then just apply the same principle. What do you think about this?Thanks for your help and time!

 

[mark726]

Your approach is correct for the first two identities. For the third identity, you can indeed obtain expressions for Q and P from the given expressions for p and q, and then apply the same principle. This will show that [Q,P]_{qp} is equal to [Q,P]_{QP}. Overall, your solution is correct and you have successfully shown that all three identities are satisfied.