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Poisson brackets for simple harmonic oscillator

  1. Jun 23, 2013 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant 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]

    3. 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! :smile:
     
    Last edited: Jun 23, 2013
  2. jcsd
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