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Generating function for canonical transformation

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Given the transformation

    [tex] Q = p+iaq, P = \frac{p-iaq}{2ia} [/tex]


    2. Relevant equations
    find the generating function


    3. The attempt at a solution

    As far as I know, one needs to find two independent variables and try to solve. I couldn't find such to variables.

    I've tried expressing it in terms of F(Q,P), and F(q,p) but always had one more term in the equation that prevented me from getting to [tex] H(q,p) = -H(Q,P) + \frac{\partial F}{\partial t} [/tex]

    I'm pretty clueless as to what is needed here. Can someone help me get started?

    Thanks.
     
  2. jcsd
  3. Dec 7, 2009 #2
    Ok, this is what I did :

    [tex] Q = 2ia(P + 2q) , P = \frac{p-iaq}{2ia} [/tex]
    which means that Q and p are independent coordintes, which means the generating function will be of the third kind, [tex] F_3(Q,p) [/tex].

    for the third kind,
    [tex] q = -\frac{\partial F_3}{\partial p} = \frac{p}{ia}-2P [/tex]
    [tex] P = -\frac{\partial F_3}{\partial Q} = \frac{Q}{2ia}-2q [/tex]

    from the first equation we get
    [tex] F_3 = 2pP - \frac{p^2}{2ia} + F(Q) [/tex]
    and from the second
    [tex] F_3 = 2Qq - \frac{Q^2}{4ia} + F(p) [/tex]

    summing both I get
    [tex] F_3 = 2Qq + 2pP - \frac{1}{2ia} (p^2 + \frac{Q^2}{2}) [/tex]

    Does this seem right??
     
    Last edited: Dec 7, 2009
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