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Canonical transformation for Harmonic oscillator

  1. Oct 11, 2012 #1
    Find under what conditions the transformation from (x,p) to (Q,P) is canonical when the transformation equations are:
    Q = ap/x , P=bx2
    And apply the transformation to the harmonic oscillator.
    I did the first part and found a = -1/2b
    I am unsure about the next part tho:
    We have the hamiltonian in (x,p):
    H = 1/2m(p2+(m[itex]\omega[/itex]x)2 , [itex]\omega[/itex]2=k/m
    So transforming to (Q,P) whilst setting am=-1 you get the nice equation:
    H = P(Q2+[itex]\omega[/itex]2)
    Should I now use hamiltons equation in the new coordinate basis, i.e.;
    dH/dP = Q2+[itex]\omega[/itex]2 = dQ/dt
    dH/dQ = 2QP = -dP/dt
    And solve these differential equations for Q,P and transform back? I think so but these equations are just not very easu. I mean the second one is easy to do by means of substitution but really the first one is just a mess. What is the smartest way to do this? I can't really get the first one solved.
     
  2. jcsd
  3. Oct 14, 2012 #2

    gabbagabbahey

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    Do you mean [itex]a=-\frac{1}{2b}[/itex], or [itex]a=-\frac{1}{2}b[/itex]? Brackets are important! :wink:

    Again, brackets are needed if you meant [itex]H=\frac{1}{2m}\left(p^2+(m\omega x)^2\right)[/itex]

    (1) Why are you setting am=-1?
    (2) You don't get that when you set am=-1

    Without setting am=-1, you should get [itex]H=-\frac{1}{4m}PQ^2 -am\omega^2P[/itex]

    The second DE is only easy to solve once you know Q. The first one is seperable, so al that is needed is some striaghtforward integration... try a trig substitution if you are stuck on the integration part
     
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