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Canonical transformation

  1. Dec 31, 2005 #1

    I need to solve the Hamiltonian of a one-dimensional system:

    [tex]H(p, q) = p^2 + 3pq + q^2[/tex]

    And I've been instructed to do so using a canonical transformation of the form:

    [tex]p = P \cos{\theta} + Q \sin{\theta}[/tex]
    [tex]q = -P \sin{\theta} + Q \cos{\theta}[/tex]

    And choosing the correct angle so as to the get the Hamiltonian of an harmonic oscillator.

    Applying this transformation, I get:

    [tex]H(P, Q) = P^2 + Q^2 - 3/2 (P^2 - Q^2) \sin{2 \theta} + 3 P Q \cos{2 \theta}[/tex]

    And as far as I can see, no choice of angle will get me to an Hamiltonian of an harmonic oscillator.

    Am I correct? Can someone please check my calculation?

  2. jcsd
  3. Dec 31, 2005 #2
    I got the same expression for H(P,Q) as you, and I can't see what angle would give a nice form either. You'd want to get something like

    [tex]H(P,Q) = \frac{P^{2}}{2} + kQ^{2}[/tex]

    from which you can then get the frequency from, but that would want [tex]\sin 2\theta = \frac{1}{3}[/tex] which isn't going to drop the 3PQ term you've got.
  4. Dec 31, 2005 #3


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    All you need is to get [itex]\cos 2\theta=0[/itex]. So what about [itex]\theta=\pi/4[/itex]?
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