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Canonical Transformation / Poisson Brackets

  1. Aug 17, 2014 #1

    Zag

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    Question:

    (A) Show that the following transformation is a canonical transformation:
    [itex]Q = p + aq[/itex]
    [itex]P = (p - aq)/(2a)[/itex]

    (B) Find a generating functions for this transformation.


    Attempt of Solution:
    Alright, so this seems to be a very straight forward problem. Transformations are canonical if their Poisson Brackets satisfy:

    [itex][P_{\alpha},Q_{\beta}]_{PB} = -\delta_{\alpha\beta}[/itex]
    [itex][Q_{\alpha},Q_{\beta}]_{PB} = 0[/itex] and
    [itex][P_{\alpha},P_{\beta}]_{PB} = 0[/itex]

    For the given problem the system is one-dimensional, so the Poisson Brackets' identities reduce to a single relevant expression:

    [itex][P,Q]_{PB} = -1[/itex]

    We can now calculate this Poisson Bracket explicitly since we were given the transformation connecting the variables [itex]Q[/itex], [itex]P[/itex], [itex]q[/itex] and [itex]p[/itex]. The calculation is shown below:

    [itex][P,Q]_{PB} = \frac{\partial P}{\partial q}\frac{\partial Q}{\partial p} - \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p} = -\frac{1}{2}a - a\frac{1}{2a}[/itex]

    [itex]\therefore[/itex] [itex][P,Q]_{PB} = \frac{a}{2} - \frac{1}{2}[/itex]

    Comments and Issues:
    It is clear from this calculation that the transformation is not canonical for a general [itex]a[/itex], since the result of the Poisson Bracket is not [itex]-1[/itex]. In fact, the only value of [itex]a[/itex] for which the transformation would be canonical is [itex]a = 1[/itex].

    Now, the problem is that the question doesn't mention anything about finding the value of [itex]a[/itex] for which the transformation is canonical, it simply asks the student to show that the transformation is canonical. So I am not sure whether there is something wrong with my solution or the problem statement is formulated in an ambiguous.

    The second part of the problem could also be easily solved if one assumes [itex]a = 1[/itex], otherwise - according to my solution - it would be impossible to find a generating function for a general [itex]a[/itex] since in that case the transformation would not even canonical to begin with.

    Any thoughts or comments on this issue would be greatly appreciated!

    Thank you very much guys! Sincerely,
    Zag
     
  2. jcsd
  3. Aug 17, 2014 #2

    Orodruin

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    ##\frac{\partial Q}{\partial p}= 1 \neq a##

    The transformation is canonical.
     
  4. Aug 17, 2014 #3

    Zag

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    Nice! Thanks Orodruin, I indeed mixed my partial derivative results there. It works now! :)
     
  5. Aug 17, 2014 #4
    it's simple derivation's mistake
     
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