Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Canonical transformation

  1. Nov 6, 2003 #1
    Problem: Verify that the infinitesimal transformation generated by any dynamical variable g is a canonical transformation.

    I've worked out that an infinitesimal canonical transformation can be represented as follows:

    q_i -> q_i' = q_i + ε(∂g/∂p_i) ≡q_i + δq_i

    p_i -> p_i' = p_i - ε(∂g/∂q_i) ≡p_i + δp_i

    I also know that, for any Hamiltonian, a new set of coordinates is only canonical if

    {q_j',q_k'} = 0 = {p_j',p_k'}

    ==> {q_j',p_k'} = δ_jk

    where {} are Poisson brackets and the lower scores denote subscript for the following character.

    So my question is: Can I use these relationships to solve the problem? If so, how do I get started? The use of indices in these relationships confuses me, and I don't know if the δ_jk is supposed to be Kronecker delta, or some infinitesimal number. My intuition says that it's the Kronecker, but the indices used in the book I'm using are not explained, and I am unsure of how to apply the criterion for a canonical transformation to the general form of an infinitesimal transformation as given above. Any hints would be greatly appreciated!


  2. jcsd
  3. Nov 8, 2003 #2

    so the idea here is to prove that a certain type of transformation is canonical correct? So you need to think of what defines a canonical transformation and you have this already. The g in your equations is the quantity that is conserved if the transformation is canonical and the Hamiltonian is invariant under this transformation. You are correct that the term in your last equation is the kronecker delta. So the thing to do would be to check that your new coordinates, the ones with the primes, still obey your canonical transformation rules... that is the equations that define a canonical transformation are still satisfied.
    Hope this helps. If this is unclear let me know.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook