Prove that infinitesimal transf. is canonical

[ibreakkidsleg]

Homework Statement

Prove that the infinitesimal transformation generated by any dynamical variable g(q,p) is canonical.

Homework Equations

q' = q + e{q,g}
p' = p + e{p,g} where e is some small number.

The Attempt at a Solution

Demanding that {q',p'} = 1 and that {q,q'} = {p,p'} = 0 yields the result. However, I'm unable to find or produce any proof that these conditions guarantee that active transformations are canonical. In other words, all proofs that I've seen (in class, in Shankar) assume that the Hamiltonian H(q,p) = H(q',p').

Any help is appreciated.

 

[mark726]

Thank you for your question. The proof that the infinitesimal transformation generated by any dynamical variable g(q,p) is canonical can be shown using the Poisson bracket notation. The Poisson bracket of two functions A and B is defined as {A,B} = ∂A/∂q ∂B/∂p - ∂A/∂p ∂B/∂q.

Using this notation, we can rewrite the infinitesimal transformation equations as:

q' = q + e{q,g} = q + e∂qg ∂pg - e∂pg ∂qg = q + e{q,g} = q + e{q,g}
p' = p + e{p,g} = p + e∂pg ∂qg - e∂qg ∂pg = p + e{p,g} = p + e{p,g}

Now, to show that this transformation is canonical, we need to prove that the Poisson bracket of q' and p' is equal to 1, and that the Poisson bracket of q and q' and p and p' is equal to 0.

{q',p'} = ∂q'∂p' - ∂p'∂q' = (1 + e{q,g})(1 + e{p,g}) - (1 + e{p,g})(1 + e{q,g}) = 1 + e{q,g} + e{p,g} + e^2{q,g}{p,g} - (1 + e{p,g} + e{q,g} + e^2{p,g}{q,g}) = 1

{q,q'} = ∂q∂q' - ∂q'∂q = 0

{p,p'} = ∂p∂p' - ∂p'∂p = 0

Therefore, the infinitesimal transformation generated by any dynamical variable g(q,p) is canonical, as the Poisson bracket of q' and p' is equal to 1 and the Poisson bracket of q and q' and p and p' is equal to 0.

I hope this helps to clarify the proof for you. Let me know if you have any further questions.