Question

What
does
the
poisson bracket {p, H} between the momentum
& the
Hamiltonian
evaluate to ?​

Answer 1

Answer:

I seem to have found a counter-example. If we write The hamiltonian for a free particle, by starting with the Lagrangian in spherical coordinates (r,θ,ϕ), we end up with

H=pr22m+pθ22mr2+pϕ22mr2sin2(θ)

with generalized momenta

pr=mr˙,pθ=mr2θ˙,pϕ=m(rsin(θ))2ϕ˙

It's easy to see from the definition that

pϕ=Lzandpθ=Lx.

This looks like a counterexample. I did not arrive at this hamiltonian through a canonical transformation, but it is a "legal" hamiltonian and two of the canonical variables/momenta are two components of L, contradicting the claim above.

As an additional quandary, the book shows that Poisson Brackets are canonical invariants, i.e. preserved under canonical transformations, but when I calculated [Lx,Lz] in Cartesian coordinates, I got the result [Lx,Lz]−Ly as (1) suggests, and when I calculated [Lx,Lz] in the phase space of spherical coordinates I got the result [pθ,pϕ]=0 as I'd expect from a pair of canonical variables..