Consider the gauge transformation treated. Let the p0 , p before the gauge transformation be the p..

Consider the gauge transformation treated.
Let the p0 , p before the gauge transformation be the p variables, and the p 0
, p after the gauge transformation be the transformed P variables, where q = t,
x, y,z and the p0 and p 0 are the “standard” momenta discussed at the end of
Section 16.9. Assume that a gauge transformation makes no change in the
coordinates, so that Qk = qk .

(a) Use the Poisson bracket conditions to
prove that this transformation is canonical.

(b) Find an F2 generating function, and
demonstrate that it does generate the transformation.

(c) Use the standard Hamiltonian K(q, p)
and the simple substitution

(Recall that o (δa)2 means that the
dropped terms are of order smaller than (δa)2 as δa → 0. See
the definitions in Section D.11.)

(b) Define q(1,2) k = Qk − qk where
Qk is the final value in eqn (18.121). Define q(2,1) k similarly, but with the
order of application of G1 and G2 reversed. Make similar definitions for the p
variables. Show that

Thus two successive canonical
transformations produce a different result when their order of application is
reversed.

(c) Compare eqn (18.123) with eqn (18.100).
Show that, not only do the canonical transformations in reversed order produce
different changes, but (when the calculation is carried to second order) the
difference itself is a differential canonical transformation generated by a
generating function G = [G1, G2] equal to the Poisson bracket of G1 and G2.

(d) Write a short paragraph justifying the
following statement (or demolishing it if you disagree): Since Lx generates
rotations about the x-axis, Ly generates rotations about the y-axis, and Lz
generates rotations about the z-axis, the Poisson bracket of Lx and Ly must be
Lz (as is shown by direct calculation in eqn (12.75)), due to the structure of
the rotation group as demonstrated.