§ 1. The algebra of observables in classical mechanics 3
For simplicity the exposition to follow is conducted using the example
of a system with one degree of freedom. The Hamiltonian equations
in this case have the form
(6) ˙ q =
∂H
∂p
, ˙ p =
∂H
∂q
, H = H(q, p).
The Cauchy problem for the system (6) and the initial conditions
(7) q|t=0 = q0, p|t=0 = p0
has a unique solution
(8) q = q(q0, p0, t), p = p(q0, p0, t).
For brevity of notation a point (q, p) in phase space will sometimes
be denoted by µ, and the Hamiltonian equations will be written in
the form
(9) ˙ µ = v(µ),
where v(µ) is the vector field of these equations, which assigns to each
point µ of phase space the vector v with components
∂H
∂p
,
−∂H
∂q
.
The Hamiltonian equations generate a one-parameter commuta-
tive group of transformations
Gt : M M
of the phase space into
itself,2
where Gtµ is the solution of the Hamil-
tonian equations with the initial condition Gtµ|t=0 = µ. We have the
equalities
(10) Gt+s = GtGs = GsGt, Gt
−1
= G−t.
In turn, the transformations Gt generate a family of transformations
Ut : A A
of the algebra of observables into itself, where
(11) Utf(µ) = ft(µ) = f(Gtµ).
In coordinates, the function ft(q, p) is defined as follows:
(12) ft(q0, p0) = f(q(q0, p0, t), p(q0, p0, t)).
2We
assume that the Hamiltonian equations with initial conditions (7) have a
unique solution on the whole real axis. It is easy to construct examples in which
a global solution and, correspondingly, a group of transformations Gt do not exist.
These cases are not interesting, and we do not consider them.
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