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Now, let R^2 n^ be the phase space for a particle in Rn. A point (q, p) ∈ R^2 n^ describes the particle’s position q ∈ Rn^ and momentum p ∈ Rn. The algebra of smooth real-valued functions C∞(R^2 n) becomes a Poisson algebra with
{F, G} =
∑^ n i=
∂F
∂pi
∂G
∂qi
∂G
∂pi
∂F
∂qi
- Given A ∈ so(n), let φ: R × R^2 n^ → R^2 n
be given by φ(t, q, p) = (exp(tA)q, exp(tA)p),
then φ is a flow. Proof: Note that φ 0 (q, p) = (exp(0)q, exp(0)p) = (Iq, Ip) = (q, p)
and
φt(φs(q, p)) = φt(exp(sA)q, exp(sA)p) = (exp(tA) exp(sA)q, exp(tA) exp(sA)p) = (exp((t + s)A)q, exp((t + s)Ap) = φt+s(q, p).
Since φ is clearly smooth, φ is a flow.
(For example, in 3 dimensions, this flow would rotate both the position and the momentum about some axis.)
- Given A ∈ so(n), define an observable F ∈ C∞(R^2 n^ by
F (q, p) =
∑^ n
i,j=
Aij (qj pi − qipj ),
then F generates the flow φ defined above. Proof: For simplicity of notation let ∂/∂q and ∂/∂p) represent the tangent vectors (∂/∂q 1 ,... , ∂/∂qn, 0 ,... , 0) and (0,... , 0 , ∂/∂p 1 ,... , ∂/∂pn), respectively. It then follows from the definition that {q, ·} = −∂/∂p and {p, ·} = ∂/∂q. Note that Aij = −Aji, so
F (q, p) =
∑^ n i,j=
Aij (qj pi − qipj )
∑^ n i,j=
−Ajiqj pi −
∑^ n i,j=
Aij qipj )
∑^ n i,j=
Aij qipj
= −q∗Ap.
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Therfore, by the Leibnitz rule we have
vF = {−q∗Ap, ·} = −({p, ·}∗q∗)A − Ap{q, ·} = −A∗q
∂q
∂p = Aq
∂q +^ Ap
∂p.
Let ψ be the flow gernerated by vF , and write ψt(q, p) = (q(t), p(t)). We must have
q˙(t) = Aq(t) p˙(t) = Ap(t)
and we see that ψ(q, p) = (exp(tA)q, exp(tA)p) is a solution. By uniquness, it follows that ψ = φ. The moral: The observable that generates the flow φ is called angular momentum in the A direction. But beware: A is not a vector in Rn! It’s a matrix in so(n)! For n = 3 we have an isomorphism so(n) ∼= Rn
so we can talk about angular momentum in some direction v ∈ Rn. But, this is not true in any other dimension (except n = 0)!
- When n = 3, the observable F (q, p) = q 1 p 2 − q 2 p 1
is usually called angular momentum in the z direction and denoted Jz. What flow does this observable generate? Solution: With a fixed basis (∂/∂q 1 ,... , ∂/∂p 3 ) for the tangent space the vector field generated by F is:
vF = {q 1 p 2 − q 2 p 1 , ·} = (−q 2 , q 1 , 0 , −p 2 , p 1 , 0).
Let φt(q, p) = (q(t), p(t)) be the flow generated by vF. When viewing R^3 as C × R, the above says:
d dt (q 1 (t) + iq 2 (t)) = i(q 1 (t) + iq 2 (t)) d dt (p 1 (t) + ip 2 (t)) = i(q 1 (t) + ip 2 (t)) d dt q 3 (t) = 0 d dt p 3 (t) = 0,
from which we can immediately conclude that
q 1 (t) + iq 2 (t) = eit(q 1 + iq 2 ) p 1 (t) + ip 2 (t) = eit(p 1 + ip 2 )
and q 3 (t) = a and p 3 (t) = b are constants. In otherwords,
φt(q 1 + iq 2 , q 3 , p 1 + ip 2 , p 3 ) = (eit(q 1 + iq 2 ), a, eit(p 1 + ip 2 ), b).
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