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1 How Hamiltonians Generate Time Evolution
We have seen how any observable H on a Poisson manifold X gives rise a vector field vH which, if in- tegrable, generates a flow φ. When H has the physical meaning of ‘energy’, we call it a Hamiltonian and call this flow time evolution. Let’s see how this works in some examples we’ve already studied.
Example: The simple harmonic oscillator. Here the configuration space is R and the phase space is X = T ∗R ∼= R × R 3 (q, p). We have Newton’s second law
m
d^2 dt^2
q(t) = −kq(t)
Let us set m = k = 1 to simplify the math a bit:
d^2 q dt^2
= −q
Since p = m q˙, with m = 1 this equation gives Hamilton’s equations
dq dt
= p
dp dt
= −q
with solution: q(t) = A cos t + B sin t p(t) = −A sin t + B cos t
where A = q(0) and B = p(0). So we get a flow:
φ: R × R^2 → R^2
defined by (t, q(0), p(0)) 7 → (q(t), p(t))
describing the time evolution of a point (q, p) in the phase space R^2. This flow is rotation clockwise by an angle t over time t.
picture of R^2 with flow lines and rotated point.
Now let’s obtain this flow using the Poisson approach. So, we will remember the Hamiltonian for the harmonic oscillator and work out the vector field vH generated by that, and see that it does indeed generate this flow. The energy of the harmonic oscillator is:
E =
m q˙^2 +
k 2
q^2
or with m = k = 1
E =
(q^2 + ˙q^2 )
so the Hamiltonian H: R^2 → R is
H(q, p) =
(q^2 + p^2 )
(Note: flow moves along level sets of H)
Our prescription says: see what vector field H generates and then see what flow that vector field generates. H generates the vector field
vH = {H, ·}
=
∂H
∂p
∂q
∂H
∂q
∂p
(or
∂H ∂p ,^ −^
∂H ∂q
in the old-fashioned notation for vector fields on R^2 .)
With our H this is:
vh = p
∂q
− q
∂p picture of this vector field
If φ: R × R^2 → R^2
is the flow generated by vH , then by definition
d dt
φt(q, p) = vH (φt(q, p))
Notation: φt(q, p) := φ(t, q, p). For short, let’s write
φt(q, p) = (q(t), p(t)) ∈ R^2.
So, our equation becomes:
( ˙q(t), p˙(t)) = vH (q(t), p(t)) = (p(t), −q(t))
and this is Hamilton’s equations: q˙(t) = p(t) p˙(t) = −q(t)
so we must have: (q(t), p(t)) = (cos(t)q + sin(t)p, − sin(t)q + cos(t)p)
which is the flow we had before.
Example: A particle in a potential in Rn. Now our configuration space is Rn^ and so the phase space is
X = T ∗Rn^ ∼= Rn^ × Rn^3 (q, p)
The potential energy is V (q) where V ∈ C∞(Rn). The kinetic energy is 12 mv^2 = p
2 2 m since^ p^ =^ mv. So the Hamiltonian is
H(q, p) =
p^2 2 m
whereas
vF = {F, ·}
=
∑^ n
i=
∂F
∂pi
∂qi
∂F
∂qi
∂pi
so ∂F ∂qi
= 0 and
∂F
∂pi
= ui
So F (q, p) = p · u + c
for any c ∈ R, which we’ll take to be 0, so we say “spatial translations in the u dirction are generated by F ”, where F is “momentum in the u direction”.
Example: What observable generates Galilei boosts by an amount s in the u direction (where u ∈ Rn)? Here the flow is: φs(q, p) = (q, p + msu)
i.e., φs does not change the position but it adds su to the velocity! This is generated by:
v =
d ds
φs(q, p)
s= = (0, mu)
=
mui
∂pi
and this equals
vF =
∑ ∂F
∂pi
∂qi
∂F
∂qi
∂pi
for F = −mu · q.
So, there’s a nice pattern: for a single particle in Rn: Galilei boosts in the direction u ∈ Rn^ are generated by −mq · u. Translations in the direction u ∈ Rn^ are generated by p · u.
For a bunch of particles in Rn, this generalizes: Galilei boosts in the direction u ∈ Rn^ are generated by −
miqi · u, i.e. −u · (total mass × center of mass). Translations in the direction u ∈ Rn^ are generated by p · u, where p =
pi is the total momentum (sum over particles: pi ∈ Rn^ is the momentum of the ith^ particle).
