Student Seminar: Classical and Quantum Integrable Systems

Let us show that the variables (I i , θ i ) are canonically conjugate. For that we need
to construct a canonical transformation (p i , q i ) → (I i , θ i ). Consider a generating
function depending on I i and q i :
We see that
Let us introduce
S(I, q) =
∫ m
m 0
α =
∫ q
q 0
p i (q ′ , I)dq ′ i .
p j = ∂S
∂q j
=⇒ p = p(q, I).
θ j = ∂S
∂I j
=⇒ θ = θ(q, I).
and show that θ j are indeed coincide with the properly normalized angle variables.
We have
∮
1
dθ i = 1 ∮
d ∂S = ∂ ( ∮ 1
)
dS = ∂ ( ∮ 1 ∂S
dq k +
∂S dI k
2π C j
2π C j
∂I i ∂I i 2π C j
∂I i 2π C j
∂q k ∂I
} {{ k
}
Furthermore,
= ∂ ( ∮ 1
)
α = δ ij .
∂I i 2π C j
=0 on C j
)
( ∂S
) (
dI i ∧ dθ i = −d(θ i dI i ) = −d dI i = −d dS − ∂S )
dq i = d(p i dq i ) = dp i ∧ dq i .
∂I i ∂q i
Problem. Find action-angle variables for the harmonic oscillator.
We have
E = 1 2 (p2 + ω 2 q 2 ) =⇒ p(E, q) = ± √ 2E − ω 2 q 2 .
and, therefore,
I = 1 ∮
dq √ 2E − ω
2π
2 q 2 = 2
E
2π
∫ √ 2E
ω
− √ 2E
ω
The generating function of the canonical transformation reads
while for the angle variables we obtain
S(I, q) = ω
∫ q
dx √ 2I − x 2 ,
dq √ 2E − ω 2 q 2 = E ω .
θ = ∂S
∂I = ω ∫ q
dx
√
2I − x
2 = ω arctan
q
√
2I − q
2
=⇒ q = √ 2I sin θ ω .
Finally, we explicitly check that the transformation to the action-angle variables is canonical
(
dI
dp ∧ dq = ω √ −
2I − q
2
qdq
)
√ ∧ dq =
2I − q
2
ω
√
2I − q
2 dI ∧ d(√ 2I sin θ ω
)
= dI ∧ dθ .
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