• F4(~p, ~P , t)

F4(~p, ~P , t) = F1(~q, ~Q, t) − ~q · ~p + ~Q · ~P ,

Legendre transforms ~q → ~p, ~Q → ~P . So

dF4 = ~p · d~q − ~P · d ~Q − (H − K)dt − ~q · d~p − ~p · d~q + ~Q · d~P + ~P · d ~Q

= −~q · d~p + ~Q · d~P − (H − K)dt

≡ ∂F4

∂~p· d~p +

∂F4

∂ ~P· d~P +

∂F4

∂tdt .

This gives

qi = −∂F4

∂pi

, Qi =∂F4

∂Pi

, K = H +∂F4

∂t, F3 ≡ F3(~p, ~P , t) .

4.2.3 Examples of canonical transformations

Identity transformation: F2(~q, ~P , t) = ~q · ~P

pi =∂F2

∂qi

= Pi , Qi =∂F2

∂Pi

= qi , K = H .

Point transformation: F2(~q, ~P , t) = ~f(~q, t) · ~P

Qi =∂F2

∂Pi

= fi(~q, ) .

[Point transformations are transformations where the new coordinates are deter-mined solely by the old coordinates.]

Exchange transformation F1(~q, ~Q, t) = ~q · ~Q

pi =∂F1

∂qi

= Qi , Pi = −∂F2

∂Qi

= −qi , K = H ,

interchanges coordinates and momenta.

Note that we have to choose carefully which generating function you use. For ex-ample for the exchange transformation cannot use F2, as pi = ∂F2(~q, ~P )/∂qi is not

a function of ~Q.

49

One dimensional Harmonic Oscillator

H(q, p) =p2

2m+

1

2mω2q2 .

We want to guess a time independent transformation to make q cyclic. Try

p = f(P ) cos Q

q = f(P )mω

sin Q

}

⇒ K =f(P )2

2mie Q cyclic .

Now taking ratio gives

p = mωq cotQ ≡ ∂

∂QF1(q, Q) ,

giving

F1(q, Q) =1

2mωq2 cot Q ,

(if added a ‘constant’ h(Q) then find that need to take h′ = 0 for consistency; alsocanonical transformations only defined up to a constant, hence h = 0). So

P = −∂F1

∂Q=

1

2mωq2cosec2Q ,

giving

q =

√

2P

mωsin Q =⇒ f(P ) =

√2mωP .

Thus

K = ωP ,

and so

P = −∂K

∂Q= 0 or P = const. =

E

ω

Q =∂K

∂P= ω or Q(t) = ωt + Q(0) ,

or

q(t) =

√

2E

mω2sin(ωt + Q(0)) .

We have solved the problem by solving only trivial differential equations – canonicaltransformations are an algebraic approach.

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4.2.4 Infinitesimal canonical transformations

These are transformations close to the identity. Suppose (~q, ~p) → ( ~Q, ~P ) is generatedfrom

F2(~q, ~P , t) = ~q · ~P + ǫ G(~q, ~P , t) ,

which implies that

pi =∂F2

∂qi

= Pi + ǫ∂G

∂qi

Qi =∂F2

∂Pi

= qi + ǫ∂G

∂Pi

,

or

Pi = pi + O(ǫ)

Qi = qi + O(ǫ) ,

so

δqi(~q, ~p, t) = Qi − qi = ǫ∂G(~q, ~P , t)

∂Pi

= ǫ∂G(~q, ~p, t)

∂pi

+ O(ǫ2)

δpi(~q, ~p, t) = Pi − pi = −ǫ∂G(~q, ~P , t)

∂qi

= −ǫ∂G(~q, ~p, t)

∂qi

+ O(ǫ2) .

G(~q, ~p, t) is the generator of the infinitesimal canonical transformation.

For example consider the infinitesimal coordinate transformation defined by Hamil-ton’s equations

Qi ≡ qi(t + δt) = qi(t) + qiδt = qi + δt ∂H∂pi

Pi ≡ pi(t + δt) = pi(t) + piδt = pi − δt, ∂H∂qi

}

G = H where ǫ = δt .

4.2.5 Matrix form for infinitesimal canonical transforma-

tions

In general writing f + f = 2f dimensional objects,

~η =

(

~q~p

)

,

and a 2f × 2f dimensional matrix

J =

(

0 I−I 0

)

JT = −J ,

51

This is a more compact notation.

For example Hamilton’s equations are now given by

ηi = Jij

∂H

∂ηj

.

Now let

~ζ =

(

~Q~P

)

,

and then infinitesimal transformations are

ζi = ηi + δηi = ηi + ǫJij

∂G

∂ηj

.

Define

Mij =∂ζi

∂ηj

= δij + ǫ JikGkj Gij ≡∂2G

∂ηi∂ηj

= Gji

or M = I + ǫJG .

Now consider

MJMT = (I + ǫJG) J (I + ǫJG)T

= J + ǫ(JGJ + J(JG)T ) + O(ǫ2)

= J + O(ǫ2) , as GT = G , JT = −J ,

and thus

M J MT = J Symplectic Condition ,

So infinitesimal canonical transformations obey the symplectic condition, including(~q(t), ~p(t)) → (~q(t + δt), ~p(t + δt))

Volumes in phase space are invariant as first we have

d2fζ = det

(

∂ζi

∂ηj

)

d2fη = det M d2fη .

Also

MJMT = 1 =⇒ det M det J det MT = det J =⇒ (det M)2 = 1 ,

52

and so

det M = +1 ,

as can get continuously to the identity (by definition). So

V =

∫

d2fη =

∫

dfq dfp is a canonical or a Poincare invariant .

Note that for generator G = H , then this is just Liouville’s theorem again.

Now consider two infinitesimal transformations

ξ → η M1JMT1 = J (M1)ij = ∂ηi

∂ξj

η → ζ M2JMT2 = J (M2)ij = ∂ζi

∂ηj

then the combined transformation has

Mij =∂ζi

∂ξj

=∂ζi

∂ηk

∂ηk

∂ξj

= (M2M1)ij

so

MJMT = M2M1J(M2M1)T

= M2M1JMT1 MT

2

= M2JM2

= J ,

ie combined transformation again obeys the symplectic condition.

4.2.6 Finite Canonical Transformations

• Also obey symplectic condition

• Direct proof is tedious (according to Goldstein)

Problem is that not all canonical transformations can be continuously connected theidentity (eg exchange transformation ~Q = ~p, ~P = −~q). So use a trick:

Consider (~q, ~p) → ( ~Q, ~P ) as paths A → B → A′.

• A, A′

Apply succession of infinitesimal transformations with H as the generator.Since ζ(t) → ζ(t + δt) obeys the symplectic condition, so does ζ(t) → ζ(t′),∀t, t′; just divide the time interval [t, t′] into steps of size δt.So Hamilton’s equations may be regarded as a gradual succession of infinites-imal canonical transformations with generator H .

53

time

B

A′

Q, P

q, p

t0

A

~q, ~p

• Now at fixed time t0 we make the canonical transformation ~η(t0) → ~ζ(~η, t0) so

ζi =∂ζi

∂ηj

ηj no explicit time dependence

= MijJjk

∂H

∂ηk

= MijJjk

∂ζl

∂ηk

∂H

∂ζl

no explicit time dependence, so K = H

= (MJMT )ij

∂H

∂ζj

≡ Jij

∂H

∂ζj

K = H .

Hence we have MJMT = J again with Mij = ∂ζi/∂ηj .

A → B → A′ =⇒

All canonical transformations obey symplectic condition

(even those that are time dependent). The inverse is also true: if the transformationobeys the sympletic condition then it is canonical [no proof]

Hence we have the result

sympletic ⇐⇒ canonical

Miscellaneous points:

• Useful identities:since J2 = −I or J−1 = −J then

MJMT = J =⇒ MJ = JMT−1 ⇒ JMJ2 = J2MT−1J ⇒ JM = MT−1J ,

so

MT JM = J , or M−1JMT−1 = J .

54

• Canonical transformations form a group – Sympletic group, Sp(2f) – withelements M , where MJMT = J :

– the identity transformation is canonical

– if a transformation is canonical then so is its inverse (M−1JMT−1 = J)

– two successive canonical transformations (product) define a canonicaltransformation (proof identical to given infinitesimal case)

– the product is associative

• It can be shown that the symplectic condition ⇔ det M = +1 ie there is no

det M = −1 in distinction to eg O(3), which has reflections as well as rota-tions.As an example of a transformation which cannot be built from infinitesimaltransformations close to the identity (which obviously have det M = +1) con-sider the exchange transformation, which interchanges coordinates and mo-menta, Qi = pi, Pi = −qi. This gives M = J so det M = det J = +1.

• Symplectic condition ⇔ transformation preserves the metric Jij .

4.3 Poisson Brackets

[u, v]q,pdef=

∂u

∂qi

· ∂v

∂pi

− ∂u

∂pi

· ∂v

∂qi

, u = u(~q, ~p, t) , v = v(~q, ~p, t) .

4.3.1 Fundamental relations

The fundamental Poisson brackets are

[qi, qj]q,p = 0 = [pi, pj]q,p ,

and

[qi, pj]q,p = δij = −[pi, qj]q,p .

Using again ~η = (~q, ~p) we may re-write the Poisson bracket as

[u, v]η =∂u

∂ηi

Jij

∂v

∂ηj

.

55

4.3.2 Invariance under canonical transformations

Poisson brackets are invariant under canonical transformations

ProofAlso writing ~ζ = ( ~Q, ~P ) then

[u, v]η =∂u

∂ηi

Jij

∂v

∂ηj

=∂ζk

∂ηi

∂u

∂ζk

Jij

∂v

∂ζl

∂ζl

∂ηj

=∂u

∂ζk

Mki Jij Mlj

∂v

∂ζl

Mij =∂ζi

∂ηj

=∂u

∂ζk

Jkl

∂v

∂ζl

= [u, v]ζ .

So we do not need to specify with respect to which set of canonical variables Poissonbrackets are defined. Equations expressed entirely in terms of Poisson brackets areinvariant in form under canonical transformations.

Invariance of fundamental Poisson brackets ⇐⇒ symplectic condition

[ηi, ηj]η = Jij

[ζi, ζj]η = ∂ζi

∂ηkJkl

∂ζj

∂ηl= (MJMT )ij

}

So if under the transformation ~η → ~ζ we find that [ζi, ζj]η = Jij then

⇔ symplectic condition MJMT = J

⇔ transformation is canonical .

(Note, test not involving the generating function.)

4.3.3 Algebraic Properties

1. Antisymmetry

[u, v] = −[v, u]

(so [u, u] = 0)

2. Bilinearity

[au + bv, w] = a[u, w] + b[u, w]

(a, b constants)

56