Canonical Transformations

Adrian Down

October 25, 2005

1 Motivation

We saw previously that the physical path of a system can be found by mini-mizing the action integral,

S =

t2t1

L(q, q, t)dt = 0

We also defined the Hamiltonian last time.

H(q, p, t) = piqi L(q, q(q, p), t)The action integral can then be written

t2t1

(piqi L(q, q(q, p), t)

)dt = 0

We devised a method for computing the Hamiltonian. However, theHamiltonian obtained in this method may not be the simplest or most trans-parent. The goal of the canonical transformation is to find a new Hamiltonianthat still satisfies Hamiltons equation of motion. The quality of the motionis the same, but the appearance in phase space may be different.

See Goldstein, chapter 9, as a reference.

2 Transformations in general

We seek some transformations to new variables Q and P of the form

Qi = Qi(q, p, t) Pi = Pi(q, p, t)

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We insist that these Q and P satisfy Hamiltons equations of motion. Callthe transformed Hamiltonian

H = K(Q,P, t)

To satisfy Hamiltons equations of motion,

Qi =K

PiPi = K

Qi

The action integral becomes

t2t1

(PiQi K(Q,P, t)

)dt = 0

There are two possible forms of such transformations.

2.0.1 Scale transformations

Definition (Scale transformation). A scale transformation is one of the form

PiQi K = (piqi H)We used a transformation of this form previously when dealing with the

harmonic oscillator.

2.0.2 Canonical transformation

Definition (Canonical transformation). A canonical transformation is oneof the form

PiQi K(Qi, Pi, t) + dFdt

= piqi H(qi, pi, t)

The action integral becomes

(PiQi K

)dt =

t2t1

(piqi H(q, p, t) dF

dt

)dt = F |t2t1

Note that

F = F (q, p,Q, P, t) F = 0Definition (Generating function). A function F of the form above is calleda generating function

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3 Examples

3.1 Identity Transformation

Let F = qiPi QiPi dF

dt= qiPi + qiPi QiPi QiPi= (qi Qi) Pi PiQi + qiPi

Substituting the canonical transformation,

piqi H = PiQi K + dFdt

=PiQi K + (qi Qi) Pi PiQi + qiPi

= (qi Qi) Pi + qiPi KComparing terms on both sides of the equation,

qi = Qi pi = Pi K = H

Definition (Identity Transformation). A transformation such as the oneabove that leaves the Hamiltonian unchanged is called an identity transfor-mation

3.2 Coordinate swap

We would like to find the conditions on a canonical transformation that leavesthe Hamiltonian is unchanged. We require that

K(Q,P, t) = H(q, p, t)

From the equation of the canonical transformation,

PiQi K(Qi, Pi, t) + dFdt

= piqi H(qi, pi, t)

dFdt

= piqi PiQi

We choose a generator of the form F (q,Q), which is called a type 1 generator.There are four types of generators F which satisfy this condition (see below).

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Assuming this form of F , the chain rule gives

dF

dt=F

q

q

t+F

Q

Q

t

=F

qiqi +

F

QiQi

Comparing the the condition on dFdt

obtained by requiring the F leave theHamiltonian unchanged,

piqi PiQi = Fqi

qi +F

QiQi

Comparing terms on either side of the equation gives the desired conditionson F .

pi =F

qiPi = F

Qi

3.3 General Type 1 generator

The generator could include time dependence, which we did not considerabove..

F (q,Q, t)

The conditions found above then become

pi =

qiF (q,Q, t) Pi =

QiF (q,Q, t)

Taking the time derivative,

dF

dt=F

qiqi +

F

QiQi +

F

t=piqi H PiQi +K

K = H + Ft

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3.4 Simple Harmonic oscillator

3.4.1 Find the Lagrangian

The Lagrangian for the simple harmonic oscillator is

L =1

2mq2 1

2kq2

The momentum is

p =L

q= mq

q = pm

The Hamiltonian is then

H = pq L = p2

m(1

2mq2 1

2kq2

) H(q, p) = p

2

2m+

1

2kq2

=1

2m

(p2 +m22q2

)3.4.2 Choose a particular form for F

Assume a solution of the form

p = f(P ) cos (Q) q =f(P )

msin (Q)

This choice is motivated by past experience with the problem.The Hamiltonian becomes

2mH = p2 +m22q2

= f(P )2 cos2 (Q) +f(P )2

m22m22 sin2 (Q)

= f(P )2

We use a type 1 generator, so that

K = H =f(P )2

2m

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3.4.3 Find f(P )

The explicit form of the generator can be found by integration,

F1(q,Q, t) p =F

qP = F

Q

Dividing the definitions for f(P ) in terms of q and p gives

p

q= m cot(Q)

p = mq cot(Q) = Fq

F = mq2

2cot(Q)

We can now find the explicit form of P and Q.

P = FQ

=mq2

2

1

sin2 (Q)

q2 = 2Pm

sin2(Q)

q(Q,P ) =

2P

msin(Q) p(Q,P ) =

2Pm cos(Q)

3.4.4 Transformed Hamiltonian

The Hamiltonian becomes

H =1

2m

(p2 +m22q2

)=

1

2m

(2Pm cos2(Q) +

m222P

m22sin2(Q)

)Since we required that H = K,

K(P ) = P

We also require that the transformed Hamiltonian satisfy Hamiltons equa-tions of motion.

Q =K

pP = K

Q

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Thus,

Q =

This equation can be easily integrated,

Q = t+Q0

Also,

K

Q= 0

P = 0 P = constant

Because there is no dissipation,

Etot = K = P

P = E

3.4.5 Phase space diagrams

Transformed The graph of Q as a function of t increases linearly. P is ahorizontal line.

The phase space diagram is a rectangle with height E. For a single

oscillation, the width of the rectangle is 2pi. Thus the area enclosed by therectangle is

A =2piE

Original system The phase space diagram for the original system is anellipse. The area is given by

A = piab =2piE

The system traces the ellipse once per oscillation, so this is the area enclosedby the ellipse during a single oscillation.

That the areas enclosed by the two phase space diagrams is the same isa demonstration of Louisvilles theorem.

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4 General method (H & F p. 211)

4.1 Recipe

1. Choose a specific generating function F

2. Use

P = FQ

p =F

q

to give specific equations of the transformation

3. Find K(Q,P, t) from

PiQi K + dFdt

= piqi H

expressing p and q in terms of P and Q

4. Apply Hamiltons equations of motion.

Type Function Derivatives Example

1 F1(q,Q, t)pi =

F1qi

Pi = F1Qi

F1 = qiQiQi = piPi = qi

2 F2(q,Q, t)pi =

F2qi

Pi =F2Qi

F2 = qiPiQi = qiPi = pi

3 F3(q,Q, t)pi = F3qiPi = F3Qi

F3 = qiQiQi = qiPi = pi

4 F4(q,Q, t)pi = F4qiPi =

F4Qi

F4 = piPiQi = piPi = qi

Table 1: Types of generating functions

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