Canonical Transformations
Adrian Down
October 25, 2005
Motivation
We saw previously that the physical path of a system can be found by minimizing the action integral,
Z t2
S =
L(q, q,
t)dt = 0
t1
We also defined the Hamiltonian last time.
H(q, p, t) = pi qi L(q, q(q,
p), t)
The action integral can then be written
Z t2
pi qi L(q, q(q,
p), t) dt = 0
t1
We devised a method for computing the Hamiltonian. However, the
Hamiltonian obtained in this method may not be the simplest or most transparent. The goal of the canonical transformation is to find a new Hamiltonian
that still satisfies Hamiltons equation of motion. The quality of the motion
is the same, but the appearance in phase space may be different.
See Goldstein, chapter 9, as a reference.
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)
1
We insist that these Q and P satisfy Hamiltons equations of motion. Call
the transformed Hamiltonian
H = K(Q, P, t)
To satisfy Hamiltons equations of motion,
K
Q i =
Pi
K
Pi =
Qi
The action integral becomes
Z t2
Pi Q i K(Q, P, t) dt = 0
t1
There are two possible forms of such transformations.
2.0.1
Scale transformations
Definition (Scale transformation). A scale transformation is one of the form
Pi Q i K = (pi qi 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 one
of the form
dF
Pi Q i K(Qi , Pi , t) +
= pi qi H(qi , pi , t)
dt
The action integral becomes
Z
Z t2
dF
dt = F |tt21
Pi Q i K dt =
pi qi H(q, p, t)
dt
t1
Note that
F = F (q, p, Q, P, t) F = 0
Definition (Generating function). A function F of the form above is called
a generating function
2
3
3.1
Examples
Identity Transformation
Let F = qi Pi Qi Pi
dF
= qi Pi + qi Pi Q i Pi Qi Pi
dt
= (qi Qi ) Pi Pi Q i + qi Pi
Substituting the canonical transformation,
dF
dt
i K + (qi Qi ) Pi
i + qi Pi
=
Pi
Q
Pi
Q
= (qi Qi ) Pi + qi Pi K
pi qi H = Pi Q i K +
Comparing terms on both sides of the equation,
qi = Qi
p i = Pi
K=H
Definition (Identity Transformation). A transformation such as the one
above that leaves the Hamiltonian unchanged is called an identity transformation
3.2
Coordinate swap
We would like to find the conditions on a canonical transformation that leaves
the Hamiltonian is unchanged. We require that
K(Q, P, t) = H(q, p, t)
From the equation of the canonical transformation,
dF
= pi qi H(qi , pi , t)
Pi Q i K(Qi , Pi , t) +
dt
dF
= pi qi Pi Q i
dt
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).
3
Assuming this form of F , the chain rule gives
dF
F q F Q
=
+
dt
q t Q t
F
F
qi +
=
Qi
qi
Qi
dF
dt
Comparing the the condition on
Hamiltonian unchanged,
obtained by requiring the F leave the
F
F
Qi
qi +
pi qi Pi Q i =
qi
Qi
Comparing terms on either side of the equation gives the desired conditions
on F .
pi =
3.3
F
qi
Pi =
F
Qi
General Type 1 generator
The generator could include time dependence, which we did not consider
above..
F (q, Q, t)
The conditions found above then become
pi =
F (q, Q, t)
qi
Pi =
F (q, Q, t)
Qi
Taking the time derivative,
dF
F
F
F
i + K
=
qi +
Q i +
=
pi
qi H
Pi
Q
dt
qi
Qi
t
F
K=H+
t
3.4
3.4.1
Simple Harmonic oscillator
Find the Lagrangian
The Lagrangian for the simple harmonic oscillator is
1
1
L = mq2 kq 2
2
2
The momentum is
L
= mq
q
p
q =
m
p=
The Hamiltonian is then
1 2 1 2
p2
mq kq
H = pq L =
m
2
2
2
p
1
H(q, p) =
+ kq 2
2m 2
1
=
p2 + m2 2 q 2
2m
3.4.2
Choose a particular form for F
Assume a solution of the form
p = f (P ) cos (Q)
q=
f (P )
sin (Q)
m
This choice is motivated by past experience with the problem.
The Hamiltonian becomes
2mH = p2 + m2 2 q 2
= f (P )2 cos2 (Q) +
f (P )2 2 2 2
m sin (Q)
m2 2
= f (P )2
We use a type 1 generator, so that
K=H=
f (P )2
2m
3.4.3
Find f (P )
The explicit form of the generator can be found by integration,
F1 (q, Q, t)
p=
F
q
P =
F
Q
Dividing the definitions for f (P ) in terms of q and p gives
p
= m cot(Q)
q
F
p = mq cot(Q) =
q
2
mq
F =
cot(Q)
2
We can now find the explicit form of P and Q.
mq 2
F
1
=
P =
2
Q
2 sin (Q)
2P
q2 =
sin2 (Q)
m
r
2P
q(Q, P ) =
sin(Q)
p(Q, P ) = 2P m cos(Q)
m
3.4.4
Transformed Hamiltonian
The Hamiltonian becomes
1
H=
p2 + m2 2 q 2
2m
2
1
m
2 2P
2
2
sin (Q)
=
2P m cos (Q) +
2m
m2 2
Since we required that H = K,
K(P ) = P
We also require that the transformed Hamiltonian satisfy Hamiltons equations of motion.
K
K
Q =
P =
p
Q
6
Thus,
Q =
This equation can be easily integrated,
Q = t + Q0
Also,
K
=0
Q
P = 0
P = constant
Because there is no dissipation,
Etot = K = P
E
P =
3.4.5
Phase space diagrams
Transformed The graph of Q as a function of t increases linearly. P is a
horizontal line.
The phase space diagram is a rectangle with height E . For a single
oscillation, the width of the rectangle is 2. Thus the area enclosed by the
rectangle is
A=
2E
Original system The phase space diagram for the original system is an
ellipse. The area is given by
A = ab =
2E
The system traces the ellipse once per oscillation, so this is the area enclosed
by the ellipse during a single oscillation.
That the areas enclosed by the two phase space diagrams is the same is
a demonstration of Louisvilles theorem.
7
General method (H & F p. 211)
4.1
Recipe
1. Choose a specific generating function F
2. Use
P =
F
Q
p=
F
q
to give specific equations of the transformation
3. Find K(Q, P, t) from
Pi Qi K +
dF
= pi qi H
dt
expressing p and q in terms of P and Q
4. Apply Hamiltons equations of motion.
Type
Function
Derivatives
F1 (q, Q, t)
1
pi = F
qi
F1
Pi = Q
i
F2 (q, Q, t)
2
pi = F
qi
F2
Pi = Q
i
F3 (q, Q, t)
3
pi = F
qi
F3
Pi = Q
i
F4 (q, Q, t)
4
pi = F
qi
F4
Pi = Q
i
Example
F1 = qi Qi
Qi = pi
Pi = qi
F2 = qi Pi
Qi = qi
Pi = p i
F3 = qi Qi
Qi = qi
Pi = pi
F4 = pi Pi
Qi = pi
Pi = qi
Table 1: Types of generating functions
