8/8/2019 PHYS2100_HM_week2

1/16

Chapter 3

Canonical transformations

3.1 Volume preserving ows

The concept of the conservation of volume in phase space is very important in Hamiltonian me-chanics. For a system with n degrees of freedom, volume is dened as

V = dq dp , (3.1)where is the integration volume. Thus for a system with one degree of freedom, the volume isan area with units kg m2 s 1

J s .

Liouvilles theorem: The volume of a phase space ow is conserved for Hamiltonian systems.

We can show this explicity for a 1D conservative Hamiltonian system.

Consider a small rectangular region bounded by two paths in phase space, with energies given byH and H + H respectively as depicted in Fig. 3.1. The magnitude of the ow vector eld is givenby

|v

|=

H

p,

H

q=

H

p

2

+H

q

2 1/ 2

=

|H

|, (3.2)

So the length x of the area A is simply

x |v |t = |H |t. (3.3)As H is perpendicular to curves of constant H then it follows that

|H | H y

. (3.4)

21

8/8/2019 PHYS2100_HM_week2

2/16

q

p

H+ H

H

x

y

Figure 3.1: Volume preserving ows in phase space.

Thus the area A is given by

A = xy = |H |t y =H y

t y = H t. (3.5)So what this means is that for any point along the two trajectories, the area will be given by H t,and these quantities are chosen by us. Thus the area of the element is conserved as it moves throughphase space.

This result also holds of non-conservative Hamiltonian systems, and for any number of degrees of freedom.

3.2 Change of coordinates

There are many choices of coordindates to represent a particular Hamiltonian system. A goodchoice will

1. Reduce Hamiltons euqations to the simplest possible form.

2. Preserve the integrity of Hamiltons equations.

So given a system described by coordinates q and p with Hamiltonian H (q, p) where

q =H (q, p)

p, p =

H (q, p)q

, (3.6)

22

8/8/2019 PHYS2100_HM_week2

3/16

AR p

q Q

P AS

Figure 3.2: Coordinate transform of area AR in (q, p) to area AS in (Q, P ).

we wish to nd a transformation (q, p) (Q, P ) that simplies the analysis of the system andalso ensures that

Q = H (Q, P )

P , P =

H (Q, P )Q

, H (q, p) = H (Q, P ). (3.7)

Such a transformation is called a canonical transformation .

To determine if a transformation is canonical we exploit the area preserving nature of a Hamil-tonian system. Consider a region R with area AR in (q, p) space. Suppose it is transformed to a

region S with area AS in the (Q, P ) system, as depicted in Fig. 3.2. The area of the region R is

AR = R dq dp, (3.8)and the area of region S is

AS = S dQdP, (3.9)In general there is no reason to assume that AR and AS will be the same. In fact, it can be shownthat

S dQdP = R dq dp (Q, P ) (q, p) . (3.10)where (Q,P ) (q,p) is known as the Jacobian of the transformation. It is dened by

(Q, P ) (q, p)

=Qq

Qp

P q

P p

=Qq

P p

Qp

P q

. (3.11)

23

8/8/2019 PHYS2100_HM_week2

4/16

Example

We again consider a freely rotating mass. We may choose our set of generalised coordinates as(q, p) where q is the distance around the circumference of the pivot from a xed point, and p =

mq. An alternative set of coordinates is (, ) as considered in Section 2.5.3. The transformationbetween the coordinates is

=qa

, = ma 2 = ma q = ap. (3.12)

The Jacobian is

(, ) (q, p)

=q

p

q

p

=1/a 0

0 a=

1a a 0 0 = 1. (3.13)

Thus the areas in phase space are identical in the two coordinate systems.

An important result of classical mechanics is that for a ow to be area preserving then the areamust be independent of the coordinate system. This can be compactly expressed as

{Q, P }(q,p) = 1 . (3.14)where we have introduced the Poisson bracket notation.

Poisson brackets

For any two functions f (q, p) and g(q, p) the Poisson bracket is dened as

{f, g }(q,p) =f q

gp

f p

gq

. (3.15)

Another important result is that if the Jacobian of a transformation is unity then the Poisson bracketof any two functions f and g is also independent of the representation. That is

(Q, P ) (q, p)

= 1 {f, g }(q,p) = {f, g }(Q,P ) . (3.16)Proof of this fact is left to an exercise.

The Poisson bracket also provides a means of determing time dependence. For example, we seethat

{q, H }(q,p) =qq

H p

qp

H q

= 1 H p 0

H q

=H q

= q, (3.17)

where we have used the fact that q and p are independent variables and Hamiltons equations.Similarly it can be shown that

p = { p,H }. (3.18)

24

http://-/?-http://-/?-

8/8/2019 PHYS2100_HM_week2

5/16

For any function f (q,p,t) it can be shown that

f = {f, H }+f t

, (3.19)

and you are asked to prove this in your tutorial sheet.

So far we have shown that a ow is area preserving if

{Q, P }(q,p) = 1 . (3.20)We now show that if Eq. ( 3.20 ) is satised then the transformation (q, p) (Q, P ) is canonicali.e. Hamiltons equations remain valid under the transformation according to Eq. ( 3.7 ).

Beginning with Q, we have

Q(q, p) =Qq

q +Qp

p, (3.21)

= Qq

Qq

+ Qp H q . (3.22)

As we have H = H , then it follows that

H p

= H Q

Qp

+ H P

P p

, (3.23)

H q

= H Q

Qq

+ H P

P q

, (3.24)

Subsituting Eqs. ( 3.23 ) and ( 3.24 ) back into Eq. ( 3.22 ) gives

Q(q, p) = Qq

H Q

Qp

+ H P

P p Qp H Q Qq + H P P q , (3.25)

= H Q

Qq

Qp

Qp

Qq

+ H P

Qq

P p

Qp

P q

, (3.26)

= H P {Q, P }(q,p) =

H P

, (3.27)

where in the nal line we have made use of Eq. ( 3.20) . Similarly it can be shown that (see tutorialsheet)

P (q, p) =

H

Q . (3.28)Thus in summary, we have shown that if {Q, P }(q,p) = 1 then the transformation (q, p) (Q, P )preserves Hamiltons equations and is thus known as a canonical transformation.

Aside: A lot of the formalism of classical Hamiltonian mechanics carries through to quantummechanics. In particular, the Poisson bracket is replaced by the commutator. When it comes toconsidering the dynamics of an operator A (in what is called the Heisenberg picture), we nd that

dAdt

= [A, H ] + At

, (3.29)

25

8/8/2019 PHYS2100_HM_week2

6/16

where H is the Hamiltonian operator for the system, and we have the denition of the commutatoris

[A, H ] = AH H A. (3.30)

This is equivalent to the classical expressiondAdt

= {A, H }+At

. (3.31)

where now A and H are functions on phase space and H is the Hamiltonian of the system.

3.3 Action-angle variables

One of the main uses of a canonical transformation is to transform to a set of coordinates for whichthe equations of motion take on a particularly simple form.

For example, suppose we nd a transformation (q, p) (Q, P ) such that H Q

= 0 . (3.32)

This has two important consequences:

1. Since P =

H/Q then P is a constant of the motion, and H = H (P ) only.

2. Since Q = H/P , then Q = constant as H/P is a function of P alone. Thus thesolution for Q has the simple form

Q(t) = a1t + a2. (3.33)

for some constants a1 and a2 that are determined by the boundary conditions.

The action-angle variables (I , ) are designed specically to be a transformation of this type. Theyare particularly powerful for application to systems displaying periodic motion.

Consider a conservative system with Hamiltonian

H = H (q, p) =p2

2m+ V (q). (3.34)

We have already shown that H is a constant of the motion. This can be rearranged to give

p(q, H ) = [2m(H V (q)]1/ 2 (3.35)where you should think of H as a constant value that dened a particular trajectory. The sign of pdepends on the region of phase space the solution is in.

26

8/8/2019 PHYS2100_HM_week2

7/16

We now dene the action variable

I =1

2 p dq (3.36)where the integral is over a complete period of motion. From Eqs. ( 3.35 ) and ( 3.36 ) we can see

that I is a function of H alone i.e. I = I (H ). Since H is a constant of motion, then so too will I .

Inverting the relationship, we can also see that H will be a function of I alone i.e. H = H (I ). If we den the generalised coordinate conjugate to I as , then from Hamiltons equations we have

=H I

= (t) = t + , (3.37)

for some (as yet undetermined) constants and .

Note that the action I is simply the area A enclosed by the trajectory with the particular value of H divided by 2 . Thus, we can equivalently write

I =1

2 A dq dp = 12 p(q, H ) dq = 12 q( p,H ) dp. (3.38)To determine and its connection to the parameters of periodic motion consider the function W known as Hamiltons characteristic function

W = q

0 p(I, q) dq. (3.39)

To determine we need only impose the area preserving condition of a canonical transformation

Area W = q0 pdq = I . (3.40)As the area is preserved (see Fig. 3.3)

W = I =W I

, (3.41)

i.e.

=

I q

0 p(I, q) dq. (3.42)

To obtain a physical interpretation of (t), consider the change in over one period of motion

= q dq = 2W

qI dq = W q dq, (3.43)

where we have used Eq. ( 3.41 ) and the fact that I is constant during the motion. From the denitionof the characteristic function Eq. ( 3.39 ) we can see that

W q dq = p dq = 2 I . (3.44)

27

8/8/2019 PHYS2100_HM_week2

8/16

q

p

p(I+ I,q)

p(I,q)

I

I+ I

I

Figure 3.3: Determination of the variable in action-angle coordinates

So this means that over the period of motion changes by 2 . But from Eq. ( 3.37 ) this means that must be 2 multiplied by the inverse of the period T . That is, is the frequency of periodicmotion

=2T

=H I

. (3.45)

Thus, one of the advantages of the action-angle coordinate system becomes clear: without solvingthe equations of motion we can nd the angular frequency of the motion.

Furthermore, since changes by 2 through one period of motion then for any other set of coordi-nates that we may wish to use to describe the system (e.g. , it follows that

( + 2 , I ) = (, ), ( + 2 , I ) = (, ). (3.46)

Example: simple harmonic oscillator

Hamiltonian is

H =p2

2m+

1

2m2q2, (3.47)

and so

p = (2mH m22q2)1/ 2. (3.48)The action variable is

I =1

2 pdq, (3.49)=

12

q1

q1(2mH m22q2)1/ 2dq +

q1

q1 (2mH m22q2)1/ 2dq . (3.50)

28

8/8/2019 PHYS2100_HM_week2

9/16

However, because

a

af (x)dx =

a

af (x)dx, (3.51)

then Eq. ( 3.50 ) becomes

I =1

q1

q1(2mH m22q2)1/ 2dq. (3.52)

The limit of integration q1 is a turning point of the motion where p = 0 . So from the HamiltonianEq. ( 3.48 ) we see that

q1 = 2H m2 (3.53)If we make the variable substitution

q = 2H m2 sin dq = 2H m2 cos d, (3.54)then Eq. ( 3.52 ) becomes

I =1

/ 2

/ 22mH m22

2H m2

sin2 1/ 2 2H m2 cos d, (3.55)

=2H

/ 2

/ 2cos2 d, (3.56)

=H

, (3.57)

where we have used the result that

/ 2

/ 2cos2 =

2

.

The frequency of oscillation is given by H/I and thus it is clear that represents the frequencyof oscillation as before. Also

t = t + , (3.58)

where is determined by the boundary conditions. Thus we have solved the SHO in the action-angle coordinate system.

3.4 Application to the pendulum

We now use action-angle coordinates to nd the general solutions for the dynamics of the pendu-lum. Remember the Hamiltonian is

H (, ) =2

2J mga cos . (3.59)

29

8/8/2019 PHYS2100_HM_week2

10/16

We dene a new parameter

=21 + H

221

1/ 2

, 1 = J0 = mga. (3.60)

where H represents the energy of a particular trajectory. Note that from Eq. ( 2.75 ) 21 is the energyof the system on the separatrix. It follows that

= 1, on the separatrices, < 1, libration (within separatrices), > 1, rotation (outside separatrices).

We have already found the solution for motion on the separatrices. We now consider motion insideand outside.

Librations: < 1

We begin by calculating the action

I (H ) =1

2 pdq. (3.61)For the pendulum the generalised coordinate and momentum are and and therefore

I (H ) =1

2 1

1 d +

1

1 d , (3.62)

= 12

1

1 2J (H + 21 cos ) d + 1

1 2J (H + 21 cos ) d , (3.63)where we have divided the integral over a complete period into two integrals for which > 0 and < 0. Due to the symmetry of the integrals this reduces to

I (H ) =1

1

1 2J (H + 21 cos ) d. (3.64)The limit of integration 1 is the angle at the turning point of the motion, for which = 0 . Bymaking this substituion into the Hamiltonian we nd that

1 = cos 1 H 21 = 2sin

1 . (3.65)

The evaluation of the integral is rather lengthy, and instead we simply quote the result

I (H ) =8J0

E (2) (1 2)K (2) , < 1, (3.66)

where

K (m) = / 2

0(1 m sin2 )

1/ 2d, (3.67)

30

http://-/?-http://-/?-

8/8/2019 PHYS2100_HM_week2

11/16

0 0.5 11

2

3

4

5

m

K ( m

)

0 0.5 11

1.2

1.4

1.6

1.8

m

E ( m

)

Figure 3.4: Complete elliptic integrals of the rst K (m) and second E (m) kind.

is a complete elliptic integral of the rst kind, and

E (m) = / 2

0(1 m sin2 )1/ 2d, (3.68)

is a complete elliptic integral of the second kind. (See http://mathworld.wolfram.com for a handyreference to special functions such as these.) These functions are plotted in Fig. 3.4.

Earlier we showed that the angular frequency of motion is given by

= H I = I (H )H

1

. (3.69)

Using the properties of elliptic integrals we can differentiate Eq. ( 3.66 ) with respect to H to nd

(H ) =2

0K (2)

, < 1. (3.70)

This result is plotted in Fig. 3.5 .

Rotations: > 1

As before we have

I (H ) =1

2 p dq = 12

d. (3.71)

where now the pendulum explores all values of . We shall assume that the motion is in theanticlockwise direction, and so > 0. Thus

I (H ) =1

2

2J (H + 21 cos ) d. (3.72)

31

8/8/2019 PHYS2100_HM_week2

12/16

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

( ) /

0

Figure 3.5: Angular frequency of a pendulum as a function of (H ).

Using the fact that cosine is an even function cos = cos() and the result cos = 1 2sin2(/ 2), we can write Eq. ( 3.72 ) asI (H ) =

1

02J (H + 21 221 sin2 / 2)

1/ 2d, (3.73)

=2J

(H + 21)

1/ 2

01

221H + 21

sin22

1/ 2

d (3.74)

Making the change of variable = / 2 we nd

I (H ) = 2J 221 H + 21

221

1/ 2

2 / 2

01

12

sin2 1/ 2

d , (3.75)

=41J

E (1/ 2). (3.76)

We nd the angular frequency again according to Eq. ( 3.69) and nd

(H ) =1J

K (1/ 2)

, > 1. (3.77)

Combining this result with the frequency for < 1 from Eq. ( 3.70) we get Fig. 3.5 . As well asproducing this plot on a computer, we can get it from the series expansion of K (k)

K (m) =2

1 +12

2

m +1 32 4

2

m2 +1 3 52 4 6

2

m3 + . . . . (3.78)

and so we see K (m) / 2 as m 0. Thus we have the three limiting cases for the frequency 0 as 0, 0 as 1, 20 as .

(3.79)

32

8/8/2019 PHYS2100_HM_week2

13/16

You should try to derive these results yourself.

Thus, we have exploited one of the advantages of action-angle coordinates: we have found theangular frequency of the pendulum without actually solving the equations of motion.

Full solution: (t)

We can now work out the full solution to Hamiltons equations for (t) and (t). We applyEq. ( 3.42 ) to the pendulum to nd

() =

I

0( , H ) d . (3.80)

If we solve this for (, H ) and substitute in

(t) = (H )t + (0), (3.81)

then the solutions are

(t) = 2cos 1[dn(0t, 2)], 1,2sin 1[sn(0t, 1/ 2)], > 1, (3.82)

where we have taken the boundary condition to be (0) = 0 . Here we have introduced the Jacobielliptic functions dn and sn.

Aside: Jacobi elliptic functions

Adapted from http://mathworld.wolfram.com/JacobiEllipticFunctions.html (11/10/2006):

The Jacobi elliptic functions are standard forms of elliptic functions. The three basic functions aredenoted cn( u, m ), dn( u, m ), and sn( u, m ), where m = k2 and k is known as the elliptic modulus.They arise from the inversion of the elliptic integral of the rst kind

u = K (, m) =

0

dt1 m sin2 t

, (3.83)

where 0 < m < 1, and = am (u, m ) = am(u) is the Jacobi amplitude, giving = K 1(u, m ) =am (u, m ).

From this, it follows that

sin = sin( am(u, m )) sn(u, m ), (3.84)cos = cos( am(u, m )) cn(u, m ), (3.85)

1 k2 sin2 = 1 m sin2(am (u, m )) = dn(u, m ). (3.86)These functions are doubly periodic generalizations of the trigonometric functions satisfying

sn(u, 0) = sin u, cn(u, 0) = cos u, dn(u, 0) = 1 . (3.87)

33

8/8/2019 PHYS2100_HM_week2

14/16

0 5 103

2

1

0

1

2

3

t / 0

( t )

(a)

0 5 102

1

0

1

2

t / 0

( t )

(b)

0 5 100

5

10

15

20

t / 0

l ( t ) /

0 J

(c)

0 5 100

0.5

1

1.5

2

2.5

t / 0

l ( t ) /

0 J

(d)

Figure 3.6: Solutions for the pendulum. (a) (t) and (b) (t) for = 0.2 (solid), = 0.6 (dashed),and = 0.99 (dotted). (c) (t) and (d) (t) for = 1 (solid), = 1.02 (dashed), and = 1.2(dotted).

34

8/8/2019 PHYS2100_HM_week2

15/16

Full solution: (t)

With the boundary condition (0) = 0 , then we can see that at t = 0

H =2(0)2J mga =

(0)2

2J 21 . (3.88)

Inserting this into the denition of from Eq. ( 3.60 ) we have

2 =21 + (0)2/ 2J 21

221, (3.89)

(0) = 2J 1. (3.90)Using this as the boundary condition we can then show that the angular momentum is

(t) =20J cn(0t, 2)],

1,

20J dn(0t, 1/ 2)], > 1. (3.91)

This can be derived by, for example, substituting the solution Eq. ( 3.82 ) into the Hamiltonian andsolving for (t). Examples of the solutions are shown in Fig. 3.6.

Limiting cases

The Jacobi elliptic functions can be written in terms of innite expansions which allow us to ndsimple expressions for (t) and (t) in limiting cases.

sn(u, m ) = u (1 + m)u3

3!+ (1 + 14 m + m2)

u5

5!+ . . . , (3.92)

cn(u, m ) = 1 u2

2!+ (1 + 4 m)

u4

4! (1 + 44m + 16 m2)

u6

6!+ . . . , (3.93)

dn(u, m ) = 1 mu2

2!+ m(4 + m)

u4

4! m(16 + 44m + m2)

u6

6!+ . . . . (3.94)

Case: 1

In this limit we nd that

cn(u, 0) = 1 u2

2!+

u4

4! u6

6!+ . . . = cos u, (3.95)

and so we nd for the angular momentum

(t) = 20J cos(0t), (3.96)

which is the same as the SHO solution in the limit of small oscillations.

35

8/8/2019 PHYS2100_HM_week2

16/16

Case: = 1

Here we nd

cn(u, = 1) = 1 u2

2! + 5u4

4! 61u6

6! + . . . =1

cosh u , (3.97)

and so we have

(t) =20J

cosh(0t), (3.98)

as we have already derived for the separatrix.

Case: 1

Here we have

dn(u, 1/ 2 0) 1 1

2u2

2!+

14

u4

4! 1

6u6

6!+ . . . 1. (3.99)

Ignoring the terms of O( 1/ 2) and higher we have

(t) =20J

, (3.100)

which is constant rotation as we might have expected.

36