MechanicsPhysics 151

Lecture 21Canonical Transformations

(Chapter 9)

What We Did Last Time

Canonical transformationsHamiltonian formalism isinvariant under canonical + scale transformationsGenerating functions define canonical transformationsFour basic types of generating functions

They are all practically equivalent

Used it to simplify a harmonic oscillatorInvariance of phase space

i i i idFPQ K p q Hdt

− + = −

1( , , )F q Q t 2 ( , , )F q P t 3 ( , , )F p Q t 4 ( , , )F p P t

Four Basic Generators

Trivial CaseDerivativesGenerator

1( , , )F q Q t

2 ( , , ) i iF q P t Q P−

3 ( , , ) i iF p Q t q p+

4 ( , , ) i i i iF p P t q p Q P+ −

1i

i

Fpq

∂=

∂1

ii

FPQ

∂= −

∂ 1 i iF q Q= i iQ p=

i iP q= −

2i

i

Fpq

∂=

∂2

ii

FQP

∂=

∂ 2 i iF q P=i iP p=i iQ q=

3i

i

Fqp

∂= −

∂3

ii

FPQ

∂= −

∂

4i

i

Fqp

∂= −

∂4

ii

FQP

∂=

∂

3 i iF p Q=i iP p= −i iQ q= −

4 i iF p P= i iQ p=

i iP q= −

Goals for Today

Dig deeper into Canonical TransformationsInfinitesimal Canonical Transformation

Very small changes in q and pDefine generator G for an ICT

Direct Conditions for Canonical TransformationNecessary-and-sufficient conditions for any CT

Poisson BracketInvariant of any Canonical TransformationConnect to Infinitesimal Canonical Transformation

Infinitesimal CT

Consider a CT in which q, p are changed by small (infinitesimal) amounts

ICT is close to identity transf.Generating function should be

i i iQ q qδ= + i i iP p pδ= + Infinitesimal Canonical Transformation (ICT)

2 ( , , ) ( , , )i iF q P t q P G q P tε= +

Identity CT generator Small

2i i

i i

F Gp Pq q

ε∂ ∂= = +

∂ ∂2

i ii i

F GQ qP P

ε∂ ∂= = +

∂ ∂Look at the

generator table

ii i

G GqP p

δ ε ε∂ ∂= ≈

∂ ∂ ii i

G Gpq Q

δ ε ε∂ ∂= − ≈ −

∂ ∂Since ε is

infinitesimal

Generator of ICT

An ICT is generated by

G is called (inaccurately) the generator of the ICTSince the CT is infinitesimal, G may be expressed in terms of q or Q, p or P, interchangeably

For example:

2 ( , , ) ( , , )i iF q P t q P G q P tε= +

i ii

GQ qP

ε ∂= +

∂ i ii

GP pq

ε ∂= −

∂

( , , )G G q p t= i ii

GQ qp

ε ∂= +

∂ i ii

GP pq

ε ∂= −

∂

Hamiltonian

Consider

What does ε look like? Infinitesimal time δt

Hamiltonian is the generator of infinitesimal time transformation

In QM, you learn that Hamiltonian is the operator that represents advance of time

( , , )G H q p t=

i ii

Hq qp

δ ε ε∂= =

∂ i ii

Hp pq

δ ε ε∂= − =

∂

i iq q tδ δ= i ip p tδ δ=

Direct Conditions

Consider a restricted Canonical TransformationGenerator has no t dependence

Q and P depends only on q and p

0Ft

∂=

∂( , ) ( , )K Q P H q p= Hamiltonian

is unchanged

( , )i iQ Q q p= ( , )i iP P q p=

i i i ii j j

j j j j j j

Q Q Q QH HQ q pq p q p p q

∂ ∂ ∂ ∂∂ ∂= + = −

∂ ∂ ∂ ∂ ∂ ∂

i i i ii j j

j j j j j j

P P P PH HP q pq p q p p q

∂ ∂ ∂ ∂∂ ∂= + = −

∂ ∂ ∂ ∂ ∂ ∂

Hamilton’s equations

Direct Conditions

On the other hand, Hamilton’s eqns say

i ii

j j j j

Q QH HQq p p q

∂ ∂∂ ∂= −

∂ ∂ ∂ ∂

i ii

j j j j

P PH HPq p p q

∂ ∂∂ ∂= −

∂ ∂ ∂ ∂

j ji

i j i j i

q pH H HQP q P p P

∂ ∂∂ ∂ ∂= = +

∂ ∂ ∂ ∂ ∂

j ji

i j i j i

q pH H HPQ q Q p Q

∂ ∂∂ ∂ ∂= − = − −

∂ ∂ ∂ ∂ ∂

Direct Conditions

for a Canonical Transformation

,,

ji

j i Q Pq p

pQq P

⎛ ⎞ ∂⎛ ⎞∂=⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠ ,,

ji

j i Q Pq p

qQp P

⎛ ⎞ ∂⎛ ⎞∂= −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠

,,

ji

j i Q Pq p

pPq Q

⎛ ⎞ ∂⎛ ⎞∂= −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠ ,,

ji

j i Q Pq p

qPp Q

⎛ ⎞ ∂⎛ ⎞∂=⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠

Direct Conditions

Direct Conditions are necessary and sufficient for a time-independent transformation to be canonical

You can use them to test a CT

In fact, this applies to all Canonical TransformationsBut the proof on the last slide doesn’t work

,,

ji

j i Q Pq p

pQq P

⎛ ⎞ ∂⎛ ⎞∂=⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠ ,,

ji

j i Q Pq p

qQp P

⎛ ⎞ ∂⎛ ⎞∂= −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠

,,

ji

j i Q Pq p

pPq Q

⎛ ⎞ ∂⎛ ⎞∂= −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠ ,,

ji

j i Q Pq p

qPp Q

⎛ ⎞ ∂⎛ ⎞∂=⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠

Infinitesimal CT

Does an ICT satisfy the DCs?2( )i i i

ijj j i j

Q q q Gq q P q

δ δ ε∂ ∂ + ∂= = +

∂ ∂ ∂ ∂

2( )j j jij

i i i j

p P p GP P P q

δδ ε

∂ ∂ − ∂= = +

∂ ∂ ∂ ∂

ii i

G GqP p

δ ε ε∂ ∂= ≈

∂ ∂

ii i

G Gpq Q

δ ε ε∂ ∂= − ≈ −

∂ ∂

2( )i i i

j j i j

Q q q Gp p P p

δ ε∂ ∂ + ∂= =

∂ ∂ ∂ ∂

2( )j j j

i i i j

q Q q GP P P p

δε

∂ ∂ − ∂= = −

∂ ∂ ∂ ∂2( )i i i

j j i j

P p p Gq q Q q

δ ε∂ ∂ + ∂= = −

∂ ∂ ∂ ∂

2( )j j j

i i i j

p P p GQ Q Q q

δε

∂ ∂ − ∂= =

∂ ∂ ∂ ∂2( )i i i

ijj j i j

P p p Gp p Q p

δ δ ε∂ ∂ + ∂= = −

∂ ∂ ∂ ∂

2( )j j jij

i i i j

q Q q GQ Q Q p

δδ ε

∂ ∂ − ∂= = −

∂ ∂ ∂ ∂

Yes!

Successive CTs

Two successive CTs make a CT

Direct Conditions can also be “chained”, e.g.,

1i i i i

dFPQ K p q Hdt

− + = − 2i i i i

dFY X M PQ Kdt

− + = −

1 2( )i i i i

d F FY X M p q Kdt+

− + = − True for unrestricted CTs

,,

ji

j i Q Pq p

pQq P

⎛ ⎞ ∂⎛ ⎞∂=⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠ ,,

ji

j i X YQ P

PXQ Y

⎛ ⎞ ∂⎛ ⎞∂=⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠

,,

ji

j i X Yq p

pXq Y

⎛ ⎞ ∂⎛ ⎞∂=⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠

Easy to prove

Unrestricted CT

Now we consider a general, time-dependent CT

Let’s do it in two steps

First step is t-independent Satisfies the DCsWe must show that the second step satisfies the DCs

( , , )i iQ Q q p t= ( , , )i iP P q p t=FK Ht

∂= +

∂

,q p 0 0( , , ), ( , , )Q q p t P q p t ( , , ), ( , , )Q q p t P q p t

Time-independent CT Time-only CT

Fixed time

Unrestricted CT

Concentrate on a time-only CTBreak t – t0 into pieces of infinitesimal time dt

Each step is an ICT Satisfies Direct Conditions“Integrating” gives us what we needed

The proof worked because a time-only CT is a continuous transformation, parameterized by t

( ), ( )Q t P t0 0( ), ( )Q t P t

0 0( ), ( )Q t P t 0 0( ), ( )Q t dt P t dt+ + ( ), ( )Q t P t

All Canonical Transformations satisfies the Direct Conditions, and vice versa

Poisson Bracket

For u and v expressed in terms of q and p

This weird construction has many useful featuresIf you know QM, this is analogous to the commutator

Let’s start with a few basic rules

[ ] ,,

q pi i i i

u v u vu vq p p q

∂ ∂ ∂ ∂≡ −

∂ ∂ ∂ ∂Poisson Bracket

[ ]1 1, ( )u v uv vui i

≡ − for two operators u and v

Poisson Bracket Identities

For quantities u, v, w andconstants a, b[ , ] 0u u =

[ ] ,,

q pi i i i

u v u vu vq p p q

∂ ∂ ∂ ∂≡ −

∂ ∂ ∂ ∂

[ , ] [ , ]u v v u= −

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

[ , ] [ , ] [ , ]uv w u w v u v w= +

[ ,[ , ]] [ ,[ , ]] [ ,[ , ]] 0u v w v w u w u v+ + =

Jacobi’s Identity

All easy to prove

This one is worth trying.See Goldstein if you are lost

Fundamental Poisson Brackets

Consider PBs of q and p themselves

Called the Fundamental Poisson Brackets

Now we consider a Canonical Transformation

What happens to the Fundamental PB?

[ , ] 0j jk k

i ij k

i i

q qqp q

q q qq p

∂ ∂∂ ∂= −

∂ ∂ ∂=

∂[ , ] 0j kp p =

[ , ] j jk k

i i i ij k jk

q qp pq p q

q pp

δ∂ ∂∂ ∂

=∂ ∂

=−∂ ∂

[ , ]j k jkp q δ= −

, ,q p Q P→

Fundamental PB and CT

Fundamental Poisson Brackets are invariant under CT

,[ , ] 0j j j j jk k i ij k q p

i i i i i k i k k

Q Q Q Q QQ Q q pQ Qq p p q q P p P P

∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂= − = − − = − =

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

,[ , ] 0j j j j jk k i ij k q p

i i i i i k i k k

P P P P PP P q pP Pq p p q q Q p Q Q

∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂= − = + = =

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

,[ , ] j j j j jk k i ij k q p jk

i i i i i k i k k

Q Q Q Q QP P q pQ Pq p p q q Q p Q Q

δ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂

= − = + = =∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

,[ , ] [ , ]j k q p k j jkP Q Q P δ= − = − Used Direct Conditions here

Poisson Bracket and CT

What happens to a Poisson Bracket under CT?For a time-independent CT

[ ] ,

, ,

,

[ , ] [ , ] [

j j j jk k k k

j i j i k i k i j i j i k i k i

j k j k j k

Q Pi i i i

j k Q P j k Q P

q p q pq p q pu u v v u u v v

q Q p Q q P p P q P p P q Q p Q

u v u v u v

q q q p p q

u v u vu vQ P P Q

q q q p

∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + − + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂≡ −

∂ ∂ ∂ ∂

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

= + +

[ ]

, ,

,

, ] [ , ]

,

j k

j k j k

j k Q P j k Q P

jk jk

q p

u v

p p

u v u v

q p p q

p q p p

u v

δ δ

∂ ∂

∂ ∂

∂ ∂ ∂ ∂

∂ ∂ ∂ ∂

+

= −

= Poisson Brackets are invariant under CT

Invariance of Poisson Bracket

Poisson Brackets are canonical invariantsTrue for any Canonical Transformations

Goldstein shows this using “simplectic” approach

We don’t have to specify q, p in each PB[ ] ,

,q p

u v [ ],u v good enough

ICT and Poisson Bracket

Infinitesimal CT can be expressed neatly with a PB

For a generator G,

On the other hand

We can generalize further…

i ii

GQ qp

ε ∂= +

∂ i ii

GP pq

ε ∂= −

∂

[ , ] i ii i

j j j j i

q qG G Gq G qq p p q p

ε ε ε δ⎛ ⎞∂ ∂∂ ∂ ∂

= − = =⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠

[ , ] i ii i

j j j j i

p pG G Gp G pq p p q q

ε ε ε δ⎛ ⎞∂ ∂∂ ∂ ∂

= − = − =⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠

ICT and Poisson Bracket

For an arbitrary function u(q,p,t), the ICT does

That is

[ , ]

ICTi i

i i

i i i i

u u uu u u u q p tq p tu G u G uu tq p p q t

uu u G tt

δ δ δ δ

ε ε δ

ε δ

∂ ∂ ∂⎯⎯⎯→ + = + + +

∂ ∂ ∂∂ ∂ ∂ ∂ ∂

= + − +∂ ∂ ∂ ∂ ∂

∂= + +

∂

[ , ] uu u G tt

δ ε δ∂= +

∂

Infinitesimal Time Transf.

Hamiltonian generates infinitesimal time transf.Applying the Poisson Bracket rule

Have you seen this in QM?

If u is a constant of motion,

That is,

[ , ] uu t u H tt

δ δ δ∂= +

∂[ , ]du uu H

dt t∂

= +∂

[ , ] 0uu Ht

∂+ =

∂

[ , ] uH ut

∂=

∂u is a constant of motion

Infinitesimal Time Transf.

If u does not depend explicitly on time,

Try this on q and p

[ , ] [ , ]du uu H u Hdt t

∂= + =

∂

[ , ] i ii i

j j j j i

p pH H Hp p Hq p p q q

∂ ∂∂ ∂ ∂= = − = −

∂ ∂ ∂ ∂ ∂

[ , ] i ii i

j j j j i

q qH H Hq q Hq p p q p

∂ ∂∂ ∂ ∂= = − =

∂ ∂ ∂ ∂ ∂ Hamilton’sequations!

Summary

Direct ConditionsNecessary and sufficientfor Canonical Transf.

Infinitesimal CTPoisson Bracket

Canonical invariantFundamental PB

ICT expressed by

Infinitesimal time transf. generated by Hamiltonian

,,

ji

j i Q Pq p

pQq P

⎛ ⎞ ∂⎛ ⎞∂=⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠ ,,

ji

j i Q Pq p

qQp P

⎛ ⎞ ∂⎛ ⎞∂= −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠

,,

ji

j i Q Pq p

pPq Q

⎛ ⎞ ∂⎛ ⎞∂= −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠ ,,

ji

j i Q Pq p

qPp Q

⎛ ⎞ ∂⎛ ⎞∂=⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠

[ ],i i i i

u v u vu vq p p q

∂ ∂ ∂ ∂≡ −

∂ ∂ ∂ ∂

[ , ] [ , ] 0i j i jq q p p= = [ , ] [ , ]i j i j ijq p p q δ= − =

[ , ] uu u G tt

δ ε δ∂= +

∂