MechanicsPhysics 151

Lecture 20Canonical Transformations

(Chapter 9)

What We Did Last Time

Hamilton’s Principle in the Hamiltonian formalismDerivation was simple

Additional end-point constraints

Not strictly needed, but adds flexibility to the definitionof the action integral

This connects to: Canonical Transformations

Principle of Least Action

1 2 1 2( ) ( ) ( ) ( ) 0q t q t p t p tδ δ δ δ= = = =

( )2

1

( , , ) 0t

i itI p q H q p t dtδ δ≡ − =∫

2

1

0t

i itp q dt∆ =∫

Got into this a bit

Canonical Transformation

Goal: To find transformations

that satisfy Hamilton’s equation of motion

K is the transformed Hamiltonian

Hamilton’s principle requires

1 1( , , , , , , )i i n nQ Q q q p p t= … … 1 1( , , , , , , )i i n nP P q q p p t= … …

ii

Hpdq∂

= −ii

Hqdp∂

= ii

KPdQ∂

= −ii

KQdP∂

=

( , , )K K Q P t=

( )2

1

( , , ) 0t

i itp q H q p t dtδ − =∫ ( )2

1

( , , ) 0t

i itPQ K Q P t dtδ − =∫and

General Transformation

Two types of transformations are possible

Both satisfy Hamilton’s principle

Combined, we find

( )2

1

( , , ) 0t

i itp q H q p t dtδ − =∫ ( )2

1

( , , ) 0t

i itPQ K Q P t dtδ − =∫and

( )i i i iPQ K p q Hλ− = −

i i i idFPQ K p q Hdt

− + = −

Scale transformation

Canonical transformation

( )i i i idFPQ K p q Hdt

λ− + = − Extended Canonicaltransformation

Scale Transformation

We can always change the scale of (or unit we use to measure) coordinates and momenta

To satisfy Hamilton’s principle, we can define

This is trivial

We now concentrate on – Canonical transformations

( , , ) ( , , )K P Q t H p q tµν=

i iP pν= i iQ qµ=

( )i i i iPQ K p q Hµν− = − Scale transformation

Canonical Transformation

Hamilton’s principle

Satisfied if δp = δq = δP = δQ = 0 at t1 and t2F can be any function of pi, qi, Pi, Qi and t

It defines a canonical transformationCall it the generating function of the transformation

i i i idFPQ K p q Hdt

− + = −

( ) [ ]2 2 2

11 1

0t t t

i i i i tt t

dFPQ K dt p q H dt Fdt

δ δ δ⎛ ⎞− = − − = − =⎜ ⎟⎝ ⎠∫ ∫

or generator

Simple Example [1]

Try a generating function:Canonical transformation generated by F is

OK, that was too simpleLet’s push this one step further…

i i i iF q P Q P= −

( )i i i i i i i i idFPQ K K q Q P Pq p q Hdt

− + = − + − + = −

i iQ q= i iP p= Identity transformation

K H=

i i i idFPQ K p q Hdt

− + = −

Simple Example [2]

Let’s try this one:fi are arbitrary functions of q1…qn and t

We can do all what we could do before

( ) i ii i i i i i j i i i

j

f fdFPQ K K f Q P P q P p q Hdt q t

∂ ∂− + = − + − + + = −

∂ ∂

1( , , , )i n i i iF f q q t P Q P= −…

1( , , , )i i nQ f q q t= …

ji j

i

fp P

q∂

=∂

All “point transformations” of generalized coordinates are covered

ii

fK H Pt

∂= +

∂

Must invert these n equations to get Pi

i i i idFPQ K p q Hdt

− + = −

Arbitrarity

Generating function F a canonical transformationOpposite mapping is not unique

There are many possible Fs for each transformatione.g. add an arbitrary function of time g(t) to F

F is arbitrary up to any function of time onlySo is the Hamiltonian

( )i i i i

dF dF dg tPQ K PQ Kdt dt dt

− + → − + + Does not affect the action integral

( )dg tK Kdt

→ + Just modifies the Hamiltonian without affecting physics

Finding the Generator

Let’s look for a generating functionSuppose for simplicity

Easiest way to satisfy this would be

Trivial example:

i i i idFPQ K p q Hdt

− + = −

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

i i i idF p q PQdt

= −

( , )F F q Q= ii

F PQ∂

= −∂i

i

F pq∂

=∂

( , ) i iF q Q q Q=

i ip Q= i iP q= − In the Hamiltonian formalism, you can freely swap the

coordinates and the momenta

Type-1 Generator

is not very generalIt does not allow t-dependent transformationFix this by extending to

This affects the Hamiltonian

( , )F F q Q=

1( , , )F F q Q t= Call it Type-1

1( , , )i

i

F q Q tPQ

∂= −

∂1( , , )

ii

F q Q tpq

∂=

∂

i i i idFPQ K p q Hdt

− + = −

1 1 1i i i i i i

i i

F F FdF q Q p q PQ K Hdt q Q t

∂ ∂ ∂= + + = − + −∂ ∂ ∂

1FK Ht

∂= +

∂

Harmonic Oscillator

Consider a 1-dimensional harmonic oscillator

Sum of squares Can we make them sine and cosine?

Suppose

Trick is to find f(P) so that the transformation is canonicalHow?

( )2 2

2 2 2 21( , )2 2 2p kqH q p p m qm m

ω= + = + 2 km

ω ≡

( ) cosp f P Q=( ) sinf Pq Q

mω=

{ }2( )2

f PK H

m= = Q is cyclic P is constant

Harmonic Oscillator

Let’s try a Type-1 generator

Express p as a function of q and Q

Integrate with q

( ) cosp f P Q=( ) sinf Pq Q

mω=

1( , , )F q Q t 1Fpq

∂=∂

1FPQ∂

= −∂

cotp m q Qω=

2

1 cot2

m qF Qω=

2

1 cot2

m qF Qω=

21

22sinF m qPQ Q

ω∂= − =

∂We are getting somewhere

Harmonic Oscillator

We need to turn H(q, p) into K(Q, P)Solve the above equations for q and p

Now work out the Hamiltonian

Things don’t get much simpler than this…

1 cotFp m q Qq

ω∂= =∂

21

22sinF m qPQ Q

ω∂= − =

∂

2 sinPq Qmω

= 2 cosp Pm Qω=

( )2 2 2 212

K H p m q Pm

ω ω= = + =

Harmonic Oscillator

Solving the problem is trivialK P Eω= =

KQP

ω∂= =∂

Q tω α= +const EPω

= =

2 cos 2 cos( )p Pm Q mE tω ω α= = +

2

2 2sin sin( )P Eq Q tm m

ω αω ω

= = +Finally

Oscillator moves in the p-q and P-Q phase spaces

One cycle draws the same area in both spaces

The area swept by a cyclic system in the phase space is invariant

Phase Space

p

q

22E

mω

2mE

Q

PEω

2π2 Eπω

Will come back to this in Lecture 23

Other Types of Generators

Type-1 generator is still not so generalJust try to find a generator for

We need generating functions of different set of independent variables

In fact, we may have 4 basic types of them

We can derive them using the now-familiar rulei.e. we can add any dF/dt inside the action integral

1( , , )F F q Q t=

i iQ q= i iP p=

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

Type-2 Generator

In the last lecture, I used to convert

Switch the definition of canonical transformations

To satisfy this

( )2

1

( , , ) 0t

i itp q H q p t dtδ − =∫

i iF q p= −

( )2

1

( , , ) 0t

i itp q H q p t dtδ − − =∫

i i i idFPQ K p q Hdt

− + = − i i iidFK p q Hd

Qt

P − + = −−

i i i idF p q Q P K Hdt

= + + −

2 ( , , )F F q P t= 2i

i

F QP

∂=

∂2

ii

F pq

∂=

∂2FK Ht

∂= +

∂

Type-2 Generator

If we go back to the original definition of generating

function

Trivial case:

We push the same idea to define the other 2 types

i i i idFPQ K p q Hdt

− + = −

2 ( , , ) i iF F q P t Q P= − 2i

i

F QP

∂=

∂2

ii

F pq

∂=

∂2FK Ht

∂= +

∂

2 i iF q P=

i iQ q=i ip P= Identity transformation

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∂

=∂

1i

i

FPQ∂

= −∂ 1 i iF q Q= i iQ p=

i iP q= −

2i

i

Fpq

∂=∂

2i

i

FQP

∂=∂ 2 i iF q P=

i iP p=i iQ q=

3i

i

Fqp∂

= −∂

3i

i

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= −

Four Basic Generators

The 4 types of generators are almost equivalentIt may look as if F1 is special, but it isn’t

1i i i i

dFPQ K p q Hdt

− + = −

2i i i i

dFPQ K p q Hdt

− − + = −

3i i i i

dFPQ K p q Hdt

− + = − −

4i i i i

dFPQ K p q Hdt

− − + = − −

There is no reason to consider any of these 4 definitions to be more fundamental than the others

We arbitrarily chose the first form (which happens to be the Lagrangian form) to write the generating functions in the table

Four Basic Generators

Some canonical transformations cannot be generated by all 4 types

e.g. identity transf. is generated only by F2 or F3

This does not present a fundamental problemOne can always swap coordinate and momentum

One can always change sign by scale transformation

These transformations make the 4 types practically equivalent

i iQ p= i iP q= −

i iQ q= ± i iP p= ±

One More Example

1-dim system withTryLet’s use Type-2

Step 1: Express p with q and PStep 2: Integrate with q to get

Step 3: Differentiate to getNow we have a canonical transformation

2

2

12 2pH

q= +

2

12

Vq

=

P pq=

2 ( , , )F F q P t= 2F QP

∂=

∂2F p

q∂

=∂

Ppq

=

2 logF P q=

logQ q= Qq e=

E

assuming q > 0

One More Example

Now rewrite the Hamiltonian

Equation of motion:

QPp Peq

−= =2 logF P q= Qq e=

2 22

2

1 12 2 2

Qp PH e Eq

−+= + = =

2 2( 1) 2QP P e E−= + =

2P Et C= +2 2

21 12 22 2

Q P Cq e Et CtE E+ +

= = = + +

constant

Summary

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 area

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