Mechanics
Physics 151
Lecture 20
Canonical Transformations
(Chapter 9)
What We Did Last Time
Hamilton’s Principle in the Hamiltonian formalism
Derivation was simple δ I ≡ δ ∫
t2
t1
( pi qi − H (q, p, t ) ) dt = 0
Additional end-point constraints
δ q (t1 ) = δ q(t2 ) = δ p(t1 ) = δ p(t2 ) = 0
Not strictly needed, but adds flexibility to the definition
of the action integral
This connects to: Canonical Transformations
t2
Principle of Least Action ∆ ∫t pi qi dt = 0
1
Got into
this a bit
Canonical Transformation
Goal: To find transformations
Qi = Qi (q1 ,… , qn , p1 ,… , pn , t ) Pi = Pi (q1 ,… , qn , p1 ,… , pn , t )
that satisfy Hamilton’s equation of motion
∂H
qi =
dpi
∂H
pi = −
dqi
∂K
Pi = −
dQi
∂K
Qi =
dPi
K is the transformed Hamiltonian K = K (Q, P, t )
Hamilton’s principle requires
δ∫
t2
t1
( pi qi − H (q, p, t ) ) dt = 0
and δ ∫
t2
t1
( PQ − K (Q, P, t ) ) dt = 0
i
i
General Transformation
δ∫
t2
t1
and δ ∫
t1
( PQ − K (Q, P, t ) ) dt = 0
i
i
Two types of transformations are possible
( pi qi − H (q, p, t ) ) dt = 0
t2
PQ
i i − K = λ ( pi qi − H )
Scale transformation
dF
PQ
= pi qi − H
Canonical transformation
i i −K +
dt
Both satisfy Hamilton’s principle
Combined, we find
dF
PQ
= λ ( pi qi − H )
i i −K +
dt
Extended Canonical
transformation
Scale Transformation
We can always change the scale of (or unit we use to
measure) coordinates and momenta
Pi = ν pi Qi = µ qi
To satisfy Hamilton’s principle, we can define
K ( P, Q, t ) = µν H ( p, q, t )
PQ
i i − K = µν ( pi qi − H )
Scale transformation
This is trivial
We now concentrate on – Canonical transformations
Canonical Transformation
dF
= pi qi − H
dt
Hamilton’s principle
PQ
i i −K +
δ∫
t2
t1
( PQ − K ) dt = δ ∫
i
i
t2
t1
dF ⎞
t2
⎛
⎜ pi qi − H −
⎟ dt = −δ [ F ]t1 = 0
dt ⎠
⎝
Satisfied if δp = δq = δP = δQ = 0 at t1 and t2
F can be any function of pi, qi, Pi, Qi and t
It defines a canonical transformation
Call it the generating function of the transformation
or generator
Simple Example [1]
dF
PQ
= pi qi − H
i i −K +
dt
Try a generating function: F = qi Pi − Qi Pi
Canonical transformation generated by F is
dF
PQ
= − K + (qi − Qi ) Pi + Pq
i i −K +
i i = pi qi − H
dt
Qi = qi Pi = pi
Identity transformation
K=H
OK, that was too simple
Let’s push this one step further…
Simple Example [2]
dF
PQ
= pi qi − H
i i −K +
dt
Let’s try this one: F = fi (q1 ,… , qn , t ) Pi − Qi Pi
fi
are arbitrary functions of q1…qn and t
∂fi
∂fi
dF
PQ
= − K + ( f i − Qi ) Pi + Pi
qj +
Pi = pi qi − H
i i −K +
dt
∂q j
∂t
Qi = fi (q1 ,… , qn , t )
pi =
∂f j
∂qi
Pj
All “point transformations” of
generalized coordinates are covered
Must invert these n equations to get Pi
∂fi
K=H+
Pi
∂t
We can do all what we could do before
Arbitrarity
Generating function F Æ a canonical transformation
Opposite mapping is not unique
There are many possible Fs for each transformation
e.g. add an arbitrary function of time g(t) to F
dF
dF dg (t )
Does not affect
PQ
−
K
+
→
PQ
−
K
+
+
i i
i i
the action integral
dt
dt
dt
dg (t )
K→K+
dt
Just modifies the Hamiltonian
without affecting physics
F is arbitrary up to any function of time only
So is the Hamiltonian
Finding the Generator
PQ
i i −K +
dF
= pi qi − H
dt
Let’s look for a generating function
Suppose K (Q, P, t ) = H (q, p, t ) for simplicity
dF
= pi qi − PQ
i i
dt
Easiest way to satisfy this would be
∂F
∂F
= pi
= − Pi
F = F ( q, Q )
∂qi
∂Qi
Trivial example: F (q, Q) = qi Qi
pi = Qi Pi = −qi
In the Hamiltonian formalism,
you can freely swap the
coordinates and the momenta
Type-1 Generator
PQ
i i −K +
dF
= pi qi − H
dt
F = F (q, Q) is not very general
It does not allow t-dependent transformation
Fix this by extending to F = F1 ( q, Q, t )
Call it Type-1
pi =
∂F1 (q, Q, t )
∂qi
Pi = −
∂F1 (q, Q, t )
∂Qi
This affects the Hamiltonian
∂F1
∂F1
dF ∂F1
=
qi +
Qi +
= pi qi − PQ
i i +K −H
dt ∂qi
∂Qi
∂t
∂F1
K=H+
∂t
Harmonic Oscillator
Consider a 1-dimensional harmonic oscillator
p 2 kq 2
1
k
2
2 2 2
2
H ( q, p ) =
+
=
p +m ω q )
ω ≡
(
2m 2
2m
m
Sum of squares Æ Can we make them sine and cosine?
f ( P)
Suppose p = f ( P ) cos Q q =
sin Q
mω
2
f
(
P
)
{
}
Q is cyclic Æ P is constant
K=H =
2m
Trick is to find f(P) so that the transformation is canonical
How?
Harmonic Oscillator
Let’s try a Type-1 generator
F1 (q, Q, t )
∂F1
p=
∂q
∂F1
P=−
∂Q
Express p as a function of q and Q
f ( P)
sin Q
p = f ( P) cos Q q =
mω
Integrate with q
p = mω q cot Q
mω q2 q2
cot Q
F1 =
2
∂F1
mω q 2
P=−
=
∂Q 2sin 2 Q
We are getting somewhere
Harmonic Oscillator
∂F
p = 1 = mω q cot Q
∂q
We need to turn H(q, p) into K(Q, P)
Solve the above equations for q and p
2P
q=
sin Q
mω
∂F1
mω q 2
P=−
=
∂Q 2sin 2 Q
p = 2 Pmω cos Q
Now work out the Hamiltonian
1
K=H =
p 2 + m 2ω 2 q 2 ) = ω P
(
2m
Things don’t get much simpler than this…
Harmonic Oscillator
K = ωP = E
Solving the problem is trivial
P = const =
E
ω
∂K
= ω Q = ωt + α
Q=
∂P
p = 2 Pmω cos Q = 2mE cos(ω t + α )
Finally
2P
2E
q=
sin Q =
sin(ω t + α )
2
mω
mω
Phase Space
Oscillator moves in the p-q and P-Q phase spaces
p
P
2mE
2E
mω 2
E
ω
q
One cycle draws the same area
2π E
ω
2π
Q
in both spaces
The area swept by a cyclic system in the phase space is
invariant
Will come back to this in Lecture 23
Other Types of Generators
Type-1 generator F = F1 (q, Q, t ) is still not so general
We need generating functions of different set of
independent variables
Just try to find a generator for Qi = qi Pi = pi
In fact, we may have 4 basic types of them
F1 (q, Q, t ) F2 (q, P, t ) F3 ( p, Q, t ) F4 ( p, P, t )
We can derive them using the now-familiar rule
i.e. we can add any dF/dt inside the action integral
Type-2 Generator
In the last lecture, I used F = −qi pi to convert
δ∫
t2
t1
( pi qi − H (q, p, t ) ) dt = 0
δ∫
t2
t1
( − pi qi − H (q, p, t ) ) dt = 0
Switch the definition of canonical transformations
dF
dF
PQ
= pi qi − H
− Pi Qi − K +
= pi qi − H
i i −K +
dt
dt
dF
= pi qi + Qi Pi + K − H
dt
To satisfy this
∂F2
= pi
F = F2 (q, P, t )
∂qi
∂F2
= Qi
∂Pi
∂F2
K=H+
∂t
Type-2 Generator
If we go back to the original definition of generating
dF
= pi qi − H
dt
∂F2
∂F2
∂F2
=
p
=
=
+
Q
K
H
F = F2 (q, P, t ) − Qi Pi
i
i
∂Pi
∂qi
∂t
function PQ
i i −K +
Trivial case: F2 = qi Pi
pi = Pi Qi = qi
Identity transformation
We push the same idea to define the other 2 types
Four Basic Generators
Generator
Derivatives
Trivial Case
Qi = pi
∂F1
∂F1
F1 (q, Q, t )
F1 = qi Qi
pi =
Pi = −
Pi = −qi
∂Qi
∂qi
Qi = qi
∂F2
∂F2
F2 (q, P, t ) − Qi Pi
F2 = qi Pi
pi =
Qi =
Pi = pi
∂Pi
∂qi
Qi = −qi
∂F3
∂F3
F3 ( p, Q, t ) + qi pi
F3 = pi Qi
qi = −
Pi = −
Pi = − pi
∂pi
∂Qi
Qi = pi
∂F4
∂F4
F4 ( p, P, t ) + qi pi − Qi Pi qi = −
F4 = pi Pi
Qi =
Pi = −qi
∂pi
∂Pi
Four Basic Generators
The 4 types of generators are almost equivalent
It may look as if F1 is special, but it isn’t
dF1
PQ
i i −K +
dt
dF2
− PQ
i i −K +
dt
dF3
PQ
i i −K +
dt
dF4
− PQ
i i −K +
dt
= pi qi − H
= pi qi − H
= − pi qi − H
= − pi qi − H
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
This does not present a fundamental problem
e.g. identity transf. is generated only by F2 or F3
One can always swap coordinate and momentum
Qi = pi Pi = −qi
One can always change sign by scale transformation
Qi = ± qi Pi = ± pi
These transformations make the 4 types practically
equivalent
One More Example
p2
1
H
=
+
1-dim system with
2 2q 2
Try P = pq
Let’s use Type-2
∂F2
∂F2
=p
F = F2 (q, P, t )
=Q
∂q
∂P
E
P
p=
q
Step 1: Express p with q and P
Step 2: Integrate with q to get
F2 = P log q
assuming q > 0
Step 3: Differentiate to get Q = log q
q = eQ
Now we have a canonical transformation
V=
1
2q 2
One More Example
P
= Pe − Q
q
Now rewrite the Hamiltonian
p2
1
P 2 + 1 −2Q
+ 2 =
H=
e =E
2 2q
2
F2 = P log q q = eQ p =
constant
Equation of motion: P = ( P 2 + 1)e −2Q = 2 E
P = 2 Et + C
q=e =
Q
P2 + 1
C2 +1
2
= 2 Et + 2Ct +
2E
2E
Summary
Canonical transformations
They are all practically equivalent
Used it to simplify a harmonic oscillator
dF
= pi qi − H
dt
Hamiltonian formalism is
invariant under canonical + scale transformations
Generating functions define canonical transformations
Four basic types of generating functions
F1 (q, Q, t ) F2 (q, P, t ) F3 ( p, Q, t ) F4 ( p, P, t )
PQ
i i −K +
Invariance of phase space area
