Canonical Transformations
For a given system of particles, we can
different sets of generalized coordinates.
define
A transformation between two sets of generalized
coordinates is called a canonical transformations if it
leaves the Hamilton’s Equations invariant.
Suppose {qi , pi }is one set of generalized coordinates and
generalized momenta. H is the Hamiltonian of the system
then
H
H
 qi and
  p i .
pi
qi
Suppose {Qi , Pi } is another set of generalized
coordinates and generalized momenta.
The transformation
{qi, pi }
{Qi , Pi }
Is called a canonical transformation if there exists a
function Ho , called Hamiltonian of the system with the
coordinates {Qi , Pi }, satisfying the equations
H o 
H o
 Qi and
  Pi
Pi
Qi
.
Theorem
The transformation {qi , pi }
 p dq  P dQ
i
i
i
i
{Qi , Pi } is canonical if
is an exact differential.
i
E.g. Show that the following transformation of one degree
of freedom is canonical. 
1 2
2

 P  2 p  q

1  q 
 Q  tan  

 p
Solution : Since Q  tan 1  q 
 p
We get dQ=

p dq  q dp p dq  q dp
dQ 

2
2
2
2
p
p

q
q

1  
 p
1

 p dq  P dQ  p dq  P dQ
i
i
i
i
i


1 2
2 pdq  qdp
 pdq  p  q
2
p2  q2
1
 pdq   pdq  qdp
2
1
  pdq  qdp
2
1 
 d  pq .
2 
So transformation is canonical.


E.g. Show that the transformation P  ln sin p  is canonical.

 Q  q tan p
Solution :
Q  q tan p
 dQ  d q tan p 
 tan p dq  q sec 2 p dp

2
p
dq

ln

sin
p

tan
p
dq

q
sec
p dp
 p dq  P dQ 

  p  ln sin p  tan p dq  q ln sin p  sec 2 p dp
d
cos p
2

p

ln

sin
p

tan
p


1

tan
p

ln

sin
p

sec
p
Since
dp
sin p
  ln sin p  sec 2 p
p dq  P dQ  d  p  ln sin p  tan p 
So transformation is canonical.
Generating Functions
The transformation {qi , pi }
{Qi , Pi } is canonical
provided that there is a function G
called generating function, satisfying
d
G  Lqi ,qi ,t   LQi ,Q i ,t  , here L’s are the
dt
Lagrangians with respect to the old and new variables.
We can express the
generating function G
using one of the two old
variables and one of the
two new variables.
qi
pi
Qi
T
U
Pi
S
V
Result I
When the generating function G is expressed using qi , Qi and t
It is denoted by T and we get
T
T
pi 
, Pi  
qi
Qi
and H o  H  T .
t
Proof of the result :
T  T qi , Qi , t 
N

 T
 T
T

dT 
dqi 
dQi  
dt
qi
Qi
t


j 1
dT
Further
 Lqi , qi , t   LQi , Q i , t 
dt

pi qi  H 
Pi Q i  H o
Eq. 1


 dT   p dq  Hdt   P dQ  H dt
i
i
i
i
o
Eq. 2
Eq. 1
and
Eq. 2
T
T
pi 
, Pi  
qi
Qi
Give us
and H o  H  T .
t
E.g. Show that the following transformation is canonical, and find
the generating function in terms of p and q.
Pq

Q   p
Solution : Canonical part is easy.
We rewrite the given transformation as
Above first equations becomes
T
p
q
 Q
Pq

 p  Q
E.g. Show that the following transformation is canonical, and find
the generating function in terms of p and q.
1 2

2


P

p

q

2

1 q
 Q  tan
p

Solution : Canonical part is easy.
By rewriting the given transformation becomes
p  q cot Q
1 2
P  q cos ec 2 Q
2
T
The equation pi 
becomes
qi
T
 q cot Q
q
By integrating partially with respect to q, we get
1 2
T  q cot Q  f Q 
2
T
T
The other equation Pi  
becomes P  
, and hence
Qi
Q
1 2
q cos ec 2 Q  P
2
T

Q
 1 2


q
cot
Q

f

Q



Q  2

1 2
d
2
 q cos ec Q 
f Q 
2
dQ
d

f Q   o 
dQ
f Q   c  T  Qq  c
1 2
So the generating function is T  q cot Q , neglecting the constant.
2
Result II
When the generating function G is expressed using qi , Pi and t
It is denoted by S and we get
S
S
pi 
, Qi 
qi
Pi
T
Ho  H  .
t
and
Proof of the result :
S  S qi , Pi , t 
 S
 S
S

dG 
dqi 
dPi  
dt
qi
Pi
 t
j 1 
dG
Further
 Lqi , q i , t   LQi , Q i , t 
dt

pi qi  H 
Pi Q i  H o
N



Eq. 1

  p dq  Q dP    Q dP  P dQ   H
i
i
i
i
i
i
i
i
o
 H dt
  Q dP  P dQ     p dq  Q dP   H
d
G
dt
 dS 
i
i
i
i
i
  p dq  Q dP   H
i
Eq. 1
i
and
i
i
Eq. 2
S
S
pi 
, Qi 
qi
Pi
o
i
i
i
 H dt
o
 H dt
Eq. 2
Give us
and H o  H  S .
t
P  ln sin p 

E.g. For the canonical transformation 
, find the
 Q  q tan p
generating function in terms
of p and q.
Solution :
Here
1
p  sin 1 e P 
Q  q tansin 1 e P 
q
eP
1 e 2 P
eP
Above first equations becomes
S
p
q
 sin 1 e P
By integrating partially with respect to q, we get
S  q sin 1 e P  f P, t 
T
The other equation Qi 
Pi
eP
q
Q
2P
1 e
S

P
q

f  P, t   o
P
eP
1  e2P
S , and hence
becomes Q 
P


f  P, t 
P
f P, t   f t 
So the generating function is
S  q sin 1 e P  f P, t  , neglecting the constant.
Result III
When the generating function G is expressed usingQi , pi and t
It is denoted by U and we get
U
U
U
qi  
, Pi 
and H o  H 
.
pi
Qi
t
Proof of the result :
U  U  pi , Qi , t 
 U
 U
U

dU 
dpi 
dQi  
dt
pi
Qi
 t
j 1 
dG
 Lqi , q i , t   LQi , Q i , t 
dt

pi qi  H 
Pi Q i  H o
N



Eq. 1
   p q     q dp  P dQ   H
d
G
dt
 dU 
i i
i
  q dp  P dQ   H
i
Eq. 1
and
i
i
i
Eq. 2
U
U
qi  
, Pi  
pi
Qi
i
o
i
i
o
 H dt
 H dt
Eq. 2
give us
and H o  H  S .
t
E.g. Does there exist a generating
of the form G=G( p, Q, t ) for the transformation
P  q, Q   p.
Result IV
When the generating function G is expressed using Pi , pi and t
It is denoted by V and we get
V
U
qi  
, Qi 
pi
Pi
and
T
Ho  H  .
t
Proof of the result :
V  V  pi , Pi , t 
 U
 U
U

dU 
dpi 
dPi  
dt
pi
Pi
 t
j 1 
dG
 Lqi , q i , t   LQi , Q i , t 
dt

pi qi  H 
Pi Q i  H o
N



Eq. 1
   p Q  p q     q dp  Q dP   H
d
G
dt
 dU 
i
i
i i
i
  q dp  Q dP   H
i
Eq. 1
and
i
i
i
Eq. 2
V
V
qi  
, Qi 
pi
Pi
o
i
i
i
 H dt
o
Eq. 2
give us
and H o  H  S .
t
E.g. Find the generating function for the transformation
P  q, Q   p.
 H dt