Classical Mechanics
Examples (Canonical Transformation)
Dipan Kumar Ghosh
Centre for Excellence in Basic Sciences
Kalina, Mumbai 400098
November 10, 2016
1
Introduction
In classical mechanics, there is no unique prescription for one to choose the generalized
coordinates for a problem. As long as the coordinates and the corresponding momenta
span the entire phase space, it becomes an acceptable set. However, it turns out in practice that some choices are better than some others as they make a given problem simpler
while still preserving the form of Hamilton’s equations. Going over from one set of chosen
coordinates and momenta to another set which satisfy Hamilton’s equations is done by
canonical transformation.
For instance, if we consider the central force problem in two dimensions and choose Cartek
. However, if we choose (r, θ) coordinates,
sian coordinate, the potential is − p
x2 + y 2
k
the potential is − and θ is cyclic, which greatly simplifies the problem. The number of
r
cyclic coordinates in a problem may depend on the choice of generalized coordinates. A
cyclic coordinate results in a constant of motion which is its conjugate momentum. The
example above where we replace one set of coordinates by another set is known as point
transformation.
In the Hamiltonian formalism, the coordinates and momenta are given equal status and
the dynamics occurs in what we know as the phase space. In dealing with phase space
dynamics, we need point transformations in the phase space where the new coordinates
(Qi ) and the new momenta (Pi ) are functions of old coordinates (pi ) and old momenta
(pi ):
Qi = Qi ({qj }, {pj }, t)
Pi = Pi ({qj }, {pj }, t)
c D. K. Ghosh, IIT Bombay
2
Such transformations are called contact transformations.
However, in Hamiltonian mechanics, only those transformations are of interest for
which the quantities Qi and Pi are canonically conjugate pairs. This means that there
exists some Hamiltonian K(Q, P, t) with respect to which P and Q satisfy Hamilton’s
equations:
∂K
∂Pi
∂K
Ṗi = −
∂Qi
Q̇i =
Note that Q and P are not specific to a particular mechanical system but are common
to all systems having the same degree of freedom. We may, for instance, obtain P and
Q for a plane harmonic oscillator and use the same set to solve Kepler’s problem. Such
transformations, which preserve the form of Hamilton’s equations, are known as canonical transformation. Though the terms contact transformation is used synonymously
with the term canonical transformation, not all contact transformations are canonical, as
the following example shows. Let us consider a free particle system with the Hamiltonian
p2
H=
. Since q is cyclic, the conjugate momentum p is a constant of motion, ṗ = 0,
2m
p
and we have q̇ = . Consider now, a contact transformation
m
P = pt; Q = qt
(1)
Does a Hamiltonian K(P, Q) exist for which P and Q satisfy the Hamilton’s equations:
∂K
∂Q
∂K
Q̇ =
∂P
Ṗ = −
∂ Ṗ
∂ Q̇
= −
as each of the expressions
∂P
∂Q
∂ 2K
∂ Ṗ
∂ Q̇
1
equals −
. However, Since
=
= , this condition is not satisfied and the
∂P ∂Q
∂P
∂Q
t
modified Hamiltonian K does not exist. Thus the contact transformations (1) are noncanonical, they do not preserve the volume of the phase space.
It may be noted that this does not imply that one cannot find an equation of motion
using these variables. Indeed, since p is constant, dP/dt = p = P/t, which gives P = kt,
where k is a constant. Likewise, we have,
dq
p
Pt
k
d Q
=
=
=
=
dt t
dt
m
m
m
Existence of such a Hamiltonian would imply
c D. K. Ghosh, IIT Bombay
3
k
so that Q = t2 + C, where C is a constant.
m
To illustrate a canonical transformation, consider the free particle Hamiltonian again for
which ṗ = 0 and q̇ = p/m, as before. Consider a linear transformation of (p, q) to a new
set (P, Q), which is given by
P = αp + βq
Q = γp + δq
(2)
If the matrix of transformation is non-singular, i.e. if ∆ =
α β
6= 0, then the above
γ δ
transformation is invertible and we have
1
(δP − βQ)
∆
1
q = (−γP + αQ)
∆
p=
(3)
It can be checked that
p
β
=
(δP − βQ)
m
m∆
p
δ
Q̇ = γ ṗ + δ q̇ = δ q̇ = δ =
(δP − βQ)
m
m∆
Ṗ = αṗ + β q̇ = β q̇ = β
(4)
Is there a Hamiltonian K for which P and Q are canonical? It is seen that
βδ
∂ Ṗ
=
∂P
m∆
βδ
∂ Q̇
=−
∂Q
m∆
so that K exists. To determine K note that Ṗ = −
Z
K=−
and Q̇ =
Ṗ dQ = −
∂K
which gives
∂Q
βδ
β 2 Q2
PQ +
+ f (P )
m∆
m∆ 2
∂K
gives
∂P
Z
K=−
βδ
δ2 P 2
Q̇dP = −
PQ +
+ f (Q)
m∆
m∆ 2
These two expressions give
1
(δP − βQ)2
2m∆
Except for the factor ∆, this is just the old Hamiltonian written in terms of the new variables. For ∆ = 1, the two Hamiltonians are identical and the transformation corresponds
K=
c D. K. Ghosh, IIT Bombay
4
to a point transformation of Lagrangian mechanics.
Consider a second example where, once again we take the free particle Hamiltonian of
the previous example. Consider a transformation
P = p cos q,
Q = p sin q
(5)
The transformation is invertible and it is easily seen that
p=
p
P 2 + Q2 ,
q = tan−1
Q
P
(6)
Using the same algebra as for the previous example, we get
Ṗ = ṗ cos q − q̇p sin q = −
and
Q̇ =
p2
Qp 2
sin q = −
P + Q2
m
m
Pp 2
P + Q2
m
∂ Q̇
∂ Ṗ
=−
, showing that the Hamiltonian exists. It is easy to show
∂P
∂Q
1 p 2
that the Hamiltonian is given by K =
P + Q2 .
3m
It can be seen that
However, instead of considering the free particle Hamiltonian, consider the Hamiltonian of a particle in a uniform gravitational field.
H=
p2
+ mgq
2m
p
In this case ṗ = −mg and q̇ = . We had seen that the transformation is invertible. In
m
this case, we get
Ṗ = ṗ cos q − q̇p sin q
p2
= −mg cos q − sin q
m
P
1
Q
= −mg p
− p
2
2
2
m P + Q2
P +Q
and
Q̇ = ṗ sin q + q̇p cos q
p2
cos q
P 2 + Q2 m
Q
1 p
= −mg p
− P P 2 + Q2
P 2 + Q2 m
= −mg p
Q
+
c D. K. Ghosh, IIT Bombay
5
∂ Ṗ
∂ Q̇
6= −
, so that the transformation is not canonical with respect to
∂P
∂Q
the given Hamiltonian. This shows that (5) is not a valid canonical transformation. This
is because, in order that a transformation may be canonical, it must satisfy Hamilton’s
equation for all systems having the same number of degrees of freedom.
One can check
2
Canonical Transformation & Generating Function
We start with an observation that Hamilton’s equations of motion can be derived from
R
R
the Hamilton’s principle by writing action written in the form S = Ldt = (pq̇ − H)dt
and optimizing the action. In deriving the Euler Lagrange equation, we had optimized
action subject to the condition that δq(ti ) = δq(tf ) = 0, i.e. there is no variation at the
end points. We have
Z
tf
(pi q̇i − H)dt = 0
δS = δ
t0
with implied summation on repeated index and we have denoted the initial time as t0 so
as not to confuse with the index i. The variation in the action is then given by
Z tf ∂H
∂H
δS =
q̇i δpi + pi δ q̇i −
δqi −
δpi dt
∂qi
∂pi
t0
Z tf ∂H
d
∂H
δpi δpi + (pi δqi ) − ṗi δqi −
δqi dt
=
q̇i −
∂pi
dt
∂qi
t0
Z tf ∂H
∂H
t
δpi δpi − ṗi +
dt + [pi δqi ]tf0
q̇i −
=
∂pi
∂qi
t0
We have seen earlier that setting δS = 0 for all variations of qi gave us Euler Lagrange
equation and we expect that setting δS = 0 for all variations of qi and pi will lead to
Hamilton’s equations. However, there is a subtle difference between the two cases. In the
Lagrangian formulation qi and q̇i were not independent variables and hence we only needed
variation with respect to qi . In the Hamiltonian formalism, however, we need to vary both
qi and pi . THe last term of the above equation vanishes if δqi (t0 ) = δqi (tf ) = 0 and it
does not requires the corresponding relation for vanishing of the momentum coordinates.
Thus, we have an additional degree of freedom and we could define the action as
Z
dF (q, p)
)dt
S = (pi q̇i − H +
dt
so that for all variations of pi and qi , the Hamilton;’s equations would be valid.
The variational principle requires the function pq̇ − H to satisfy the Euler Lagrange
equations. Thus we have
d
∂
∂
(pq̇ − H) − (pq̇ − H) = 0
dt ∂ q̇
∂q
d
∂
∂
(pq̇ − H) − (pq̇ − H) = 0
dt ∂ ṗ
∂p
c D. K. Ghosh, IIT Bombay
6
∂H
∂H
d
p+
= 0, i.e. ṗ = −
and likewise, since
dt
∂q
∂q
∂H
∂H
= 0, i.e., q̇ =
. The variational porinciple
pq̇ − H no dependence on ṗ, −q̇ +
∂p
∂p
satisfied by the old pair of variables (q, p) is
Z
δS = δ (pq̇ − H)dt = 0
Since H = H(p, q), we get from above,
with no variation at the end points. With respect to the new variables, one then must
have
Z
0
δS = δ (P Q̇ − H)dt = 0
These two forms of S and S 0 are completely equivalent if the two integrands differ either
by a scale factor or by total time differential of a function F , i.e. if
X
X
Pi Q̇i − K = λ
(pi q̇i − H)
(7)
i
i
known as the scale transformation or,
X
Pi Q̇i − K +
i
X
dF
=
pi q̇i − H
dt
i
(8)
known as the canonical transformation or by a combination of both
X
Pi Q̇i − K +
i
X
dF
=λ
(pi q̇i − H)
dt
i
(9)
which is usually referred to as extended canonical transformation If we rewrite
(8) as
X
X
pi dqi − Hdt −
Pi dQi + Kdt = dF
(10)
i
i
we notice that the difference between the two differential forms must be an exact differential. Analogous statement can be made regarding (9) as well. It turns out that
this condition of exact differentiability is both necessary and sufficient condition for a
transformation to be canonical.
2.1
Scale Transformation
Scale transformation is achieved by multiplying the old coordinates and momenta with
some scale factors to get the corresponding new quantities, Pi = νpi and Qi = µqi . The
scale factors can be different for momenta and the coordinates but must be the same for
all coordinates (and all momenta). P and Q are obtainable from the new Hamiltonian K
through Hamilton’s equations
Ṗi = −
∂K
∂Qi
Q̇i =
∂K
∂Pi
c D. K. Ghosh, IIT Bombay
7
Thus we have, using Pi = νpi and Qi = µqi ,
ν ṗi = −
∂K
∂µqi
which gives
ṗi = −
1 ∂K
µν ∂qi
∂H
, we get K = µνH. This gives
∂qi
X
X
Pi Q̇i − K(P, Q, t) = µν
(pi q̇i − H(p, q, t))
Comparing this with the equation ṗi =
i
i
Comparing this with (7) we get λ = µν.
2.2
Generating Function
The function F in (8) or (9) is, in principle, a function of pi , qi , Pi , Qi and t. It is known as
the generating function of canonical transformation. We will illustrate it with a couple
of examples. We will use Einstein summation convention according to which a repeated
P
index in a single term implies a sum, i.e. fi gi = i fi gi .
Example : Identity Transformation:
Consider a transformation F = qi Pi − Qi Pi in which the dependence on time is implicit.
We have
Pi Qi − K +
dF
= Pi Qi − K + q̇i Pi + qi Ṗi − Q̇i Pi − Qi Ṗi
dt
= −K + (qi − Qi )Ṗi + Pi q̇i
≡ pi q̇i − H
The equation is satisfied by identity transformation pi = Pi , qi = Qi and K = H.
Example: Swapping Coordinates and Momenta
We can generalize the above example a consider a generating function of the following
type (summation convention used)
F = fi (q1 , q2 , . . . , qn ; t)Pi − Qi Pi
where fi is a function of the arguments q1 , q2 , . . . , qn and t. We then have
dF
∂fi
∂fi
= Pi Q̇i − K +
q̇j Pi +
Pi + fi Ṗi − (Q̇i Pi + Qi Ṗi )
Pi Q̇i − K +
dt
∂qj
∂t
∂fi
∂fi
q̇j Pi +
Pi
= −K + (fi − Qi )Ṗi +
∂qj
∂t
≡ pi q̇i − H
c D. K. Ghosh, IIT Bombay
8
∂fi
Pi and
The set of equations above can be satisfied by taking Qi = fi , K = H +
∂t
∂fi
requiring pi q̇i =
q̇j Pi which can be rearranged as follows:
∂qj
pi q̇i =
∂fi
∂fj
q̇j Pi =
q̇i Pj
∂qj
∂qi
where in the last term we have interchanged the sum over i and j. This gives
pi =
X ∂fj
j
∂qi
Pj
where we have explicitly reintroduced the sum for clarity. These are n equations which
must be inverted to get Pi . It may be noted that the choice of F is not unique corresponding to a particular canonical transformation. For instance, if we add any g(t) to
dF
dg
dF
→
+
does not change the
F it dopes not affect the action integral, because
dt
dt
dt
Lagrangian equations.
Let us illustrate how to find generator. Let K(Q, P, t) = H(q, p, t). Plugging this into
(8), we get
X
dF
=
(pi q̇i − Pi Q̇i )
dt
i
This is not very interesting because there is no time dependence. The simplest way to
satisfy it is
F = F (q, Q)
We have
X ∂F
dF
∂F
=
q̇i +
Q̇i
dt
∂q
∂Q
i
i
i
X
=
(pi q̇i − Pi Q̇i )
i
∂F
∂F
This gives with
= pi ;
= −Pi . A trivial way to satisfy the equations is to take
∂Qi
P ∂qi
F (qi , Qi ) = i qi Qi so that
∂F
= Qi = pi
∂qi
∂F
= qi = −Pi
∂Qi
which simply requires us to swap the coordinates and the momenta in the Hamiltonian
formalism (but for a sign).
c D. K. Ghosh, IIT Bombay
3
9
Different types of Generating Functions
We have remarked that the generating function F is a function of the old coordinates (and
momenta) q, p, the new coordinates Q, P and time t. However, since Q and P themselves
are functions of q and p (and vice-versa) , only two of the four variables are needed in the
description of the generating function. We need one each from the old pair and the new
pair to describe the generating function, in addition to time. There are four possibilities
and depending on the combination that we choose, the generating functions are classified
as type 1, 2, 3 or 4.
3.1
Type 1 Generator
Type 1 generator, denoted by F1 is a function of qi and Qi . We denote
F = F1 (qi , Qi , t)
(11)
dF
∂F1
∂F1
∂F1
=
Q̇i +
q̇i +
dt
∂q
∂Qi
∂t
Xi
X
≡
pi q̇i −
Pi Q̇i + K − H
(12)
Since F1 = F1 (q, Q, t), we can write
i
dF1
where the last expression is from (8). We can compare the last two expressions for
dt
to get
∂F1
∂qi
∂F1
Pi = −
∂Qi
∂F1
K =H+
∂t
pi =
(13)
Note that (12) can be rewritten as
dF1 = pi dqi − Pi dQi + (K − H)dt
(14)
If the Hamiltonian has no explicit time dependence, the generating function will also
not have an explicit time dependence and in such a case K = H. In such a situation (8)
can be written as (summation convention used)
dF
= pi q̇i − Pi Q̇i
dt
For instantaneous transformation, we can write the above as
dF = pi δqi − Pi δQi
c D. K. Ghosh, IIT Bombay
10
Since the left hand side is a perfect differential, so is the right hand side. In order that
this may be so, we require
∂pi
∂Pi
=
(15)
∂Qi
∂qi
When the functional dependence are given, one can always use (15) to test whether the
transformation is canonical.
As shown earlier, one of the simplest examples of a F1 type transformation is F1 (qi , Qi ) =
qi Qi , which gives the swapping transformation pi = Qi and Pi = −qi . Let us see what it
p2
1
does to the Hamiltonian of the Harmonic oscillator for which H =
+ kq 2 . Since the
2m 2
Hamiltonian has no explicit time dependence, we have K = H. In such a case
K=
1
Q2
+ kP 2
2m 2
The new equations of motion are
∂K
= kP
∂P
∂K
Q
Ṗ = −
=−
∂Q
m
Q̇ =
Thus we get
Q̈ = k P̈ = −
Q
m
Q̇
which has the solution Q = Q0 cos(ωt + δ) . We can obtain P from the equation P =
k
Q0 ω
which gives P = −
sin(ωt + δ). Since this is equal to −q, we can identify q0 = Q0 ω/k.
k
We then can write
q = q0 cos(ωt + δ 0 )
where δ 0 − δ = π/2.
3.2
Type 2 Generator
Type 2 generators, donoted by F2 (q, P, t) can be obtained from the type 1 generator
F1 (q, Q, t) by means of a Legendre transformation. Since we have, in view of (13)
Pi = −
∂F1
∂Qi
and we wish to replace the variable Qi by the new variable Pi , we define
F2 (qi , Pi , t) = F1 (qi , Qi , t) +
X
i
Pi Qi
(16)
c D. K. Ghosh, IIT Bombay
11
Note that starting with the equation (14) for F1 can be written (use summation convention) by observing
dF1
= pi q̇i − Pi Q̇i + K − H
(17)
dt
Using (16), we have
dF1
dF2
=
+ Pi Q̇i + Ṗi Qi
dt
dt
= pi q̇i + Ṗi Qi + (K − H)
(18)
Since F2 depends on q and P , we can write
∂F2
∂F2
∂F2
dF2
=
q̇i +
Ṗi +
dt
dt
∂qi
∂Pi
∂t
Comparing it with (17), we get
∂F2
= pi
∂qi
∂F2
= Qi
∂Pi
∂F2
=K −H
∂t
3.3
(19)
Type 3 Generator
In a similar way as above, we can define F3 (p, Q, t) through a Legendre transform on F1 .
∂F1
Referring to (13), we have, since the variable qi is being replaced by pi , and, pi =
∂qi
F3 (p, Q, t) = F1 (q, Q, t) − pi qi
We then have
dF1
dF3
=
− pi q̇i − qi ṗi
dt
dt
= −qi ṗi − Pi Q̇i + (K − H)
(20)
On the other hand, since F3 depends on (pi , Qi , t), we can write,
dF3
∂F3
∂F3
∂F3
=
ṗi +
Q̇i +
dt
∂pi
∂Qi
∂t
Comparing this with (20) we get
∂F3
= −Pi
∂Qi
∂F3
= −qi
∂pi
∂F3
=K −H
∂t
(21)
c D. K. Ghosh, IIT Bombay
3.4
12
Type 4 Generator
Finally we would like to define F4 (p, P, t). In this case it is better to start with F2 or F3
and do a Legendre transformation. Starting with F3 , we have to replace Qi with Pi . In
view of (21), we have
F4 (p, P, t) = F3 (p, Q, t) − Qi Pi
We then have
dF4
dF3
=
+ Qi Ṗi + Pi Q̇i
dt
dt
= −qi ṗi + Qi Ṗi + (K − H)
Since F4 = F4 (p, P, t), we have
∂F4
∂F4
∂F4
dF4
=
ṗi +
Ṗi +
dt
∂pi
∂Pi
∂t
Comparing with the preceding relation, we have we get
∂F4
= −qi
∂pi
∂F4
= Qi
∂Pi
∂F4
=K −H
∂t
3.5
(22)
Summary of the different types of Generators
The following table gives a summary of relationships for different types of generators.
Generators Derivatives
Relationship
F1 (q, Q, t)
pi =
∂F1
∂F1
; Pi = −
∂qi
∂Qi
∂pi
∂Pj
=−
∂Qj
∂qi
F2 (q, P, t)
pi =
∂F2
∂F2
; Qi =
∂qi
∂Pi
∂pi
∂Qj
=
∂Pj
∂qi
F3 (p, Q, t)
Pi = −
F4 (p, P, t)
qi = −
∂F3
∂F3
; qi = −
∂Qi
∂pi
∂Pi
∂qj
=
∂pj
∂Qi
∂F4
∂F4
; Qi =
∂pi
∂Pi
∂Qi
∂qj
=−
∂pj
∂pi
c D. K. Ghosh, IIT Bombay
4
13
Invariance of Poisson’s Bracket
A canonical transformation preserves the Poisson’s bracket. This property can be used
to verify whether a transformation is indeed canonical. Let (q, p) → (Q, P ) be a transformation. The set (q, p) satisfies Poisson’s bracket relationship, i.e.
{qi , pj } = δij
(23)
What we need to show is that Poisson’s bracket calculated using either basis is the same,
i.e.,
{Qi , Pj }QP = {Qi , Pj }qp
(24)
Clearly, since {Qi }, {Pi } are canonically conjugate pairs, they satisfy
{Qi , Pj }QP = δij
(25)
To prove (24), we note
{Qi , Pj }qp
X ∂Qi ∂Pj
∂Qi ∂Pj
=
·
−
·
∂qk ∂pk
∂pk ∂qk
k
X ∂Qi ∂qk
∂Qi
∂pk
∂Qi
=
·
−
· −
= δij
=
∂q
∂Q
∂p
∂Q
∂Q
k
j
k
j
j
k
(26)
In proving the above, we have used (15).
In a similar way, we can prove that the Poisson’s brackets of the old variables are
preserved in the new basis,
{qi , pj }qp = {qi , pj }QP
In general, for any two dynamical variables X and Y , one can show that the Poisson’s
bracket relationship remains the same in either basis
{X, Y }QP = {X, Y }qp
5
(27)
Illustrative Problems
Example 1: Harmonic Oscillator The Hamiltonian for the Harmonic oscillator is
p2
1
H(p, q) =
+ mω 2 q 2 . Let us carry out a transformation
2m 2
p = f (P ) cos Q
f (P
q=
sin Q
mω
(28)
f (P )2
. Note that
2m
the Hamiltonian is cyclic in Q so that P is constant. The problem is to find f (P ) such
This converts the Hamiltonian to a function of P alone with K = H =
c D. K. Ghosh, IIT Bombay
14
that the transformation in canonical.
Let us try an F1 type transformation for which,as per eqn. (13)
∂F1
∂q
∂F1
P =−
∂Q
∂F1
=H
K =H+
∂t
p=
(29)
the last equality because the Hamiltonian is time independent. From (28), we get,
p = mωq cot Q =
∂F1
∂q
Integrating, we get
mωq 2
cot Q
2
The momenta conjugate to Q is then given by
F1 =
∂F1
∂Q
mωq 2
cosec2 Q
=
2
p2
1
=
2mω cos2 Q
P =−
which gives
√
p=
2P mω cos Q
√
f (P )2
2P mω. This in turn gives the new Hamiltonian to be K =
= ωP .
2m
Thus the transformation (28) is equivalent to
√
p = 2P mω cos Q
r
2
q=
sin Q
mω
so that f (P ) =
Inverting, we get
1
(q 2 m2 ω 2 + p2 )
2mω mωq
−1
Q = tan
p
P =
c D. K. Ghosh, IIT Bombay
15
Note that
∂Q
∂q
∂P
∂p
∂Q
∂p
∂P
∂q
=
=
p2
mωp
+ m2 ω 2 q 2
p
mω
=−
p2
mωq
+ m2 ω 2 q 2
= qmω
which shows that
{Q, P }qp = 1
i.e. as expected, the transformation preserves Poisson’s bracket.
E
As the energy is constant we have P = . The Hamilton’s equations for the new
ω
variables then gives
∂K
=ω
Q̇ =
∂P
yielding Q = ωt + α The original problem is now solved,
√
√
p = 2mE cos Q = 2mE cos(ωt + α)
r
r
2P
2E
q=
sin Q =
sin(ωt + α)
mω
mω 2
Example 2:
1
1
Consider the Hamiltonian H = (p2 +q 2 ). Determine if the transformation Q = (q 2 +p2 )
2
2
and P = − tan−1 (q/p) is canonical. If so, find a generating function of type F1 .
In order that the given transformation is canonical, pδq − P δQ = dF must be an exact
differential. Note that
q q 2 + p2
pδq − P δQ = pδq + tan−1 δ(
)
p
2
q
−1 q
= p + q tan
δq + p tan−1 δp
p
p
In order that the above is an exact differential,
∂
∂
−1 q
−1 q
p + q tan
=
p tan
∂p
p
∂q
p
It can be easily seen that both sides of the above condition equals
transformation is canonical.
p2
so that the
p2 + q 2
c D. K. Ghosh, IIT Bombay
16
The Poisson’s brackets can be shown to be preserved:
∂ 1 2
∂
∂ 1 2
∂
2
−1 q
2
−1 q
{Q, P } =
(q + p )
− tan
−
(q + p )
− tan
∂q 2
∂p
p
∂p 2
∂q
p
2
(q/p )
1/p
=1
=q
−
p
−
1 + q 2 /p2
1 + q 2 /p2
We can obtain F by integrating either of the factors. For instance, a bit of exercise in
integration gives
Z −1 q
F =
p + q tan
dq
p
q
pq q 2 + p2
+
tan−1
=
2
2
p
However, the above form cannot be a generating function as it has no dependence on the
new variables. Sincep
we wish it to be of form F1 , we need to eliminate p. This is done
q
q
q
by substitution p = 2Q − q 2 so that tan−1 = tan−1 p
= sin−1 √ . With
2
p
2Q
2Q − q
these substitutions, we get
q
1 p
F = Q sin−1 √
+ q 2Q − q 2
2Q 2
∂K
1
= 0, i.e. Q is constant, and
The Hamiltonian K = (q 2 + p2 ) = Q, so that Q̇ =
2
∂P
∂K
Ṗ = −
= −1. Since Q is constant, so is the Hamiltonian and P decreases with time.
∂Q
Example 3:
sin p
Prove that the transformation Q = ln
and P = q cot p is canonical and derive all the
q
different forms of generating functions.
We have, for the given teansformation
1
δQ = cot pδp − δq
q
Thus
1
pδq − P δQ = pδq − q cot p cot p δp − δq
q
= (p + cot p) δq − q cot2 p δp
This is an exact differential if
∂
∂
(p + cot p) =
(−q cot2 p)
∂p
∂q
c D. K. Ghosh, IIT Bombay
17
It can be seen that both sides equal − cot2 p and hence the transformation is canonical.
It is also easy to see that Poisson’s bracket is preserved.
The structure of generating function is
Z
F = (p + cot p)dq = pq + q cot p
(We can also integrate the other factor over p and show that both the forms are the
same if the constant integration is taken to be zero). We will now express this in terms
of different forms of the generating
the given transformation,
p function. Note that for p
Q
2
2Q
sin p = qe which gives cos p = 1 − q e and q cot p = e−2Q − q 2 . Substituting
these, we get
p
p
F1 = q cos−1 1 − q 2 e2Q + e−2Q − q 2
Suppose we wish to find the generator form F2 . In such a case, we need to replace p by
P , which is quite straightforward. We can find F2 by a Legendre transformation. Since
∂F1
P =−
, we have F2 (q, P ) = F1 (q, Q) + P Q. Thus
∂Q
F2 (q, P ) = F1 (q, Q) = q sin−1 (qeQ ) +
1p
1 − q 2 e2Q + P Q
eQ
However, as F2 must be expressed as a function
of q and P , we must accordingly express
p
2
1
1 − q e2Q
, which gives eQ = p
the variables. We had shown that P =
. Thus
Q
e
P 2 + q2
F2 (q, P ) = q sin−1
q
p
P 2 + q2
!
+P −
P
ln(P 2 + q 2 )
2