Download Poisson Brackets and Constants of the Motion (Dana Longcope 1/11

Poisson Brackets and Constants of the Motion (Dana Longcope 1/11/05)
Poisson brackets are a powerful and sophisticated tool in the Hamiltonian formalism of Classical
Mechanics. They also happen to provide a direct link between classical and quantum mechanics.
A classical system with N degrees of freedom, say a set of N/3 particles in three dimensions,
is described by 2N phase space coordinates. These are the N generalized coordinates q1 , q2 , . . . qN
and N conjugate momenta p1 , p2 , . . . pN . The system’s Hamiltonian depends on these 2N variables
and possibly on time t as well
H(q1 , q2 , . . . qN , p1 , p2 , . . . pN , t) ≡ H(qi , pi , t) ,
where the second expression is a more concise way of expressing the fact that H depends on all
coordinates, qi , and all momenta pi (implicitly we take i = 1, 2, . . . N ).
The Poisson bracket is an operation which takes two functions of phase space and time, call
them F (qi , pi , t) and G(qi , pi , t) and produces a new function. It is defined1
{F, G} ≡
N
X
j=1
=
∂F ∂G
∂F ∂G
−
∂qj ∂pj
∂pj ∂qj
!
(1)
∂F ∂G
∂F ∂G
∂F ∂G
∂F ∂G
∂F ∂G
∂F ∂G
−
+
−
+ ··· +
−
∂q1 ∂p1 ∂p1 ∂q1 ∂q2 ∂p2 ∂p2 ∂q2
∂qN ∂pN
∂pN ∂qN
That’s all there is to it. To take the Poisson bracket of two functions you need only evaluate all
partial derivatives and assemble them in a sum of products. Each term in the sum contains one
derivative of F and one derivative of G; one of the derivatives is with respect to a coordinate qj
and the other is with respect to the conjugate momentum pj . The terms change sign depending on
which function is differentiated w.r.t. the coordinate and which w.r.t. the momentum.
In the case of a single degree of freedom, N = 1, phase space is 2-dimensional: (q, p) and the
Poisson bracket has only two terms
{F, G} =
∂F ∂G
∂F ∂G
−
.
∂q ∂p
∂p ∂q
Taking, for example, F (p, q) = p − q and G(p, q) = sin(q), the Poisson bracket is
{p − q, sin(q)} =
∂(p − q) ∂ sin(q)
∂(p − q) ∂ sin(q)
−
= − cos(q)
∂q
∂p
∂p
∂q
|
{z
=−1
} | {z }
=0
|
{z
=+1
} | {z }
=cos(q)
To see why the Poisson bracket is useful recall the form of Hamilton’s equations
q̇i =
∂H
∂H
, ṗi = −
, i = 1, 2, . . . N .
∂pi
∂qi
(2)
These describe how the phase space coordinates qi (t) and pi (t) of a particular particle vary in time
as a result of the Hamiltonian H(qi , pi , t). That particle will see a particular value of the function
1
Many authors use square brackets to write the Poisson bracket: [F, G]. This is the same way we will later write
the commutator in quantum mechanics. We will use curly brackets for the Poisson bracket to avoid confusion later
on.
1
F (qi , pi , t) at each time. This value can vary for two reasons: 1. the phase space coordinates of the
particle are changing and 2. the function F depends explicitly on time. The total time derivative
accounts for both effects through the magic of the chain rule:
dF
dt
=
∂F dq1
∂F dq2
∂F dqN
∂F dp1
∂F dpN
∂F
+
+ ··· +
+
+ ··· +
+
,
∂q1 dt
∂q2 dt
∂qN dt
∂p1 dt
∂pN dt
∂t
=
N
X
j=1
∂F
∂F
q̇j +
ṗj
∂qj
∂pj
!
+
∂F
.
∂t
We can use Hamilton’s equations (2) to eliminate q̇j and ṗj
N
X
dF
∂F ∂H
∂F ∂H
=
−
dt
∂qj ∂pj
∂pj ∂qj
j=1
!
+
∂F
.
∂t
The sum in this expression is (not coincidentally) the Poisson bracket. This means that the time
evolution of an arbitrary function of phase space is given by
dF
∂F
= {F, H} +
.
dt
∂t
(3)
In fact, after using Hamilton’s equations to arrive at expression (3) we can, if we like, forget
about them entirely and use equation (3) as the definition of all time derivatives. Taking the
particular choice F (qi , pi , t) = q1 we can quickly verify that
q̇1 = {q1 , H} +
∂q1
∂H
,
=
∂t
∂p1
|{z}
=0
since ∂q1 /∂pj = 0 always, and ∂q1 /∂qj = 0 as long as j 6= 1. This is the first of Hamilton’s
equations. This same trick will work for ṗ1 to give the second equation:
ṗ1 = {p1 , H} +
∂p1
∂H
.
= −
∂t
∂q1
|{z}
=0
And similarly for q̇i and ṗi for all other i’s. Thus Hamilton’s equations follow from (3).
There are two very useful properties of the Poisson bracket which can be easily check by manipulating the definition (1).
1. The Poisson bracket is anti-symmetric in its two arguments
{G, F } = − {F, G} .
An immediate consequence of this is that {F, F } = 0 for any function at all.
2. The Poisson bracket is linear in either of its arguments
{F1 + F2 , G} = {F1 , G} + {F2 , G} ,
{F, G1 + G2 } = {F, G1 } + {F, G2 } .
2
Constants of the Motion
The Poisson bracket displays its true power in the search for constants of the motion. A constant of
the motion is some function of phase space, independent of time, F (qi , pi ), whose value is constant
for any particle. In other words, F (qi , pi ) is a constant of the motion if dF/dt = 0. Since we
specified that F does not depend explicitly in time it follows that ∂F/∂t = 0. This means that
F is a constant of the motion if and only if {F, H} = 0 for all points in phase space.
All of the familiar constants of the motion can be checked using this one simple prescription
• Energy: First of all {H, H} = 0 always due to the anti-symmetry of the Poisson bracket.
Using this in (3) we quickly find that
dH
∂H
=
.
dt
∂t
(4)
If you’re not clear on the difference between total and partial derivatives this equation appears trivial; it is not. The derivative on the right is taken at a fixed point in phase space,
(q1 , q2 , . . . , pN ), while the one on the left is taken by following a particle along as it moves
through phase space. The fact that these two derivatives give the same result should strike
you as truly remarkable.
In those cases where Hamiltonian does not depend on time explicitly ∂H/∂t = 0. Using
this in (4) gives the immediate result that H(qi , pi ) is a constant of the motion: Energy is
conserved in cases where the Hamiltonian is time-independent.
• Linear Momentum: In a case where the Hamiltonian does not contain a particular coordinate, qk , explicitly it is said to be cyclic in that coordinate. Applying the definition (1)
directly we find that
∂H
{pk , H} = −
= 0
∂qk
so pk is a constant of the motion: Momentum is conserved if it is conjugate to a cyclic
coordinate.
• Angular Momentum: Consider a particle in three dimension, (x, y, z), subject to a centralforce potential
q
x2 + y 2 + z 2
V (x, y, z) = V
,
p
where r = x2 + y 2 + z 2 is distance from the origin. The Hamiltonian can be written as the
sum of the kinetic and potential energy
H(px , py , pz , x, y, z) =
p2y
p2x
p2
+
+ z + V
2m
2m}
|2m
{z
q
T (px ,py ,pz )
Angular momentum about the z axis is defined by the function
Lz = xpy − ypx .
3
x2 + y 2 + z 2
.
Linearity allows us to break the Poisson bracket with the Hamiltonian into two smaller pieces
{Lz , H} = {Lz , T + V } = {Lz , T } + {Lz , V } .
Since T depends only on momenta the first bracket on the right has only two non-vanishing
terms:
{Lz , T } =
∂Lz ∂T
∂Lz ∂T
px
py
+
= py − p x
= 0 .
∂x ∂px
∂y ∂py
m
m
(5)
Other terms, such as −(∂Lz /∂px )(∂T /∂x) vanish since derivatives of T w.r.t. coordinates,
such as x, will always be zero. V depends only on coordinates so the second bracket has two
different non-vanishing terms:
{Lz , V } = −
∂Lz ∂V
∂V
∂V
∂Lz ∂V
−
= y
−x
.
∂px ∂x
∂py ∂y
∂x
∂y
Since V depends on x and y only in the combination r =
rule to write
(6)
p
x2 + y 2 + z 2 we can use the chain
∂V
x
∂V
y
0
p
= p 2
V
(r)
,
=
V 0 (r) .
∂x
∂y
x + y2 + z2
x2 + y 2 + z 2
Using these in (6) yields {Lz , V } = 0 regardless of what V 0 (r) actually is. Combined with (5)
this means that {Lz , H} = 0 and the z-angular momentum is a constant of the motion.
Following similar steps it can be shown that the other two components of angular momentum
Lx = ypz − zpy
,
Ly = zpx − xpz ,
are also constants of the motion: {Lx , H} = 0 and {Ly , H} = 0. Therefore for a particle
moving in a central force potential all three components of angular momentum are
conserved.
For the Hamiltonian (1.13) in Liboff Lz is a constant of the motion. Without recourse to Poisson
brackets, however, the book can only show this by transforming into spherical coordinates to get
the new Hamiltonian (1.20). Only then is it apparent that φ is a cyclic coordinate and therefore
that pφ is a constant of the motion. (A little further manipulation, see problem 1.5, shows that
pφ = Lz .) The power of the Poisson bracket is that it allows one to identify constants of the motion
regadless of which coordinates you choose to use. In this case we can quickly see this even in
cartesian coordinates:
{Lz , V } =
Lz ,
qΦ0
z
d
4
= −
∂Lz qΦ0
= 0 .
∂pz d