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 1 them F (qi, pi, t) and G(qi, pi, t) and produces a new function. It is deﬁned

N ∂F ∂G ∂F ∂G F,G ! (1) X ∂q ∂p ∂p ∂q { } ≡ j=1 j j − j j ∂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 − = 1 =0 =+1 =cos(q) | {z− } | {z } | {z } | {z } To see why the Poisson bracket is useful recall the form of Hamilton’s equations ∂H ∂H q˙i = , p˙i = , i = 1, 2,...N. (2) ∂pi − ∂qi

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 1Many 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 ∂F dq ∂F dq ∂F dq ∂F dp ∂F dp ∂F = 1 + 2 + + N + 1 + + N + , dt ∂q1 dt ∂q2 dt ··· ∂qN dt ∂p1 dt ··· ∂pN dt ∂t N ∂F ∂F ∂F = q˙ + p˙ ! + . X ∂q j ∂p j ∂t j=1 j j

We can use Hamilton’s equations (2) to eliminateq ˙j andp ˙j

dF N ∂F ∂H ∂F ∂H ∂F = ! + . dt X ∂q ∂p ∂p ∂q ∂t j=1 j j − j j 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 + . (3) dt { } ∂t 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

∂q1 ∂H q˙1 = q1,H + = , { } ∂t ∂p1 =0 |{z} since ∂q1/∂pj = 0 always, and ∂q1/∂qj = 0 as long as j = 1. This is the ﬁrst of Hamilton’s 6 equations. This same trick will work forp ˙1 to give the second equation:

∂p1 ∂H p˙1 = p1,H + = . { } ∂t − ∂q1 =0 |{z} And similarly forq ˙i andp ˙i 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 ma- nipulating 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 = . (4) dt ∂t If you’re not clear on the difference between total and partial derivatives this equation ap- pears 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 coor- • dinate, 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 central- • force potential 2 2 2 V (x, y, z) = V qx + y + z ,

where r = x2 + y2 + z2 is distance from the origin. The Hamiltonian can be written as the p sum of the kinetic and potential energy

2 p2 2 px y pz 2 2 2 H(px, py, pz, x, y, z) = + + + V x + y + z . 2m 2m 2m q | T (px{z,py,pz) } Angular momentum about the z axis is deﬁned by the function

Lz = xpy ypx . − 3 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 px py Lz,T = + = py px = 0 . (5) { } ∂x ∂px ∂y ∂py m − m

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 . (6) { } − ∂px ∂x − ∂py ∂y ∂x − ∂y

Since V depends on x and y only in the combination r = x2 + y2 + z2 we can use the chain p rule to write ∂V x ∂V y = V 0(r) , = V 0(r) . ∂x x2 + y2 + z2 ∂y x2 + y2 + z2 p p

Using these in (6) yields Lz,V = 0 regardless of what V (r) actually is. Combined with (5) { } 0 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:

qΦ0 ∂Lz qΦ0 Lz,V = Lz, z = = 0 . { } d − ∂pz d