Poisson Brackets

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Let's imagine that we have a quantity f that is a function of the coordinates, momentaand possibly time as well. Let's take its total time derivative to get

\frac{d f}{d t} = \frac{\partial f}{\partial t} + \sum_k \left ( \frac{\partial f}{\partial q_k} {\dot q}_k + \frac{\partial f}{\partial p_k} {\dot p}_k \right ) = \frac{\partial f}{\partial t} + \sum_k \left ( \frac{\partial f}{\partial q_k} \frac{\partial H}{\partial p_k} - \frac{\partial f}{\partial p_k} \frac{\partial H}{\partial q_k} \right ) = \frac{\partial f}{\partial t} + \left [ H , f \right ].

There is lots of different notation for the Poisson brackets. To wit, it is often written with braces { } rather than brackets [ ], probably to avoid confusion with the quantum mechanical commutator to which it is intimately related. Since we call them Poisson brackets it seems natural to use brackets rather than braces.

If the function f is an integral of the motion we have 
{\partial f}/{\partial t} + \left [ H , f \right ] = 0.
In particular if f doesn't explicitly depend on time, we find that that 
\left [ H , f \right ] = 0.

Definition of the Poisson Brackets

We can define the Poisson brackets for any two quantities

[f , g ] =\sum_k \left ( \frac{\partial f}{\partial q_k} \frac{\partial g}{\partial p_k} - \frac{\partial f}{\partial p_k} \frac{\partial g}{\partial q_k} \right )

Properties of the Poisson Brackets

The brackets have the following properties

\left . \right . [ f , g ] = - [g, f], [ f, c] = 0, [ f_1 + f_2 , g ] = [f_1, g ] + [f_2 , g], [f_1 f_2, g] = f_1 [f_2, g] + f_2 [f_1, g]

We can also look at the time derivative of the brackets

\frac{\partial}{\partial t} [ f, g]  = \left [ \frac{\partial f}{\partial t}, g \right ] + \left [ f, \frac{\partial g}{\partial t} \right ]

If one of the functions f and g is a momentum or a coordinate we have

[ f , q_k ] = \frac{\partial f}{\partial p_k}, [ f , p_k ] = -\frac{\partial f}{\partial q_k},

Furthermore we can look at the Poisson brackets among the momenta and coordinates themselves

\left [q_i, q_k \right ] = 0, \left [p_i, p_k \right ] = 0, \left [p_i, q_k \right ] = \delta_{ik}

We also have the important relation

\left [ f, [g , h] \right ] + \left [ g, [h , f] \right ] + \left [ h, [f , g] \right ] = 0

known as Jacobi's identity.

Integrals of the Motion

Let's say the functions f and g are integrals of the motion, then we can prove that [f, g] is also an integral of the motion.

\frac{d}{dt} \left [ f, g \right ] = \frac{\partial}{\partial t} [f, g] + \left [ H, [f, g ] \right ]


\frac{d}{dt} \left [ f, g \right ] = 
  \left [ \frac{\partial f}{\partial t}, g \right ] + \left [ f, \frac{\partial g}{\partial t} \right ]
- \left [ f, [g,H] \right ] - \left [ g, [H, f ] \right ]

where I have used Jacobi's identity to replace the final term in the first equation. Rearranging yields

\frac{d}{dt} \left [ f, g \right ] = \left [ \frac{\partial f}{\partial t} + [H,f] , g \right ] + \left [ f, \frac{\partial g}{\partial t} + [H, g] \right ] = \left [ \frac{d f}{d t}, g \right ] + \left [ f, \frac{d g}{d t} \right ] .

The right-hand side vanishes because f and g are integrals of the motion. This result is known as Poisson's theorem.

An Example with Angular Momentum

The angular momentum about the x-axis and y-axis are given by

\left . \right .
L_x = y p_z - z p_y, L_y = z p_x - x p_z

Let's calculate [Lx,Ly]

[L_x, L_y] = [ y p_z - z p_y, z p_x - x p_z ] = [ y p_z, z p_x - x p_z ] - [z p_y, z p_x - x p_z ]

= [ y p_z, z p_x ] - [y p_z, x p_z ] - [z p_y, z p_x] + [z p_y, x p_z ]

= y [ p_z, z p_x ] + p_z [ y, z p_x] - y [ p_z, x p_z ] - p_z [ y, x p_z]  - z [z p_y, z p_x] - p_y [z , z p_x] + z [ p_y, x p_z ] 
+ p_y [z , x p_z ]

= y p_x - p_y [x p_z , z ] =y p_x - x p_y = L_z

where we only used the properties of the brackets themselves and the results for the brackets among the momenta and coordinates.

Quantum Mechanics

The Poisson bracket defined here is connected to the commutator of quantum mechnanics. Specifically, the commutator is

\left . \right .
[A , B ] = A B - B A.

Classically the commutator would vanish because real numbers commute. Quantum mechanically think of A and B as "operators" that act on vectors or functions (you can imagine these as matrices that don't necessarily commute). We take

[{\hat p}_x, {\hat x}] = -i \hbar

and scale all of the results from the algebra of the Poisson brackets. For example

[{\hat p} _x, f]  = -i \hbar \frac{\partial f}{\partial x}

One way to make sense of these relations is to think of the operators acting on a particular function of position and time ψ(x,t) (the wave function), so we have 
[{\hat p} _x, f] \psi  = {\hat p}_x ( f \psi ) - f ({\hat p}_x \psi ) = -i \hbar \frac{\partial f}{\partial x} \psi

which tells us that the momentum operator is

{\hat p}_x = -i \hbar \frac{\partial}{\partial x}

Classically the Hamiltonian of a system often looks like

H = \frac{p^2}{2 m} + V(x)

so we have in quantum mechanics

{\hat H} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x)

Classically, the Hamiltonian tells quantities how to evolve forward in time. It is similar in quantum mechanics we have

-i \hbar \frac{d f}{d t} = -i\hbar \frac{\partial f}{\partial t} + [{\hat H}, f].

so if the Hamiltonian commutes with a particular quantity and that quantity is not an explicit function of time, the quantity is conserved by the motion even in quantum mechanics.

Again let's assume that these operators act on a function ψ and that f is an integral of the motion

[{\hat H}, f] \psi = -i \hbar \frac{d f}{d t} \psi + i\hbar \frac{\partial f}{\partial t} \psi =  i\hbar \frac{\partial f}{\partial t} \psi.

As with the momentum we have that

{\hat H} = i \hbar \frac{\partial}{\partial t}


i \hbar \frac{d \psi}{d t} = {\hat H} \psi = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V(x) \psi

This result is known as the Schrödinger equation.

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