References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.7; Exercises 2.7.1 – 2.7.2.
Hamilton’s canonical equations are:
If a coordinate is missing in the Hamiltonian (that is, is indepedent of ), then
Thus the conjugate momentum is conserved. Such a missing coordinate is known as a cyclic coordinate. [I’m not sure of the origin of this term. Again Google doesn’t provide a definitive answer.]
There is a general method for calculating the rate of change of some function that depends on the momenta and coordinates, but not explicitly on the time ( is allowed to depend implicitly on time since and/or can depend on time). The time derivative can then be written using the chain rule:
Since automatically, the total energy (represented by the Hamiltonian) is conserved, provided there is no explicit time dependence. Such a time dependence can arise if the system is subject to some external force, for example.
From the definition 5 we can derive a few fundamental properties of Poisson brackets. We’ll consider a general Poisson bracket between two arbitrary functions and . Then
A Poisson bracket is distributive, in the sense that
One more identity is useful, which we can derive using the product rule:
The Poisson brackets involving the coordinates and momenta turn up frequently, so it’s worth deriving them in detail. We have
This follows because, in the Hamiltonian formalism, the s and s are independent variables, so for all and . For the same reason, we have
The mixed Poisson bracket is a different story, however:
Example In two dimensions, we have a Hamiltonian:
If , then in polar coordiantes, the only coordinate appearing in is the radial distance from the origin , which means that the polar angle is a cyclic coordinate. This means that the conjugate momentum must be conserved. That is,
However, is the angular momentum , so this just says that angular momentum is conserved.
To see this explicitly, it’s easier to convert to polar coordinates. From Hamilton’s equations
where in the fourth line, is the linear velocity and in the fifth line we converted this to polar coordinates. Thus the Hamiltonian becomes, in the case where :
To find the conjugate momenta in polar coordinates, we can write out the Lagrangian. We use and and get
The conjugate momenta are thus
From this we can see that is indeed angular momentum as it’s proportional to the product of and the tangential velocity . (‘Real’ momentum and angular momentum must, of course, also contain a factor of a mass, but from the definition of the Hamiltonian above, we see that the mass has been incorporated into the momentum parameters.)
Plugging these back into 32 we get
We can now calculate the Poisson brackets easily:
Thus (the angular momentum) is conserved, while , so that the object is always being pulled in towards the origin.