References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.8; Exercise 2.8.5.
If the Hamiltonian is invariant under a regular canonical transformation and we can find a generator such that an infinitesimal version of this transformation is given by
then is conserved.
If we are dealing with a finite regular canonical transformation where we go from , and the Hamiltonian is invariant under this transformation, then it turns out that if a trajectory satisfies Hamilton’s equations of motion:
then the trajectory obtained by transforming every point in the original trajectory to the barred system is also a solution of Hamilton’s equations in the sense that
The proof of this is a bit subtle, but goes as follows. To begin, review the derivation of the conditions for a transformation to be canonical. This derivation applied to a passive transformation, in which the two sets of parameters refer to the same point in phase space. The transformation we’re considering here is an active transformation, in which actually moves the point in phase space. The original derivation (for passive transformations) relied on the fact that the numerical value of the Hamiltonian is the same in both coordinate systems, since both and refer to the same point in phase space. However, for our active transformation, we’re assuming that the Hamiltonian is invariant under the transformation, that is , where and now refer to different points in phase space. Since the assumption that the Hamiltonian satisfies was all that we used in the original derivation, the same derivation works both for passive transformations (always) and for active transformations (if the Hamiltonian is invariant under the active transformation). We therefore end up with the equations
Since the transformation is specified to be canonical, the conditions on the Poisson brackets apply here:
We can now revisit the 2-d harmonic oscillator to show that a noncanonical transformation violates these results. The Hamiltonian is
and we consider the transformation where we rotate the coordinates but not the momenta. The transformation is
As we’ve seen, this is a noncanonical transformation. To see what happens, we’ll consider the initial conditions
The mass is started off at a point on the axis with a momentum only in the direction. In this case, the mass behaves like a one-dimensional harmonic oscillator, moving along the axis only. To be precise, we can work out Hamilton’s equations of motion:
The equations for and are the same, with replaced by everywhere. We can solve these ODEs in the usual way, by differentiating the first one and substituting the second one into the first to get
This has the general solution
We can do the same for and get
Applying the initial conditions, we get
Now suppose we start off with , , and . That is, we have rotated the coordinates through , but not the momenta. We now begin with the mass on the axis, but moving in the direction, so as time progresses, it will have components of momentum in both the and directions. Although it’s fairly obvious that this motion will not be simply the motion in the first case rotated through , let’s go through the equations. By the same technique as above, we can solve the equations to get
If we look at the system at, say, , then and . The mass that started off on the axis will be at position and so will the mass that started off on the axis. Since the two masses are in the same place, obviously one is not the rotated version of the other.
Another, probably easier, way to see this is that since the first mass moves only along the axis, if the rotated version of the trajectory was also to be a solution, the rotated trajectory would have to lie entirely along the axis, which is certainly not true for the mass that starts off on the axis, but with a momentum .
In the general case, if the transformation is noncanonical, then the Poisson brackets in 7 and 8 don’t satisfy the conditions 9 and 10, with the result that Hamilton’s equations aren’t satisfied in the coordinates. (There may be a deeper, physical interpretation that I’ve missed, but from a mathematical point of view, that’s what goes wrong.)