Principal Bundle

Angles in Quantum Mechanics

Posted in Physics, Quanum Mechanics by principalbundle on February 28, 2009

We know from intro quantum mechanics the positions x and y of a particle should be interpreted as operators. What happens when we transform to polar coordinates, x=r \cos\theta, y=r\sin\theta? Since \theta classically becomes a dynamical variable, we expect it to become an operator quantum mechanically. What happens when we do this? The result is a very basic consequence of operators on a Hilbert space, but comes as a bit of a surprise, since these types of issues are not normally discussed in quantum mechanics classes or texts.

We can define the angular momentum operator, L_z = -i \frac{\partial}{\partial\theta}. And we can find the commuter [L_z,\theta]=-i. We also know L_z has integer eigenvalues. Hitting each side of the commuter with eigenstates, we have

(m-m') \langle m | \theta | m' \rangle = -i\delta_{m,m'}.

Simple, right? No! This equation is false. When m=m', it reads 0=1.

What has gone wrong?

To answer this, let’s look at another question. From the commutation relation above, we can, by using the generalized uncertainty principal, deduce,

\Delta L_z \Delta\theta \geq \hbar/2.

However, eigenstates of L_z have \Delta L_z = 0. Another contradiction! How to we resolve this?

Let’s recall our definition of L_z. It is an operator in a Hilbert space which acts on functions. But which functions does it act on? It only acts on functions which are periodic in \theta.1 So we have to be careful when acting on a function like \theta, which is either multi-valued, or nonanalytic.

Because multivalued and nonanalytic operators are not allowed as observables, \theta should not be thought of an observable over its whole range.

The obvious resolution to this is to only consider the analytic functions made from \theta. So we can find the commutation relations

[L_z,\sin\theta] = -i \cos\theta
[L_z,\cos\theta] = +i \sin\theta

We can then find the uncertainty relations,

\Delta L_z \Delta \sin\theta \geq \frac{1}{2} \langle\cos\theta\rangle
\Delta L_z \Delta \cos\theta \geq \frac{1}{2} \langle\sin\theta\rangle

which is not a contradiction, because eigenstates of L_z have \langle\sin\theta\rangle = \langle\cos\theta\rangle = 0.

1 How do we phrase this in terms of compactness? Functions that L_z acts on take their values from a compact set, however the domain isn’t the set of functions whose domain is compact. So something else about the topology of the domain of functions that L_z acts on is needed, but I don’t know enough functional analysis to have an obvious more elegant way to write this. We can phrase this in terms of when L_z is self-adjoint, but I don’t know that this leads to a nicer description offhand.