# Angular momentum in a rectangular box

1. Nov 27, 2007

### Mjolnir

This is more of a conceptual question. I'm looking for a "physical argument" as to why the angular momentum of a classical particle about the center of a 3d box wouldn't be conserved, as opposed to spherical well in which orbital angular momentum is a constant.

2. Nov 27, 2007

### nrqed

It depends what you mean by this. What are the forces on the particle? I am not sure if you mean a physical 3 d box which has a a gravitational pull on the particle, if the particle is inside or outside of the box, etc.

(nice handle...Thor's hammer if I remmeber correctly)

3. Nov 27, 2007

### Mjolnir

by a 3d box I mean a potential well such that V(x,y,z) = 0 for -a/2 < x < a/2, -b/2 < y < b/2, and -c/2 < z < c/2; or V -> infinity otherwise

edit: glad you like the screen name :-)

4. Nov 28, 2007

### nrqed

But then the angular momentum is conserved as long as the particle does not touch the sides of the box, trivially.
If the particle bounces on the sides, then angular momentum is not conserved because the force exerted by the walls is not a central force (it's not along th eradius). Therefore, the change of momentum is not along the radius an dthe angular momentum is not conserved.

You can see this easily if you imagine the box to have a finite mass. when the particle bounces, the box would start to spin. Of course, the angular momentum of the combined system (box plus particle) would be conserved.

5. Nov 28, 2007

### Mjolnir

To expand on this a bit, how would one go about using the time-independent Schrodinger equation to prove that there are no states with both definite energy and definite orbital angular momentum magnitude?

6. Nov 28, 2007

### nrqed

Well, one way is to write the energy eigenfunctions (which are easy to write down since they are products of sine function) and then to apply the angular momentum square operator and show that these states are not eigenstates of L^2.