Angular momentum in a rectangular box

AI Thread Summary
The discussion centers on the conservation of angular momentum for a classical particle in a 3D box potential. It is noted that angular momentum is conserved as long as the particle does not collide with the box's walls. When the particle bounces off the walls, angular momentum is not conserved due to the non-central forces involved. The conversation also touches on the implications of a finite mass box, which would result in the box spinning upon particle collisions, conserving the total angular momentum of the system. Additionally, there is a query about using the time-independent Schrödinger equation to demonstrate the absence of states with both definite energy and orbital angular momentum magnitude.
Mjolnir
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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.
 
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Mjolnir said:
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.

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)
 
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 :-)
 
Mjolnir said:
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 :-)

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.
 
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?
 
Mjolnir said:
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?

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.
 
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