# Orbital Angular Momentum Origin

by widderjoos
Tags: angular momentum
 P: 21 We know from classical mechanics that angular momentum $L = r \times p$ depends on your choice of origin. My question is: How does this work quantum mechanically? We know we get certain eigenvalues, but does this apply only in a certain choice of origin? How do we calculate angular momentum at some other point? I had a similar problem concerning torque on a magnetic dipole, $\tau = \mu \times B = r \times F$. About what point do we measure the moment arm? Do we just assume our origin is at the "center" of the orbit? Thanks for the help.
P: 3,260
 Quote by widderjoos We know from classical mechanics that angular momentum $L = r \times p$ depends on your choice of origin. My question is: How does this work quantum mechanically? We know we get certain eigenvalues, but does this apply only in a certain choice of origin? How do we calculate angular momentum at some other point? I had a similar problem concerning torque on a magnetic dipole, $\tau = \mu \times B = r \times F$. About what point do we measure the moment arm? Do we just assume our origin is at the "center" of the orbit? Thanks for the help.
Angular momentum depends on position like in classical mechanics. However, when speaking of L e.g. for an atom, one always refers to the angular momentum of the atom in the rest frame of its center of mass so that L becomes independent of position. In principle it would be more appropriate to talk of a contribution to the spin of the compound particle than of angular momentum.
 P: 343 In quantum mechanics, orbital angular momentum usually describes electron orbitals, where orbitals are located in an atom. When we regarding to an atom, I don't think choosing an arbitrary axis, say the tree trunk outside, would mean any thing to solving problems. Therefore, it has no necessity to specify the axis, since we all know what we are referring to
P: 21

## Orbital Angular Momentum Origin

Yeah, that's what I was thinking, but just wanted to make sure since I couldn't find it explicitly stated anywhere. Thanks!

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