# Is angular momentum something that is characteristic for a particle?

1. Dec 26, 2011

### Oww2

Hi,

I've got some questions about angular momentum. I hope they aren't too stupid, but I can't see the wood for the trees.

Is angular momentum something that is characteristic for a particle? I know that spin is characteristic (for example, the spin of a pi- is always 0) if I'm correct, but is the spin of the combination pi+pi- always 0?

And if I want to know the angular momentum of the combination of particles 123, can I group them the way I want? If I know the angular momentum of the combination 12, can I group those two particles and determine the angular momentum of (12)3? Is that the same as the angular momentum of 1(23)?

Thanks!

2. Dec 26, 2011

### EmpaDoc

>Is angular momentum something that is characteristic for a particle?
Certainly not. First of all, just as in classical physics, a.m. is always defined relative to a point in space. r cross p depends on what the origin of r is! And the angular momentum of a particle about this point can certainly change as time goes on, just as it does in classical physics. (Example: An electron is in orbit around a nucleus, then is excited to become a free particle) All the conservation laws from classical physics apply also in quantum physics, just remember that angular momentum is quantized.

>I know that spin is characteristic (for example, the spin of a pi- is always 0) if I'm correct, but is the spin of the combination pi+pi- always 0?
No, for example they could be spinning around their common center of mass.

>And if I want to know the angular momentum of the combination of particles 123, can I group them the way I want? If I know the angular momentum of the combination 12, can I group those two particles and determine the angular momentum of (12)3? Is that the same as the angular momentum of 1(23)?
Yes. The total angular momentum of a system is uniquely defined and does not depend on how you calculate it.

3. Dec 26, 2011

### Oww2

Thanks! :)

I forgot to ask something:

There is always conservation of total angular momentum, but spin doesn't have to be conserved, is that correct? And if there's a reaction: a + b -> c + d, and I want to check if the reaction is possible, one of the things I have to do is determine the total angular momentum of a + b and c + d and see if it's the same? If the spatial angular momentum is 0 at both sides of the reaction, then you just have to add the spins of a + b and c + d and see if they overlap (for example: spin of a + b = 0,1, spin of c + d = 0, then the reaction is possible because the spins both can be 0?)? But how do you add them when the spatial angular momentum isn't 0?

I know this are a lot of questions, I'd be very thankful if someone answered them :)

4. Dec 28, 2011

### EmpaDoc

Of course, you'll have all these questions answered when you enter an undergraduate course on quantum mechanics. Including lots of problem solving, that's the proper way to learn this!
But OK, here goes. :)

The answer to your question is that spin and orbital angular momentum are treated on an equal footing when adding angular momenta. For the observables and operators,
J = S + L
(S is spin, L is orbital a.m., and J is total a.m.)
and for the quantum numbers you thus may obtain one of the cases
|s-l| <= j <= s+l

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook