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Is angular momentum something that is characteristic for a particle?

  1. Dec 26, 2011 #1
    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. jcsd
  3. Dec 26, 2011 #2
    >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.
     
  4. Dec 26, 2011 #3
    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 :)
     
  5. Dec 28, 2011 #4
    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
     
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