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Angular momentum quantization

  1. Oct 23, 2006 #1
    You can define angular momentum for a free particle, with respect to another particle? i.e.

    L = v x r?

    This kind of angular momentum, would it be quantized?

  2. jcsd
  3. Oct 23, 2006 #2


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    Staff: Mentor

    Actually, in vectors it's

    [tex]\vec L = \vec r \times \vec p = \vec r \times m \vec v[/tex]

    For a free particle, L is not quantized, just as neither E nor p are quantized. For a bound system such as the hydrogen atom, L is indeed quantized as is of course the energy.
  4. Oct 23, 2006 #3


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    L, that is "orbital" angular momentum, -i X * d/dX, is always quantized -- it is a Hermitean operator, and thus has a spectrum, a discrete one in fact. The fact is that, often, we don't use L for free particles -- except, for example, when we are interested in partial wave scattering amplitudes, or in the multipole expansion for photons, or when we are using the Jacob. and Wick formalism, or when we follow Weinberg's approach to free single particle wave functions. (In both of these, you define the state's momentum along the z axis, and generate all other states by means of rotations.)
    Reilly Atkinson
    Last edited: Oct 23, 2006
  5. Oct 24, 2006 #4

    Meir Achuz

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    L is always quantized.
  6. Oct 28, 2006 #5
    And the quantization of am is because rotations don't commute, in quantum and classical worlds. (EDIT: or why is it?)

    Whereas the quantization of energy comes from confinement in space. When you confine a wave you get standing waves which only allow a certain set of frequency. Due to energy=Planck constant x frequency you get discrete energy values in quanum physics.
    Last edited: Oct 28, 2006
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