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Quantum angular momentum

  1. Nov 27, 2008 #1
    There exists quantized angular momentum for orbiting electrons of the atom. What would be the quantized angular momentum of a free electron undergoing non orbital angular momentum.
     
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  3. Nov 27, 2008 #2

    jambaugh

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    In the usual treatment free electrons will have non-negative energy and the angular momentum (about some point) will have a continuous spectrum. Remember that angular momentum must be defined relative to some center even if the particle is not in a bound orbit.

    Now if you want to get into some more advanced cases, suppose that the spatial universe is a compact 3-sphere (in the same way that the surface of the earth is a 2-sphere) then linear and angular momentum are unified (moving in a "straight" line is just rotating on the 3-sphere just as walking on the Earth is just rotating about it's center)....

    ...if you assume this then quantum mechanics dictates that both linear and angular momentum will be quantized.
     
  4. Nov 27, 2008 #3
    i"f you assume this then quantum mechanics dictates that both linear and angular momentum will be quantized"

    Is this the general assumption of quantum mechanics?
     
  5. Nov 27, 2008 #4

    Avodyne

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    No, angular momentum is always quantized, even for a free particle. A plane wave can be written as a superposition of "spherical waves", each with a definite, integer value of the orbital angular momentum (around some arbitrarily chosen point in space that serves as the origin of coordinates). This is usually explained in discussions of "scattering theory" in QM books.
     
  6. Nov 27, 2008 #5

    jambaugh

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    Not an assumption i.e. axiom but rather a direct calculation from a mathematical theorem about representations of Lie groups.
    If the particle is confined to a compact manifold then its energy and momenta will be quantized. You can do the calculation directly from the Schrodinger equation.

    A simpler way to see it though is in the same way you see quantized angular momentum for a bound electron... in rotating around the atom (as a wave) its wavelength must match up with the circumference of its orbit. So too in an orbit around the universe if the universe is compact and periodic.
     
  7. Dec 1, 2008 #6
    there seems to be two conflicting answers which one of you guys is the most right?
     
  8. Dec 1, 2008 #7

    jambaugh

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    I'm likely the one mistaken...
     
  9. Dec 1, 2008 #8

    jambaugh

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    Ahhhhrrrrrrg!

    Avodyne is quite correct and I am terribly mistaken. I was mis-remembering. It is the energy spectrum which is quantized in the bound modes (negative values) and continuous in the free modes. (This due to the continuous spectrum of linear momentum.) Again if you consider the closed universe case then you again get a discrete spectrum...

    And that is precisely because linear momentum in this spherical universe case acts just like more components of an angular momentum and so has a discrete spectrum!

    Ugggg! I knew this . . . i.e. even was relying on it but mixed it up in my tired memory.

    My deepest apologies I jumped the gun with my shoes untied and tripped onto my face.
     
  10. Dec 1, 2008 #9
    I'm inclined to disagree. Although I agree that the plane wave for the electron can be a superposition of spherical waves with integer values of angular momentum, we need to remember that it's now a superposition of several wavefunctions. each of the wavefunctions have quantized angular momentum, the coeficents in front of each wave function are not.

    For example, let's say the new plane wave is a superposition of two of the old spherical waves. The expectation value for L would be...

    L total = A^2 <psi1 | L1 | psi1> + B^2 <psi2 | L2 | psi2>

    Where each of the expectation values in braket notation are quantized, but the coeficients are determined by the physical setup of the problem. Anyone agree? If not where did I go wrong?
     
  11. Dec 1, 2008 #10
    the main problem i seem to come across when trying to understand this is that i can't find any information on the subject. Everywhere i look for quantum angular momentum, orbital quantum angular momentum comes up ie. electrons of the atom. Can anyone tell me where to find more on the subject?
     
  12. Dec 1, 2008 #11
    Well, let's say you have a free electron in orbit around the sun. There's no reason to expect that orbital angular momentum to be quantized.
     
  13. Dec 1, 2008 #12

    Avodyne

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    This is not correct. First of all, if it's in orbit, it's not free, it's being acted on by the sun's gravity. In fact, since gravity and the electrostatic force are both [itex]1/r^2[/itex] forces, this is, mathematically, just the same problem as the hydrogen atom, where angular momentum is definitely quantized.

    Second, it doesn't matter whether there is a potential or not, or whether space is compact or not; the eigenvalues of any component of the angular momentum operator must be integer multiples of [itex]{1\over2}\hbar[/itex]. This can be proven directly from the angular momentum algebra,
    [tex][L_x,L_y]=i\hbar L_z,[/tex]
    and cyclic permutations. This is done in many QM texts.

    Quantization still allows for continuous expectation values; quantization means that the eigenvalues of an operator form a discrete set. Again, this is true of angular momentum whether or not there is a potential.
     
    Last edited: Dec 2, 2008
  14. Dec 1, 2008 #13

    jtbell

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    The coefficients in front of each wave function are not what? :confused:

    The superposition state has an indeterminate angular momentum. When something "measures" the angular momentum, you always get one of the quantized values, chosen at random according to probabilities that are determined by the coefficients.
     
  15. Dec 1, 2008 #14

    Avodyne

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    Hey, happens to us all!
     
  16. Dec 2, 2008 #15

    jambaugh

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    It is quantized but this quantum is so small compared to the magnitude that it is difficult to measure. However if you could fly a large enough S-G magnet past the whole system... and do this repeatedly enough times to get a distribution ... and cool the system down enough to eliminate thermal noise (i.e. use a solar sized cold neutron star)...

    Then in principle you'd see a discretized spectrum of angular momenta.

    Ultimately you need only look at the coordinate dual to the generalized momentum observable in the canonical treatment. If the coordinate is compact (in this case angle) then the observable has a discrete spectrum. If the coordinate is not compact then the observable may have a continuous spectrum.

    This to me argues for a compact spatial universe, otherwise you run into infinite information encoded even a small interval of a particle's continuous spectrum.
     
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