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Angular Momentum the Full Operator Approach

  1. Jan 7, 2012 #1
    Greetings all,

    In going through an operator approach to deriving the rules of angular momentum I find myself asking a curious question. Is it possible to fully derive the Angular momentum algebra relying on only a minimal set of classical relations.

    That is to say, if we take the only classical relation (which I call classical in the sense that we relate it to a physical picture in classical mechanics) that [itex]L^2[/itex] and [itex]L_z[/itex] (the z component being arbitrarily chosen as our quantization axis) are well defined simultaneously at any given moment in time and hence [itex]\left[L^2,L_z\right]=0[/itex]. Then can we find the commutation relations [itex][L_i,L_j]=i\epsilon_{ijk}L_k[/itex] (natural units) without ever referring to the classical definition of angular momentum? (I realize that perhaps the well defined nature of these two is intrinsic to the classical definition so perhaps that answers my question...)

    Essentially we have the following relations to work with:

    [itex]L^2=L_x^2+L_y^2+L_z^2[/itex] by definition (this is just a vector relation, so I don't consider it a classical physics relation, though perhaps it is... what do you think?)

    [itex][L^2,L_z]=[L^2,L_x]=[L^2,L_y]=0[/itex] from the arbitrary nature of choosing the quantization axis (we can always find a rotation that moves our system into a basis where any of the three components can be (up to a rotation) the [itex]L_z[/itex] component).

    [itex](L^2-L_z^2)\left|\lambda,\mu\right>=(\lambda^2 - \mu^2)\left|\lambda,\mu\right>[/itex] which can be shown to be a re-expression of the commutative nature of [itex]L^2[/itex] and [itex]L_z[/itex].

    And finally, by the first relation we can find
    [itex]L^2-L_z^2=L_x^2+L_y^2=\left(L_x+iL_y\right)\left(L_x-iL_y\right)+i[L_y,L_x]=L_+L_-+i[L_y,L_x][/itex]
    where we have chosen to identify arbitrarily the two operators [itex]L_+[/itex] and [itex]L_-[/itex] with the two terms of the product in the first term of the second part of the equality.

    Now the question is, can we derive using only this information what the commutator [itex][L_y,L_x][/itex] is?

    I have tried several times to find an intrinsic relation that will give the correct answer, However I keep coming up short as it seems without referring to the classical definition of angular momentum [itex]\vec{L}=\vec{r}\times\vec{p}[/itex] and letting [itex]\vec{p}\rightarrow i\nabla[/itex] there is no way of incorporating the uncertainty principle into the angular momentum relations.

    What do you think? Is it necessary to revert to a classical relation to find this quantity? Perhaps it is because we have started at the classical level and worked down to the quantum level that we have this requirement. Though that doesn't mean there isn't a more fundamental starting point from which we could work up to quantum mechanics and its simply that I do not know what that more fundamental description is...

    If anyone knows a way to get the commutation relations without the classical relation I would be most interested to see it. Either way I am curious what your thoughts are on this.
     
  2. jcsd
  3. Jan 7, 2012 #2

    kith

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    Science Advisor

    You can't derive the commutation relations of angular momentum from these two assumptions alone, because they also hold for many other observables (X and P for example).

    However, you can derive them without using the classical definition of angular momentum by using the fact that classical rotations in different directions don't commute. This is done in Sakurai for example.
     
  4. Jan 7, 2012 #3
    I'd recommend looking through chapters 2 and 3 of Ballentine's book on quantum mechanics, I think he fully discusses pretty close to what you want.
     
  5. Jan 8, 2012 #4
    Excellent! Thank you for the references!
     
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