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Particle physics J=L+S

  1. May 18, 2014 #1

    Matterwave

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    I've never really been...convinced...of the statement

    $$\bf{J}=\bf{L}+\bf{S}$$

    I've always just gone along with it, but I've never seen why this is "right". So I guess now's as good a time as any to ask.

    Thinking about this from a "classical" perspective (which obviously is not correct, but perhaps I can at least show where my doubt comes from), if the L stands for the angular momentum of the particle with respect to the center of mass, and the S stands for the angular momentum of the particle "spinning" around (again, obviously not right), then the two should be measured from different coordinate origins (e.g. L measured from the proton in a Hydrogen nucleus if we are looking at the electron, and S is measured from the "center" of the electron). So, I can not motivate the correctness of this statement from naive classical analyses.

    Looking at this mathematically (e.g. from the analysis in Ballentine chapter 7), we have that the ##\bf{L}## operators act on the physical space while the ##\bf{S}## operators act on the internal space.

    Ballentine then says something along the lines of, the total rotation operator ##e^{in_\alpha J_\alpha }## must be in the form:

    $$e^{in_\alpha J_\alpha }=e^{in_\alpha L_\alpha }e^{in_\alpha S_\alpha }$$

    From which the statement ##\bf{J}=\bf{L}+\bf{S}## is true if the L's and S's commute. But I'm not convinced that the above formula is a simple product, and not a direct product. The two different operators operate in different spaces, and so, shouldn't it be a direct product? If I express my state function as a 2 component vector e.g. ##\Psi_i (x,t), i=1,2##, for example, the rotation dealing with ##\bf{L}=-i\hbar \bf{x}\times\nabla## must be applied to each component individually, while the rotation dealing with S applies to my 2 component vector as a whole. The whole J=L+S thing doesn't make sense to me taken as an operator equation since L is a differential operator, and S is a matrix. What's the sum of a derivative and a matrix supposed to mean? Unless I am now constructing a 2x2 diagonal matrix for ##\bf{L}##? I'm confused. =/


    This has always bothered me.
     
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  3. May 18, 2014 #2
    The correct form of the equation is
    [tex] \hat{\bf{J}} = \hat{\bf{L}}\otimes\hat{\bf{I}} + \hat{\bf{I}}\otimes\hat{\bf{S}}[/tex]
    where the [itex]\hat{\bf{L}}[/itex] and [itex]\hat{\bf{S}}[/itex] generate rotations in different subspaces.
     
  4. May 19, 2014 #3

    Matterwave

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    And so should there be a direct product in:

    $$e^{i\theta_\alpha J_\alpha}=e^{i\theta_\alpha L_\alpha}\otimes e^{i\theta_\alpha S_\alpha}$$

    ?
     
  5. May 19, 2014 #4
    Yes, mathematically (and intuitively), there should be a direct product.
    I would suggest Modern Quantum Mechanics, J.J. Sakurai, section 3.7 for further reading.
    Edit:
    There is a little discrepancy in the exponential operators. Please check Sakurai.
     
  6. May 19, 2014 #5

    dextercioby

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    Well, something like this:

    $$e^{i\theta_\alpha J_\alpha}=e^{i\theta_\alpha L_\alpha}\otimes \hat{1} + \hat{1}\otimes e^{i\theta_\alpha S_\alpha}$$
     
  7. May 19, 2014 #6

    Matterwave

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    Ah, that's helpful, thanks.
     
  8. May 20, 2014 #7
    The infinitesimal rotation operator that affects the Hilbert space (made from tensor product of position space and spin space) is
    [tex]
    \hat{I}-\frac{i(\hat{\bf{L}}\otimes\hat{I}_2+\hat{I}_1\otimes\hat{\bf{S}}).\bf{n} d\theta}{\hbar} =\left(\hat{I}_1-\frac{i\hat{\bf{L}}.\bf{n} d\theta}{\hbar}\right)\otimes\left(\hat{I}_2-\frac{i\hat{\bf{S}}.\bf{n} d\theta}{\hbar}\right).
    [/tex]

    For a fine angle rotation, the equation takes the form
    [tex]
    \exp\left(\frac{-i\hat{\bf{J}}.\bf{n} \theta}{\hbar}\right)=\exp\left(\frac{-i\hat{\bf{L}}.\bf{n} \theta}{\hbar}\right)\otimes\exp\left(\frac{-i\hat{\bf{S}}.\bf{n} \theta}{\hbar}\right).
    [/tex]
     
  9. May 20, 2014 #8

    Matterwave

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    Uh...isn't that what I had in post #3?
     
  10. May 20, 2014 #9

    ChrisVer

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    well of course there is a direct product, since we are talking about operators on the exponentials...
     
  11. May 20, 2014 #10
    Yes you were right (I am sorry for creating the confusion). It was that you wrote the rotation operator as [itex]e^{in_\alpha J_\alpha}[/itex] in the first post and then you changed it to [itex]e^{i\theta_\alpha J_\alpha}[/itex].
     
  12. May 20, 2014 #11

    Matterwave

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    Indeed, I changed notations around and did not explain them. I get your point.

    But the direct product of the two exponentials look a little different than what Dexter wrote, or are they equivalent?
     
  13. May 20, 2014 #12
    Yes. They are equivalent. The exponential operators, that dextercioby mentions, give superimposition of the seperate rotations in position and spin space. This is the same rotation which [itex]\hat{\bf{J}}[/itex] generates in whole of position [itex]\otimes[/itex] spin space.
     
    Last edited: May 20, 2014
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