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Lz and L^2 are commuting operators

  1. Feb 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Using the definitions of Lz and L^2, show that these two operators commute.


    2. Relevant equations
    Lz = -ih_bar * d/d(phi)
    L^2 = -(h_bar)^2 {1/sin(theta) * d/d(theta) * [sin(theta) * d/d(theta)] + 1/sin^2(theta) d^2/d(phi)^2}


    3. The attempt at a solution
    I'm actually not too sure how to begin this. If someone can show me an example of commuting operators that is a bit complex, that would definitely help. My professor only gave us a simple example, which is why I'm confused with this one. Thanks for your help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 23, 2009 #2

    gabbagabbahey

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    If [itex][L^2,L_z]=0[/itex] (that is L^2 and L_z commute), what can you say about [itex]L^2(L_z\psi)[/itex] and [itex]L_z(L^2\psi)[/itex] for any wavefunction [itex]\psi[/itex]?
     
  4. Feb 23, 2009 #3

    CarlB

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    [tex]x\partial_x[/tex] and [tex]y\partial_y[/tex] commute because
    [tex]x\partial_x ( y\partial_y ( f) ) = y\partial_y ( x\partial_x (f)) [/tex] for any function f.

    On the other hand, [tex]x\partial_y[/tex] does not commute with [tex]y\partial_x[/tex]
     
  5. Feb 23, 2009 #4
    gabba, I would say that LaTeX Code: L^2(L_z\\psi) and LaTeX Code: L_z(L^2\\psi) for any wavefunction LaTeX Code: \\psi would be equivalent.

    My problem is more with doing the problem. Do I do the product rule when it comes to -h_bar^2 * 1/sin(theta) * d/d(theta), or is -h_bar^2 pulled out to the front and I just take the derivative of 1/sin(theta)?
     
  6. Feb 23, 2009 #5

    gabbagabbahey

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    Well, -h_bar^2 is a constant isn't it.....don't you always pull the constants out when taking a derivative?

    Why don't you post an attempt and I'll see where you might be going wrong....
     
  7. Feb 23, 2009 #6

    lanedance

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    constants shouldn't matter in differentiation and keep in mind a phi partial derivative means theta is kept constant for teh dervative

    might help to remember that the partial derivative cross terms are the same for continuous & differentiable functions ie

    for f(u,v) then fuv = fvu

    (what are the exact conditions for this to apply?)

    So maybe make some assumtions about the wavefunction continuity & its derivatives etc...
     
  8. Feb 23, 2009 #7
    this is what i have...
    Lz(L^2) = ih_bar^3 * {[1/cos(theta) (sin(theta)) + (cos(theta))*1/sin(theta)] + 1/sin^2(theta)}
    simplified...ih_bar^3 [sin(theta)/cos(theta) + cos(theta)/sin(theta) + 1/sin^2(theta)]

    I did the exact same thing for L^2(Lz), but I realize that making a mistake for each of these will result in the same answer.
     
  9. Feb 23, 2009 #8

    gabbagabbahey

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    This makes no sense :confused:

    [itex]L_z[/itex] and [itex]L^2[/itex] are differential operators, so their product should be as well. You should have [tex]\frac{\partial^2}{\partial \phi \partial \theta}[/tex] terms in your expression for [itex]L_z(L^2)[/itex].
     
  10. Feb 23, 2009 #9
    how do i work with 1/sin(theta)d/d(theta) * (sin(theta)d/d(theta)) part?
     
  11. Feb 23, 2009 #10
    do you use the product rule or are they two totally different derivations?
     
  12. Feb 23, 2009 #11

    gabbagabbahey

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    Use the product rule!

    For example,

    [tex]\frac{\partial}{\partial x}\left[f(y)\frac{\partial}{\partial y}\left(g(y)\frac{\partial}{\partial y}\right)\right]=\frac{\partial}{\partial x}\left[f(y)g'(y)\frac{\partial}{\partial y}+f(y)g(y)\frac{\partial^2}{\partial y^2}\right]=f(y)g'(y)\frac{\partial^2}{\partial x \partial y}+f(y)g(y)\frac{\partial^3}{\partial x \partial y^2}[/tex]
     
  13. Feb 24, 2009 #12
    2nd time's a charm?
    -i*(h_bar)^3 [ cos(theta)/sin(theta) * d^2/d(theta)^2 + d^3/d(theta)^3 + 1/sin^2(theta) * d^3/d^3(theta)

    why is it that it's d/dx [ f(y)g'(y) d/dy + f(y)g(y)d^2/dy^2] instead of
    d/dx [ f(y)g'(y) d/dy + f'(y)g(y)d^2/dy^2]? (i added in an f ' to the second part. it might be hard to tell with this font.) i thought you said product rule. could you explain?
     
  14. Feb 24, 2009 #13

    gabbagabbahey

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    Because d/dy acts only on the product g(y) d/dy....it doesn't act on f(y)-which comes before the operator.

    And 2nd time is definitely not the charm....sorry::frown:....theta and phi are independent variables....but all I see is theta derivatives in your expression...what happened to the phi derivatives?
     
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