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For Angular Momentum Operator L, prove [Lx,Ly] = ihLz

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data

    For an angular momentum operator ~L =ˆiLx +ˆjˆLy + ˆkˆLz = ˆr × ˆp, prove that
    [ˆLx, ˆLy] = i[itex]\hbar[/itex]ˆLz,
    [ˆLx, ˆLz] = −i[itex]\hbar[/itex]ˆLy,
    [ˆL2, ˆLx] = 0,
    [[itex]p^{2}[/itex], ˆLx] = 0,
    [[itex]r^{2}[/itex], ˆLx] = 0,
    [ˆLx, ˆy] = i[itex]\hbar[/itex]ˆz,
    [ˆLx, ˆpy] = i[itex]\hbar[/itex]ˆPz.

    **Note: I'm really only looking for help for the first part here, but any suggestions on how to approach the others would be appreciated. I figure I can't approach those, until I understand fully how to work through the first one.**

    2. Relevant equations

    Px = −i[itex]\hbar[/itex] d/dx
    Py = −i[itex]\hbar[/itex] d/dy
    Pz = −i[itex]\hbar[/itex] d/dz

    Lx = yPz - zPy
    Ly = zPx - xPz
    Lz = xPy - yPx

    3. The attempt at a solution

    I'm simply looking for help on [Lx,Ly] = −i[itex]\hbar[/itex] Lz right now:

    Take [Lx,Ly] = [yPz - zPy, zPx - xPz] = [yPz, zPx]-[yPz, xPz]-[zPy, zPx]+[zPy, xPz]

    Looking at the first term:
    [yPz,zPx] = (yPz zPx - zPx yPz)
    ==> first term here: yPz zPx = y(-i[itex]\hbar[/itex])d/dz z(-i[itex]\hbar[/itex])d/dx
    ==> second term here: zPx yPz = z(-i[itex]\hbar[/itex])d/dx y(-i[itex]\hbar[/itex])d/dz

    ==> first term simplified: -[itex]\hbar^{2}[/itex]yz d/dz d/dx ????
    ==> second term simplified: -[itex]\hbar^{2}[/itex]zy d/dx d/dz ????

    Subtracting the second term from the first term, I get:
    [ -[itex]\hbar^{2}[/itex]yz d/dz d/dx + [itex]\hbar^{2}[/itex]zy d/dx d/dz ] ????

    The "????" is to represent where I'm having trouble with the algebra and differentiation.
    I'm not sure that these operations actually work out like this. Also, the next step I have in my notes here says:

    [yPz,zPx] = -[itex]\hbar^{2}[/itex]yz [itex]\partial^{2}[/itex]/dzdx - [itex]\hbar^{2}[/itex]y [itex]\partial[/itex]/dx + [itex]\hbar^{2}[/itex]yz [itex]\partial^{2}[/itex]/dxdz = -[itex]\hbar^{2}[/itex]y [itex]\partial[/itex]/dx

    It could be a simple chain rule issue that I'm just not seeing, but would somebody please explain to me how you obtain each of these three terms? The first and third terms I seem to have no problem coming up with, but the second term is throwing me, especially since there isn't a matching term to follow before the [itex]\hbar^{2}[/itex]yz [itex]\partial^{2}[/itex]/dxdz term. I think the second term is actually more common sense than the first and third terms since d/dz z = 1.... As I said, anyone who can explain how to arrive at each of these terms, your assistance would be greatly appreciated.
    Last edited: Oct 17, 2011
  2. jcsd
  3. Oct 17, 2011 #2
    Don't bother with actual representation in space (that momentum become a partial derivative operator).

    Since [itex]L_z = x p_y - y p_x[/itex] etc, just write out the commutator in Cartesian coordinates:
    [itex][L_x, L_y] = [y p_z - z p_y, z p_x - x p_z][/itex] and use properties of the commutator to get rid/factor out anything that commutes.

    You should end up with [itex][L_x, L_y] = y p_x [p_z, z] + x p_y [z, p_z][/itex]. Then apply the canonical commutation relation.
    Last edited: Oct 17, 2011
  4. Oct 17, 2011 #3


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    You're not evaluating the products correctly. You need to apply the product rule, not the chain rule.
    [tex]y\frac{\partial}{\partial z} z\frac{\partial}{\partial x}
    = y \left(\frac{\partial}{\partial z} z\right) \frac{\partial}{\partial x} + y z \left(\frac{\partial}{\partial z}\frac{\partial}{\partial x}\right) = y\frac{\partial}{\partial x} + \frac{\partial^2}{\partial z\partial x}[/tex]
    But as mathfeel implied, this is the long, tedious way of doing it. I suggest you follow his suggestion.
  5. Oct 18, 2011 #4
    Thank you both. I was able to solve that problem, I have moved forward in this problem and am stuck on the [r^2, Lx] portion. I know that the x[x,Lx] + [x,Lx]x = 0 and that I'm left with the y and z terms, but I'm having trouble with those. Do you have any suggestions towards that approach?

    I've got [y,Lx] = [y, xPz-zPy] ==> [y,yPz] - [y, zPy]
    and following [A,BC] = [A,B]C + B[A,C] this expands out to be

    [y,y]Pz + y[y,Pz] - [y, z]Py + z[y,Py]

    If I understand this correctly, the [y,y] term goes to zero, as it must, and so must [y,z]
    so I'm left with two terms y[y,Pz] - z[y,Py]

    Is this correct?
  6. Oct 18, 2011 #5


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    That's correct, but [y, pz] also vanishes.
  7. Oct 18, 2011 #6
    so I'm left only with -z[y,Py] huh?

    I've worked it out a few times, and if I remember correctly [y,Pz] clearly vanishes when operating on some form of dummy function, but without actually working it out, can you explain briefly why exactly it vanishes?
  8. Oct 18, 2011 #7
    "The only operator in Lz that does not commute with x, is Px..." but why?
  9. Oct 18, 2011 #8


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    Because when you take the partial with respect to z, you hold y constant.
  10. Oct 18, 2011 #9
    Most definitely, but forgive my lack of understanding at 1:00am , but is there a physical explanation, or is this just one of those things that is only mathematical?
  11. Oct 18, 2011 #10
    Never mind, Vela, you're amazing and awesome, and I really appreciate all the help you've given me. You answer just about every question I ask, and for that I thank you. I totally just woke up, and now things are making sense again. I think I need a short break.
  12. Oct 18, 2011 #11
    I do have one last question for you, does [Px, Py] commute? (Do each part of [Pi,Pj] commute?
  13. Oct 18, 2011 #12


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    P_x and P_y commute, of course. This is postulated in a tensorial notation [p_i,p_j] = 0, [itex]\forall[/itex] [itex] i,j=\bar{1,3} [/itex]. When i=1. j=2, it means exactly [p_x,p_y]=0.
  14. Oct 18, 2011 #13
    If you are in doubt, think intuitively back to differential operator. Can you change the order of taking partial derivative in the x and in the y direction?
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