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Problem with angular momentum operator math

  1. Oct 24, 2008 #1
    1. The problem statement, all variables and given/known data
    I am basically trying to show that LxLy-LyLx=i(hbar)Lz

    2. Relevant equations

    3. The attempt at a solution

    I get to the end where I have i(hbar)Lz-z*y*PxPz+z*xPyPz. How do I get these last two terms to cancel out? I am not too strong in operator math (it hasnt been taught to me, jsut assumed I can do it), so I dont know if the d/dz part of the Pz operator can still apply to a z that is to the left of it, but if it does then I get LxLy-LyLx=2*i(hbar)Pz which is wrong according to my book.... Can anyone help?
    Last edited: Oct 24, 2008
  2. jcsd
  3. Oct 24, 2008 #2
    Do you know how to calculate z*P_z-P_z*z?

    The equation you're trying to prove is wrong anyway, so that will be quite hard...
  4. Oct 24, 2008 #3
    Is the solution to your calculation 0 or zP_z+i(hbar)? This is where my operator math gets fuzzy.

    And I extracted that equation from a larger problem that asks me to prove that the angular momentum operator (L) squared commutes with each component of the angular momentum. The book goes about solving this by showing that:


    It does not explain how or why this is equivalent, so seeing how the left has to equal the right, I extracted my question. Can you explain why this is not correct or what I am doing wrong?

    AH! I apologize, I had a P instead of an L in my first post, sorry if that caused confusion.
    Last edited: Oct 24, 2008
  5. Oct 24, 2008 #4
    Neither: you should apply the operator z*P_z -P_z *z to a dummy wavefunction:

    (z*P_z -P_z *z) psi = S psi

    where S turns out to be a very simple operator (so, you'll have to find what that simple operator S is). That gives you

    z*P_z -P_z *z=S

    Then you can use that to figure out what L_x L_y-L_yL_x is.
  6. Oct 24, 2008 #5
    Im sorry but I dont understand what you did. It seems like you threw up an arbitrary operator S and used that to define the solution to (z*P_z -P_z *z) as itself. If it is not 0 or i(hbar), then what is its exact solution?
  7. Oct 24, 2008 #6
    It is i\hbar (but that's not what you wrote in your question). That is, I wanted you to find the simple operator S: the answer is indeed S=i\hbar I (with I the identity, which is usually left out)
  8. Oct 24, 2008 #7
    Now substitute zP_z-P_z z=i\hbar in:


    what do you get?
  9. Oct 25, 2008 #8
    Before I do that, can you show me how it ends up being i/hbar? I cannot for the life of me get to that answer which is probably the root of my difficulties haha.

    Ill show my steps I guess so you can point out where I am wrong:
    zP_z-P_z z=(z*-i*hbar*d/dz)-(-i*hbar*d/dz*z)=-i*z*hbar*d/dz+ihbar=zP_z+ihbar (if the d/dz operator cannot apply to z since z is to its left)

    OR -ihbar+ihbar=0 (if it can apply to z)
    Last edited: Oct 25, 2008
  10. Oct 25, 2008 #9
    I tried to tell you how to calculate it: apply the operators to a (dummy) function. The operator d/dz then applies to everything to its right: for one term, the z*P_z term, it only applies to the function, but for the term P_z*z it will apply to the product of z and the function.
  11. Oct 25, 2008 #10
    Ok, so even if I do put a function in I dont see how you get your result:

    (zP_z-P_z z)psi(z)=(z*(-i*hbar*d/dz*psi(z))-(-i*hbar*d/dz*z*psi(z))=-i*z*hbar*Psi(dz)+ihbar*Psi(dz)=(-i*z*hbar+i*hbar)Psi(dz)

    Where is i/hbar coming from?
    Last edited: Oct 25, 2008
  12. Oct 25, 2008 #11
    First, it's not i/hbar, it's i\hbar, which is LaTeX for i*hbar.

    Second, from your manipulations I get the impression you don't know how to apply d/dz to a function.
  13. Oct 25, 2008 #12
    Well I have a hard time typing it out, but it is just Psi(dz) instead of Psi(z) when d/dz is applied to it correct?

    And can you show me where I am going wrong or how much further I have to go to get to that result because that will essentially enable me to solve this problem since I keep running into the same issues that I am having with your simple sample.

    Ill edit the above post to reflect this.
  14. Oct 25, 2008 #13
    Psi(dz)?? I have no idea where you would get this from.

    Do you know what d/dz stands for?
  15. Oct 25, 2008 #14
    Yes it is the derivative with respect to z. So if I have a dummy function psi and I take the derivative of it with respect to z, then I get Psi'(z) (prime of z) or I was calling it Psi(dz) in the event that it had other variables (d/dz psi(x,y,z)=psi (x,y,dz)). If this is incorrect notation I apologize and I do not know the correct way to express it.
  16. Oct 25, 2008 #15
    Okay, good: let's go with Psi' then. What is d/dz (z*Psi)? Then subtract z*Psi', and you have your answer.
  17. Oct 25, 2008 #16
    d/dz (z*Psi)=Psi(z)+zPsi'(z)

    Ahhhh ok I totally forgot about product rule and was stopping the derivative operator before psi. Thank you so much!
  18. Oct 25, 2008 #17
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