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Commutator with Tensor Notation

  1. Apr 6, 2013 #1
    Greetings,

    I would like to find the commutator [itex]\left[Lx^2,Ly^2\right][/itex] and prove that
    [itex]\left[Lx^2,Ly^2\right][/itex]=[itex]\left[Ly^2,Lz^2\right][/itex]=[itex]\left[Lz^2,Lx^2\right][/itex] I infer from the cyclic appearance of the indices that using the index notation would be much more compact and insightful to solve the problem. However due to summation convention I do not know how to write a component squared instead of the whole vector squared. What is the remedy? Any help suggestion is welcome.
     
  2. jcsd
  3. Apr 6, 2013 #2

    atyy

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  4. Apr 6, 2013 #3
    Thanks for the reply. Yes, I am sure the index is squared the question is from Liboff Introduction to Quantum Mechanics 4th edition, and it is given as I stated. I do not know the physical insight behind this question but I want to prove that identity using index notation. How can I do so?
     
  5. Apr 6, 2013 #4

    strangerep

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    I don't have a copy of Liboff, but I guess the ##L##'s are so(3) generators? If so, you don't need index notation. Just apply the Leibniz product rule for commutators, i.e., ##[a,bc] = b[a,c] + [a,b]c##, to the first commutator ##[L_x^2, L_y^2]## twice. Then cyclically permute x,y,z to get another expression and use the so(3) commutation relations to show it's the same as the original.

    If you need more help than this, don't forget to show your work -- like in the homework forums.

    (Actually, this probably should be in the homework forums. :devil: )
     
  6. Apr 7, 2013 #5
    I do not have a solid background in the mathematical aspect of the quantum theory so I may not say that if the operator is a so(3) generator or not. I applied the same approach as you illustrated to the problem and it is tractable that way but I insist on using the index notation.
     
  7. Apr 8, 2013 #6
    if you really want to invoke index notation then note that for example Li2 can be written as
    Li2ijkrjpkεilmrlpm=(δjlδkmjmδkl)rjpkrlpm
    after it you will have to use the first quantized form for momentum operators.
     
  8. Apr 8, 2013 #7
    andrien I think your suggestion also sums over all components of angular momentum from the way you expand the product of two Levi Civita Tensors.
     
  9. Apr 9, 2013 #8
    yes,that is what I wanted from you to recognise.It is the total angular momentum square and so you will get the famous operator L2 from it i.e.
    L2=xijxij-xijxji=(r22+r.∇)-{3r.∇+(r.∇)2-(r.∇)}=r22-r.∇-(r.∇)2,which ultimately gives the L2.you can see it will be of no use to determine the components separately.
     
  10. Apr 9, 2013 #9

    Fredrik

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    The first equality isn't true. The expression after the first equality is equal to ##L_iL_i##, which is equal to ##\mathbf L^2##, not ##L_i^2##.

    There's no need to use the definition of L in terms of r and p here. It's sufficient to use the commutation relations for the components of L. There's also no need to use tensor index notation, because what strangerep said completely solves the problem. However, if you really want to use tensor index notation for some reason, then you need to start with ##[L_i^2,L_j^2]##. Ah, I see the problem. When you use the rule [AB,C]=A[B,C]+[A,C]B, the same index will appear twice, and you don't want this to imply summation. This just means that you can't use the summation convention in this problem.

    So if you insist on using the index notation, you need to start by saying that in this calculation, you're not using the summation convention. Then if you need to sum over some index, you just write a summation sigma.
     
  11. Apr 9, 2013 #10
    Li2 means LiLi where the summation is understood (no use of metric tensor) which is equal to L2.It should not pose any ambiguity.Also with the components of say Lx2 really poses problem because indices should not be summed even if it appears twice.So it is better not use index notation.
     
  12. Apr 9, 2013 #11

    Fredrik

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    This is where we disagree. Obviously, if you just write ##L_iL_i## without any comments, then it should be interpreted as ##\mathbf L^2## because of the summation convention, but ##L_i^2## simply means ##L_1{}^2## if i=1, ##L_2{}^2## if i=2 and ##L_3{}^2## if i=3. That's why you can't rewrite ##L_i^2## as ##L_iL_i## without saying that we're not using the summation convention.

    I obviously can't rule out that there's some book that uses the notation ##L_i^2## to mean ##L_iL_i##, but I would consider that a bad notation, because an index that doesn't explicitly appear twice shouldn't be summed over. Also, we already have the notation ##\mathbf L^2## for ##L_iL_i##. We don't need another.
     
  13. Apr 9, 2013 #12
    Okay thanks I also think I cannot make any contractions so tensor notation is not that suitable for this problem at hand.
     
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