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Poisson brackets and angular momentum

  1. Aug 24, 2012 #1

    Rhi

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    1. The problem statement, all variables and given/known data

    Let f(q, p), g(q, p) and h(q, p) be three functions in phase space. Let Lk =
    εlmkqlpm be the kth component of the angular momentum.
    (i) Define the Poisson bracket [f, g].
    (ii) Show [fg, h] = f[g, h] + [f, h]g.
    (iii) Find [qj , Lk], expressing your answer in terms of the permutation symbol.
    (iv) Show [Lj , Lk] = qjpk−qkpj . Show also that the RHS satisfies qjpk−qkpj =
    εijkLi. Deduce [Li, |L|2] = 0.
    [Hint: the identity εijkεklm = δilδjm − δimδjl may be useful in (iv)]



    2. Relevant equations n/a

    3. The attempt at a solution


    i) [f,g]=[itex]\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}[/itex]

    ii) easy to show from the definition in i)

    iii) after a bit of working, I get εlmkql

    iv) my working is quite long, but I get [Lj,Lk]=qjpk-qkpjijkLi as required.

    The bit I'm having trouble with is the very last bit of the question, to deduce [Li, |L|2] = 0.

    Since it's only a small part of the question, it seems as though this part should be fairly simple so maybe I'm overlooking something, but I don't get 0. This is my working:

    [Li, |L|2]=[Li, LjLj]=Lj[Li, Lj]+[Li, Lj]Lj=2Lj[Li, Lj]

    I'm not entirely sure where to go from here so any help (or pointing out of any glaring errors) would be great.
     
  2. jcsd
  3. Aug 24, 2012 #2
    I don't really see how in the last line you get to 2L_j[L_i,L_j] if this commutator contains a L_k and the L_subscripts don't commute.
     
  4. Aug 25, 2012 #3

    Rhi

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    I'm not really sure what you mean, there's no L_k involved in the last line? And I'm not really sure what you mean by a commutator, either..
     
  5. Aug 25, 2012 #4
    Haha I was being infinitely stupid. I forgot you were talking about Poisson brackets. I have a lame excuse for it though namely that I usually use { , } for poisson and [ , ] for commutator. Now to redeem myself I will actually look at this last exercise. be back shortly!
     
  6. Aug 25, 2012 #5
    So here it goes. Leave the last equality out and when you get

    [itex] L_j[L_i,L_j]+[L_i,L_j]L_j = L_j\epsilon_{kij} L_k + \epsilon_{kij} L_kL_j = \epsilon_{kij}L_jL_k - \epsilon_{jik} L_kL_j = \epsilon_{kij}L_jL_k - \epsilon_{kij} L_jL_k = 0 [/itex]

    Where in the equality before last I just relabel j to k and vice versa in the second summand.
     
  7. Aug 25, 2012 #6

    Rhi

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    Ah, I get it. That makes a lot of sense, cheers :)
     
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