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Poisson brackets and angular momentum |
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| Aug24-12, 08:31 AM | #1 |
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Poisson brackets and angular momentum
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-qkpj=εijkLi 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. |
| Aug24-12, 05:26 PM | #2 |
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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.
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| Aug25-12, 03:38 AM | #3 |
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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..
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| Aug25-12, 09:39 AM | #4 |
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Poisson brackets and angular momentum
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!
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| Aug25-12, 09:53 AM | #5 |
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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. |
| Aug25-12, 10:33 AM | #6 |
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Ah, I get it. That makes a lot of sense, cheers :)
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