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Poisson brackets and angular momentum 
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#1
Aug2412, 08:31 AM

P: 10

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 = ε_{lmk}q_{l}p_{m} 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 [q_{j} , L_{k}], expressing your answer in terms of the permutation symbol. (iv) Show [L_{j} , L_{k}] = q_{j}p_{k}−q_{k}p_{j }. Show also that the RHS satisfies q_{j}p_{k}−q_{k}pj = ε_{ijk}L_{i}. Deduce [L_{i}, L^{2}] = 0. [Hint: the identity ε_{ijk}ε_{klm} = δ_{il}δ_{jm} − δ_{im}δ_{jl} 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 ε_{lmk}q_{l} iv) my working is quite long, but I get [L_{j},L_{k}]=q_{j}p_{k}q_{k}p_{j}=ε_{ijk}L_{i} as required. The bit I'm having trouble with is the very last bit of the question, to deduce [L_{i}, 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: [L_{i}, L^{2}]=[L_{i}, L_{j}L_{j}]=L_{j}[L_{i}, L_{j}]+[L_{i}, L_{j}]L_{j}=2L_{j}[L_{i}, L_{j}] I'm not entirely sure where to go from here so any help (or pointing out of any glaring errors) would be great. 


#2
Aug2412, 05:26 PM

P: 117

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.



#3
Aug2512, 03:38 AM

P: 10

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..



#4
Aug2512, 09:39 AM

P: 117

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!



#5
Aug2512, 09:53 AM

P: 117

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. 


#6
Aug2512, 10:33 AM

P: 10

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



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