Poisson brackets and angular momentum

  • Thread starter Rhi
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  • #1
Rhi
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Homework Statement



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)]



Homework Equations

n/a

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.
 

Answers and Replies

  • #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.
 
  • #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..
 
  • #4
133
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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
133
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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.
 
  • #6
Rhi
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Ah, I get it. That makes a lot of sense, cheers :)
 

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