Poisson brackets and angular momentum

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Homework Help Overview

The discussion revolves around the properties of Poisson brackets in the context of angular momentum in phase space. The original poster presents a series of tasks related to defining Poisson brackets and demonstrating specific relationships involving angular momentum components.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to define the Poisson bracket and show relationships between functions and angular momentum. Some participants question the steps taken in the derivation of the last part of the problem, particularly regarding the treatment of commutators versus Poisson brackets.

Discussion Status

Participants are actively engaging with the problem, clarifying terminology, and correcting misunderstandings. There is a productive exchange of ideas, with some participants providing insights into the calculations and reasoning behind the Poisson brackets and angular momentum relationships.

Contextual Notes

There is a noted confusion between the notation for Poisson brackets and commutators, which may affect the clarity of the discussion. The original poster expresses uncertainty about the final deduction involving the Poisson bracket of angular momentum components.

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

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