Poisson brackets and angular momentum

Click For Summary
The discussion focuses on the application of Poisson brackets in the context of angular momentum. The Poisson bracket is defined as [f,g]=∂f/∂q_i ∂g/∂p_i - ∂f/∂p_i ∂g/∂q_i. The participants successfully demonstrate the properties of Poisson brackets, including showing that [Lj, Lk] = qjpk - qkpj and that this expression can be related to the permutation symbol. A key point of confusion arises regarding the deduction of [Li, |L|^2] = 0, which is clarified through manipulation of the Poisson brackets and recognizing the symmetry in the expressions. Ultimately, the participants resolve their misunderstandings and confirm the validity of their calculations.
Rhi
Messages
8
Reaction score
0

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]=\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}

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.
 
Physics news on Phys.org
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

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

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 :)
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
7
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
571
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K