# Homework Help: Prove the basic rule's of Poisson's Brackets?

1. Oct 17, 2011

### zhillyz

1. The problem statement, all variables and given/known data
Show that:
(i){pi,pj}=0
(ii){qi,qj}=0
(iii){qi,pj}=δij

2. Relevant equations

{x,Y} = [(∂x/∂q * ∂Y/∂p) - (∂Y/∂q * ∂x/∂p)]

3. The attempt at a solution

Right I know first of all that qij and pij are generalized coordinates so that we dont need to worry about what units etc the answers are in.

I know that if the Poisson Bracket is equal to zero then the point you have used it on is a conserved quantity.

I think (i) and (ii) are ok but stuck on what to do on (iii). I have a feeling it has something to do with the Levi Civita Tensor as that is the last place I came across Kronecker Delta.

(i/ii){qi,qj} = [(∂qi/∂q)*(∂qj/∂p) - (∂qj/∂q)*(∂qi/∂p)]

The ∂qi/∂p and the ∂qj/∂p on either side are just '0' I think as it is a partial derivative of a q component with respect to p and it does not have any p component so QED 0 - 0 = 0

And that would be basically the same solution for parts (i) and (ii). Now...

(iii) {qi,pj} = [(∂qi/∂q) * (∂pj/∂p) - (∂pj/∂q)*(∂qi/∂p)]

Last part is equal to zero for same reasons as part (i) and (ii) leaving me with

(∂qi/∂q)*(∂pj/∂p)

Now I dont really have a clue what to do like I said before I think it is something to do with Levi.... any help is appreciated.

2. Oct 17, 2011

### grzz

Is not q$_{i}$ independent of q$_{j}$ for i being not equal to j?

So the partial derivative of one wrt the other is 1 if i = j and 0 otherwise.
Similarly for the p.