Prove the basic rule's of Poisson's Brackets?

[zhillyz]

Homework Statement

Show that:
(i){p_i,p_j}=0
(ii){q_i,q_j}=0
(iii){q_i,p_j}=δ_ij

Homework Equations

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

The Attempt at a Solution

Right I know first of all that q_ij and p_ij are generalized coordinates so that we don't 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){q_i,q_j} = [(∂q_i/∂q)*(∂q_j/∂p) - (∂q_j/∂q)*(∂q_i/∂p)]

The ∂q_i/∂p and the ∂q_j/∂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) {q_i,p_j} = [(∂q_i/∂q) * (∂p_j/∂p) - (∂p_j/∂q)*(∂q_i/∂p)]

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

(∂q_i/∂q)*(∂p_j/∂p)

Now I don't really have a clue what to do like I said before I think it is something to do with Levi... any help is appreciated.

 

[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.