Canonical Transformation/Poisson Brackets

[roeb]

Homework Statement

I am trying to show that [q_j, p_k] = \delta_{jk} (this is part of exercise 2.7.3 from Shankar's QM). I'm having difficulties with the summation notation.

Homework Equations

The Attempt at a Solution

[q_j, p_k] = \sum_{k} (\frac{\partial q_j}{\partial q_k} \frac{\partial p_k}{\partial p_k} - \frac{\partial q_j}{\partial p_k} \frac{\partial p_j}{\partial q_k} = \sum_{k} - \delta_{jk} = \delta_{jk} ??
I'm not so confident on my choice of 'k' as the summation variable. It seems to me the summation should not disappear like that. If I am interpreting this correctly, the negative sign isn't such a big deal... Can anyone check my work, I don't think I am doing it correctly

 

[xepma]

Yes, you have to be careful about that. Left, the p_k carries an index k. That means you shouldn't use k as a summation dummy index on the right hand side. Try working it out starting from the following:

[q_j,p_k] = \sum_n \left( \frac{\partial q_j}{\partial q_n}\frac{\partial p_k}{\partial p_n} - \frac{\partial q_j}{\partial p_n}\frac{\partial p_k}{\partial q_n}\right)

EDIT: mixed up the indices myself...

 

[Cyosis]

There is more amiss here. The indices of the numerators are wrong as well. (we're talking about Poisson brackets here right?)

<br /> [q_j,p_k]=\sum_n \left( \frac{\partial q_j}{\partial q_n}\frac{\partial p_k}{\partial p_n} - \frac{\partial q_j}{\partial p_n}\frac{\partial p_k}{\partial q_n}\right)<br />