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## Homework Statement

So we have infinitesimal transformations from ##q_i## to ##\bar{q_i}## and ##p_i## to ##\bar{p_i}## ( where ##p_i## represents the canonical momentum conjugate of ##q_i##) given by $$\bar{q_i} = q_i + \epsilon \frac{\partial g}{\partial p_i}$$ $$\bar{p_i} = p_i - \epsilon \frac{\partial g}{\partial q_i}$$ where ##g(q,p)## is any dynamical variable. We must show that this is a conanical transformation. The hint is to work in first order in ##\epsilon##.

## Homework Equations

[/B]For canonical transformations we have ##\{ \bar{q_i} , \bar{q_j} \} = 0, \{ \bar{p_i}, \bar{p_j} \} = 0, \{ \bar{q_i}, \bar{p_j} \} = \delta_{ij} ##

## The Attempt at a Solution

Summing over ##j##, we have

$$\{ q_i,q_k \} = \left ( \frac{\partial q_i}{\partial q_j} + \epsilon \frac{\partial^2 g}{\partial q_j \partial p_i} \right ) \left ( \frac{\partial q_k}{\partial p_j} + \epsilon \frac{\partial^2 g}{\partial p_j \partial p_k} \right ) - \left ( \frac{\partial q_i}{\partial p_j} + \epsilon \frac{\partial^2 g}{\partial p_j \partial p_i} \right ) \left ( \frac{\partial q_k}{\partial q_j} + \epsilon \frac{\partial^2 g}{\partial q_j \partial p_k} \right ).$$

Any ##\epsilon^2## terms will be disregarded, so only the following remains:

$$\frac{\partial q_i}{\partial q_j} \frac{\partial q_k}{\partial p_j} - \frac{\partial q_i}{\partial p_j} \frac{\partial q_k}{\partial q_j} + \left( \epsilon \frac{\partial^2 g}{\partial q_j \partial p_i} \frac{\partial q_k}{\partial p_j} + \epsilon \frac{\partial^2 g}{\partial p_j \partial p_k} \frac{\partial q_i}{\partial q_j} \right ) - \left ( \epsilon \frac{\partial^2 g}{\partial p_j \partial p_i} \frac{\partial q_k}{\partial q_j} + \epsilon \frac{\partial^2 g}{\partial q_j \partial p_k} \frac{\partial q_i}{\partial p_j} \right ).$$

However, it does not look like these cancel to me. What have I done wrong?