# Fundamental Poisson Bracket under Canonical Transformation

We have $H(q,Q,t)$. There is unique solution $Q$ for $p_i = \frac{\partial H}{\partial q_i}$ and we have $P_i = - \frac{\partial H}{\partial Q_i}$.

We want to prove from only this that the fundamental Poisson Bracket equations are fulfilled.

These are: $$\{Q_i, P_j\} = \delta_{i,j}$$ $$\{Q_i, Q_j\} = 0$$ $$\{P_i, P_j\} = 0$$

where $$\{f, g\} = \sum_{k=1}^{n} \left(\frac{\partial f}{\partial q_k}\frac{\partial g}{\partial p_k} - \frac{\partial f}{\partial p_k}\frac{\partial g}{\partial q_k}\right)$$

The solution:

I think I have proven $\{P_i, P_j\} = 0$ by using the definition for $P$, then using the unique solution and then applying the multi-dimensional chain rule (here the sum is eliminated). As in all of this, as the system is continuously differentiable, we exchange partial derivatives, I get $$...= \frac{\partial H}{\partial Q_i \partial Q_j} - \frac{\partial H}{{\partial Q_j \partial Q_i}} = 0$$

I'm having trouble with $$\{Q_i, Q_j\} = \sum_{k=1}^{n} \left(\frac{\partial Q_i}{\partial q_k}\frac{\partial Q_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial Q_j}{\partial q_k} \right)= ?$$

There doesn't seem to be any relation I can use.

My question: What underlying principal or relationship exists here that I can use to proof this and the Dirac delta identity?

2nd question: May I use the chain rule in the first proof and is this way correct or how is it usually done?

To continue with $\{Q_i, Q_j\} = 0$, write out the following:
$$0 = \frac{\partial p_i}{\partial q_k} = \dots \quad\text{and}\quad \delta_{ik} = \frac{\partial p_i}{\partial p_k} = \dots$$