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?


1 Answer 1


It sounds like your first step is correct. But I would have to see the steps to be sure. The chain rule is definitely the tool of choice, I don't know any other relations you can utilise. This is pretty calculation heavy so I hope these tips will suffice.

To continue with $ \{Q_i, Q_j\} = 0$, write out the following:

$$ 0 = \frac{\partial p_i}{\partial q_k} = \dots \quad\text{and}\quad \delta_{ik} = \frac{\partial p_i}{\partial p_k} = \dots$$

by using a again the chain rule. Then combine those two equation, by treating them as matrices. There are some subtleties regarding the unique solution part, but otherwise the rest is pretty straight forward.

  • $\begingroup$ Thank you very much. I think I used the chain rule correctly and will now try to use your suggestions to go on. I just wasn't sure if the multidimensional chain rule is applicable here at all due to some dependecies but if you say that's the tool of choice here I'm pretty confident. I will post my results later. $\endgroup$ Jul 21, 2015 at 20:28

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