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I have just started learning about Poisson Brackets, and came across the following property

$$\{q_i,q_j\}=0$$

And

$$\{p_i,p_j\}=0.$$

Where $p$ and $q$ are respectively the momentum and position coordinates i.e. phase space coordinates.

Now Poisson Brackets are defined as

$$\{F,G\}=\frac{\partial F}{\partial q_i}\frac{\partial G}{\partial p_i}-\frac{\partial G}{\partial q_i}\frac{\partial F}{\partial p_i}$$ $i$ and $j$ here stand for the $i$'th and $j$'th spatial coordinates.

$$\{q_i,q_j\}=0$$ $$\Rightarrow \{q_i,q_j\}=\frac{\partial q_i}{\partial q_i}\frac{\partial q_j}{\partial p_i}-\frac{\partial q_j}{\partial q_i}\frac{\partial q_i}{\partial p_i} =0$$ But I am having a hard time proving it. I know that the second term $(\frac{\partial q_j}{\partial q_i})$ is zero because the i'th and j'th spatial coordinates are orthogonal and hence, there is no change in $q_i$ on changing $q_j$. However I don't know how to prove the first term to be zero, and that is where I need help.

To summarise, my question is prove that $$\frac{\partial q_i}{\partial q_i}\frac{\partial q_j}{\partial p_i}=0$$ Any help will be deeply appreciated.

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As $p$ and $q$ do not depend functionally on one another $$ \frac{\partial q_i}{\partial p_j} = 0$$ and also $$ \frac{\partial p_i}{\partial q_j} = 0$$ for all $i,j$

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I think there is an misunderstanding on your side. $$\frac{\partial q_j}{\partial q_i} = \delta_{ij},$$ where $\delta_{ij}$ is the so-called Kronecker delta.

I hope this helps so far.

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  • $\begingroup$ I understand that $\frac{\partial q_i}{\partial q_j}=\delta_{ij}$ and also that $\delta_{ij}=0$ for $i\neq j$ but my question doesn't involve the $\frac{\partial q_i}{\partial q_j}$ term, I want the proof for the other term in the Poisson Bracket being zero, more specifically I need the proof for the following statement $\frac{\partial q_j}{\partial p_i}=0$ $\endgroup$ – SK Dash Mar 20 at 9:10

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