# Confusion regarding properties of Poisson Brackets

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.

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 $$i,j$$
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 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 $\frac{\partial q_j}{\partial p_i}=0$ Mar 20, 2020 at 9:10