Why are positions and momenta independent variables in the Hamiltonian formulation?

Question

Lifshitz and Landau's Vol. $1$ explicitly states that

$$ \cfrac{\partial{q_k}}{\partial{p_i}} = 0$$

And seems to imply also that $$ \cfrac{\partial{p_k}}{\partial{q_i}} = 0.$$

I guess that whenever $k \neq i$ the two quantities are independent of each other, but for $k=i$ it seems that they do depend on each other since $p_i \text{ is defined in terms of } \dot{q_i}$.

Answer 1

There are various ways to see this:

1. Consider the Hamiltonian formulation. Let there be given an arbitrary but fixed instant of time $t_0\in [t_i,t_f]$. The (instantaneous) Hamiltonian $H(q(t_0),p(t_0),t_0)$ is a function of both the instantaneous position $q^i(t_0)$ and the instantaneous momentum $p_k(t_0)$ at the instant $t_0$. Here $q^i(t_0)$ and $p_k(t_0)$ are independent variables. The point is that since Hamilton's equations are $2n$ coupled first-order ODEs one is entitled to make $2n$ independent choices of initial conditions.

2. Alternatively, in the Lagrangian formulation with Lagrangian $L(q,v,t)$ the generalized positions $q^i$ and the generalized velocities $v^j$ are independent variables, cf. e.g. this Phys.SE post.

   A Legendre transformation of the Lagrangian wrt. the variables $v^j\leftrightarrow p_k$ yields the Hamiltonian $H(q,p,t)$, where positions $q^i$ and momenta $p_k$ are independent variables.