This question might be a replication but I could not find an answers till now.
Why are $(x,p)$ independent in the Hamiltonian formulation?
I'm interested in the independence of $(x,p)$ in the Hamiltonian, specifically under symmetry transformations.
A symmetry transformation should leaf the equations of motion, the Euler-Lagrange equations ($ELG$), form invariant. Lets say if $\phi$ is a symmetry transformation then $x'=\phi(x,\dot{x},t)$ we demand that $ELG(x) = ELG(x'(x)$). This leads us to the invariance condition state in Noether theorem that the Lagrangian should transform as
$$ \frac{dt'}{dt} L(x' , \frac{d x'}{d t'}, t') |_{x'=\phi(x)} = L(x,\dot{x},t) + \frac{d}{dt} W $$
Now let us look at the Hamiltonian. I understand that the first Hamilton equation giving the connection between $p$ and $\dot{x}$ is a result of the Legendre transformation. Therefore the variation to derive the Hamilton equations is "trivial" for $\delta p$.
If i demand that $(x,p)$ are independent i can transform them independently. Lets consider now a free particle with $H(x,p,t)=p^2/2m$. Under a coordinate transformation of $x'=\phi(x,t)$ this Hamiltonian is invariant, $H(x',p,t)=H(\phi(x),p,t) = H(x,p,t) $.
So i would get that $\dot{x} = \frac{d}{dt} \phi(x) =p/m$, which look wrong. Therefore i would conclude that $p$ can not be transformed independently of $x$, due to the condition given by the Legendre transformation.
Another point of confusion is that the action we consider for the Lagrangian gets curves in real space. These curves are functions $x(t): \mathbb{R} \rightarrow \mathbb{R}^d $. The Legrendre transformation between Lagrangian and Hamiltonian should be involutive. But if $(x,p)$ are independet then the curves are $(x,p)(t): \mathbb{R} \rightarrow \mathbb{R}^d \times \mathbb{R}^d$. So the transformation should not be involutive anymore, since we need to map this again to $\mathbb{R}^d$ functions.
