How to derive the fact that $p\sim d/dx$ and $H\sim d/dt$ from classical mechanics?

Question

I am trying to understand Noether's conserved quantities to shifts in time and or position. I have seen the derivation of the operators for Schrodinger's equation but not for classical mechanics. Is it true or perhaps obvious that the derivative of the action $S=∫(KE-PE)dt$ with respect to time is Energy since it is the $d/dt$ of a $∫dt$ that contains energy? Also is it equally obvious that the derivative of the action with respect to position, is the momentum $dL/dx=$ momentum? Or does it require some extra steps and logic to derive the fact that $d/dx=$ momentum and $d/dt=$ Hamiltonian?

Answer 1

These observables-are-derivatives equations carelessly obscure what's really happening, viz.$$\langle x|\hat{p}_k|\psi\rangle=-i\hbar\frac{\partial}{\partial x_k}\langle x|\psi\rangle,\,\hat{H}|\psi\rangle=i\hbar\frac{\partial}{\partial t}|\psi\rangle$$or something like that. Since these statements are in quantum-mechanical Hilbert space language, they cannot be derived in classical mechanics. Instead, quantum mechanics is constructed to represent certain truths about classical mechanics in these equations. The Poisson brackets $\{x_j,\,p_k\}=\delta_{jk}$ give rise to $[x_j,\,p_k]=i\hbar\delta_{jk}\Bbb I$ ($\Bbb I$ the Hilbert space's identity operator) because momentum and the Hamiltonian are the generators of infinitesimal space and time translations (see here and here, respectively).

Answer 2

The correspondence with classical mechanics is best seen in the Heisenberg picture by acting on operators $\hat{f}(\hat{x},\hat{p},t)$ rather than on kets $|\psi\rangle$.

First of all, recall that the commutator $\frac{1}{i\hbar} [\cdot,\cdot]$ corresponds to the Poisson bracket $\{\cdot,\cdot\}$, cf. e.g. this Phys.SE post.

Now let us answer OP's title question:

1. Differentiation of the operator $\hat{f}$ wrt. time $t$ is given by Heisenberg's EOM $$ \frac{d\hat{f}}{dt}~=~ \frac{1}{i\hbar} [\hat{f},\hat{H}]+\frac{\partial \hat{f}}{\partial t}, \tag{1Q}$$ which corresponds to Hamilton's EOM $$ \frac{df}{dt}~=~\{f,H\}+\frac{\partial f}{\partial t}. \tag{1C}$$

2. Differentiation of $\hat{f}$ wrt. $\hat{x}$ is given by $$ \frac{1}{i\hbar} [\hat{f},\hat{p}], \tag{2Q}$$ which corresponds to $$ \frac{\partial f}{\partial x}~=~\{f,p\}.\tag{2C} $$