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

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?

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).

• I think the best way to phrase it is "we can go from QM to classical mechanics by the identification of commutators with Poisson brackets" and not the other way around, especially because of ambiguity of operator ordering. That's why quantum mechanics is sometimes presented from an axiomatic viewpoint.
– MSha
Oct 12, 2021 at 19:59
• @MSha It's a bit of both. Whatever QM becomes in the direction you describe has to be CM, but we worked out how QM had to look based on knowing that, so the historical theory development order is "the other way around".
– J.G.
Oct 12, 2021 at 20:56

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 $$\frac{\partial f}{\partial x}~=~\{f,p\}.\tag{2C}$$