How to derive the Momentum and Energy Operators from first principles?

[Svend]

TL;DR

Wondering about the origins of the Momentum and Energy Operators, i want to know how they are derived, assuming we have no knowlegde of the Schrødinger Equation.

So we all know that the form of the momentum operator is: iħd/dx. And for energy it is iħd/dt. But how do we derive these operators?

The only derivations of the i have seen is where the schrødinger equation was used, but that makes the logic circular, because the Schrødinger-Equation is derived from the momentum operator applied to the classical hamiltonian?

So i am interested in knowing other ways to derive the momentum and energy operators, from first principles. Can anyone help?I have read about cannonical quantization, where you postulate that [p,x]=iħ, and this should appereantly give the form of p=ih*d/dx, if we insist that x is diagonal. Has anyone heard of this?
Any contribution is deeply appreciated!

 

[anuttarasammyak]

As you say the commutation relation
xp-px=i\hbar
is often used to derive
p=-i \hbar \frac{d}{dx}
with x just a number as it is in classical physics.
You can confirm it by direct calculation of
(xp-px)f=i\hbar f
for any function f(x). For f(p) you get instead
x=i \hbar \frac{d}{dp}
with p just a number as it is in classical physics.

 

[Gere]

I assume it can be derived from something like Noether's law from the invariance of physical laws in space and time. But I don't know the exact derivation in QM. I think I saw an argument about space invariance and momentum operator in QM somewhere.
For energy it's roughly like: you need a continuous change of the wavefunction and it has to stay normalized. This already says d/dt psi=A*psi. Why not just call A the energy? The imaginary unit is just convenient to make A Hermitian and the eigenvalues real (otherwise A would be skew Hermitian and eigenvalues imaginary). So it's easy to show than it has to be the form of Schrödinger equation, but you still need to argue why the usual trick of taking classical energy and making it an operator works.
Maybe someone else spell out these ideas more precisely.

 

[PeterDonis]

So we all know that the form of the momentum operator is: iħd/dx.

Yes.

And for energy it is iħd/dt.

No. The energy operator is the Hamiltonian. $i \hbar \partial / \partial t$ is not an operator in QM (it does not act on the Hilbert space).