# What is the origin of the quantum operators for $p$ and $E$ in standard QM?

It is always stated the quantum operators for $${p}$$ and $$E$$ are the ones we´re familiar with (the operator for energy, $$E=i\hbar\frac{\partial}{\partial t}$$ and the momentum operator, $$p=-i\hbar\nabla$$). But where do these operators come from?

I understand that substitution of these operators in the standard momentum-energy relation amounts to the appearance of the Schrödinger equation (or the Klein-Gordon equation in the relativistic case, where, by the way, the $$E$$ and $$p$$ operators stay the same).

But in the time of the development of quantum mechanics nobody knew the (by now known) form of a wave equation, so why replace the $$p$$ and $$E$$ by their now well-known operator-form?

Did they find the right operators, corresponding to $$p$$ and $$E$$ by educated guessing, by trial and error, or some other way, from which the Schrödinger equation emerged when applied to the standard energy-momentum relation?

Did they try all kinds of combinations of operators, $$i$$, and the Planck-constant until the right operators were found, from which the equation (the Schrödinger equation) followed which fitted the data? Or what?

• Comment to the post (v6): Note that $i\hbar\frac{\partial}{\partial t}$ is not the Hamiltonian operator, cf. this Phys.SE post. Jun 13 '16 at 8:32

Before quantum mechanics, we were familiar with particles of momentum $\mathbf{p}$ and plane waves $\propto \exp i\left(\mathbf{k}\cdot\mathbf{x}-\omega_\mathbf{k} t\right)$ with $\mathbf{k}$ the wavevector and $\omega_\mathbf{k}$ a $\mathbf{k}$-dependent angular, not linear, frequency. (The formula relating $\omega$ to $\mathbf{k}$ is called a dispersion relation.) Although electromagnetic radiation was regarded as a wave, it was known to have momentum. After Planck hypothesised electromagnetic radiation has quanta of energy $E=hf=\hbar\omega$ for some constants $h>0,\,\hbar=\tfrac{h}{2\pi}$, de Broglie rearranged to give $\lambda = \tfrac{c}{f} = \frac{h}{p}$, since speed-$c$ matter in special relativity satisfies $p:=\left|\mathbf{p}\right|=\tfrac{E}{c}$. In fact, it was supposed this condition held even for massive matter; indeed, early particle wave duality experiments supported this. But it was realised the wavenumber $k=\tfrac{2\pi}{\lambda}=\tfrac{p}{\hbar}$ is perhaps more helpful, since we can immediately generalise to a vector hypothesis $\mathbf{p}=\hbar\mathbf{k}$. Partnering this with $E=\hbar\omega$, we see that quantum mechanics can achieve the correct eigenvalues of momentum's components and energy from a plane wave using the operators $\hat{\mathbf{p}}=-i\hbar\boldsymbol{\nabla},\,E=i\hbar\partial_t$. Moreover, this approach also relates energy-momentum relations to dispersion relations. For example, $E=\tfrac{p^2}{2m}$ becomes $\omega=\tfrac{\hbar k^2}{2m}$, while $E^2=c^2p^2+m_0^2c^4$ becomes $\omega^2=c^2\left(k^2+m^2\right)$ with $m:=\tfrac{m_0 c}{\hbar}$.