# What are origins of the definitions for the momentum and position operators in quantum mechanics?

For that matter what is origin for the energy operator? I now they are defined as: $\hat{x} = x$ $\hat{p} = i\hbar\frac{d}{dx}$ $\hat{E} = -i\frac{d}{dt}$

I understand they might not have strict derivations as such but I am just looking for some sort of justification for them. Also I wanted to understand the origin of the commutation relations between them?

Thank you

• Commutation relation can be traced back to classical mechanics (Hamilton mechanics) in which there is something called Poisson bracket. – K_inverse Sep 12 '18 at 12:01
• I don't believe that $\hat{E}$ is defined to be $-i\frac{d}{dt}$ – Alfred Centauri Sep 12 '18 at 12:16
• – Alfred Centauri Sep 12 '18 at 12:23
• – Frobenius Sep 12 '18 at 15:03

1. For a particle with a well-defined wavelength $\lambda$, the spatial dependence goes as $\psi \sim e^{i kx}$ for $k=2\pi/\lambda$, and therefore $$-i\frac{\mathrm d}{\mathrm dx} \psi = k \,\psi.$$
2. The de Broglie relationship between momentum and wavelength reads $$p = \frac{h}{\lambda} = \frac{h}{2\pi/k} = \hbar k,$$ and if you substitute that into the above, you have $$-i\hbar\frac{\mathrm d}{\mathrm dx} \psi = \hbar k \,\psi = p \,\psi.$$
3. Finally, this needs to be extended from wavefunctions that have well-defined wavelengths to wavefunctions that don't, and this is done by postulating that this extension be done in a linear way, i.e. that if $\psi = e^{ik_1x}+ e^{ik_2x}$ is a superposition of two matter waves with different momenta, then $\hat p\psi = p_1 e^{ik_1x}+ p_2e^{ik_2x}$, i.e. the momentum acts separately on the two components. Depending on how much of the theory you've already laid down, this is probably a brand-new postulate ("matter waves are linear") and it doesn't "come from" anywhere - you just postulate and see if it works to explain experiments. (Which, of course, it does.)
As for the energy operator, it's the same thing, but now you swap in the Planck relation $E=h\nu$ where the de Broglie relation used to sit.