# Why we cannot describe operator for force $F$ in quantum mechanics?

In quantum mechanics we describe operators corresponding to momentum but we don't define operator for force what is the reason behind it?

Jimmy's answer is true, but the OP should be careful that the expectation value of this force over the entire wave-packet $\left\langle \vec{F}\right\rangle$ is what determines the acceleration of the center of the wave packet $m\frac{d^2\left\langle \vec{X}\right\rangle}{dt^2}$ according to Ehrenfest's theorem, and not the value of the force operator at the center of the wave wave packet $\left.\vec{F}\right|_{\vec{X} = \left\langle \vec{X}\right\rangle}$.

It is only in the classical limit that it's approximately true that $\left\langle\vec{F}\right\rangle\approx \left.\vec{F}\right|_{\vec{X} = \left\langle \vec{X}\right\rangle}$

This is in general why the Force operator even though can be defined, has very limited usefulness, if at all. Unless you're are interested in a quasi-classical situation.

• OP might also be interested in knowing why we shift from a force-centred (Newton's laws) discussion in classical mechanics to a potential/energy-centred one in quantum mechanics (Schrödinger equation). – Demosthene Mar 8 '15 at 11:16
• ...and via an energy-centred classical mechanics in the Lagrangian/Hamiltonian formalism :) – danimal Mar 8 '15 at 12:34

You can: i[H,p]/ℏ is force in QM. It is not used often because we normally need to find the entire Hamiltonian, not just its commutators with other interactions.

• It might be more useful if you included a reason why this is or is not used. – Kyle Kanos Mar 8 '15 at 14:58
• Okay, I will add it. – Jimmy360 Mar 8 '15 at 20:56