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

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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.

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  • $\begingroup$ 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). $\endgroup$ – Demosthene Mar 8 '15 at 11:16
  • $\begingroup$ ...and via an energy-centred classical mechanics in the Lagrangian/Hamiltonian formalism :) $\endgroup$ – danimal Mar 8 '15 at 12:34
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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.

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

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