# Force through quantum mechanics

In classical physics force is: $$F=\frac {dp}{dt}$$ How about quantum mechanics? In Old Quantum Mechanics momentum is: $p=\hbar \cdot k$ so force will be: $$F=\hbar \frac {dk}{dt}$$ what does $\frac {dk}{dt}$ mean?

It's still true in quantum mechanics – it's a "Heisenberg equation of motion" – but both sides are operators. However, because we often want to find the spectrum of the Hamiltonian in quantum mechanics – a set of possibilities – and the properties of the energy eigenstates, the importance of the force is diminishing. The force may be written as $i[H,p]/\hbar$, so it's the commutator of the momentum with the Hamiltonian, but we often need to study the whole Hamiltonian and not just its commutators with other operators. That's why the potential energy $V$ appears much more often in quantum mechanical calculations than the force – even though the latter may be written as $-V'(x)$ if a potential is known. Note that $-V'(x)$ is the result of the calculation of the commutator $i[H,p]/\hbar$ for Hamiltonians $H=p^2/2m+V(x)$.

Just to add to Luboš' excellent answer: if you haven't seen this sort of thing before it can be helpful to think in terms of expectation values – that is, the average of the possible outcomes of a measurement – rather than the whole spectrum of possibilities. In these terms Newton's second law (in one dimension) becomes $$\frac{\mathrm{d}}{\mathrm{d}t}\langle p\rangle = \big\langle-\frac{\mathrm{d}V}{\mathrm{d}x}\big\rangle$$ where the right-hand side is more or less $\langle F\rangle$.

$$\frac{\mathrm{d} \mathbf{p}}{\mathrm{d}t}=\mathbf{F}$$
The split $p=\hbar k$ where $k$ is the wave vector (or wavevector) is also valid in relativistic quantum theory. Wave vectors $k$ are also used in classical mechanics but the definition is different and does not use $\hbar$ of course [*].
Since $\hbar$ is an universal constant the meaning of $\frac {\mathrm{d}k}{\mathrm{d}t}$ can be obtained directly from $\frac {\mathrm{d}p}{\mathrm{d}t}$. The former is the variation of momentum in units or 'packets' of $\hbar$
$$\frac {\mathrm{d}k}{\mathrm{d}t} = \frac {1}{\hbar} \frac {\mathrm{d}p}{\mathrm{d}t}$$