Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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 $H=p^2/2m+V(x)$.

share|cite|improve this answer

The quantum analogue is the Heisenberg equation where the bold font denote matrices (Heisenberg and co-workers developed the matrix formulation of quantum mechanics)

$$\frac{\mathrm{d} \mathbf{p}}{\mathrm{d}t}=\mathbf{F}$$

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 $\hbar$

$$ \frac {\mathrm{d}k}{\mathrm{d}t} = \frac {1}{\hbar} \frac {\mathrm{d}p}{\mathrm{d}t}$$

[*] The wave vector in classical mechanics is introduced using Fourier transforms.

share|cite|improve this answer

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 $\langle F\rangle$.

If you're interested in finding out more about this the keyword to Google is the Ehrenfest theorem.

share|cite|improve this answer
Thanks for the useful addition, +1. – Luboš Motl Dec 11 '12 at 11:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.