## Are newton's laws also an approximation?

So are newton's laws also an approximation to quantum phenomena. Can it be derived from quantum laws?
 Recognitions: Gold Member You may be interested in reading about the Correspondence Principle: http://en.wikipedia.org/wiki/Correspondence_principle
 Recognitions: Gold Member To point to a few specifics, there seems to be a few possible interpretations, though I imagine someone else here could tell you more about the current consensus: "Once the Schrödinger equation was given a probabilistic interpretation, Ehrenfest showed that Newton's laws hold on average: the quantum statistical expectation value of the position and momentum obey Newton's laws." "Because quantum mechanics only reproduces classical mechanics in a statistical interpretation, and because the statistical interpretation only gives the probabilities of different classical outcomes, Bohr has argued that classical physics does not emerge from quantum physics in the same way that classical mechanics emerges as an approximation of special relativity at small velocities. He argued that classical physics exists independently of quantum theory and cannot be derived from it. His position is that it is inappropriate to understand the experiences of observers using purely quantum mechanical notions such as wavefunctions because the different states of experience of an observer are defined classically, and do not have a quantum mechanical analog."

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## Are newton's laws also an approximation?

Ehrenfest's theorem is not saying that the average values obey Newton's laws, which is wrong! This is only the case for the motion in a harmonic-oscillator potential or in a constant force field. Ehrenfest's theorem says
$$\frac{\mathrm{d}}{\mathrm{d} t} \langle A \rangle =\frac{1}{\mathrm{i} \hbar}\langle [\hat{A},\hat{H}] \rangle,$$
where $A$ is a not explicitly time dependent observable. For momentum you find
$$\frac{\mathrm{d}}{\mathrm{d} t} \langle \vec{p} \rangle =-\langle \vec{\nabla} V(\hat{\vec{x}}) \rangle.$$
Except for a constant force or a force that is linear in $\vec{x}$ the expectation value on the right-hand side is not the same as $-\vec{\nabla} V(\langle x \rangle)$!
 Recognitions: Gold Member Ahh, I stand corrected. The wikipedia quote was a bit misleading, so thanks for the insight vanhees!
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