Just a short query, given an electron at rest at the origin in the presence of a magnetic field whose magnitude is constant but whose direction is rotating around a cone at constant angular velocity $\omega$: $$\vec{B}(t) = B_0[\sin \alpha \cos (\omega t)\hat{x} + \sin \alpha \sin (\omega t) \hat{y} + \cos \alpha \hat{z}],$$ where the corresponding Hamiltonian is $$H(t) = \frac{e}{m}\vec{B} \cdot \vec{S} = \frac{\hbar \omega_1}{2} \dbinom{\cos \alpha~~~e^{-i \omega t}\sin \alpha}{e^{i \omega t}\sin \alpha~~~-\cos \alpha}.$$ It is considered in a text I am using that the characteristic time for changes in the Hamiltonian is $\frac{1}{\omega}$, why is this the chosen characteristic time for change of the Hamiltonian? Is this standard?

  • $\begingroup$ The Hamiltonian oscillates with period $2 \pi / \omega$. So there's a timescale of $1/\omega$ here, i.e. a quantity with dimensions of time that relates to how the Hamiltonian is changing. $\endgroup$ – knzhou May 4 '17 at 17:44
  • $\begingroup$ @knzhou Yes I understand that. But the characteristic time is not just any measurement of time, it is the time taken for something to change appreciably. For example the characteristic time for wave functions to change appreciably is given as $t \approx \frac{\hbar}{\Delta E}$. I'm asking why is this particular time considered to be a time where the Hamiltonian changes appreciably? $\endgroup$ – user101311 May 6 '17 at 13:40
  • $\begingroup$ Would you agree that the Hamiltonian 'changed appreciably' after time $\pi/\omega$, since this is the time it takes for the transverse field to flip sign? $\endgroup$ – knzhou May 6 '17 at 19:37
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    $\begingroup$ Characteristic times are always defined just 'up to an order of magnitude'. After all, you can't make a strong argument for why its' $\pi/\omega$ rather than, say, $2\pi/\omega$ or $\pi/2\omega$, so might as well throw out the prefactors and call it $1/\omega$. $\endgroup$ – knzhou May 7 '17 at 18:19
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    $\begingroup$ It's the same way with radioactive decay, where the characteristic time is the time to decay to $1/e$. Why not $1/2$ or $1/4$? It doesn't really matter. $\endgroup$ – knzhou May 7 '17 at 18:20

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