Timeline for Correlation Amplitude in QM
Current License: CC BY-SA 3.0
4 events
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| Apr 11, 2017 at 20:27 | comment | added | Amara | @JohnDoe, $\omega:= \frac{E-E_0}{\hbar}$ This makes the units work out. Now $\omega$ is related to the period of your oscillations for $ \sin \omega t$. For times of the order $ \frac{1}{\omega} \approx T $ you are beginning to see the function oscillate and so when you integrate you begin to pick up negative and positive areas under the graph. So either don't evolve all the way to times of order the period (T) or make $\omega$ small or your period large | |
| Apr 11, 2017 at 14:34 | comment | added | user100411 | I think I understand, if we consider a simple plane periodic wave: $\sin(\omega t)$ then in analogy with this example, in the integrand $\omega := \frac{t}{\hbar}$ and $t := (E - E_0)$ (with this is the $t$ from $\sin (\omega t)$). Hence $\omega = 2 \pi f$ implies $\frac{t}{\hbar} = 2 \pi f$ implies that the period $T = \frac{2 \pi \hbar}{t}$. Hence if the period $T = \frac{2 \pi \hbar}{ t}$ is very large compared to $E - E_0$ then the wave has not traversed sufficiently to have enough oscillations to cancel sufficiently. Is this basically the idea? | |
| Apr 11, 2017 at 14:18 | comment | added | user100411 | Thanks I understand much of your answer, but could you elaborate on why if $|E - E_0| << \frac{\hbar}{t}$ that we would not get rapid oscillations? How does it imply long period mathematically? | |
| Mar 26, 2017 at 14:19 | history | answered | Amara | CC BY-SA 3.0 |