Timeline for How can the expected value $\left<x\right>$ of a given state in an harmonic oscillator differ from $0$?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 6, 2018 at 10:14 | vote | accept | Mr. Nobody | ||
| Mar 6, 2018 at 10:14 | |||||
| Mar 6, 2018 at 8:38 | comment | added | The Vee | @Mr.Nobody You won't find the sign of the wave function, but the relative phase of $|0\rangle$ and $|1\rangle$. This is because it contributes to the mod square. Consider in a simple case: $|e^{ix}+e^{iy}|^2=2+e^{i(x-y)}+e^{i(y-x)}=2(1+\cos(x-y))$. The global phase of the wavefunction is unphysical, but relative phases can be identified by measurements. In the above examples we always happened to keep the phase of $|0\rangle$ at zero, so the difference between the two concepts was not evident. But the measurement at an instant only revealed $\theta_1-\theta_0$. There's no contradiction. | |
| Mar 5, 2018 at 14:06 | comment | added | Mr. Nobody | Yes, thanks for the explanation, but anyway in a particular instant, the wave functions will each have a particular sign, so if you could measure just an instant you could actually know how the wave function is, besides the square of it. I know I'm probably wrong, I just don't know where. | |
| Mar 5, 2018 at 12:36 | comment | added | J.G. | In short, adding an $e^{i\theta}$ factor to $|1\rangle$ obtains $\langle x(t)\rangle = \sqrt{\frac{\hbar}{2m\omega}}\cos(\omega t -\theta)$, which again time-averages to $0$. However, we can average over $\theta$ instead to get the same result. | |
| Mar 5, 2018 at 12:27 | history | answered | The Vee | CC BY-SA 3.0 |