Timeline for The classical limit of quantum mechanics through Ehrenfest's theorem
Current License: CC BY-SA 4.0
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| Oct 15, 2023 at 13:16 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Oct 14, 2023 at 23:21 | comment | added | dennis | Actually, if we use Euclidean time evolution, we're in better shape: take an arbitrary initial state $\psi(x)$. Now use the Euclidean time evolution operator $e^{-\hbar^{-1}H\tau}$ where $H$ is the Hamiltonian: $H=-\frac{\hbar^2}{2}\partial_x^2+V(x)$. When $\hbar \to 0$, $V(x)$ is the dominant term, so the wavefunction just gets multiplied by $e^{-\hbar^{-1}V(x)\tau}$. Now $V(x)$ has a global minimum at some point $x_0$. At late times, $e^{-\hbar^{-1}V(x)\tau}$ is damping, and it localises the wavefunction around $x_0$. The spread is low and classical evolution ensues. | |
| Oct 14, 2023 at 22:29 | comment | added | Cosmas Zachos | Indeed, you may introduce states which will not have a reasonable classical limit, to the extent that there is no absolutely no action restriction on states. You may also describe classical mechanics in the KvN formulation. However the TDSE does have its scale set by ℏ, which controls the type of of state one is ultimately discussing. Classical states are the ones whose characteristic actions are much larger than ℏ, otherwise they are quantum. These limits are most visible in deformation quantization. Bog swill redux. | |
| Oct 14, 2023 at 21:39 | comment | added | dennis | In fact, I can easily construct an initial state with large spread in $x$, so that initially there is large deviation from classical evolution. The best you can hope for is that at late times the spread becomes small, and classical evolution is followed, but there is no good reason for that to actually happen. (And notice, none of what I said had to do with the size of $\hbar$! Actually, $\hbar$ enters in the time-evolution operator but it's size is not going to dictate whether the state becomes more or less spread.) | |
| Oct 14, 2023 at 21:29 | comment | added | dennis | I could, however, arbitrarily introduce a length-scale $\ell$. In that case $\langle x^n \rangle\propto \ell^n$ (if I take the state $e^{-x^2/\ell^2}$ for example). | |
| Oct 14, 2023 at 21:18 | history | answered | Cosmas Zachos | CC BY-SA 4.0 |