Timeline for Inaccuracies of an energy level's degeneracy
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 19, 2020 at 23:32 | comment | added | Phy | Judging by the fact that 1 cubic volume unit is counted as 1 state, while in fact a cubic volume has 8 lattice points (1 on each corner), would this mean that for large volumes, one volume unit would have on average 1 lattice point? Otherwise the volume won't be approximately equal to the number of lattice points, i.e. the number of discrete quantums states. This seems to be the case for a circle area as stated here | |
| Apr 19, 2020 at 12:47 | comment | added | Roger V. | Indeed, you cannot make such a direct comparison. However the number of states in a large volume should be approximately equal in the continuous and the discrete approaches. | |
| Apr 19, 2020 at 12:17 | comment | added | Phy | Is this correct? | |
| Apr 17, 2020 at 18:39 | comment | added | Phy | Ok, this has helped me quite a bit. So does this mean that I can not compare the outcome of $4\pi n^2 dn/8$ to the amount of integer solutions for a particular $n^2=n_x^2+n_y^2+n_z^2$? I.e. they will never be the same purely because the approach is discrete vs continuous? | |
| Apr 17, 2020 at 17:50 | comment | added | Roger V. | There is exactly one state for every combination $n_x, n_y, n_z$, that is one state occupies volume 1, and their density is uniform, i.e. volume in $n$-space is literally the number of states in it. Then the number states in any shell of radius $n$ and thickness $dn$ is $4\pi n^2dn/8$. This can be converted to the energy units using that $n^2=n_x^2+n_y^2+n_z^2$. | |
| Apr 17, 2020 at 16:02 | comment | added | Phy | I'm trying to comprehend how the number of lattice points within the shell would be approximately the same as its volume but I don't know how. | |
| Apr 17, 2020 at 12:59 | comment | added | Roger V. | I think that talking about an infinitely thin shell doesn't make much sense here - this is a kind of coarse-graining procedure, where you have always deal with the volumes containing lots of points. Otherwise, one shouldn't be doing such an approximation. | |
| Apr 17, 2020 at 12:56 | comment | added | Phy | Thanks for your explanation. There is a big issue I have when talking about a shell in n-dimensions. I would deduce that the true number of quantum states within an infinitely thin n-shell would be achieved by counting the number of n-grid lattice points that is within/on the n-shell. But when one calculates it using n-shell volume, thus by multiplying the surface of the shell by an infinitely small thickness $0.00000...00001$, I would deduce that this thickness would make the outcome way less than the number of lattice points. What am I reasoning wrong here? | |
| Apr 17, 2020 at 5:28 | history | answered | Roger V. | CC BY-SA 4.0 |