Timeline for Why is wave-number $k$ taken real & not complex when dealing with infinite line of atoms in a lattice?
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| Oct 3, 2015 at 15:59 | comment | added | user36790 | Yours is correct; Feynman takes $k$ as $ik'$; if I replace it with $k$ in $e^{ikx_n}$, what should I get? $$e^{ikx_n}= e^{i(ik')x_n}= e^{i^2 k'x_n}= e^{-k'x_n} .$$ So, it's not giving $e^{ik'x_n}.$ There is a '$-$' sign which Feynman omitted. | |
| Oct 3, 2015 at 13:08 | comment | added | garyp | No, Feynman is ok. The wave function blows up either to the left or to the right depending on the sign of $k'$. Another way to look at it is that I could just as well have written $k = k_r - ik_i$ | |
| Oct 3, 2015 at 13:06 | comment | added | garyp | @Ari That's correct. But the $k$ vectors in a solid have a size that's roughly the order of magnitude $1/a$ where $a$ is the lattice constant, the distance between unit cells in the solid. That means that the wave function would become exponentially large after only several lattice constants. For a macroscopic solid this is not an acceptable solution. But it does become important for small structures such as thin atomic layers. In those cases you can't declare that the wave function is strictly real. | |
| Oct 3, 2015 at 9:11 | vote | accept | CommunityBot | moved from User.Id=36790 by developer User.Id=2911 | |
| Oct 3, 2015 at 9:10 | comment | added | user36790 | Didn't Feynman wrongly write $e^{k'x_n}$? Wouldn't it be $e^{-k'x_n}$? | |
| Oct 3, 2015 at 4:36 | comment | added | Ari | but does that mean that for a finite crystal we may take "k" to be complex? | |
| Oct 3, 2015 at 3:36 | history | answered | garyp | CC BY-SA 3.0 |