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Dec 17, 2014 at 7:07 history bounty ended Xin Wang
Dec 17, 2014 at 7:07 vote accept Xin Wang
Dec 17, 2014 at 4:53 comment added Ruslan for given $k$, $E$ as a function of $n$ is discrete — the bands are still discrete entities. This is because of the finite length of lattice cell. You can see this if you apply Bloch's theorem and solve the problem for the case of $k=0$. It'd be the same as original problem, but constrained to one lattice cell.
Dec 17, 2014 at 4:51 comment added Ruslan See also Kronig-Penney model for introduction in simple models in solid state physics. 4) Your initial domain can already have multiple cells (say, $N$). You just apply the Bloch theorem and reduce the problem to the domain of length $L/N$. After that you may consider the problem for any $N$ solved — just use the corresponding values of $k$. 5) The energy spectrum is always in the form $E_n(k)$ where $k$ is quasiwavevector, and $n$ is number of band. If $k$ varies continuously, then of course $E$ is also continuous. But
Dec 17, 2014 at 4:46 comment added Ruslan 1) Right 2) Because we only require physically preparable states to be in Hilbert space. Eigenstates of continuous spectrum are usually treated using rigged Hilbert space formalism. Without these states you'd find that there're no eigenstates at all in these problems, and even if there were, you'd not be able to expand most of the wavefunctions in that basis, so it's not complete without continuous spectrum states. 3) Take $A \cos(x)$ for example. With it Schrödinger equation is solvable in terms of Mathieu functions.
Dec 16, 2014 at 20:08 comment added Xin Wang 4.) Assume that you solved the finite-interval problem with periodic boundary conditions( so $H \psi = E\psi$ on some finie domain $[0,L]$ with periodic Boundary conditions). How do you construct from this the solution for mutiple cells? 5.) So if we are on the full real lines, do we get continuity only in the $k-$values or is also the energy spectrum for each individual $k$ itself continuous? ( I guess in the finite-interval case are both discrete)
Dec 16, 2014 at 20:07 comment added Xin Wang I like the mathematical approach that you take here: There are just a few things that I want to ask you now in order to be sure that I got this: 1.) In principle we could show that Bloch's theorem even holds if we take as the domain the whole real line and just demand that our solutions are bounded everywhere? 2.) Why aren't we looking for proper $L^2$ eigenfunctions instead, if we are looking at the full real line, as bounded functions do not have to be an element of any proper Hilbert space? 3.) Do you know how these periodic potentials look like in practice?
Dec 16, 2014 at 19:59 history edited Ruslan CC BY-SA 3.0
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Dec 16, 2014 at 19:51 history answered Ruslan CC BY-SA 3.0