Timeline for Central Equation and the Formal Proof of Bloch's Theorem by Kittel
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 15, 2023 at 16:20 | comment | added | 蕭力諶 | Also great thanks to @JacobStuligross | |
| Jun 15, 2023 at 16:19 | comment | added | 蕭力諶 | @naturallyInconsistent Thank you very much. I finally get it. | |
| Jun 15, 2023 at 8:20 | comment | added | naturallyInconsistent | @蕭力諶 Yes, and that is always the case, not just in the Bloch–translation case. We are almost always interested in some maximally commuting set of operators that include the Hamiltonian, and if you swap some of them for others, the old complete basis will span the same space as the new basis but will appear as superpositions that will violate the eigenvalue equation for the operators that are swapped out. | |
| Jun 15, 2023 at 7:41 | comment | added | 蕭力諶 | @naturallyInconsistent I see. Is it proper to say that a solution is not necessarily a Bloch state and an eigenstate of Hamiltonian is not necessarily an eigenstate of translation operators, and Bloch's theorem is merely stating the existence of a complete basis taking the form of Bloch states? | |
| Jun 15, 2023 at 1:37 | comment | added | naturallyInconsistent | @蕭力諶 No, it will not be, and that is the point. If you want to be a Bloch state, in particular, if you want to be a simultaneous eigenfunction of Hamiltonian and Translation operators, then you need to only have one single $k$ in the 1BZ and no more. The concept that is appearing here is the same as that in single atoms: The different $m_j$ in H atom for the same $n,\ell,s,j$ have the same energy, but if you superpose them, they will no longer be $J_z$ eigenstates. Nobody is saying that every solution of TISE is in $J_z$ eigenstates, but rather that they span the whole Hilbert space. | |
| Jun 14, 2023 at 19:24 | vote | accept | 蕭力諶 | ||
| Jun 14, 2023 at 19:23 | comment | added | 蕭力諶 | And while the solution is a superposition of different of Bloch state basis functions, it is still a Bloch state, right? | |
| Jun 14, 2023 at 18:47 | comment | added | Jacob Stuligross | @蕭力諶 yes, that's correct | |
| Jun 14, 2023 at 18:04 | comment | added | 蕭力諶 | I see. Does it mean that solutions to in which there are coefficients $C(k'\neq k_0-G)\neq0$ can actually exist, and just imply superposition of different degenerate states? | |
| Jun 14, 2023 at 14:25 | history | edited | Jacob Stuligross | CC BY-SA 4.0 |
corrected a statement about which wavevectors can be represented
|
| Jun 14, 2023 at 14:22 | comment | added | Jacob Stuligross | On the other hand, while Kittel is justified in looking for wavefunctions of this form, the statement I quoted is not true. The earlier proof was in the absence of degeneracy. In this proof, the statement that "if $k$ is in $\psi$ then all other wavevectors in the Fourier expansion will be of the form $k+G$" is not correct. You can have a wavevector with a well-defined energy which has $k$ and $k'\neq k+G$ in its Fourier expansion (like the $\psi$ I constructed at the end). | |
| Jun 14, 2023 at 14:21 | comment | added | Jacob Stuligross | That's totally correct that not all $k$ values have the right energy. I'll edit the answer today | |
| Jun 14, 2023 at 8:31 | comment | added | naturallyInconsistent | Of course, I think you already know these. Just felt that a good answer deserves to be made perfect. | |
| Jun 14, 2023 at 8:26 | comment | added | naturallyInconsistent | "There's nothing actually stopping us from having all values of $k$ represented." Actually, just because you need a constant eigen-energy $\epsilon$, it is clear that there will only be a handful of $k$ points inside the 1BZ that would have the exact same eigen-energy. You can thus only pick from those points. | |
| Jun 14, 2023 at 8:24 | comment | added | naturallyInconsistent | "The answer is that, well, Charles Kittel is lying here." No, he is not. IIRC, earlier in the book he derives Bloch theorem using translation operator, and thereby already knows that the wavefunction should look like $\psi_k=e^{ikr}u(r)$ where $u(r)$ has lattice symmetry. So he is totally justified to only seek the wavefunctions that have this form. | |
| Jun 14, 2023 at 6:40 | history | answered | Jacob Stuligross | CC BY-SA 4.0 |