Timeline for Time reversal of scalar Bloch states
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 6, 2021 at 20:25 | comment | added | Valter Moretti | I think so. An issue would be if that phase depends on the eigenvector or not... | |
| Feb 6, 2021 at 20:13 | vote | accept | Jamin | ||
| Feb 6, 2021 at 20:10 | comment | added | Jamin | Right, I realize my mistake now. Fundamentally, each Bloch state is labelled by the quantum number $\mathbf{k}$, but it is only defined modulo a shift by reciprocal lattice vectors. I guess I should go back to Ashcroft & Mermin to review, but hopefully my understanding is correct now. In that case, shouldn't we really be saying that $K\psi_{n\mathbf{k}} = e^{i\phi}\psi_{n,-\mathbf{k}}$ for some unknown phase factor? | |
| Feb 6, 2021 at 19:08 | comment | added | Valter Moretti | It is just the Bloch theorem I think. The form of wavefunction is $e^{ikx}u(X)$, where is translationally invariant (with respect to the lattice). $k$ is such that the exponential changes with a constant phase when $x$ varies in the lattices. This condition defines the reciprocal lattice. | |
| Feb 6, 2021 at 18:58 | comment | added | Jamin | Thanks for your answer. I'm still a little confused trying to understand this mathematically. After eqn 1.17 in this reference, the author mentions that for a reciprocal lattice vector G, two Bloch states |n, k> and |n, k+G> can differ only by a phase, as you also pointed out. How could we prove that? We would need to show that |n, k> and |n, k+G> satisfy the same Schrodinger equation, but I don't understand how we can demonstrate that. I'm guessing there is a simple linear algebra result that I'm not realizing applies here. | |
| Feb 6, 2021 at 12:08 | history | edited | Valter Moretti | CC BY-SA 4.0 |
added 166 characters in body
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| Feb 6, 2021 at 12:02 | history | edited | Valter Moretti | CC BY-SA 4.0 |
added 166 characters in body
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| Feb 6, 2021 at 8:15 | history | answered | Valter Moretti | CC BY-SA 4.0 |