Timeline for How to prove Bloch function is periodic in reciprocal lattice?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 8, 2020 at 9:55 | comment | added | Floyd4K | I didn't and wouldn't say must, but can, in the following sense: The two functions will "look" transform the same way under translations, so they won't be able to represent states differing in the translation properties. Of course this doesn't mean you cannot have more than one physical state at a given $k$. On the contrary, that is why we have a band structure. But all wavefunctions at this $k$ will transform similarly under translation by $a$. | |
| Nov 21, 2020 at 1:49 | comment | added | Who | Thank you for your response. Sorry, it is not clear yet to me. So the argument is that the operator T has only one set of eigenvalues, and since both states have the same eigenvalues then they must be the same state. But how we know this must be the case? what let us conclude this? how are we sure they arent different states with the same eigenvalues? | |
| Nov 19, 2020 at 15:53 | comment | added | Floyd4K | It is a property of the operator $T$ which only has $N$ eigenvalues and therefore eigenfunctions labelled by the $N$ values of $k$. Does that help? | |
| Nov 6, 2020 at 11:21 | comment | added | Who | I have the same doubt. Thank you much for your help. I have a doubt. I hope you can answer. You wrote: "Ψk′ yields the same eigenvalues of T as Ψk and is therefore not distinguishable from Ψk" Is this a property of the eigenfunctions? or it is part of group theory? | |
| Aug 20, 2020 at 16:42 | history | edited | Floyd4K | CC BY-SA 4.0 |
probably less confusing to use T instead of T^n
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| Aug 20, 2020 at 16:00 | review | Late answers | |||
| Aug 20, 2020 at 17:09 | |||||
| Aug 20, 2020 at 15:41 | review | First posts | |||
| Aug 20, 2020 at 17:22 | |||||
| Aug 20, 2020 at 15:40 | history | answered | Floyd4K | CC BY-SA 4.0 |