[QE-developers] eigenstate calculated by QE

[Paolo Giannozzi]

Hi

if I understand correctly your question: yes, ALL degenerate eigenstates
are random (but orthonormal) linear combinations inside the degenerate
subspace. And no, typically they do not resemble  the eigenstates one would
find in textbooks.

Paolo

On Thu, Dec 9, 2021 at 10:59 AM 曾梓萌 <zengzm20 at mails.tsinghua.edu.cn> wrote:

> dear develpoers
>
> For Kramer's degenerate system, there are two energy degenerate
> eigenstates on each k and E, and any linear combination of these two
> degenerate states is still the eigenstate of the system. Is the eigenstate
> calculated by QE this random linear combination eigenstate?
>
> I will appreciate any helps in this subject.
> Truly yours,
> Zeng Zimeng
> Tsinghua university
>
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-- 
Paolo Giannozzi, Dip. Scienze Matematiche Informatiche e Fisiche,
Univ. Udine, via delle Scienze 206, 33100 Udine, Italy
Phone +39-0432-558216, fax +39-0432-558222
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