[QE-developers] eigenstate calculated by QE

[曾梓萌]

I'm very sorry that I still don't understand this question. This is my understanding: when I try to calculate the eigenstates of the system, QE will randomly select a set of orthonormal linear combinations inside the degenerate subspace, so the QE selection method is also different during repeated calculations(Because even if the input parameters are exactly the same, the choice of the orthonormal linear combinations  is still random).
But in fact the eigenstates I calculated are the same in repeated calculations, so maybe QE has some way to artificially select a special orthonormal linear combinations inside the degenerate subspace?
Is my understanding correct? If so, what is this special way of selection?

ZengZimeng

在2021-12-10 16:01:13，Paolo Giannozzip.giannozzi at gmail.com写道：

If you repeat the same calculation, you get the same eigenvectors. Slightly different calculations may actually give very similar eigenvectors, but there is no guarantee.


Paolo


On Thu, Dec 9, 2021 at 8:55 PM 曾梓萌 <zengzm20 at mails.tsinghua.edu.cn> wrote:

dear Paolo

Thank you very much for your reply. I have done some tests on this issue and got some incomprehensible results. For a system with Kramer degeneracy, perform multiple calculations and draw the spatial distribution of an eigenstate. In repeated calculations, the spatial distribution of the states is almost the same. If in multiple calculations, the eigenstates calculated by QE are random (but orthonormal) linear combinations inside the degenerate subspace, then it seems that a consistent spatial distribution should not be obtained？

Zeng Zimeng

在2021-12-09 18:14:45，Paolo Giannozzip.giannozzi at gmail.com写道：

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






--

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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