[QE-developers] eigenstate calculated by QE

[曾梓萌]

dear develpoersFor 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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