 Re: Different Z2 Invariants with and without Band.dispersion 'on' ( No.1 ) |
- Date: 2021/06/04 08:32
- Name: T. Ozaki
- Hi,
Could you show us your input file?
Then, we may have a more chance to respond your question properly.
Regards,
TO
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| Re: Different Z2 Invariants with and without Band.dispersion 'on' ( No.2 ) |
- Date: 2021/06/04 17:11
- Name: Simba <pshubham2805@gmail.com>
- How should I share input files with you?
Should I copy-paste here? There is no attachment option here?
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| Re: Different Z2 Invariants with and without Band.dispersion 'on' ( No.3 ) |
- Date: 2021/06/04 17:56
- Name: Simba <pshubham2805@gmail.com>
- Okay, I found the source of the error. There was a slight change in my atomic positions.
I got the Z2= (1, 1, 0, 0) for both cases.
But now my question is that if my system is 2D what does it mean by (1; 1, 0, 0)?
If it was (1; 0, 0, 0) we say it is a strong topological Insulator and if it is (0; 0, 0, 1) we say it is weak TI in 3D while strong in 2D.
But it is (1, 1, 0, 0), what do we call this one. In literature, it is only written Z2=1 (just single 1) which is also correct for 2D systems. But I am getting two 1. Please explain this and correct me if I am wrong?
sincerely
Thank you
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| Re: Different Z2 Invariants with and without Band.dispersion 'on' ( No.4 ) |
- Date: 2021/06/09 22:48
- Name: T. Ozaki
- Hi,
By the definition of Z2, (1; 1, 0, 0) implies x0=0, x_pi=1, y_0=1, y_pi=0, z_0=1, and z_pi=0.
I guess that the k-path perpendicular to the 2D plane may not have physical meaning, but we see
that Z2=1 appears along the other two directions, respectively.
So, we may be able to call it strong topological Insulator.
Regards,
TO
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Different Z2 Invariants with and without Band.dispersion 'on'