Timeline for Topology of Fermi surface
Current License: CC BY-SA 3.0
13 events
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| Apr 12, 2015 at 20:34 | comment | added | hehuan0430 | Hi, @MengCheng, I may still have a question of this issue. I read Horava's paper. The thing I don't understand is what the winding number tells us? What is the physical difference of systems with different winding numbers N=1,2? | |
| Apr 2, 2015 at 16:39 | comment | added | Meng Cheng | It seems that the Volovik thing pre-assumes the existence of Fermi surface. I don't think it makes sense trying to apply it to insulators anyway. | |
| Apr 2, 2015 at 11:46 | comment | added | FraSchelle | @MengCheng Ok, I understand now your previous comment. Indeed, the comment I made to user:079 was perfectly unclear... sorry for that. So let me state it correctly in the answer. Please check the edit, and tell me if you agree. | |
| Apr 2, 2015 at 11:45 | history | edited | FraSchelle | CC BY-SA 3.0 |
add the last section
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| Apr 2, 2015 at 11:34 | history | edited | FraSchelle | CC BY-SA 3.0 |
add the last section
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| Apr 2, 2015 at 9:45 | comment | added | FraSchelle | @MengCheng You're perfectly right, but for the Volovik construction, the integral is defined in momentum, and so the poles are the poles with respect to p, not z. The spectral properties correspond to the poles in z indeed, as you say. I developed the above (boring) machinery to be sure to say no stupidity in the case of Volovik $N_{1}$ invariant (it has been done by Horava in fact, see arxiv.org/abs/hep-th/0503006 ). But any way to prove the stability of the Fermi surface in a better way is indeed welcome :-) Thanks for your comment | |
| Apr 2, 2015 at 5:53 | comment | added | Meng Cheng | The Green function, as a function of frequency, have poles corresponding to single-particle excitations. In your example, the pole is $z=\sigma(p^2+\Delta)-\mu$. The difference between metal and insulator is whether that the location of the pole can be arbitrarily small frequency (just gapless v.s. gapped). | |
| Apr 2, 2015 at 5:19 | comment | added | FraSchelle | @079 The Green function clearly has no pole along the real axis in the insulating region, see edit about a simple model to see how this happens. In fact you're right the Green function gives you knowledge about the spectrum of the Schrödinger equation. More precisely the Green function has a pole along the real axis for each discrete energy, eventually it has branch cuts for bands. So how could a Green function has a pole when there is no associated state, as for a trivial insulator ? | |
| Apr 2, 2015 at 5:13 | history | edited | FraSchelle | CC BY-SA 3.0 |
a few math typos
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| Apr 2, 2015 at 5:05 | history | edited | FraSchelle | CC BY-SA 3.0 |
add the edit, as a try to complete the answer
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| Apr 2, 2015 at 4:52 | history | edited | FraSchelle | CC BY-SA 3.0 |
add the edit, as a try to complete the answer
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| Apr 2, 2015 at 2:03 | comment | added | 079 | I think Green function will also have poles for an insulator because the poles of the green function gives you only knowledge about the spectrum of the system it doesn't tell you about the system is insulating or not?@ FraSchelle am i right? | |
| Apr 1, 2015 at 22:40 | history | answered | FraSchelle | CC BY-SA 3.0 |