1
$\begingroup$

The above figure is Rashba-split free electron surface state. The below on is TI surface state figure.

The above figure is Rashba-split free electron-like surface state in a projected bulk band gap. The below figure is the band structure of a topological insulators. x axis is the wave vector, y axis is the energy. red and blue lines are surface states with one spin up one spin down. What is the difference between the two figures why TI has this linear one but the above is parabolic? They all have Rashba spin orbit interaction.

$\endgroup$
3
  • 1
    $\begingroup$ C'mon, give us a bit more to work with here! What do these figures show? What are the axes, what are the lines? And, the difference between the figures is rather obvious (one is linear, the other is parabolic as you say) - are you asking for the reason for that difference? $\endgroup$
    – ACuriousMind
    Apr 13, 2015 at 18:53
  • $\begingroup$ The above figure is Rashba-split free electron-like surface state in a projected bulk band gap. The bellow figure is the band structure of Topological insulators. x axis is the wave vector, y axis is the energy. red and blue lines are surface states with one spin up one spin down $\endgroup$
    – Michelle
    Apr 13, 2015 at 22:06
  • $\begingroup$ Please edit that information into the question instead of commenting. $\endgroup$
    – ACuriousMind
    Apr 13, 2015 at 22:19

2 Answers 2

5
$\begingroup$

The difference is obvious: In the second figure, the blue and red line connect the valence and conduction bands. These are actually surface states. So regardless of where your chemical potential lies, there will be low-energy excitations on the surface, i.e. the surface is conducting. This is what happens in a topological insulator. In the first figure, you can put your Fermi energy between the valence band and the surface states (there is a whole energy window where nothing happens), and the whole system, both bulk and surface included, is insulating.

You should also notice that the red/blue lines have opposite spins. Imagine there are impurities on the surface. Normally electrons just backscatter by the impurities, causing dissipation of electric current. But in topological insulators, backscattering means you have to go from the red to the blue, since one is right-moving and the other is left-moving. But in the process the spin has to flip, which breaks time-reversal symmetry. So if the impurities do not break time-reversal symmetry (i.e. non-magnetic), they do not cause backscattering (to the leading order). This is why the surface states are robust: they are protected by time-reversal symmetry.

$\endgroup$
2
  • $\begingroup$ Thank you for your answer. What about the above figure? Is this figure protected by time reversal symmetry too? Why these two band structure all have spin split according to Rashba model but only the bottom one become TI? Also, what condition makes material become TI besides strong spin orbital coupling? $\endgroup$
    – Michelle
    Apr 14, 2015 at 19:01
  • $\begingroup$ The above figure is not. As mentioned already in the answer, you can choose to put the Fermi energy below the surface bands and the whole system is insulating. Even when the Fermi energy intersects the surface bands, for a given spin there are both left and right moving electrons, so they can backscatter and dissipate current without breaking time-reversal symmetry. This is completely different from the lower figure. $\endgroup$
    – Meng Cheng
    Apr 15, 2015 at 15:27
0
$\begingroup$

The two pictures show different situation, you just see path of $\Gamma$ to M, and edge states in first picture cross Fermi surface even times, while second are odd times. So the first edge states could be gapped by perturbation. You could refer Colloquium: Topological insulators

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.