Physics, asked by farhanmkhan3888, 8 months ago

What is the relation between the quantum hall effect and the quantum spin hall effect?

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Answered by Anonymous
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Answer:

This has been one of the special research area in condensed matter physics which was proposed first theoretically and then later verified experimentally. In quantum Hall effect, External huge magnetic field and low temperature  are the uttermost requirement but QSH is the special case of quantum hall effect without the application of external magnetic field. Here, Spin orbit coupling plays the crucial part and the resulting current on the surface we get is spin currents, not the normal electron current. Relativistically, charged particles of velocity v sees the electric field partly as a magnetic field . Since electrons carry spin and spin experience this magnetic field , which actually lifts the degeneracy and splits the energy levels.Thus,SO coupling plays the role of magnetic field in a superficial manner.This state is insulating in the bulk having gapless surface states. What give rise to surface states? Its not at all straight forward. It is actually the consequence of topology that leads to the zero energy modes present in the bulk. It has been explained well from Dirac equation 's negative and postive energy states.But, Simple Dirac equation won't help much because of the sheer symmetry present between these two states. In mathematical language, I can say that there wont be any topological distinction between them in Dirac explanation. To get the surface states, people have tried different corrections in the Dirac equation and found out the eigen states that let them know the presence of zero energy in the gap .Now, There are quite heavy theoretical models that can explain numerous incredible properties of these interesting systems. In laymen terms,Topologically invariant means closing and opening the band gap should be continuous without disturbing the system. Closing  the band gap in solid state means going towards conductor and opening the band gap means  getting an insualtor. So, Basically this special case sets up a connection between the conduction Band and Valence Band that leads to the surface states. We can think as if something is going from negative to positive , it has to pass through zero somewhere. So, Those zero modes are the proof for the existence of states. Also, These surface states/edge states (edge states in 2D, surface states in 3 D) are time reversal invariant which states that for every energy eigenstate,the time reversed state is also an eigen state of same energy. In classical mechanics specially for spin 1/2 systems , if we flip the arrow of time twice, everything should go back to itself  But, in quantum systems,for half-integer spins, a rotation of

Answered by Anonymous
0

This has been one of the special research area in condensed matter physics which was proposed first theoretically and then later verified experimentally. In quantum Hall effect, External huge magnetic field and low temperature  are the uttermost requirement but QSH is the special case of quantum hall effect without the application of external magnetic field. Here, Spin orbit coupling plays the crucial part and the resulting current on the surface we get is spin currents, not the normal electron current. Relativistically, charged particles of velocity v sees the electric field partly as a magnetic field . Since electrons carry spin and spin experience this magnetic field , which actually lifts the degeneracy and splits the energy levels.Thus,SO coupling plays the role of magnetic field in a superficial manner.This state is insulating in the bulk having gapless surface states. What give rise to surface states? Its not at all straight forward. It is actually the consequence of topology that leads to the zero energy modes present in the bulk. It has been explained well from Dirac equation 's negative and postive energy states.But, Simple Dirac equation won't help much because of the sheer symmetry present between these two states. In mathematical language, I can say that there wont be any topological distinction between them in Dirac explanation. To get the surface states, people have tried different corrections in the Dirac equation and found out the eigen states that let them know the presence of zero energy in the gap .Now, There are quite heavy theoretical models that can explain numerous incredible properties of these interesting systems. In laymen terms,Topologically invariant means closing and opening the band gap should be continuous without disturbing the system. Closing  the band gap in solid state means going towards conductor and opening the band gap means  getting an insualtor. So, Basically this special case sets up a connection between the conduction Band and Valence Band that leads to the surface states. We can think as if something is going from negative to positive , it has to pass through zero somewhere. So, Those zero modes are the proof for the existence of states. Also, These surface states/edge states (edge states in 2D, surface states in 3 D) are time reversal invariant which states that for every energy eigenstate,the time reversed state is also an eigen state of same energy. In classical mechanics specially for spin 1/2 systems , if we flip the arrow of time twice, everything should go back to itself  But, in quantum systems,for half-integer spins, a rotation of

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