The inversion time operator on the spin-orbit can be obtained directly without more difficulties

T(S)= -S

T(L)= -L

which can be applied straighforwardly to so its invariance under the T operator

T(L.S)=T(L).T(S)=(-L)(-S)= LS

The Dirac equation is invariant under T if the term of mass is zero, which is the case of the materials with linear dispersion law and therefore the Dirac point doesn't change due to the LS terms.

First, spin-orbit coupling shows splitting due to this effect (don't confuse with zeeman splitting).

since SL coupling has the form : -l(r) L.S =1/2[ J2-L2-S2] where J2, L2 or S2 both don't break time reversal symmmetry, while the field must cause breaking of time symmetry!

Dear Prof. Mohamed El Hafidi ,

Thank you for the explanation. My confusion is, SOC can shift the electrons energy in atomic energy levels. Hence this may also cause the shift in energy at Dirac point which may break time reversal symmetry.

If a charge is moving in one diredtion through a magnetic field, you reverse the path of the charge, then

you also change the direction of the lorenz force. So both paths the direct and the time reversed are not

the same for the charged particle.

The inversion time operator on the spin-orbit can be obtained directly without more difficulties

T(S)= -S

T(L)= -L

which can be applied straighforwardly to so its invariance under the T operator

T(L.S)=T(L).T(S)=(-L)(-S)= LS

The Dirac equation is invariant under T if the term of mass is zero, which is the case of the materials with linear dispersion law and therefore the Dirac point doesn't change due to the LS terms.

Dear Sir, Thank you very much

I see that the question contains more parts which are not answered. The spin-orbit doesn't depend of the time reversal (shown in two different forms) but you also ask why the magnetic field do it. There are many different forms to prove it but one inmediate is to take the Ampere electromagnetic equation

curl B = μ0 J

where obviously the density of the current J breaks time invertion and therefore the magnetic field must do it also because the other terms of the equation are independent of time.

The splitting of energy in Zeeman is due to one general external magnetic field B which acts on the magnetic moments associated to the spins μs, while the spin-orbit energy (fine structure or Larmor energy) has a magnetic field Bl due to the orbital angular momentum of the electron. In both cases the energy is the scalar product of

E=-μs.B (Zeeman)

E=- μs.Bl (spin-orbit)

where all the above magnitudes break time reversal (individually) and therefore their scalar product doesn’t do it, preserving the invariance of time reversal symmetry for the spin-orbit and the Zeeman energies. Thus their energies do not distinguish the time invariance T operator behaviour although the magnetic field has a very different origen.

Thank you all for the explanation!

@ Daniel Baldomir

Could you please explain how current density J breaks TRS?

Does it mean that electric field also breaks TRS?

I think, the generator of time is total energy operator i.e. Hamiltonian itself. So anything which commutes with hamiltonian like total angular momentum J, Jz will not break time reversal symmetry. And if TRS is present in the system, any angular momentum operator should have degenerate kramer's doublet.

However by applying magnetic field when J,L,S are no longer a good quantum number, TRS is broken.

why it is important to break inversion symmetry for valleytronics application?

is it possible to dope bismuthene with magnetic impurities ? Mohamed El Hafidi