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What are the consequences for the bandgap if the conduction band edge and valence band edge are contributed by different spins? Are there any driving forces that prevent an excitation from spin up to spin down for example? Will this be significantly affected by spin-orbit coupling?

I suspect this gets much more complicated considering the noncolinear spin case, but I only wish to consider the colinear case currently.

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2 Answers 2

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I don't think we need to talk about driving forces (this sounds esoteric) instead calculating the selection rule:

$$P_T = \int \psi_1^* \mu \psi_2 d\tau$$

Here $\psi_1$ and $\psi_2$ are the wavefunctions of the two states involved in the transition and $\mu$ the transition operator.

This integral represents the transition probability ($P_T$). When the result is zero, the transition is called forbidden.

As the spin is directional, the corresponding spin wavefunctions are odd. This directly implies that a transition where the spin "directions" are different, will result in $P_T = 0$ and then, are considered as forbidden transition.

Relaxation of this rule (i.e. $P_T \neq 0$) can occur with the presence of spin-orbit coupling, vibronic coupling and/or $\pi$-acceptor/donnor mixing with d-orbitals.

Answering your questions:

What are the consequences for a bandgap if the conduction band and valence band are on different spins?

If you have an odd number of electrons or anything else that break the system symmetry, you will/can have separated bands for the spin-up and spin-down configurations. This implies having two different bandgaps. You can explore this feature using your material as the core of spin-filter in spintronics, for example.

The image bellow correspond to a boron-nitride nanotube interacting with a Ni atom. enter image description here

Are there any driving forces that prevent an excitation from spin up to spin down for example?

Nothing here about driving forces, just selection rules.

Will this be significantly affected by spin-orbit coupling?

Yes.

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    $\begingroup$ Excellent answer, my choice of words for "driving forces" was intentionally vague as I figured a selection rule or some physical process involving relaxation would be required. The quick takeaway from this should effectively be that the material effectively will have two band gaps and should be considered independently if I understand correctly? $\endgroup$ Jan 4, 2021 at 21:50
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What are the consequences for a bandgap if the conduction band and valence band are on different spins?

The energy band of materials when the spin is involved (magnetic/nonmagnetic) are usually classified as follows:

enter image description here

in which :

  • (a): ferromagnetic metals
  • (b): half metals
  • (c): topological insulator
  • (d): half semiconductor
  • (e): spin gapless semiconductor
  • (f): bipolar magnetic semiconductor

Here I assume you are talking semiconductors, namely (d) and (f) in the previous screenshot. Essentially, the bandgap is decided by the position of the Fermi level. Interestingly, you will meet some materials in which the spin-up channel is metallic whereas the spin-down channel is a semiconductor. This is an important consequence of spin-involved band structure.

Are there any driving forces that prevent an excitation from spin up to spin down for example?

This question is explained by @Camps. The transitions from the valence band to the conduction band are related to the properties of bandstructure and wavefunctions, such as minimal direct gap, the overlap of the electron and the hole wavefunction, and symmetry of wavefunction. The information from the spin will also be reflected in the wavefunction.

Will this be significantly affected by spin-orbit coupling?

As for the bandgap, the spin-orbit coupling (SOC) will host a significant influence (more than 0.5eV) in some materials with heavy atoms, such as BiI$_3$ and BiTeI:

enter image description here

In addition, the SOC will influence the optical excitation in some nonmagnetic materials with heavy atoms. For example in TMDC monolayers, the spin-$z$ component is a good quantum number due to the interplay of lattice inversion symmetry and strong SOC, which will lead to the selective photoexcitation of carriers with various combinations of the valley and spin indices.

enter image description here

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