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12

As you mentioned in the question, excitons are indeed bound electron-hole pairs. They are often considered to be the signature of optical spectra in insulating solids. Adding onto the comment by tmph, there are two types of excitons: opposite spins of electrons will lead to a dark exciton with S=0 (since it doesn't allow for momentum conservation). Same spin ...


9

This has already been asked in several forms in Physics Stack Exchange. Within the semi-classical 'electric dipole' approximation, only the electric field of light interacts with the electron. The effect of magnetic field is usually very weak except for a few cases like highly concentrated pulses etc. Magnetic field can flip spins, in a process called 'spin-...


7

The model you are describing corresponds to a Wannier-Mott exciton and the binding energy is approximated by: $$ \tag{1} E_{\mathrm{B}}=\frac{\mu}{\epsilon^2_{\infty}}R, $$ where $\mu$ is the reduced mass of the electron and hole, $\epsilon_{\infty}$ is the high frequency dielectric constant, and $R$ is the Rydberg constant. You can in principle construct ...


6

Not a direct answer to your proposed approximation to the exciton group velocity as $\mathbf{Q}\to0$, but just wanted to point out that the latest version of Yambo supports the calculation of exciton dispersions, from which you can then calculate the exciton velocity at any $\mathbf{Q}$. Here are the details: Yambo 5: exciton dispersion.


6

well, you certainly can change the spin of an electron through acting on its orbital motion with the electric field. It is called "spin-orbit coupling" (SOC) and a lot of magneto-optical and opto-magnetic (inverse) effects completely rely on it. You might want to look at papers discussing all-optical magnetization switching with circularly ...


5

I think the claim assumes linear polarization of the light which has zero angular momentum. For circularly polarized light, the spin of the electron should be flipped. (e.g. Appl. Phys. Lett. 114, 041104 (2019)) Indeed, it has a probability issue as explained by @Xivi76.


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