# Tag Info

## Hot answers tagged spin

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 ...

11

Often when one is asked for the number of spin states, we are interested in determining the spin state of metal complexes containing these ions, rather than the lone ion itself. If you are interested in just the lone ion itself, Nike Dattani's answer goes into depth about that. My answer will focus on the determination of spin states of ions within metal ...

11

$\ce{Cu}^+$ This ion has 28 electrons. If all of them are up (i.e. aligned with the +z axis), then since each electron has a spin of magnitude 1/2, we would have a total spin of +14. If all of them are down we would have a total spin of -14. We could also have 1 up and 27 down, meaning $+\frac{1}{2} - \frac{27}{2} = -\frac{26}{2} = -13$. The number of total ...

10

For computing the spin-state in a metal complex there are many methods available. I'll focus in DFT as one of the most used methodologies. In a single atom cluster (sometimes described as mononuclear complexes), DFT offers good results compared with wave-function multideterminantal methods. On the other hand, if the complex is formed by two or more ...

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-...

9

Unfortunately, without performing spin polarized calculations you cannot know if it will matter. Further, you must perform different magnetic configurations (AFM, FM, PM) to know which is stable. If you can experimentally show if its magnetic or not this will help. Certain elements tend to be magnetic as well. If you have Ni, Co, or Fe then you ...

8

This took some digging, but I can supply at least a partial answer. Assuming you can copy the list of magnetic moments to a list, you could trick whatever software you are using for visualization into outputting it indirectly. A typical POSCAR can look like the following. H 1.00000000000000 25 0 0 0 25 0 0 0 25 H 2 Selective dynamics ...

8

I agree with Tristan that the fool proof method is to do a spin-polarized self consistent calculation and look at the final magnetic moment. If the final magnetic moment is non-zero you should include spin polarization in your calculations. However, from a practical point of view you can have a look at the existing databases. In particular if you are working ...

7

The word "splitting" seems to occur only in one sentence, over the course of the entire paper that you referenced: "The d-orbital spin-splitting energy is stronger than the weak crystal field splitting energy of the S ligand around the TM atoms" I can see why you had to ask this question here, because when I search "d-orbital spin-...

6

With metals, they can often exist as different multiplicities depending on the compound they are in and it is not always simple to predict the correct multiplicity. More often than not, determining the multiplicity of the ground state is either done experimentally or by comparing the energies of likely multiplicities computationally. In your case, it seems ...

6

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 ...

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

To obtain the spin multiplicity of the ground electronic state of a molecule, can be extremely hard. In your question you mentioned $\ce{UF6}$ which has 7 atoms, and not all of them being of the same element. But even for a very simple homonuclear (all atoms being of the same element) diatomic molecule like $\ce{Fe2}$, my answer to "Total spin and/or ...

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.

5

Yes, see e.g. the hydrogen molecule cation.

5

How to interpret this figure below? And what is the meaning of the blue and yellow color around atoms? This figure basically tells you the probability that you can find spin-up and spin-down electrons. If I assume the yellow color represents the spin-up electrons, then the blue color (seems like cyan) will represent spin-down electrons. From your figure, we ...

4

I see now how your initial questions were related, as they all fall under the scope of crystal field theory. I wrote a bit about this in a previous answer. At least in the context of molecular crystal field theory, you will usually see the phrase pairing energy rather than "spin-splitting". The distinction is basically just the direction, where the ...

3

I suspect this would be very difficult to do with DFT. A spin liquid is a very subtle magnetic state and DFT, as a method, is not usually good at describing the magnetic states of a system. Spin liquids are fairly difficult to understand and characterize even using methods that are specialized for strongly-correlated magnetism (like DMRG or QMC). Where DFT ...

3

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: in which : (a): ferromagnetic metals (b): half metals (c): topological insulator (d): half semiconductor (e): spin gapless semiconductor (f):...

2

I don't have much experience with QE, but in VASP, which I imagine is pretty similar, oftentimes not setting an initial magnetization on each ion results in zero magnetic moments on those ions. This is probably because, numerically, it finds a local minima that results in zero spin despite doing a spin polarized calculation. I would highly suggest ...

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