# How to understand the spin of exciton?

In very recent years, exciton physics has been intensively explored as the emergence of 2D materials. Exciton is a bound electron-hole pair and is bosonic, namely hosting an integer spin.

• My first question is what the integer spin is? We all know the spin of an electron is $$\dfrac{1}{2}$$ and the hole will hold a positive charge compared to the kicked electron. Can we consider that the spin of the hole will be $$-\dfrac{1}{2}$$ and the exciton holds an integer spin $$0$$.

• My second question is how the formation of the exciton is affected by the spin of electrons? For example, the following figure from this paper indicates that the spin is closely related to the formation of the exciton.

• My third question is how can we demonstrate these spin-related exciton landscapes with first-principles calculations?

• Interesting question. Ch 9 of Cohen and Louie's Fundamentals of Condensed Matter Physics has a brief explanation on the spin: basically it can be either a singlet ($S=0$) or triplet ($S=1$) state. Perhaps that helps address the first two questions.
– tmph
Feb 9, 2021 at 2:59
• +1 but you got lucky with Xivi76 answering all three questions right away. Otherwise I woukd have asked you to follow the "1 question per post" rule. Feb 9, 2021 at 3:40
• @NikeDattani Questions (1) and (2) are definitely inter-related. I'm not completely conversant with GW-BSE, so I wouldn't be able to do justice to question (3) - If OP wants to know how GW-BSE works, it should be a separate question for sure. Feb 9, 2021 at 4:43
• @Xivi76 I would have liked question (3) to be asked as a separate question, and your paragraph about it in your answer, could have been a comment or slightly extended answer. On some SE sites like the quantum computing stack exchange, the question would have been closed right away as being "too broad" because they are much more strict about multi-question posts. Here I would still like all posts to ask one question (within reason) but don't believe we need to close-vote to get there: I trust the users to do this one their own. The reasons why we want one question per post are large in number! Feb 9, 2021 at 5:26
• @NikeDattani I will consider your suggestion.
– Jack
Feb 11, 2021 at 7:56

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 of electrons will lead to a bright exciton with S=1 (since it allows for recombination). Therefore the electronic band-structure can often serve as an indicator for optical selection rules of excitons. The dipole strength (that dictates the relative strength of 'bright' or 'dark') is a term that is basically a superposition of dipoles from DFT (read: Fermi's golden rule). The dark exciton is not trivial though, it has been observed in experiments like this one.

Regarding the third part of your question, the method of choice to model excitons seems to be the GW-BSE method, since the exciton itself is inherently an excited state. But the mean-field starting point is Kohn-Sham DFT in most GW-BSE calculations.

Edit: Thanks to Anyon for pointing out a class of excitons with S=2 occurring in certain semiconductors, this should potentially be considered as well.

• +1 for answering all three questions! But why is GW-BSE the only option, what about TD-TDFT or excited-states using Monte Carlo or plane-wave coupled cluster? Feb 9, 2021 at 3:38
• Sometimes one also encounters $S=2$ dark excitons. This can occur in e.g. semiconductor band structures with "heavy holes" that have spin-$3/2$. Feb 9, 2021 at 3:39
• @NikeDattani Maybe I should the remove the word 'only', what I have noticed is GW-BSE is the method of choice for studying excitonic spectra. Anyon, thanks for pointing that out, I will edit it into the answer. Feb 9, 2021 at 4:38
• I think you may have the spin counting the other way around? I agree that the electron doesn't change the spin as it is excited, but this means that you go from a, say, spin up valence electron to a spin up conduction electron, leaving a hole behind with the opposite spin. So in fact, spin singlet is the bright exciton and spin triplet the dark one. This is also why, for example, singlet excitations undergo fluorescence whereas triplet excitations undergo phosphorescence. Feb 10, 2021 at 10:32
• @Jack Yes but in general in physics, it's usually not mentioned as the spin of a hole (Read: physics.stackexchange.com/questions/432044/…). That is why I said electron with opposite spin. Feb 11, 2021 at 16:15

First I mention this very similar question on Physics SE and the related answer.

Then, a couple of points which are not well addressed in the previous answer in my opinion. First of all we are here discussing excitons in extended systems. The word "exciton" is sometimes used also in isolated systems, but with a slightly different meaning.

Second we are discussing systems in which the spin is a good quantum number. This happens when spin-orbit coupling SOC is neglected. Otherwise the discussion is true only in an approximated way.

Third we have to distinguish between spin S and orbital $$L$$ angular momentum. Excitons can also have an angular momentum $$L$$, or, in presence of SOC, of their combination $$J$$. However the definition of $$L$$ is always approximated, since $$L$$ is a good quantum number in presence of $$SO(3)$$ invariance, while, the exact symmetry of the exciton will depend on the underlying crystal lattice symmetry $$O_l$$. The approximation is better for more delocalized (i.e. Wannier like) excitons.

The discussion on heavy holes is related to the angular momentum $$J$$. For localized, i.e. Frenkel excitons, one should look into the irreps of $$O_l$$, as seen in this Chem SE answer. Excitons can have $$S=0,1$$ and potentially any $$L$$ or $$J$$. (However no $$S=2$$ for obvious reasons!)

1. Bright excitons are always with $$S=0$$, $$L=0,1$$ due to electric dipole selection rules.

2. This means that the valence band from which the electron is removed and the conduction band into which the electron is added have the same spin. However, as correctly pointed out in the question, the convention is that the spin of the hole is the opposite of the one of the valence from which it is removed. Also notice that the election and the hole does not stay in a specific state in the two bands, but they can be expressed using these bands as a basis-set. So the discussion specify which state in the basis set can be used. Notice that instead S=1 excitons (i.e. magnons if $$S_z=\pm1$$) can be generated via magnetic fields. L=1 usually require instead circularly polarized light.

3. For first principles simulations, the state of the art is the Bethe-Salpeter equation (BSE). For systems with $$S=0$$ ground state, this is always constructed and solved in the singlet $$S=0$$ channel. For magnetic systems both S=0 and S=1 solutions can be obtained (but always with $$S_z=0$$). For the case with SOC all excitons are computed. This is discussed in details in one of my recent arxiv articles

Finally let me add a comment on other possible approaches beyond BSE. For extended system quantum chemistry approaches are not feasible. TDDFT cannot capture the physics of the exciton at $$q=0$$, one would need TDPolarizationDFT or do the $$q\rightarrow0$$ limit of finite momentum TDDFT. However for neither there exist good functionals able to properly describe excitons. One of the reasons is that the exciton requires a long range interaction which cannot be easily captured by these theories. Despite all this, the structure of the TDDFT equations is the same as the one of BSE. So all the above considerations for BSE apply. Instead I'm not an expert on quantum Monte Carlo and I cannot add much on this.