Why doesn't Graphene have a band gap?

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SUMMARY

Graphene lacks a band gap due to its linear energy-momentum (E-K) relationship, which results from the unique symmetry of its atomic structure. In contrast, bilayer graphene exhibits a quadratic E-K relationship, allowing an electric field to induce a band gap. The absence of strong reflections of Bloch waves along specific directions in graphene contributes to this phenomenon, as the atomic arrangement prevents the necessary interactions that would lead to band splitting. In materials like MoS2, differing atomic species in the sublattices create a band gap, highlighting the importance of atomic composition in determining electronic properties.

PREREQUISITES
  • Understanding of quantum mechanics, particularly perturbation theory.
  • Familiarity with band structure calculations and Hamiltonian formulation.
  • Knowledge of Bloch's theorem and its implications in solid-state physics.
  • Basic principles of symmetry in crystal structures.
NEXT STEPS
  • Study the Hamiltonian formulation for graphene and its implications on electronic properties.
  • Explore the concept of band structure in bilayer graphene and the effects of electric fields.
  • Research the role of symmetry in determining band gaps in materials like MoS2 and BN.
  • Learn about the application of perturbation theory in solid-state physics to analyze band gaps.
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Physicists, materials scientists, and electrical engineers interested in the electronic properties of two-dimensional materials, particularly those researching graphene and transition metal dichalcogenides like MoS2.

asheg
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Is there any simple justification about graphene having no band gap? How bout its linear E-K? Why bilayer graphene has a quadratic E-K and electric field can open a band gap there?

I do not completely understand the broken symmetry argument? Also Why MoS2 which has similar structure, do not have similar properties?

I know that we can create the Hamiltonian, plot the band-structure numerically and answer these questions. But I want to know if we can use some simple quantum theory arguments (like perturbation theory, or Fermi golden rule) or symmetry facts to get a qualitative answer for all of these questions.
 
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I don't know whether you are familiar with x-ray structure analysis. There you encounter a similar phenomenon: Systematic absences. Usually, the splitting of two bands near the Brillouin zone boundary is due to strong reflection of the Bloch waves, or, what amounts to the same, a strong mixing of counter propagating waves, leading to a sin and a cos wave with different energy. However in graphene, these reflections are absent along special directions as you have two reflecting lines of atoms which are not spaced by a lattice vector, but only half of it. You easily see this as follows: Fill the centers of the hexagons with additional C atoms: Apparently you get a primitive closest packed 2d crystal with smaller unit cell. Some of these lattice lines are also present in graphene with equal density of C atoms on them, though they are no longer connected by a lattice vector.
 
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Thanks DrDu. Do you have a reference so I can take a look into the details? Also do you have idea about the other questions?
 
No, I have no literature at hand. You can also see it like this: in graphene, each C atom has 3 nearest neighbours. At the conical point, the first C atom is in phase, 1, the second is ##\exp(\pm 2\pi/3)## out of phase and the third one ##\exp(\mp2\pi/3)## out of phase, so that the interaction of a c atom with it's nearest neighbours vanishes. As there are two possible signs of the phase factor, you see that there are two conical points in K space. Hence the two sub lattices consisting of all atoms being next-nearest neighbours don't interact at the conical points and are energetically degenerate, as the atoms are identical. In other substances, like BN, the sub lattices are formed from different elements (B and N, respectively), which have different energies, so there is a band gap.
In MoS2 and graphene double layers, similar arguments as in BN should apply.
 
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