# Tag Info

## Hot answers tagged electronic-band-theory

26

A band is essentially a (near) continuous collection of momentum eigenstates. Within the band the electrons can be treated as free to a reasonable approximation, so their eigenstates are just plane waves. The symmetry means that for every eigenstate there is another with an equal and opposite momentum. So if we populate every momentum eigenstate the net ...

22

Room-temperature superconductors are not forbidden by any known theory. However, discovery is difficult, while engineering is possible. One thing about superconductors is that they do not give off any heat. So cooling is just a function of fighting the insulation. With the discovery of super-insulators, the rest is just engineering! Here's an interesting ...

21

The answer lies in the band structure of the two materials. The band structure describes how the electrons in a solid are bound, and what other energy states are available to them. Very simply, the band gap for transparent diamonds is very wide (see this link): Normally, diamond is not a conductor: all the electrons live in the "valence band", and you ...

17

$\renewcommand{ket}{|#1\rangle}$ The basic logical connection here is $$\text{symmetry} \rightarrow \text{degeneracy} \rightarrow \text{avoided crossing} \rightarrow \text{band gap} \, .$$ $\textrm{symmetry}\rightarrow \textrm{degeneracy}$ Consider an operator $S$ and let $T(t) = \exp[-i H t / \hbar]$ be the time evolution operator. If $$[ T(t), S] = 0 ... 16 No band is special. A partially full valence band does conduct, just like a partially full conduction band. On the other hand, a perfectly full band conducts just as well as a perfectly empty band: No conduction at all. Now, nobody is surprised when you say an empty band can't conduct, but at first it seems surprising that a full band is the same way. ... 15 Even when an isolated atom has a filled shell, the electron bands in a solid crystal may be partially filled. The reason is that bands that originate from different atomic orbitals may actually overlap in energy. This is shown on the left in the figure below. When the bands overlap, the lowest-energy state has some electrons displaced from the top of the ... 12 In calculating the electron dispersion you probably obtained the diagonalized Hamiltonian in the momentum space$$ H=\sum_\mathbf{k}\left[c^{\dagger}_{\mathbf{k}A},c^{\dagger}_{\mathbf{k}B}\right]\left[\begin{array}{cc}0 & \Delta(\mathbf{k})\\ \Delta^{\dagger}(\mathbf{k}) &0\end{array}\right]\left[\begin{array}{c}c_{\mathbf{k}A} \\ c_{\mathbf{k}B}\...

12

As mentioned here, metallic hydrogen may be a conventional superconductor up to about 290 K. This is then due to the low mass of the metal ions, this leads to a strong coupling of the electrons with the lattice vibrations.

11

For any crystal, the First Brillouin Zone is found using the Wigner-Seitz construction for the reciprocal lattice. The high-symmetry points are labeled by certain letters mainly as a convention--like you said Gamma for (0,0,0) etc. The important thing to realize as far as the group theory, is that the group of the wavevector at the Gamma point has the full ...

10

(1) Your definition of strongly correlated system is correct "single-particle fails." We can still use ARPES to study strong correlated systems, we just do not see features that would be present in a weakly correlated system. The most prominent feature in a weakly correlated system is a sharp peak at certain energy and momentum. If you track this peak in ...

9

Fermi pockets (or Fermi surfaces) are contours of Fermi energy in the Brillouin zone. Depending on the effective mass $m^*$ of quasi-particles, the Fermi pockets can be divided into electron pockets (if $m^*>0$) and hole pockets (if $m^*<0$). For weakly interacting Fermion systems, according to the Fermi liquid theory, all the low-energy physics ...

9

A useful way to visualize the difference between conductors, insulators and semiconductors is to plot the available energies for electrons in the materials. Instead of having discrete energies as in the case of free atoms, the available energy states form bands. Read on the link. You ask: My question is very simple. I just want to know that is conduction ...

8

They are shown at the $\Gamma$ point in special diagrams called the reduced zone scheme in which a band will be shown folded back on itself. This way of showing the band structure is convenient for a few reasons, one of which is that it saves space on the page. If you look at that band gap at $\Gamma$ and follow the lower band down to lower energies, you ...

8

The most truthful answer, to my mind, to this is simply "because it often works in practice." It is not obvious, a priori, that band structure should apply to any realistic solid. The Coulomb interaction is typically of the order of the Fermi energy. Nonetheless, thanks to the magic of Fermi liquid theory, this strong interaction somehow only ...

8

The rule of thumb is that there are a number of bands equal to the "degrees of freedom" of the lattice. You can get additional degrees of freedom from having multiple species of atoms, multiple orbitals per atom, multiple coupling strengths, etc. These degrees of freedom all increase the dimension of your Hamiltonian matrix. In this simplest 1-atom ...

8

Kittel (at least my 5th edition) goes through this derivation. Refer to the diagram below and remember that in semiconductor physics the chemical potential $\mu$ is also the Fermi level $E_F$ below. The derivation essentially involves calculating the concentration of electrons and holes at temperature T in the conduction and valence bands respectively, ...

7

Let us take it one step at a time, when the temperature increases the vibration energy of atoms increases causing the distance between them to increase. I hope that is clear. Now we know from solid state physics that electrons exist in bands rather than discrete levels for single atoms. The electrons in the valance band are the outermost electrons, which ...

7

The quantization of energy levels appears both in quantum and classical mechanics, and it is not a consequence of the Schrödinger equation. It is a consequence of confinement. In fact, anytime that a wave equation (any quantum equation for the wavefunction, or a classical equation for a classical field, e.g., EM field) has periodic boundary conditions in ...

7

An example of a doped semiconductor might give an intuitive picture of some aspects of this topic: Consider a material like Germanium. Atoms are structured in a lattice. All valance electrons are "used" in the crystal structure to form bonds to neighbours; none are more "free" than others. source Now dope it with another atom of one higher electron number,...

7

Set aside the battery part, and think about what you get when you e.g. inject electrons into a block of insulating plastic. It looks like this: That's a "beam tree", a.k.a. a "Lichtenberg figure". It's made by using an accelerator to deposit electrons in a block of plastic. A lot of electrons. They stay where they stopped, building up charge in the ...

6

Take the solutions of the potential problem of an atom and look at the energy levels. Between the n=1 energy level and the n=2 energy level there is a forbidden gap in energy, i.e. you will not find the electron of the hydrogen atom there. Note the thick line for large n where the energy gaps become so tiny leading to a continuum , i.e. an energy "band" ...

6

Think about why the diode does conduct current when it is forward-biased. In that case, there is an electric field pointing from the P-type end to the N-type end. Negative electrons want to climb up the field. The electrons get to the junction and find holes to fall into. Now that the electrons are in the P-type material they can find their way to the ...

6

Technically, both solids and gasses have quantized energy levels. The difference is that molecules of a gas interact with other molecules very weakly, so the energy levels observed in emission or absorption of a collection of gas molecules are almost exactly the same as the energy levels that would be observed if you had a single gas molecule in isolation. ...

6

Metals are good conductors of electricity because the outer (valence) electrons of the metal atoms are only loosely bound to the nucleus and form molecular orbitals known as the conduction band. Electrons can move more or less freely through the conduction band and so metals conduct electricity generally well. When a metal is chemically oxidised its outer ...

6

Is there a nice way to look at the dispersion relation near a Dirac point? It turns out there is. Namely, since Dirac points are equivalent to having a linear dispersion relation, $E(k) \propto k$. Warm-up Consider the simple absolute-value function, $$|x| = \sqrt{x^2} = \begin{cases} -x & , ~ x < 0 \\ x & , ~ x \geq 0 \end{cases} ~,$$ ...

6

Since the velocity $v_\parallel$ parallel to the surface is given by the derivative $v_\parallel = dE/dp_\parallel$, with $p_\parallel$ the parallel momentum component, the velocity would vary discontinuously if the surface bands would not be tangential to the bulk bands. In particular, at the transition point the velocity would not exist. The reason this ...

6

It can, and does. However, the absorption of a photon with wavevector $\mathbf k$ causes an electron transition from $(\mathbf{k}_i,E_i)$ to $(\mathbf{k}_f,E_f)$ subject to the constraints that $E_f-E_i = \hbar c|\mathbf k|$ and $\mathbf{k}_f-\mathbf{k}_i = \mathbf k$. Consider a transition corresponding to an energy difference of $2$ eV. This would ...

6

This the famous Hofstadter problem. (Douglas Hofstadter is the author of Godel-Escher-Bach).

5

TB method uses the atomic orbitals as basis functions to get the matrix representation of the crystal Hamiltonian. The matrix elements of this Hamiltonian are usually computed using fitting parameters. $k \cdot p$-method is based on the matrix version of the perturbation theory derived by Lowdin. It states that the energy spectrum at some point of the ...

5

The depletion region forms due to the equilibrium between drift (field driven) and diffusion ( concentration gradient driven) currents. If you have very low doping, the depletion region will be large because a large volume of depleted semiconductor is needed to generate enough electric field to balance the diffusion current. On the other hand, if you have ...

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