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There are many schemes to make topological superconductors. Some of these schemes have restrictions on the chemical potential $\mu$. You also need to know what type of topological superconductors you are dealing with. You can refer to the periodic table to determine this: In the paper from the link you provided the authors mention two types of ...


5

I absolutely understand what you mean. Just assume that the last $\delta(E-E_n)$ is only partly occupied. It's actually a common strategy to widen such Delta functions a little bit when integrating numerically, and then the procedure is obvious. There is nothing wrong with that, you just have to know how to handle the math - and the easiest way is to use ...


5

If you want to decide whether a gas of fermions is degenerate$^*$, then you would compare the temperature of the gas with its Fermi temperature. If $T \ll T_F$ then the gas can be considered completely degenerate. If $T \sim T_F$ then the gas is partially degenerate. If $T > T_F$ then the gas is not degenerate. If the fermion gas is degenerate then the ...


5

The fact that the Fermi energy is in between two energy levels is actually general property of systems with discrete energy levels. The point is that the Fermi energy is not defined as the highest occupied energy level at zero temperature. It is the chemical potential at zero temperature. Consider a system with discrete energies, with $N$ occupied levels ...


5

So I don't know whether any of the following 3 sections is going to be review or not, so I have just typed them out as a sort of complete review of what we're talking about -- the Fermi energy is one contributor to the chemical potential and once you know that, you are prepared to deal with semiconductors. What is chemical potential? Let's take a step back ...


5

This subject causes me confusion too (as well as frustration). Below is my understanding, but take it with a grain of salt because I may be wrong about it. I think that it basically comes down to how correctly you want to define the term "Fermi level" (and to a lesser extent, if you're a physicist or engineer). Short answer: for engineering (and most ...


4

If the mass of the particles increases and the Fermi energy stays the same, then the number (density) of the particles must decrease. But the energy levels of the particles depend on their masses as well. For example, suppose we have a set of $N$ electrons in a cubical box and $N$ protons in another cubical box of the same size. The Fermi energy is ...


3

Like you said, Fermi energy corresponds to the highest energy occupied by an electron at $ T = 0 $. For the free electron gas, $E = 1/(2m)\times\left(k_x^2+k_y^2+k_z^2\right)$. If you plot the momenta on the Cartesian axes, the constant energy will be a spherical shell. Here's a more intuitive way of seeing this. Take a 3D box with $L_x\neq L_y\neq L_z$. ...


3

You can understand this in the following oversimplified model. Let us consider periodic 1D chain of $N$ atoms with lattice constant $a$. Brillouin zone boundaries are located then at $\pm n\pi/a$. Let us consider three configurations: Atoms have 1 valence electron on s-type orbital. In this case the energy level of this atomic orbital is split into $N$-...


3

What you call Fermi energy is mostly called Fermi level in semiconductor physics. This Fermi level is synonymous with the total chemical potential of the electrons in the semiconductor which is a purely thermodynamic concept. This chemical potential needs not correspond to an allowed energy level of the electrons. See Fermi level.


3

You have to keep in mind that there is also an electronic binding energy between the electrons and the potassium nuclei that tries collapse the potassium inwards. The precise density is set by the balance between the fermi pressure and the binding pressure between electrons and nuclei. The take home message in your exercise, is to me is that you need a very ...


3

The Fermi energy $\xi_F$ and the chemical potential $\mu(T)$ are two distinct but related quantities. In principle the particle number is conserved. However it is hard to deal with a system with fixed particle number $\mathcal{N}$. In order to release this constraint we introduce a Lagrange multiplier $\mu(T)$ and allow the particle number to vary. The ...


3

There is not a Fermi surface associated with every lattice point. The Fermi circle as a concept exists in k-space, and has a fixed center, which is the origin of k-space. The point is that you are filling the circle up to a surface of fixed energy.


3

It is normally situated midway between valence band and conduction band, is this how it is defined or is there other reasons this is so? Not necessarily. You're right - if the definition of $\mu$ was simply that all states with energy $E<\mu$ are occupied at $T=0$, then $\mu$ could be anywhere within the band gap. To understand exactly where it should ...


3

In general, it is close to the centre of the gap but slightly offset. This can be seen nicely by using the effective density of states approximation for conduction and valence band, $N_c$ and $N_v$ respectively. For an intrinsic semiconductor the electron density in the conduction band (that has energy $E_c$) is equal to the hole density in the valence band (...


3

This is an approximate statement, which follows from understanding the shape of the Fermi distribution: $$ n_F(E) = \frac{1}{e^{\frac{E - E_F}{k_B T}} + 1} $$ This distribution has a step-like form: it is approximately 1 for $E < E_F$ and approximately zero for $E > E_F$, except for the region around $E_F$ where $|E-E_F|\sim k_B T$. In the limit of ...


2

Undoped (intrinsic) semiconductors and insulators are identical in this picture. They only differ by the value of their electronic bandgap. Elevated temperature leads to intrinsic carriers in the semiconductor and therefore makes it conduct at T > 0 K. The bandgap of an insulator is so large that this effect can be neglected for common temperatures. The ...


2

Because the Fermy energy is not really the energy of the fastest electron(altougth for understanding how it works, it is not that bad start, it can be seen that way for metals, for example). It is more like the energy under which all the levels are filled. For semiconductors, as they have band structure, the energies that an electron can take are limited to ...


2

Fermions are indistinguishable particles, hence you cannot distibute the three fermions in different ways. The state is completely described by the occupations numbers $|1,2\rangle$ and has zero entropy. If you are talking about a Fermi system at a given temperature, that means it is (or was) in contact with the external world at that temperature (canonical ...


2

1) The Fermi function tells you the probability that a state with energy $E$ is filled. However, it does not guarantee that such a state exists! You are correct that no states exist within the band gap (assuming a perfect crystal). So, there is no contradiction: the Fermi function only applies if a state exists, and states don't exist in the gap. 2) The ...


2

The bandgap energy is the difference between two levels, the valence band edge and the conduction band edge. The Fermi level is a particular energy level, the level at which, if there were a state at that level, the state would have a 50% probability of occupancy. In order for a material to be considered a semiconductor, the Fermi level must lie within the ...


2

Let's say you have some distribution $D(x)$. If we want this to represent probabilities, then it must be that $$A=\int D(x) dx=1$$ So what we can do is just define a new function in terms of $D(x)$ and this integral over the domain of $D$: $$P(x)=\frac{1}{A}D(x)$$ So that the integral over the domain of $P$ is $1$. Now, when we want to find the ...


2

Let's begin with the Fermi-Dirac distribution function $$n_k(T) = n(\epsilon_k, T) = \frac{1}{e^{(\epsilon_k - \mu(T))/T}+1}.$$ The subscript $k$ labels the states of the system, and I have explicitly included the temperature dependence of both $n_k$ and the chemical potential $\mu$. Also, I use units such that the Boltzman constant $k_B = 1$. At $T = 0$, ...


2

There isn't an intrinsic difference between a band insulator and a semiconductor - the difference between the two is one of degree, not of kind. Generally, we use the term 'semiconductor' to refer to band insulators where either (i) the bandgap is small enough that at room temperature there will be enough thermal excitations to bring sufficient population ...


2

To quote from a note in Ashcroft and Mermin's Solid State Physics book (p. 573 in the 1976 version): It is the widespread practice to refer to the chemical potential of a semiconductor as "the Fermi level," a somewhat unfortunate terminology. Since the chemical potential almost always lies in the energy gap, there is no one-electron level whose energy is ...


2

From the Schrödinger equation, energy for a bound electron is quantized, such that only certain energy levels are allowed. Because electrons are fermions, they obey the Pauli exclusion principle, which states that no two electrons can have all their quantum numbers (such as energy level, orbital, spin) equal. This means that on each energy level, there are ...


2

One can derive the fermi-energy using quntumstatistics and is a more complex derivation. But in condensed matter you find a more vivid explanation about the meaning. Let us assume a free fermi gas or electron gas. We make the following assumptions: N>>1 conducting electrons move on a homogeneous charge background No interaction between particles Pauli-...


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