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Is it like this:

enter image description here

where one energy value which is a single line in the band can be occupied by multiple electrons - and by one energy value I mean that the exact/precise energy of all the electrons in that level (if they can) must be the same.

Or like this:

enter image description here

Where no matter how close the levels are or how continuous the band is, one single energy line in the band is occupied by only one electron. Of course since there are as many states as there are the number of atoms in the crystal these energy levels are very close.

I am getting different answers from different places. The confusion in this is troubling me in understanding the concept of density of states and that of degeneracy. The concept of density of states say that in a volume there could be different electronic states with the same energy. But this doesn't make sense to me because when bands are formed by combining the single atom states they split and no two of them have the same energy. And each such split state can only hold 2 electrons as the Pauli exclusion principle states. If that is the case then the latter image should be the right one. But the concept of density of states talk about degeneracy at a particular energy value itself, which implies that there exists multiple electrons (comparable to the no of crystal atoms) that share the same energy which if true leads to the first image being accurate.

So my question is again very simple and precise: can multiple electrons (more than two) have the same precise energy (not their energies lying in the same interval E to E + dE) or are the electrons singly occupying precise energies that are very very close and almost a continuum but nevertheless still alone in a particular line in the energy band ?

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So my question is again very simple and precise: can multiple electrons (more than two) have the same precise energy?

The answer is also simple - yes. The degeneracy of a given energy is generically infinite in the thermodynamic limit. When dealing with a finite crystal then the number of states with a given energy becomes finite, but it still may be very, very large.

The simplest possible dispersion relation is that of the free electron model, in which we have for the lowest band that $E(\mathbf k) = \frac{\hbar^2 k^2}{2m}$ for $\mathbf k$ in the first Brillouin zone. The set of states with some generic energy value $E$ is given by the sphere of radius $k = \sqrt{2mE/\hbar^2}$, which obviously constitutes an infinite set of points.

For another simple example, we might consider electrons hopping on a 2D square lattice with lattice spacing $a$, with Hamiltonian $$H = -t\sum_{m,n\in \mathbb Z} \bigg[|n,m\rangle\langle n+1,m| + |n,m\rangle\langle n,m+1| + h.c.\bigg]$$ The dispersion relation for this system takes the form $$E_{\pm}(k_x, k_y) = \pm 2t\bigg[\cos(k_xa)+\cos(k_ya)\bigg]$$ and so the set of points such that e.g. $E=-t/2$ is given by the following curve:

enter image description here

Once again, this set consists of an infinite number of points.

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In short it is both, but mostly the second option. Each fine energy level within a band is labeled by the (quasi-)momentum $q$ corresponding to this particular level. For each value $q$ there is one energy level (in fact two, if counting spin degree of freedom and ignoring spin-orbit coupling). Occasionally some states with different quasi-momentum values have same energy value $\epsilon(q_1)=\epsilon(q_2)$. There is nothing prohibiting such degeneracy.

So the answer to your question is as follows: usually, each state has its own momentum $q$ and energy level (degenerate in spin) $\epsilon(q)$ except for occasional cases of degeneracy determined by band dispersion $\epsilon(q_1)=\epsilon(q_2)$.

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    $\begingroup$ Uh? I would say there are ''infinitely'' many electrons for a given energy. A solid isn't a 1d material, usually, so the occupied states (at ordinary temperatures) form a fermi sphere. For a given E, there is a whole spherical surface of electrons, not just 2. $\endgroup$ Sep 25, 2022 at 20:48
  • $\begingroup$ As Examination12345 notes in the question, the number of electrons is proportional to the number of atoms (and the gap between the states also depends on the number of atoms). So there are not infinitely many electrons in a given energy because there aren't an infinite number of electrons. $\endgroup$
    – lnmaurer
    Sep 26, 2022 at 14:08
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So my question is again very simple and precise: can multiple electrons (more than two) have the same precise energy (not their energies lying in the same interval E to E + dE) or are the electrons singly occupying precise energies that are very very close and almost a continuum but nevertheless still alone in a particular line in the energy band ?

This may not be the question that you want to ask because technically the answer is "yes", but I'd argue that your second picture is better.

For simplicity, consider free [EDIT] electrons in a 1D lattice. Pick your favorite value of $k$ (so long as it's a reciprocal lattice vector). An electron with that $k$ will have energy edit $E_k$. It's also possible to have an electron with a wavevector of $-k$. This will also have an energy of $E_k$. So, yes, multiple electrons can have the same energy.

To be more precise: two electrons can have wavevector $k$ --- one spin up and one spin down, so four electrons can have a given energy. I'd argue that this is actually more like your second picture: there are a bunch of closely spaced energy levels with a limited number of electrons in them; it's just that 4 is the limit --- not one. Moving to 2/3D, even more states can have the same energy, so the limit is a higher value. Even that's only in an ideal crystal. Real crystals aren't ideal, and once you start adding imperfections and breaking symmetries, you won't necessarily have multiple energy levels with exactly the same energy (and spin up and down may have different energies to boot). In that case, you might be down to one electron per energy. This is your second picture.

[Edit: free electrons -> electrons in a 1D lattice per J. Murray's comment]

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  • $\begingroup$ Do you mean to require that $k$ be a reciprocal lattice vector? That dispersion relation holds for arbitrary $k\in \mathbb R$. $\endgroup$
    – J. Murray
    Sep 27, 2022 at 1:05
  • $\begingroup$ okay so now that i think about it, the pauli exclusion principle should only be applicable when the wave functions of two electrons are being described as one wave function. Two electrons have the same k when they are far apart or even when they are in different crystal directions right ? Thats why in 2D/3D it is possible to have many states with same k with different kx, ky and kz. But an electron in a particular direction (wavevector) cannot be fully delocalized across an entire macroscopic solid or is it ? $\endgroup$ Sep 27, 2022 at 5:24
  • $\begingroup$ @Examination12345 There is no notion of "far apart" for Bloch states, because they are delocalized across the entire crystal. A Bloch state $\psi_{n\mathbf k}$ is specified by a band index $n\in \mathbb Z$ and a (vector!) quasimomentum $\mathbf k$ which lies in the (first) Brillouin zone. There's no reason that two Bloch states with different quasimomenta $\mathbf k$ and $\mathbf k'$ cannot have the same energy (in the free electron model, this occurs when $|\mathbf k|=|\mathbf k'|$) which is why the degeneracy of a given energy can be arbitrarily large. $\endgroup$
    – J. Murray
    Sep 28, 2022 at 16:24
  • $\begingroup$ @J.Murray I had read your answer on the normalization of the bloch waves. But still wouldn't the fact that they are not normalizable due to their periodic nature imply that they cannot be actual states of the crystal ? However the superposition of different bloch waves with varying phases can create a localized wave packet that is normalizable within its limited length. So doesn't that mean that the electron wavefunctions are localized wavepackets and dont extend throughout the crystal (at least not very much) even though the bloch waves do ? $\endgroup$ Sep 28, 2022 at 18:02
  • $\begingroup$ @Examination12345 We typically avoid the technical issues with normalizability by putting our system in a box of length $L$ and imposing periodic boundary conditions, and then taking the limit as $L\rightarrow \infty$ to simplify expressions. It is in this limit that the energy bands become continuous, the Bloch waves become non-normalizable, and the degeneracy of a generic energy level goes to infinity. If you maintain an arbitrarily large but finite $L$, then the Bloch waves are normalizable and we don't need to worry about wave packets (at least in this context). $\endgroup$
    – J. Murray
    Sep 29, 2022 at 1:17

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