Mathematical understanding of band energy Let me first give a sketch of how I understand band energy mathematically. It is not exactly rigorous, but probably could be made rigorous under suitable conditions.
Let $H$ denote the Hamiltonian on an $d$-torus (finite volume with period boundary conditions), i.e., a self-adjoint operator $H:L^2 \rightarrow L^2$ on $L^2 = L^2(\mathbb{T}^d)$. Let $H$ be translationally invariant with respect to a Bravais lattice, i.e., if $\mathscr{A}$ is a Bravais lattice and $T:\mathscr{A} \rightarrow GL(L^2)$ denotes its naturally induced action on $L^2$ (so that it is unitary), then $T(R)HT(R)^{-1}=H$ for all $R\in \mathscr{A}$. Then each eigenspace of $H$ is invariant under $\mathscr{A}$ and thus must be decomposable into irreps of $\mathscr{A}$ (each of which 1-dim). Notice that every irrep of $T$ can be characterized by a wave vector $k$ in the first Brillouin zone $\mathscr{B}$. Therefore, the irreps are spanned be (orthonormal) vectors which can be labeled by $|E, k, (n)\rangle$, where $E$ denotes the energy, $k$ the wave vector and $(n)$ denotes possible degeneracy (in the decomposition of irreps). This is basically Bloch's theorem.
Now for simplicity, let's assume that the eigenvalues of $H$ (energy levels) are discrete $E_1,E_2,...$ and nondegenerate so that $|E_1,k_1\rangle$ is the basis for $E_1$-eigenspace and so on. If $k_1,k_2,...$ were all distinct, then we would have a single band energy, described by $\epsilon (k) = \sum_1^n E_n \delta (k-k_n)$.  If $k_1,k_2,...$ we not all distinct, then we would have multiple band energies.
Here is my question, which I divide in 2 parts:


*

*Is this understanding correct? Or is there a better way?

*Even in the case where $k_1,k_2,...$ are all distinct, it seems possible that $k_1,k_2,...$ is a proper subset of all wave vectors in the first Brillouin zone (if not, why?). Then how would such a band energy be well-defined? This problem gets worse when $k_1,k_2,...$ are not distinct, since for multiple band energies to be well-defined on the first Brillouin space, the sequence of $k_1,k_2,...$ must have exactly an integer number of each wave vector in the first Brillouin zone (which I cannot see why is true in a general setting).

 A: I think I finally understand how to mathematically interpret band energies. Please correct my if you think otherwise.
For simplicity, let me consider a 1D finite system so that the underlying single-particle Hilbert space is $\mathscr{H} \equiv L^2[0,N)$ where $N\ge 1$. Let us consider a periodic lattice (corresponding to the points $R=0,1,...,N-1$) in our finite system so that the translation group acts on $\mathscr{H}$ naturally, i.e., $(R\cdot \varphi)(r) =\varphi(r-R)$ where it's always mod $N$. Let the system be described by a Hamiltonian $H$ which commutes with this action $T$.
By Peter-Weyl, we see that $\mathscr{H}$ can be decomposed into the direct sum of (necessarily) 1-dim irreps of the translation group $T$, i.e., $\mathscr{H} = \bigoplus V$ where $V$ are irreps of $T$ characterized by some wave vector $k=2\pi/N \cdot \{0,...,N-1\}$. Let us group the 1-dim irreps corresponding to the same wave vector as a larger subspace, i.e., $W_k =\bigoplus 'V$ where the direct sum is over all $V$ that correspond to the wave vector $k$. Then we see that
$$
\mathscr{H} = \bigoplus_k W_k
$$
Since the Hamiltonian $H$ commutes with the action $T$, we see that it must map $W_k$ into $W_k$, i.e., if $1_k$ denotes the projection onto $W_k$, then $1_p H1_q=0$ for $p\ne q$. Therefore, it is "block-diagonal" and thus we can diagonal $H$ by diagonalizing its restriction on $W_k$, i.e., $H \psi_k =E_k \psi_k$ for $\psi_k\in W_k$.
In my original question, there were 2 problems with this logic.

*

*What is the guarantee that the decomposition $\mathscr{H} = \bigoplus_k W_k$ contains every $k$ in the first Brillouin zone? If not, then $E_k$ may be ill-defined for a particular $k$.

*How do we know that each $W_k$ has the same dimension, i.e., $\dim W_k = \dim W_p$ for distint $k,p$? If not, then certain bands may be ill-defined. For example, if $\dim W_k =1$ while $\dim W_p=2$ for all $p\ne k$, then it's clear that the one band of energy must be ill-defined.

I think I now know the mathemaitcally rigorous resolution to the above problem. The key insight is that $L^2[0,N)$ has an orthonormal basis consisting of plane waves, i.e., $\varphi_k (x) =e^{-ikx}$ where $k \in 2\pi \mathbb{Z}/N$. It's then easy to check that $W_k$ has the orthonormal basis $\varphi_{k+ 2\pi m}$ where $m$ varies through $\mathbb{Z}$. Therefore, the decomposition contains all possible $k$ in the first Brillouin zone (i.e., all $W_k$ are nonzero), and all the $W_k$ are of the same dimension. Thus, the band energy is rigorously defined.
A: Your symmetry argument is completely general, but the part about the discrete energy levels assumes non-interacting atoms - you need to include hopping between them in order to get a band structure. This is called tight-binding model as opposed to the nearly free electrons model. A good way of understanding the two is working through the exactly solvable Kronig-Penny model, which contains both limits.
