# Bloch's Theorem without periodic boundary condition - mathematically rigorous way

I am looking for a proof of Bloch's Theorem which does not use periodic boundary conditions. Sometimes one happens to see non-rigorous demonstrations of Bloch's Theorem without the use of periodic boundary conditions. As this matter already confused me to sufficient extent, I am interested in all the mathematical details.

## Recall: Bloch's Theorem with the use of periodic boundary conditions:

Consider a cell of $N_i$ atoms in $\mathbf a_i$-direction, $\text{Cell} = \{ \sum_{i=1}^{3} \lambda_i \mathbf a_i \vert 1\leq i \leq3: 0\leq \lambda_i \leq N_i\}$ with one atom at each element of $\left\{ \sum_{i=1}^3 (n_i+\frac{1}{2})\ \mathbf a_i \vert 1\leq i\leq 3:n_i \in \mathbb N_1^{N_i}\right\}$.

Take as your Hilbert space the space of square-integrable functions over such cell $L^2(\text{Cell})$ and employ periodic boundary conditions.

Let $T_{\mathbf R_{\mathbf n}}$ denote the translation by $n_1$ atoms in $\mathbf a_1$-direction, $n_2$ atoms in $\mathbf a_2$-direction and $n_3$ atoms in $\mathbf a_3$-driection.

As far as I understand, here, the assumption of periodic boundary conditions helps us in threefold way.

1. The spectrum of our Hamiltonian $H$ is fully discrete and to each value $E$ contained in the spectrum $\sigma(H)$ there exists a finite-dimensional eigenspace $V_E$. (Btw, what is the rigorous argument for finite-dimensionality here?)
2. These eigenspaces $V_E$ actually lie in our Hilbert space of square-integrable functions over the cell in question $L^2(\text{Cell})$.
3. The Translation group $T$ becomes a finite group.

Due to the commutation relation $[H,T_{\mathbf R_{\mathbf n}}]=0$ we know that the representation of $T$ is invariant on $V_E$. Thus on $V_E$, the representation of $T$ is a finite-dimensional complex representation. Remember that $T$ is abelian, and therefore all finite-dimensional complex irreducible representations are one-dimensional. Furthermore, now that in this case $T$ is finite, any finite-dimensional complex representation is completely reducible.

Therefore I can find a basis $\psi_1,\dots,\psi_{\dim(V_E)}$ of $V_E$ which not only is diagonal in $H$ but also in all $T_{\mathbf R_{\mathbf n}}$, i.e. $$\forall \mathbf n \in \mathbb{N}_1^{N_1} \times \mathbb{N}_1^{N_2} \times \mathbb{N}_1^{N_3}: 1\leq l \leq \dim(V_E): \exists \lambda_{\mathbf n, l} \in \mathbb{C}: T_{\mathbf R_{\mathbf n}} \psi_l = \lambda_{\mathbf n,l} \psi_l\ .$$

Which then leads quickly to Bloch's Theorem, together with the periodic boundary conditions.

## Bloch's Theorem without the use of periodic boundary conditions?

I have the feeling that we are running into trouble here. Let me sketch the problem to the point that I understand so far:

Consider as Hilbert space the space of square-integrable functions $L^2(\mathbb{R}^3)$.

Note:

1. Our Hamiltonian $H$ has no eigenvectors in $L^2(\mathbb{R}^3)$ and thus no discrete part of the spectrum $\sigma(H)$.
2. Even after extending to a rigged Hilbert space with the tempered distributions $S^*(\mathbb{R}^3)$ as extension, our generalized eigenspaces $V_E \subset S^*(\mathbb{R}^3) \setminus L^2(\mathbb{R}^3)$ to our generalized eigenvalues (values of our continuous spectrum $\sigma(H)$) seem all to be infinite-dimensional. (Is there a formal argument to make this point?)
3. While the representation of the Translation group $T$ is unitary on $L^2(\mathbb{R}^3)$, and finite-dimensional unitary representations are completely reducible, I don't see how the representation of $T$ could be taken as unitary on $S^*$. (However, $V_E$ lies entirely in $S^*\setminus L^2$.)

So it is definitively not possible to decompose the generalized eigenspaces $V_E$ along the same line as seen in the previous section.

Note:

• If all $V_E$ were finite-dimensional and in $L^2$, then by unitarity of the representation of $T$ on $L^2$, this representation would be completely reducible. But neither is $V_E$ finite-dimensional nor contained in $L^2$.

## Questions:

What is the appropriate way to state that I can decompose my infinite-dimensional eigenspaces $V_E$ in some continuous way?

I suppose that it is possible to write any generalized eigenstate to generalized eigenvalue $E$ as some integral $$\psi(\mathbf r) =\ "\int_{\text{some domain}} \mathrm{d}^3 \mathbf k\ \psi_{\mathbf k}(\mathbf r)" ,$$ where the $\psi_{\mathbf k}$ are generalized eigenstates of $H$ to generalized eigenvalue $E$ and eigenstates of all Translations.

I am interested in how to make this symmetry argument rigorous and what the right mathematical framework to employ here is. I am naively thinking of the spectral theorem here and projection-valued measures, given the integral above, is there some connection to it?

In general I am also fine with any rigorous argument that yields Bloch's Theorem without the assumption of periodic boundary conditions, or likewise an argument that shows that Bloch's Theorem is not extendable to the case of no periodic boundary conditions.

## Edit 1:

The last part of my post speculating about an integral representation of any $\psi \in V_E$ seems to be utter nonsense. (I was supposing there might be some concept of uncountable basis with then unique "uncountable sums" to represent each element...)

The only extended notion of basis there is, as far as I see, is a Schauder basis which allows us to express any element as a countable linear combination of some countable collection that we call Schauder-basis, if and only if the space in question is separable.

Well, I guess some problem to it is that $S^*$ is not even a normed space, and therefore I can't even ask whether there exists a Schauder-basis to it along the same line as I would do for a Hilbert space(?) Yet I think in the framework of rigged Hilbert spaces $S^*$ comes with some induced topology and therefore one can at least address the question of whether $S^*$ (or some $V_E$ subspace if it) is topologically separable.

Independent of $S^*$ (or some $V_E$ subspace if it) being separable, there is an uncountable Hamel-basis to the eigenspaces $V_E$ but then any state can be represented as linear combination of a finite subset of that Hamel-basis.

So not knowing whether the $V_E$ are separable, a more well-defined question one might ask is whether there exists a Hamel-basis $\{\varphi_i \}_{i\in I}$ of $V_E$ such that each $\varphi_i$ is eigenvector of all translations.

## Edit 2:

Clearly, for the existence of a Schauder-basis completeness is a necessary condition. Luckily, as $S^*$ is a topological vector space and thus a uniform space, we can talk about completeness of it. However we'd be left to show that a certain $V_E$ in question is complete and separable in order to use a Schauder-basis.

Independent of that, it's possible to ask for a Hamel-basis with the above required property. So I guess I should focus on that.

(Seems like I need to properly learn about functional analysis in order to advance by myself...)