# What is Bloch's theorem really saying (isomorphism between Bloch eigenbasis and $k$-space)?

(Throughout this development, I neglect spin.)

Bloch's theorem, strictly speaking (according to Ashcroft and Mermin), says that given a (one-electron) Hamiltonian $$\hat{H} = \frac{\hat{\mathbf{P}}^2}{2m} + \hat{U}(\hat{\mathbf{r}})$$ with a potential which is periodic with respect to some Bravais lattice ($$U(\mathbf{r} + \mathbf{R}) = U(\mathbf{r})$$ for all $$\mathbf{R} \in BL$$), there exists an eigenbasis of $$\hat{H}$$, $$E = \{|\psi \rangle \}$$, for the relevant Hilbert space which is such that, in the coordinate representation, each $$|\psi \rangle$$ obeys $$\psi(\mathbf{r} + \mathbf{R}) = e^{i\mathbf{k} \cdot \mathbf{R}} \psi(\mathbf{r})$$ for some $$\mathbf{k} \in \mathbb{R}^3$$.

The (typical) proof of this fact uses that $$\hat{H}$$ and $$\hat{T}_{\mathbf{R}}$$ commute for all $$\mathbf{R}$$ (the $$\hat{T}_{\mathbf{R}}$$ are translation operators by $$\mathbf{R}$$) and, since these are all normal operators, there therefore exists a joint eigenbasis. One can then show that it's possible to associate some $$\mathbf{k} = \sum_i x_i\mathbf{b}_i$$ for $$\mathbf{b}_i$$ the primitive vectors of the given reciprocal lattice. In effect one has shown that each $$\psi$$ corresponds to some $$\mathbf{k} \in \mathbb{R}^3$$ as stated.

However, after developing this, all of the books I have read (Ashcroft and Mermin, as well as Girvin and Yang) seem to take this as proof that there exists a bijection between k-space ($$\mathbb{R}^3$$) and the joint eigenbasis, but I can't understand why. Up to this point, we've not even shown that the map so defined is an injection, let alone a surjection. Indeed, we've just proved that there is a well-defined map from the joint eigenbasis to k-space.

• I find it hard to believe that "all the books ... take this as proof that there exists a bijection between $\mathbf k$-space and the joint eigenbasis." In my understanding, the relationship straight up isn't a bijection in real crystals. You should have multiple eigenstates for each $\mathbf k,$ each state living in a different band. Having a bijection would limit you to one band. Do you have a specific example that makes you think the authors claim otherwise?
– HTNW
Jul 3 at 1:03
• @HTNW What you say is a further addition on top; one restricts k-space to a particular primitive cell of the reicprocal lattice (usually the first Brillouin zone) and in so doing the $n$ label arises. I am talking about the structure before that step; when the choice of restriction has not yet been made there is indeed (it seems to me) a bijection with all of k-space ($\mathbb{R}^3$).
– EE18
Jul 3 at 3:24
• It simply was never about $\mathbb R^3$ k space, but rather only ever the 1BZ. The case of it being extendible to $\mathbb R^3$ is rather a special case, that makes sense for empty lattice in the extended zone scheme representation. It would not be so obvious what it means in the general case. Jul 3 at 3:59
• The $\mathbf k$ produced in the proof is, from the very start, only defined up to reciprocal lattice translation (in my copy of Ashcroft+Mermin, $x_i$ in (8.18) $c(a_i)=e^{2\pi i x_i}$ is, from the start, ambiguous up to an integer). The Bloch construction doesn't even define a function from eigenstates to $\mathbf k\in\mathbb R^3,$ much less a bijection. It's only a function once you quotient $\mathbf k$ by the reciprocal lattice. And, again, the fact is that there is not necessarily a bijection. Can you point out what specifically makes you think there is one, or that A+M think there is?
– HTNW
Jul 3 at 4:03
• If this is not answered by the end of the day (my day just started), I'll type up a big answer. I think this is a big misunderstanding. Jul 3 at 4:05

Completely forget about Bloch's theorem for a moment. Given any choice of a $$d$$-dimensional Bravais lattice $$\Lambda$$, the Hilbert space $$L^2(\mathbb R^d)$$ can be decomposed as the direct integral $$L^2(\mathbb R^d)\simeq \int_{\Gamma}^{\oplus} L^2_\mathbf k(\mathscr u) \mathrm \ d\mu(\mathbf k)$$

where $$\mathscr u$$ is a choice of primitive unit cell for $$\Lambda$$ and $$\Gamma$$ is the Wigner-Seitz unit cell of the inverse lattice $$\Lambda^{-1}$$, also called the Brillouin zone. This is a matter of kinematics, not dynamics; it doesn't matter what the Hamiltonian is, or how it relates to $$\Lambda$$; it is simply the statement that an arbitrary square-integrable function $$\psi:\mathbb R^d \rightarrow \mathbb C$$ can be expanded as

$$\psi(\mathbf x) = \int_{\Gamma} \mathrm d^dk \ e^{i\mathbf k \cdot \mathbf x}f_\mathbf k(\mathbf x)$$ where, for each $$\mathbf k\in \Gamma$$, the function $$f_\mathbf k(\mathbf x)$$ restricts to a periodic, square-integrable function on $$\mathscr u$$. To show this, we need only consider the Fourier transform. Any square-integrable function can be written as $$\psi(\mathbf x) = \int_{\mathbb R^d} \mathrm d^d\kappa \ e^{i\boldsymbol \kappa \cdot \mathbf x} \phi(\boldsymbol \kappa)$$ In turn, given any $$\Lambda$$ with Brillouin zone $$\Gamma$$, any $$\boldsymbol \kappa \in \mathbb R^d$$ can be decomposed as $$\boldsymbol \kappa = \mathbf k + \mathbf G$$, where $$\mathbf k \in \Gamma$$ and $$\mathbf G$$ is a reciprocal lattice vector. With this decomposition, we find $$\psi(\mathbf x) = \int_\Gamma \mathrm d^d k \sum_{\mathbf G} e^{i\mathbf k \cdot \mathbf x} e^{i\mathbf G \cdot \mathbf x} \phi(\mathbf k+\mathbf G)$$ $$= \int_\Gamma \mathrm d^d k \ e^{i\mathbf k \cdot \mathbf x} \left(\sum_{\mathbf G} e^{i\mathbf G \cdot \mathbf x} \phi(\mathbf k + \mathbf G)\right) \equiv \int_\Gamma \mathrm d^d k \ e^{i\mathbf k \cdot \mathbf x} f_{\mathbf k}(\mathbf x)$$

Again, I have made no reference whatsoever to a Hamiltonian; this is true for any choice of lattice $$\Lambda$$.

With this fact in mind, Bloch's theorem says that if the Hamiltonian $$H$$ commutes with the lattice translation operators corresponding to a lattice $$\Lambda$$, then it factors into the direct integral $$H = \int_\Gamma \mathrm d^dk\ e^{i\mathbf k \cdot \mathbf x} H_{\mathbf k}e^{-i\mathbf k \cdot \mathbf x}$$ such that $$H\psi = \int_\Gamma \mathrm d^d k \ e^{i\mathbf k \cdot \mathbf x} H_\mathbf k f_\mathbf k(\mathbf x)$$ where $$H_\mathbf k$$ is taken to be an operator on $$L^2_\mathbf k (\mathscr u)$$. This implies that the (generalized) eigenvectors of $$H$$ can be written in the form $$e^{i\mathbf k \cdot \mathbf x} u_\mathbf k(\mathbf x)$$ where $$u_\mathbf k$$ is an eigenvector of $$H_\mathbf k$$.

However, after developing this, all of the books I have read (Ashcroft and Mermin, as well as Girvin and Yang) seem to take this as proof that there exists a bijection between k-space ($$\mathbb{R}^3$$) and the joint eigenbasis, but I can't understand why.

You'd have to quote a specific passage for a specific explanation, but no, that is not the case. For a given $$\mathbf k\in \Gamma$$, we can label the eigenvectors of $$H_\mathbf k$$ with natural numbers $$n$$ because the spectrum of $$H_\mathbf k$$ is necessarily discrete, and so we can label a (generalized) eigenvector of $$H$$ with a pair $$(\mathbf k, n)\in \Gamma\times \mathbb N$$, so that constitutes a bijection if you'd like.

You have many misconceptions and are failing to see the rigour that is already being presented. This answer will even just cite A&M, because they have commented upon basically everything that you are having misconceptions on. The equations I am using, if they are labelled in the style of A&M, appear inside A&M.

## Impossibility of $$\mathbb R^3$$$$\vec k$$-space

In problem 1 of Chapter 8, A&M covers a direct counterexample. Figure 8.6 plots out Equation (8.76) and shows that for $$\mathbb R$$ $$k$$-space in a strong potential, there are ranges of $$k$$ where there is no possible solution. Note that Figure 8.6 explicitly plots $$k$$ on the y-axis as only available in 1BZ, whereas the horizontal x-axis is plotting $$K$$ that is really the energy eigenvalue. I take this as that it is futile to insist that there would be a bijection between $$\mathbb R^3$$ $$\vec k$$-space and the joint eigenbasis, because a too strong (or too weak) potential can kill some previously expected states.

## Authorial Intent for $$n^\text{th}$$ Band Labelling

$$\tag{8.6}\text{You have quoted}\qquad\psi(\vec r+\vec R)=e^{i\vec k\cdot\vec R}\psi(\vec r)$$ But this comes by you ignoring \begin{align} \tag{8.3}\psi_{n\vec k}(\vec r)&=e^{i\vec k\cdot\vec r}u_{n\vec k}(\vec r)\\ \tag{8.4}u_{n\vec k}(\vec r+\vec R)&=u_{n\vec k}(\vec r)\\ \tag{8.5}\psi_{n\vec k}(\vec r+\vec R)&=e^{i\vec k\cdot\vec R}u_{n\vec k}(\vec r) \end{align} and the footnotes 2 and 3 on that page. You should also look at the General Remarks About Bloch's Theorem, points 2, 3 and 4, which explicitly covers why there is a need for band indexing and where $$\vec k$$ is meant to be considered from.

I am very worried about you misunderstanding what I am going to write next, so I am going to break with A&M notation for a while. The reciprocal lattice vector $$\vec G=n_1\vec b_1+n_2\vec b_2+n_3\vec b_3$$ with $$(n_1,n_2,n_3)\in\mathbb Z^3$$ is labelled as $$\vec K$$ in A&M, which is very easy to confuse with $$\vec k\in$$1BZ i.e. strictly fractional coefficients (real-valued in the continuum limit) rather than integer coefficients. If you add any $$\vec G$$ to $$\vec k$$, bringing them out of 1BZ, then in Equation (8.5) in particular, it is very clear that the $$e^{i\vec G\cdot\vec R}=1$$ by definition. This is thus a redundant description and we do need to limit ourselves to 1BZ if we want to avoid double-counting, when we use the band indexing.

## Information from Second Proof

\begin{align} \tag{8.1 }U(\vec r)&=U(\vec r+\vec R)\\ \tag{8.31}\text{directly implies}\qquad U(\vec r)&=\sum_{\vec G}U_{\vec G}\,e^{i\vec G\cdot\vec r} \end{align} Similarly, $$\tag{8.43&44}\psi_{n\vec k}(\vec r)=e^{i\vec k\cdot\vec r}u_{n\vec k}(\vec r)\qquad\bigwedge\qquad u_{n\vec k}=\sum_{\vec G}c_{n,\vec k-\vec G}\,e^{-i\vec G\cdot\vec r}$$ openly tells you that it is actually that all $$\vec G$$ needs to contribute to each Bloch state. You really cannot distinguish which state is actually which, between the $$\vec k$$ and $$\vec k+\vec G$$, because all of these come together to make one single Bloch state.

## Extended Zone Scheme

If you look at Figure 9.4, A&M explicitly constructs the extended zone scheme, the reduced zone scheme, and the periodic zone scheme. The periodic zone scheme completely does away with even attempting to avoid double-counting. The reduced zone scheme, that all of us are telling you about, is the least confusing in practice, because the 1BZ is completely unambiguous, and you can just use the band index to go up to the energy level you are interested in.

The extended zone scheme is doable in 1D, but if you have more dimensions, it is exceedingly easy to miss or double-count. It might not even be interpretable, and the ease at which it looks in Figure 9.4 is because it is the most simple, 1D weak potential limit. Only in the case that you can guarantee that you have established a bijection from the reduced zone scheme to the extended zone scheme, then, and only then, might you say that you can drop the band index and use the extended zone scheme as the proxy for band indexing. Which, again, might not exist in strong potentials.

I hope that from here onwards, it is clear that Bloch's theorem is actually much more natural and obvious in the reduced zone scheme than in the extended zone scheme that you are looking at. Basically every author is going to want to start with a treatment in reduced zone scheme, because of how the explicit 2nd proof goes. The sum on reciprocal lattice vectors is extremely telling.

But you are also having a misconception of what it is that the argument actually is about. To illustrate that, I have to pick a detour to another system, where we can discuss the details with less confusion.

The most familiar example to try is the Hydrogen atom. The task is to get a basis for the Hilbert space. To do that, we seek a maximally commuting set of operators with the Hamiltonian. For the Schrödinger or Pauli cases, this set will be $$\{\hat{\mathcal H},\hat{L^2},\hat{S^2},\hat{J^2},\hat{J_z}\}$$ with the principal quantum numbers $$(n,\ell,\frac12,j,m_j)$$ to go with them. The equivalents in the Dirac case are $$\{\hat{\mathcal H},\hat{J^2},\hat{J_z},\hat K\}$$ and $$(n,j,m_j,\kappa)$$. In either case, spherical symmetry means that the energy eigenvalues can only depend upon $$n$$ and $$j$$, never $$m_j,\ell$$ or $$\kappa$$, at least until hyperfine corrections and above forces us to do even more work.

There is no way to do an extended zone scheme here. In order to express the full energy eigenvalues, the quantum number $$n$$ is a necessity. It is thus rather weird that you are insisting to do away with it in the case of Bloch's theorem. After all, Figure 9.4 was assuming that there was no lattice at all, just a weird periodic weak potential with one electron in it. Then, and only then, can it have the quadratic that looks so beautiful. In reality, each energy level $$n$$ of an atom, say, will spawn at least one band. It will be horrendous.

Even if we do not go that far, note that it is important to have a maximally commuting set. The consequence of not having a maximally commuting set, is that there will be a failure to distinguish degenerate states. That is, if you only have $$\{\hat{\mathcal H},\hat{J^2}\}$$, then their joint eigenbasis spans the entire Hilbert space with no problems, but then the degenerate subspaces will be confusing messes. There is never any worry that the eigenbasis will be insufficient—by postulation, any single Hermitian (observable) operator has an eigenbasis that spans the entire Hilbert space. The only problem is that it is possible for the degeneracies to cause confusion.

The relevance of the above consideration to the problem at hand, is that we are NEVER trying to establish a bijection" or whatnot. That is simply not the correct logic. The available state space is also not just something you can insist to throw all possible values into. For example, if you choose $$\ell>j$$, then there will be no possible state. Similarly, if you pick $$n=-0.2$$, then obviously there is no state, even if $$\vec k\in$$1BZ. That is, nobody proving Bloch theorem should be burdened with stating that not all combinations of the stuff is possible. That $$\vec k+\vec G$$ is 1) conditionally well-defined in the extended zone scheme by replacing band indexing or 2) illegal in the reduced zone scheme or 3) double-counted in periodic zone scheme, is actually outside the scope of Bloch's theorem.

In fact, Bloch theorem explicitly does not cover spin, and so it is manifestly not a maximally commuting set of operators with the Hamiltonian.

• A few comments: (1) Figure 8.6 shows that there are disallowed regions of the energy, not of $k$. The fact that that plot shows only $k \in BZ1$ is simply because they have performed the common restriction to $BZ1$ so as for the "band" picture to emerge. My question is about the step before this. (2) I here copy a comment from above: Consider (8.27) for example. But also from (8.20) there is clearly no restriction on $x_i$ (other than their being real since the translation operators are unitary). Further, if you examine the second proof it is clear that up to (8.39), $\mathbf{q}$...
– EE18
Jul 4 at 18:54
• ... is arbitrary (save for the macroscopically periodic boundary condition quantization). It is only later that we decide to do the conventional restriction to a reciprocal lattice primitive cell. I understand the restriction to $BZ1$ is to avoid double-counting, but that is only after we have done the restriction in the first place in order for the index $n$ to emerge. In this question I am asking about the "step" before then; when we first show that there is a correspondence between the Bloch eigenstates and k-space. Unless I am misunderstanding, it seems like a lot of people are...
– EE18
Jul 4 at 18:55
• ...skipping this step and simply pointing me to band structure with $(n,\mathbf{k})$, $\mathbf{k} \in BZ1$. I am comfortable with this but am asking about the derivation of this description which, again, I believe goes through first considering $\mathbf{k} \in \mathbb{R}^3$ (i.e. no $n$ index) and only later doing the restriction to $BZ1$ so that the $(n,\mathbf{k})$, $\mathbf{k} \in BZ1$ picture arises.
– EE18
Jul 4 at 18:57
• No, it is the other way around. You start with 1BZ and n bands, and unfold to get $\mathbb R$ and rid of n. The case of H atom makes it clear that you start with restrictions on the available state space, not the other way around. Jul 4 at 20:46

Technically, if $$\Lambda$$ is the original real lattice and $$\Lambda^*$$ its corresponding dual lattice defined by: $$\Lambda^*=\{k\in\mathbb R^3|\forall x\in\Lambda, k\cdot x \in2\pi\mathbb Z\}$$ then the lattice momentum does not lie in $$\mathbb R^3$$ but rather $$\mathbb R^3/\Lambda^*$$. Basically, it means that they are defined up to an element of the dual lattice, or equivalently, you need to restrict to the Brillouin zone (unit domain of the dual lattice).

However, there is typically a countable number of bands at each crystal momentum. At fixed crystal momentum $$k$$, you have the Hamiltonian: $$H(k) = \frac{(p-ik)^2}{2m}+U$$ with periodic boundary conditions for $$\psi$$. Your domain is compact and under some assumptions on $$U$$, the spectrum of $$H(k)$$ is discrete by applying arguments from functional analysis.

This is to be matched with the countable number of BZ to pave $$\mathbb R^3$$. Thus it is possible to construct a bijection between $$\mathbb R^3$$ and the energy eigenstates. Actually, mathematically, just by cardinality arguments, you already know that a bijection exists. The issue is whether this bijection is natural.

It is not always so. There is one special case when correspondence easily constructible which is the nearly free electron model i.e. when the potential is small. You usually treat $$U$$ as a perturbation and the energy eigenstates are constructed by perturbing momentum states. This gives you a natural correspondence between the two.

In this case, it the paving of $$\mathbb R^3$$ by BZ is natural by increasing value of $$p^2$$. The closest one to the origin is the first BZ (Wigner-Seitz cell). Then the next ones are layered around it like an onion.

While there is a natural bijection in the bulk of these BZ cells, things get tricky at the boundaries. Indeed, you need to apply degenerate perturbation theory there so the original momentum eigenstates get mixed up, typically in “equal amounts” (according to Born’s rule). Therefore, the bijection will be a arbitrary at the boundaries and loses its naturalness.

In order to extend this correspondence in general, you can always increase adiabatically the strength of the periodic potential. This will allow you to define a correspondence at finite value of the potential (that os only natural in the bulk of the BZ cells in momentum space).

While the result is true, I don’t think that it is particularly useful outside the nearly free case. I think it was probably presented as a “sanity check” to show that the number of energy eigenstates of the free Hamiltonian $$p^2/2m$$ matches the one with a periodic potential (didn’t read the book so this is speculation).

Hope this helps.

• I agree with you that eventually one arrives at the picture of discretely many states at given $\mathbb{k}$, but what I would say is that the rigorous arrival at this picture is not often painted (perhaps I need to consult a group theory text) and, to the extent that it is arrived at in Ashcroft and Mermin, I believe it is arrived at by first oging through a picture where $\mathbb{k}$ ranges throughout all space (with no $n$ label), and then the restriction to BZ1 happens and the $n$ label emerges.
– EE18
Jul 3 at 17:58
• I don't quite get what you are looking for. Do you want more details in how the energy spectrum is discrete at fixed crystal momentum? The approach you describe is the one I outlined when starting from the nearly free model, but it is not always relevant.
– LPZ
Jul 5 at 16:37

The translational symmetry $$T_R$$ that this theorem implies is not exact, but we will keep the same idea. In a perfect crystal $$\hat{H}$$ is also likely to commute with symmetry elements such as rotations $$C_n$$ and reflexions $$S_{2n}$$ or inversion in the crystal group. If the crystal has the common case $$C_2$$ symmetry, $$[\hat{H},C_2]=0$$, the local potential is also $$C_2$$ invariant $$U(\textbf{r}-\textbf{R})=U(\textbf{r})$$, $$\textbf{k}$$ and $$-\textbf{k}$$ have the same eigenvalues : $$\psi$$ is not bijective in the k-space. By using the general form $$\psi_{\textbf{k}}(\textbf{r})=e^{i\textbf{k}.\textbf{r}}u(\textbf{r})$$. If we use the map : $$r \rightarrow -r$$, $$\psi_{-\textbf{k}}(\textbf{r}) = \psi_{\textbf{k}}(\textbf{r})$$ even in a reduced scheme $$\textbf{k} \in BZ1$$.

Furthermore, there is $$\textbf{k} \in BZ1$$ / $$\textbf{k}'=\textbf{k}+\textbf{K}$$, $$\psi_{\textbf{k}'} \in BZ1$$. For a given $$\textbf{k} \in BZ1$$ there are several bands $$\psi_{n\textbf{k}}$$ with a band index $$n$$, $$\psi_{n\textbf{k}}$$ is not an injection in the k-space reduced to $$BZ1$$.

• See LPZ answer; I am quite certain that there is a bijection.
– EE18
Jul 3 at 17:58