Lecture 3: Translation Symmetry and the Bloch Eigenproblem

Quantum condensed matter becomes tractable by organizing an enormous state space before approximating its dynamics. The object-and-basis discipline established earlier now meets a spatial symmetry: lattice translations partition a one-electron problem into crystal-momentum sectors. That move supplies the kinematic scaffold for bands, but not their microscopic couplings, occupations, or experimental intensities. Later discussions of Fermi surfaces, response, quasiparticles, and topology all reuse this scaffold while changing what is computed over it. Symmetry first reduces the possible states; dynamics and preparation then decide what the material actually does.

In a calculated band plot for crystalline silicon, two curves may approach or meet near a zone boundary. The picture alone does not tell us whether the meeting is a physical degeneracy, an avoided crossing too small to resolve, or the same dispersion redrawn after a larger unit cell folded it into a smaller Brillouin zone. The distinction matters: a real gap changes which low-energy electronic states exist, whereas a change of plotting convention does not.

The ambiguity already appears in the simplest possible anchor. Take the free- electron dispersion ${E_{\mathrm{free}}(q)=\hbar^2q^2/(2m)}$ and organize its states using translations of spacing ${a}$ . The original wave number ${q}$ and its reduced-zone representative ${k}$ are different labels: a branch index must accompany ${k}$ to keep track of the original state. Distinct free states can meet in energy at a zone boundary; a microscopic coupling can then make them repel. We will follow this pair throughout the lecture rather than introducing a new example for every distinction.

We will derive the translation label and cell eigenproblem for the folded free-electron anchor, then separate a change of representation from the nonzero microscopic coupling that can open a gap.

Our governing question is

How does lattice translation produce the labels and state spaces behind a band plot, and what information beyond those labels decides whether two displayed branches truly cross or open a gap?

The core model is deliberately one-particle and spinless: one electron moves in a lattice-periodic potential. For that declared Hamiltonian, translation symmetry produces independent crystal-momentum blocks whose eigenvalues can be defined as energy bands. In an interacting crystal, a total translation label can survive even though assigning a separately conserved label and band energy to every electron generally fails; we will verify that boundary only after the one-particle object has been constructed.

For the opening figure use ${G=2\pi/a}$ , the half-open first zone ${-\pi/a\leq k<\pi/a}$ , and energy unit ${E_{\mathrm R}=\hbar^2\pi^2/(2ma^2)}$ . Before reading Figure 3.1, take the marked original wave number ${q=-1.5\pi/a}$ . Predict the representative ${k=q+G}$ , and the height at which the same state must appear if only its plotting label changes.

Figure 3.1. One selected segment of the extended free-electron parabola is relabeled by {k=q+G} inside the first zone, preserving each state’s energy. Solid and dashed amber curves are two depictions of the same selected states, not extra states. The open positive boundary excludes a representative, not a physical state. No coupling is added.

The horizontal arrow changes the plotting coordinate, not the electron’s physical momentum or energy. The marked state moves from the plotted coordinate ${q=-1.5\pi/a}$ to the representative ${k=0.5\pi/a}$ at the unchanged height ${E/E_{\mathrm R}=2.25}$ . Its branch is ${E_{-1}(k)=E_{\mathrm{free}}(k-G)}$ . The different free state with original wave number ${q=k}$ belongs to ${E_0(k)=E_{\mathrm{free}}(k)}$ ; at the same representative it has ${E_0/E_{\mathrm R}=0.25}$ . Sharing a translation label does not make two states identical or force their energies to agree.

The figure follows only the exterior segment ${-2\leq qa/\pi<-1}$ , which maps to ${0\leq ka/\pi<1}$ ; it is not the complete folded spectrum. The included negative boundary is not mapped to the excluded positive endpoint. The original parabola still continues through ${q=\pi/a}$ ; the open circle concerns the reduced representative only. Later evaluations at ${k=\pi/a}$ use the boundary limit. We will derive the branch equality and then the additional coupling that can split it, after constructing the translation blocks.

Silicon supplies the physical motivation. The toy model isolates translation kinematics; a realistic silicon calculation additionally requires its orbital content and microscopic potential. Diffraction identifies reciprocal-lattice vectors through momentum transferred to the crystal. Angle-resolved photoemission removes an electron and plots an energy against an inferred crystal-momentum representative, but its brightness also depends on orbital content and probe matrix elements. Thus the distinction between a translation label, an energy, and mechanical momentum is experimentally consequential. Within a declared one-particle model it controls whether a semiconductor has a gap and whether an avoided crossing exchanges the character seen by light, strain, or a surface probe.

The shortest path through the argument is translation sector, character, Bloch form, cell eigenproblem, and finally the distinction between folding and coupling. The finite-group and functional-analytic statements certify that path, while the changed-cell count checks that it is representation independent. A reader may return to those certifications after first following the causal spine, but none of their assumptions is optional.

Try three short predictions without looking up a formula.

A state is translated through one lattice period ${a}$ . Must its translation eigenvalue equal one, or need only its magnitude equal one?
Two diagonal levels meet in a ${2\times2}$ Hermitian matrix. Which matrix entry must be examined before deciding that the eigenvalues cross?
A wave vector is shifted by ${2\pi/a}$ in a one-dimensional lattice. Which phase measured by a lattice translation remains unchanged?

Record all three answers before the instructor comparison. Keep the reasons with the predictions: the derivations will let us test each one separately.

Translation symmetry on a finite periodic crystal

The band description begins with a constraint, not with a plotted curve. Exact lattice translation symmetry decomposes the one-particle state space into character sectors, and crystal momentum is the label of those sectors modulo reciprocal equivalence. This conclusion does not require a tight-binding picture and does not identify crystal momentum with mechanical momentum. Translation-breaking boundaries or disorder remove the exact decomposition; finite periodic samples retain it. Translation-invariant interactions preserve a total translation character rather than independent one-electron labels. Those changes mark the domain of the Bloch description instead of making it merely an approximate drawing of a solid.

Translation operators and common symmetry sectors

To decide whether the relabeling in the opening question changes a state, we need translation eigenvalues that can be counted without continuum normalization obscuring them. A finite periodic sample supplies that clean setting: it makes the translation group finite, gives exactly one allowed crystal-momentum label per unit cell, and can be enlarged after the symmetry argument is complete. This is why the boundary condition below is a teaching and counting device rather than an incidental technical assumption.

Let ${d\in\mathbb N}$ be the spatial dimension, let ${\Lambda\subset\mathbb R^d}$ be a Bravais lattice, and let ${\Omega}$ be a finite sample with Born–von Karman periodic boundary conditions compatible with ${\Lambda}$ . We neglect spin in Lectures 3–4. The one-particle Hilbert space is therefore ${\mathcal H=L^2(\Omega,\mathbb C)}$ , and a position-space wave function is represented by a map ${\psi:\Omega\to\mathbb C}$ with ${\psi\in\mathcal H}$ . Ordinary vectors are bold upright: position satisfies ${\mathbf r\in\Omega}$ , and a lattice translation satisfies ${\mathbf R\in\Lambda}$ . Hats distinguish operators acting on ${\mathcal H}$ .

A later representation check will specialize this setting to a square lattice in ${d=2}$ . Its purpose is the same as the one-dimensional opening anchor: we will subtract a reciprocal-lattice vector from a displayed point and ask whether this identifies a physical state or only its translation sector. The one-dimensional pair remains the calculation spine; the square lattice tests whether the reasoning transfers beyond an interval.

For a free particle, every continuous displacement is a symmetry and the generator of those displacements, canonical momentum, is conserved. A crystal retains only a discrete subgroup. If ${V}$ is differentiable, then for a Cartesian component ${j\in\{1,\ldots,d\}}$ ,

$[\hat p_j,\hat H] =-\mathrm i\hbar\,\partial_j V(\hat{\mathbf r}),$

which is generally nonzero. Thus an energy eigenstate in a crystal need not be a momentum eigenstate. It would nevertheless be wasteful to discard all translation information: displacement by an entire lattice vector still leaves the Hamiltonian unchanged. Crystal momentum will emerge as the label of that surviving discrete symmetry, not as a renamed expectation value of ${\hat{\mathbf p}}$ .

The modern organizing object is therefore not the plotted curve by itself. Translation symmetry decomposes the one-particle Hilbert space into representation sectors labeled by reciprocal classes ${[\mathbf k]}$ . Within each sector the Hamiltonian has eigenspaces (or, at degeneracy, spectral subspaces); an energy band records how those state spaces and their eigenvalues vary across the reciprocal quotient. A chosen vector ${\mathbf k}$ is only a representative of its class. This distinction is what allows us to say precisely why folding changes labels while a new matrix element changes the Hamiltonian.

For each ${\mathbf R\in\Lambda}$ , let ${\hat T_{\mathbf R}:\mathcal H\to\mathcal H}$ be the unitary translation operator defined by

$(\hat T_{\mathbf R}\psi)(\mathbf r) =\psi(\mathbf r-\mathbf R).$

The sign is a convention fixed by the statement that a localized packet moves from ${\mathbf r_0}$ to ${\mathbf r_0+\mathbf R}$ . Successive translations add,

$\hat T_{\mathbf R}\hat T_{\mathbf R'} =\hat T_{\mathbf R+\mathbf R'}, \qquad \hat T_{\mathbf R}^{\dagger} =\hat T_{-\mathbf R}.$

Both statements can be checked on a wave function. Composition gives ${(\hat T_{\mathbf R}\hat T_{\mathbf R'}\psi)(\mathbf r) =\psi(\mathbf r-\mathbf R-\mathbf R')}$ . For the norm, change variables to ${\mathbf s=\mathbf r-\mathbf R}$ :

$\|\hat T_{\mathbf R}\psi\|_2^2 =\int_{\Omega}|\psi(\mathbf r-\mathbf R)|^2\,\mathrm d^d r =\int_{\Omega}|\psi(\mathbf s)|^2\,\mathrm d^d s =\|\psi\|_2^2.$

The middle equality uses the periodic identification of the sample, so the shifted integration cell covers the same torus. Since ${\hat T_{-\mathbf R}}$ is the inverse, norm preservation gives unitarity and the displayed adjoint. A hard-walled finite sample would fail this boundary step; that is why its translation problem is different.

Let ${\hat{\mathbf r}}$ and ${\hat{\mathbf p}}$ be the position and canonical momentum vector operators on ${\mathcal H}$ . No vector potential is present, so canonical momentum is also mechanical momentum in this model. Let ${V:\Omega\to\mathbb R}$ be a bounded real-valued potential-energy function satisfying ${V(\mathbf r+\mathbf R)=V(\mathbf r)}$ for every ${\mathbf R\in\Lambda}$ , and let ${m>0}$ be the particle mass. The single-particle Hamiltonian is

$\hat H=\frac{\hat{\mathbf p}^{\,2}}{2m} +V(\hat{\mathbf r}), \qquad V(\mathbf r+\mathbf R)=V(\mathbf r),$

with operator domain ${\mathcal D(\hat H)=H^2_{\mathrm{per}}(\Omega)}$ . Here the subscript means that the function and its periodic traces obey the Born–von Karman boundary identifications. Under the bounded-real-potential hypothesis, ${\hat H}$ is self-adjoint on this declared domain. That conclusion is what the symmetry argument consumes.

On the finite periodic sample, the periodic Laplacian ${-\hbar^2\nabla^2/(2m)}$ is self-adjoint on ${H^2_{\mathrm{per}}(\Omega)}$ . Multiplication by the bounded real function ${V}$ is a bounded self-adjoint operator on ${L^2(\Omega,\mathbb C)}$ , since

$\lVert V\psi\rVert_2 \leq \lVert V\rVert_\infty\lVert\psi\rVert_2 \qquad \text{for every }\psi\in L^2(\Omega,\mathbb C).$

Thus ${V}$ has relative bound zero with respect to the periodic Laplacian, and the bounded-perturbation theorem makes their sum ${\hat H}$ self-adjoint on the same domain [Kuc16, arXiv v4 evidence copy, manuscript §3, pp. 17-19, Eq. (3.1) and the paragraph declaring domain H²(R^n) and self-adjointness; numbering collated by work identity, not by a line-by-line journal comparison; Tes14a, Second edition, printed pp. 157-159, §6.1, definition (6.1), Lemmas 6.2-6.3, and Theorem 6.4, official errata p. 5]. This precision statement justifies the common-eigenvector argument below; its proof is not a prerequisite for the character calculation.

Specialize to a ring of ${N\in\mathbb N}$ cells, with ${L=Na}$ , and ${V(x)=2V_c\cos(Gx)}$ , where ${V_c\in\mathbb R}$ has units of energy and ${G=2\pi/a}$ . For ${\hat H=-\hbar^2\partial_x^2/(2m)+V(x)}$ and ${(\hat T_a\psi)(x)=\psi(x-a)}$ , determine whether the Hamiltonian commutes with translation on ${H^2_{\mathrm{per}}([0,L])}$ .

Periodic translation preserves both the function and derivative boundary conditions, hence maps ${H^2_{\mathrm{per}}([0,L])}$ into itself. Both operator compositions are therefore defined on this common domain. Compare them on an arbitrary ${\psi}$ there, without solving for an energy:

$(\hat H\hat T_a\psi)(x) =-\frac{\hbar^2}{2m}\psi''(x-a)+2V_c\cos(Gx)\psi(x-a),$ $(\hat T_a\hat H\psi)(x) =-\frac{\hbar^2}{2m}\psi''(x-a)+2V_c\cos[G(x-a)]\psi(x-a).$

The derivative commutes with the fixed shift, and ${\cos[G(x-a)]=\cos(Gx-2\pi)=\cos(Gx)}$ . The two expressions agree, hence ${[\hat H,\hat T_a]=0}$ ; composition gives the result for every ${na}$ . The same calculation works for any potential satisfying ${V(x-a)=V(x)}$ . The dimensionless phase ${Ga=2\pi}$ checks the period, while the limit ${V_c=0}$ restores the free particle and its continuous translation symmetry. This calculation determines a symmetry of the state space, not the energies inside a symmetry sector. Its translated state is the input to the character construction that follows.

Characters on a finite periodic lattice

Because ${V}$ is lattice periodic, ${[\hat H,\hat T_{\mathbf R}]=0}$ for every lattice vector. Energy eigenstates may therefore be chosen as simultaneous eigenstates of all lattice translations. Write

$\hat T_{\mathbf R}|\psi_{\mathbf k}\rangle =\lambda_{\mathbf k}(\mathbf R) |\psi_{\mathbf k}\rangle.$

Here ${\mathbf k\in\mathbb R^d}$ has units of inverse length and labels the translation-eigenvalue function ${\lambda_{\mathbf k}:\Lambda\to U(1)}$ , where ${U(1)=\{z\in\mathbb C:|z|=1\}}$ . Unitarity gives ${\|\hat T_{\mathbf R}\psi_{\mathbf k}\|^2 =|\lambda_{\mathbf k}(\mathbf R)|^2\|\psi_{\mathbf k}\|^2 =\|\psi_{\mathbf k}\|^2}$ . A nonzero eigenvector therefore has ${\lvert\lambda_{\mathbf k}(\mathbf R)\rvert=1}$ . Applying two translations to that same vector and comparing their product with one combined translation gives

$\lambda_{\mathbf k}(\mathbf R+\mathbf R') =\lambda_{\mathbf k}(\mathbf R) \lambda_{\mathbf k}(\mathbf R').$

Regard the Bravais lattice ${\Lambda}$ as an additive group. A translation character is a group homomorphism ${\chi:\Lambda\to U(1)}$ ; that is, $\chi(\mathbf R+\mathbf R') =\chi(\mathbf R)\chi(\mathbf R') \quad\text{for all }\mathbf R,\mathbf R'\in\Lambda.$ It is a map assigned to an entire translation-symmetry sector, not one chosen coordinate representative or one normalized eigenvector in that sector [Zak67, APS full-text manifestation, printed p. 1385, Eqs. (1)-(3)].

Thus ${\lambda_{\mathbf k}}$ is a translation character. The definition says both more and less than “an eigenvalue is a phase”: it requires compatible phases for all lattice vectors, but it does not choose a particular state inside the common eigenspace.

Before writing the general solution, rehearse the logic on a finite one-dimensional lattice. Let the allowed translations be ${R=na}$ , where ${n\in\mathbb Z_N}$ , and let translation through the full sample satisfy ${\hat T_{Na}=\hat I}$ . The entire character is determined by the eigenvalue of one generator. If ${\lambda(a)=\mathrm e^{-\mathrm i ka}}$ , the group law forces

(3.1)

$\lambda(na)={\bigl[\lambda(a)\bigr]}^n =\mathrm e^{-\mathrm i kna}.$

Periodicity imposes ${\mathrm e^{-\mathrm i kNa}=1}$ , so the allowed representatives may be chosen as

$k_m=\frac{2\pi m}{Na}, \qquad m=0,1,\ldots,N-1.$

There are exactly ${N}$ distinct characters, not infinitely many. Moreover, their phases obey the finite Fourier orthogonality relation

$\frac{1}{N}\sum_{n=0}^{N-1} \mathrm e^{\mathrm i(k_m-k_{m'})na} =\delta_{mm'}.$

This identity will later normalize the Bloch sums and remove the sum over the absolute cell. It also gives a useful interpretation of the thermodynamic limit: as ${N}$ grows, the allowed labels become dense in a Brillouin zone, but the number of labels always equals the number of cells.

Now attach the opening anchor to this result. The two bare plane waves ${\varphi_k(x)=L^{-1/2}\mathrm e^{\mathrm i kx}}$ and ${\varphi_{k-2\pi/a}(x)=L^{-1/2}\mathrm e^{\mathrm i(k-2\pi/a)x}}$ , defined on the ring of length ${L=Na}$ , are distinct momentum eigenvectors. Yet for every lattice translation ${R=na}$ ,

$\frac{\mathrm e^{-\mathrm i(k-2\pi/a)R}} {\mathrm e^{-\mathrm i kR}} =\mathrm e^{\mathrm i2\pi n}=1.$

The discrete translation test therefore assigns the same eigenvalue function to both waves even though their canonical momenta differ by ${2\pi\hbar/a}$ . This supplies the translation test behind the two branches in Figure 3.1: a shared representative need not mean a shared physical momentum, state or energy.

Close the worked phase calculation. On a ring with ${N=12}$ , a nonzero state has ${\hat T_a\psi=\mathrm e^{-\mathrm i\pi/3}\psi}$ . Complete the compatible representative ${k=\underline{\qquad}}$ modulo ${2\pi/a}$ , and the step ${\psi(x+a)=\underline{\qquad}\psi(x)}$ . Then evaluate the ratio of the translation eigenvalues assigned to ${k}$ and ${k+2\pi/a}$ for ${R=na}$ . Check the twelfth power of the generator eigenvalue. Compare the sign and periodicity with the feedback before extending the construction to more dimensions.

Answer. The compatible representative is $k=\pi/(3a)$ modulo $2\pi/a$ , and $\psi(x+a)=\mathrm e^{+\mathrm i\pi/3}\psi(x)$ . The reciprocal-shift phase ratio is one for every integer $n$ , and the twelfth generator power is $\mathrm e^{-\mathrm i4\pi}=1$ .

After completing the phase check, keep ${N=12}$ and choose a label ${m\in\{0,\ldots,11\}}$ , with ${ka=2\pi m/12}$ . Compare the phase of ${\hat T_{na}}$ with the factor relating ${\psi(x+na)}$ to ${\psi(x)}$ : they are complex conjugates. Then replace ${k}$ by ${k+2\pi\ell/a}$ , where ${\ell}$ is an integer. The lattice phases and twelve-step closure stay unchanged. Two bare plane waves with those wave numbers nevertheless have distinct canonical momenta when ${\ell\ne0}$ ; equal characters do not identify the waves. The plot shows the real and imaginary parts of the translation character, with the selected step marked.

From generator phases to reciprocal classes

The same construction works in ${d}$ dimensions without guessing the answer. Let the primitive lattice translations be ${\mathbf a_1,\ldots,\mathbf a_d}$ . Because they generate the entire Bravais lattice, a character is fixed once we know the ${d}$ phases

$\lambda(\mathbf a_i)=\mathrm e^{-\mathrm i\phi_i}, \qquad \phi_i\in\mathbb R\ \text{modulo }2\pi.$

Choose reciprocal basis vectors satisfying ${\mathbf a_i\mathbin{\cdot}\mathbf b_j=2\pi\delta_{ij}}$ and define

$\mathbf k=\sum_{i=1}^{d}\frac{\phi_i}{2\pi}\mathbf b_i.$

For an arbitrary lattice vector ${\mathbf R=\sum_i n_i\mathbf a_i}$ , the group law then gives

$\lambda(\mathbf R) =\prod_i{\lambda(\mathbf a_i)}^{n_i} =\mathrm e^{-\mathrm i\sum_i n_i\phi_i} =\mathrm e^{-\mathrm i\mathbf k\mathbin{\cdot}\mathbf R}.$

Thus the exponential form follows from the generator phases and the group law; it is not an additional dynamical assumption. Changing any ${\phi_i}$ by ${2\pi}$ changes ${\mathbf k}$ by a reciprocal-lattice vector and leaves the character unchanged. The derivation has therefore already produced both crystal momentum and its reciprocal equivalence.

In compact form, the unit-modulus homomorphisms of the lattice translation group are the characters

$\lambda_{\mathbf k}(\mathbf R) =\mathrm e^{-\mathrm i\mathbf k\mathbin{\cdot}\mathbf R}.$

The minus sign follows from our definition of ${\hat T_{\mathbf R}}$ . Changing both conventions changes no prediction. The important point is that ${\mathbf k}$ labels how a state transforms under the discrete translation group. It need not equal the mechanical momentum of the electron. Indeed, a periodic potential generally mixes plane waves whose mechanical momenta differ by reciprocal-lattice vectors.

Reciprocal equivalence and crystal-momentum classes

Let ${\mathbf R\in\Lambda}$ denote a direct-lattice translation. A reciprocal vector ${\mathbf G}$ adds no phase to any such translation. The reciprocal lattice ${\Lambda^*\subset\mathbb R^d}$ collects these vectors:

$\Lambda^*=\left\{\mathbf G\in\mathbb R^d: \mathrm e^{\mathrm i\mathbf G\mathbin{\cdot}\mathbf R}=1 \ \text{for every }\mathbf R\in\Lambda\right\}.$

Consequently, ${\mathbf k}$ and ${\mathbf k+\mathbf G}$ give the same eigenvalue for every lattice translation:

$\mathrm e^{-\mathrm i(\mathbf k+\mathbf G) \mathbin{\cdot}\mathbf R} =\mathrm e^{-\mathrm i\mathbf k\mathbin{\cdot}\mathbf R}.$

For ${\mathbf k,\mathbf k'\in\mathbb R^d}$ , define $\mathbf k\sim\mathbf k' \quad\Longleftrightarrow\quad \mathbf k-\mathbf k'\in\Lambda^*.$ This is an equivalence relation [BSW36, APS full-text manifestation, printed pp. 59-60, Eqs. (2)-(2a), with opening priority footnote]. A crystal-momentum label is a class ${[\mathbf k]\in\mathbb R^d/\Lambda^*}$ , whereas a point ${\mathbf k}$ in a chosen first Brillouin zone is one representative of that class, subject to the boundary identifications of the chosen fundamental region.

Class and representative. The phase equation proves that equivalent vectors define the same translation character. We call the quotient ${\mathbb R^d/\Lambda^*}$ crystal-momentum space. In one dimension, ${k=\pi/a}$ and ${k=-\pi/a}$ represent the same class because their difference is ${2\pi/a}$ . Choosing a half-open zone retains only one endpoint and prevents double counting; choosing the other endpoint changes the representative, not the class or its physical predictions.

Nonexample: two bare plane waves. The functions ${\mathrm e^{\mathrm i\mathbf k\cdot\mathbf r}}$ and ${\mathrm e^{\mathrm i(\mathbf k+\mathbf G)\cdot\mathbf r}}$ are generally distinct Hilbert-space vectors and have different canonical momenta. They lie in the same translation-character fiber; they are not made into the same state by the equivalence relation. This is exactly the pair first examined on the finite ring.

For primitive direct vectors ${\mathbf a_i\in\Lambda}$ and reciprocal basis vectors ${\mathbf b_j\in\Lambda^*}$ , with ${i,j\in\{1,\ldots,d\}}$ , choose the dual normalization

$\mathbf a_i\mathbin{\cdot}\mathbf b_j=2\pi\delta_{ij}.$

On the ${d=2}$ square lattice, ${\mathbf e_x}$ and ${\mathbf e_y}$ denote orthonormal unit vectors, and ${\mathbf a_1=a\mathbf e_x}$ and ${\mathbf a_2=a\mathbf e_y}$ , so ${\mathbf b_1=(2\pi/a)\mathbf e_x}$ and ${\mathbf b_2=(2\pi/a)\mathbf e_y}$ . A convenient half-open first zone is ${-\pi/a\leq k_x,k_y<\pi/a}$ ; the opposite boundary points are identified.

Before inspecting Figure 3.2, predict the directions and magnitudes of the two reciprocal basis vectors from ${\mathbf a_i\mathbin{\cdot}\mathbf b_j=2\pi\delta_{ij}}$ . Then locate the outside point in panel (c) and decide which reciprocal vector must be subtracted to enter the displayed zone. The purpose of the figure is to test a mapping between direct-space translations and reciprocal-space labels, not to serve as a labeled lattice photograph.

Figure 3.2. Direct lattice, reciprocal lattice, first Brillouin zone, and an explicit change of reciprocal-space representative. The dashed arrow changes the representative but not the translation character. [Bri30, HAL French manifestation, printed pp. 378-384, especially reciprocal-space planes and Fig. 4]

Use the figure actively. In panel (c), the red point lies outside the displayed zone. The dashed displacement is a reciprocal-lattice vector, so the black point has the same phase under every direct-lattice translation. Nothing in this construction supplies a coupling between two energy levels. Zone folding can place two branches at the same representative ${\mathbf k}$ ; it does not make them repel.

Work through one fold explicitly. Take ${\mathbf k_{\mathrm{out}}=(5\pi/4a,-3\pi/2a)}$ . Subtracting ${\mathbf G=(2\pi/a,-2\pi/a)}$ gives

$\mathbf k_{\mathrm{BZ}} =\left(-\frac{3\pi}{4a},\frac{\pi}{2a}\right),$

which lies in the displayed square. Under ${\mathbf R=n_1a\mathbf e_x+n_2a\mathbf e_y}$ , the ratio of the two translation phases is

$\frac{\mathrm e^{-\mathrm i\mathbf k_{\mathrm{out}} \mathbin{\cdot}\mathbf R}} {\mathrm e^{-\mathrm i\mathbf k_{\mathrm{BZ}} \mathbin{\cdot}\mathbf R}} =\mathrm e^{-\mathrm i\mathbf G\mathbin{\cdot}\mathbf R} =\mathrm e^{-\mathrm i2\pi(n_1-n_2)}=1.$

Stop at this point and make the distinction operational. If the two labels had the same mechanical momentum, the plane-wave momentum eigenvalues would have to agree; they differ by ${\hbar\mathbf G}$ . What agrees is instead the outcome of every lattice-translation phase test. A response that says only “momentum is conserved modulo ${\mathbf G}$ ” has not identified which quantity is an observable and which is a symmetry-sector label.

This check is deliberately more concrete than saying “momenta are conserved modulo a reciprocal vector.” Mechanical momentum has not become equal at the two points. Rather, the discrete translation group cannot distinguish their characters.

To compute the reciprocal lattice for an arbitrary two-dimensional primitive cell, collect the direct vectors as the columns of ${A}$ and the reciprocal vectors as the columns of ${B}$ . Their duality equations give ${A^{\mathsf T}B=2\pi I}$ , hence the inverse transpose below. The areas ${S}$ and ${S^*}$ are the direct and reciprocal primitive-cell areas. Taking determinants gives their area-product identity. Here ${\ell}$ is a fixed reference length, ${r_1,r_2}$ are positive lengths in that unit, ${\phi}$ rotates the cell, and ${\theta}$ is its included angle. The construction requires an invertible ${A}$ , meaning that the two direct vectors are not collinear.

(3.2)

$\begin{gathered} A=(\mathbf a_1\ \mathbf a_2),\qquad B=(\mathbf b_1\ \mathbf b_2)=2\pi A^{-\mathsf T},\qquad A^{\mathsf T}B=2\pi I,\\ \mathbf a_1=\ell r_1(\cos\phi,\sin\phi),\qquad \mathbf a_2=\ell r_2(\cos(\phi+\theta),\sin(\phi+\theta)),\\ S=|\det A|,\qquad S^*=|\det B|,\qquad SS^*={(2\pi)}^2. \end{gathered}$

The explorer uses ${0.75\leq r_i\leq2}$ , ${-180^\circ\leq\phi\leq180^\circ}$ , and ${30^\circ\leq\theta\leq150^\circ}$ . These display limits avoid collinear vectors; the formula holds for every invertible ${A}$ .

Drag a primitive-vector head in either panel and watch both Wigner–Seitz cells. The other vector in the panel you control stays fixed; the opposite basis updates to preserve duality. Click a parameter value to type it, or open Values to check the reciprocal components, areas, and normalized dual matrix. The first reciprocal vector stays perpendicular to the second direct vector, and conversely. Predict what doubling the first length does: it halves the first reciprocal vector while leaving the second unchanged. Rigid rotation rotates both cells together. Axes stay fixed, with direct coordinates in ${\ell}$ and reciprocal coordinates in ${2\pi/\ell}$ ; plotted dots are lattice points, not atoms. Each shaded cell contains the points at least as close to the origin as to any other lattice point. Its edges lie on perpendicular bisectors between the origin and neighboring lattice points. Thus the reciprocal shaded cell is the first Brillouin zone. The arrows generate the lattice; they are not edges of these cells. Each cell still has the primitive area given above, even when its boundary changes between four and six sides.

At Reset, ${r_1=1.5}$ , ${r_2=1}$ , ${\phi=0}$ , and ${\theta=60^\circ}$ : ${\mathbf a_1/\ell=(1.5,0)}$ , ${\mathbf a_2/\ell=(1/2,\sqrt3/2)}$ , ${\ell\mathbf b_1/(2\pi)=(2/3,-2/(3\sqrt3))}$ , and ${\ell\mathbf b_2/(2\pi)=(0,2/\sqrt3)}$ . The normalized areas are ${S/\ell^2=3\sqrt3/4}$ and ${S^*\ell^2/{(2\pi)}^2=4/(3\sqrt3)}$ , whose product is one.

For the finite square sample ${\Omega}$ , let ${N_1,N_2\in\mathbb N}$ be the numbers of cells along the two primitive directions. Periodic boundary conditions make translation by ${N_i\mathbf a_i}$ the identity. The character condition then restricts ${\mathbf k\mathbin{\cdot}N_i\mathbf a_i}$ to an integer multiple of ${2\pi}$ . There are ${N_1N_2}$ distinct allowed crystal momenta in the first zone, exactly the number of unit cells. This counting will matter when a localized orbital in every cell becomes one state in each Bloch block. The continuum-looking zone is the thermodynamic-limit representation of this finite set, not an additional reservoir of states.

From translation characters to Bloch form

The phase labels now have a precise domain and reciprocal equivalence. We next construct their projectors, then translate that decomposition into a periodic factor and a cell-sized eigenproblem for the same state.

Finite character projectors

The finite sample lets us prove the decomposition algebraically before rewriting its vectors in position space.

Let ${\Lambda_\Omega\subset\Lambda}$ be the subgroup generated by the full sample translations ${N_i\mathbf a_i}$ , and let ${G_\Omega=\Lambda/\Lambda_\Omega}$ be the finite translation group. Suppose the self-adjoint Hamiltonian ${\hat H}$ on ${\mathcal H=L^2(\Omega,\mathbb C)}$ , with Born–von Karman boundary conditions, commutes with the unitary representation ${{\bigl\{\hat T_{\mathbf R}\bigr\}}_{\mathbf R\in G_\Omega}}$ . Then $\mathcal H=\bigoplus_{\chi\in\widehat G_\Omega}\mathcal H_\chi, \qquad \mathcal H_\chi= \left\{\psi\in\mathcal H: \hat T_{\mathbf R}\psi=\chi(\mathbf R)\psi \ \text{for every }\mathbf R\in G_\Omega\right\},$ and every ${\mathcal H_\chi}$ is invariant under ${\hat H}$ . Writing ${\chi(\mathbf R)=\mathrm e^{-\mathrm i\mathbf k\cdot\mathbf R}}$ , every coordinate representative in that fiber has the form $\psi_{\mathbf k}(\mathbf r) =\mathrm e^{\mathrm i\mathbf k\cdot\mathbf r} u_{\mathbf k}(\mathbf r), \qquad u_{\mathbf k}(\mathbf r+\mathbf R)=u_{\mathbf k}(\mathbf r),$ where ${\mathbf k}$ is defined modulo ${\Lambda^*}$ [Blo29, complete article facsimile, printed pp. 559-560, Eqs. (6)-(6a); Zak67, APS full-text manifestation, printed pp. 1385-1386, Eqs. (1)-(7)]. Bloch’s paper is the original periodic-potential source for the plane-wave-times-periodic-factor form; Zak supplies the later finite-translation formulation used in the character-projector proof that follows. The infinite-crystal version requires the separate direct-integral statement below.

For the finite sample, the proof is finite-dimensional representation theory applied to the translation labels, even though each common eigenspace can be infinite dimensional. A character projector keeps the component with its chosen translation character and removes components with other characters. Because the projector is a sum of translations commuting with the Hamiltonian, the Hamiltonian preserves its range: different character sectors cannot mix. The identities below establish these claims. A finite Abelian group has mutually orthogonal character projectors

$\hat P_\chi =\frac{1}{|G_\Omega|} \sum_{\mathbf R\in G_\Omega} \chi{(\mathbf R)}^*\hat T_{\mathbf R}.$

To check its range, shift the summation variable to ${\mathbf Q=\mathbf R+\mathbf S}$ . The character law gives

$\hat T_{\mathbf S}\hat P_\chi =\frac1{|G_\Omega|}\sum_{\mathbf Q} \chi{(\mathbf Q-\mathbf S)}^*\hat T_{\mathbf Q} =\chi(\mathbf S)\hat P_\chi.$

Thus every vector in the range has character ${\chi}$ . Taking the adjoint and replacing ${\mathbf R}$ by ${-\mathbf R}$ gives ${\hat P_\chi^\dagger=\hat P_\chi}$ . Applying a second projector now reduces to the already established finite Fourier sum:

$\hat P_\chi\hat P_{\chi'} =\left[\frac1{|G_\Omega|}\sum_{\mathbf R} \chi{(\mathbf R)}^*\chi'(\mathbf R)\right]\hat P_{\chi'} =\delta_{\chi\chi'}\hat P_\chi.$

For the sample generated by ${N_i\mathbf a_i}$ , this sum factorizes into one geometric sum in each primitive direction; the vectors need not be orthogonal. In one direction, with ${z=\mathrm e^{2\pi\mathrm i(m-m')/N}}$ , it is one when ${z=1}$ , and ${(1-z^N)/[N(1-z)]=0}$ otherwise. Summing over all characters instead gives ${\sum_\chi\chi{(\mathbf R)}^*=|G_\Omega|\delta_{\mathbf R,\mathbf0}}$ , so ${\sum_\chi\hat P_\chi=\hat T_{\mathbf0}=\hat I}$ . These are self-adjoint, mutually orthogonal projectors whose ranges exhaust the space, which proves the direct sum. Because every ${\hat P_\chi}$ is a linear combination of translations, ${[\hat H,\hat P_\chi]=0}$ , so ${\hat H}$ cannot couple distinct fibers. It remains only to prove the stated coordinate form, which we now do directly.

$\psi_{\mathbf k}(\mathbf r-\mathbf R) =\mathrm e^{-\mathrm i\mathbf k\mathbin{\cdot}\mathbf R} \psi_{\mathbf k}(\mathbf r).$

Equivalently, ${\psi_{\mathbf k}(\mathbf r+\mathbf R) =\mathrm e^{\mathrm i\mathbf k\mathbin{\cdot}\mathbf R} \psi_{\mathbf k}(\mathbf r)}$ . Removing the character defines

$u_{\mathbf k}(\mathbf r) =\mathrm e^{-\mathrm i\mathbf k\mathbin{\cdot}\mathbf r} \psi_{\mathbf k}(\mathbf r),$

and the two translation phases cancel explicitly:

$u_{\mathbf k}(\mathbf r+\mathbf R) =\mathrm e^{-\mathrm i\mathbf k\cdot(\mathbf r+\mathbf R)} \mathrm e^{\mathrm i\mathbf k\cdot\mathbf R}\psi_{\mathbf k}(\mathbf r) =u_{\mathbf k}(\mathbf r).$

Thus

$\psi_{n\mathbf k}(\mathbf r) =\mathrm e^{\mathrm i\mathbf k\mathbin{\cdot}\mathbf r} u_{n\mathbf k}(\mathbf r), \qquad u_{n\mathbf k}(\mathbf r+\mathbf R) =u_{n\mathbf k}(\mathbf r).$

The implication also runs backward. If ${u_{\mathbf k}}$ is any cell-periodic function, then ${\psi_{\mathbf k}(\mathbf r)=\mathrm e^{\mathrm i\mathbf k\cdot\mathbf r} u_{\mathbf k}(\mathbf r)}$ transforms with the character labeled by ${\mathbf k}$ . Bloch’s theorem is therefore an equivalence between a translation character and a plane-wave factor multiplying a periodic function. It does not say that ${u_{\mathbf k}}$ is constant, that the state is a mechanical-momentum eigenstate, or that the energy has a free-particle dispersion. All microscopic complexity has been moved into the cell-periodic factor and the eigenproblem it obeys.

Let ${\Lambda\subset\mathbb R^d}$ be a full-rank lattice with reciprocal lattice ${\Lambda^*}$ , let ${\Omega_{\mathrm{cell}}}$ be a fundamental cell, and let the real potential ${V\in L^\infty(\mathbb R^d)}$ be ${\Lambda}$ -periodic. Define $\hat H=-\frac{\hbar^2}{2m}\nabla^2+V \quad\text{on}\quad \mathcal D(\hat H)=H^2(\mathbb R^d).$ The bounded-perturbation theorem makes ${\hat H}$ self-adjoint on that domain. Equip the Brillouin torus ${\mathcal B=\mathbb R^d/\Lambda^*}$ with normalized Haar measure ${\mathrm d\mu_{\mathcal B}}$ , so that ${\mu_{\mathcal B}(\mathcal B)=1}$ . Then a correspondingly normalized unitary Bloch–Floquet transform gives

$L^2(\mathbb R^d,\mathbb C) \cong\int_{\mathcal B}^{\oplus}\mathcal H_{\mathrm{per}}\, \mathrm d\mu_{\mathcal B}(\mathbf k), \qquad \hat H\cong\int_{\mathcal B}^{\oplus}\hat H(\mathbf k)\, \mathrm d\mu_{\mathcal B}(\mathbf k),$

where ${\mathcal H_{\mathrm{per}}=L^2_{\mathrm{per}}(\Omega_{\mathrm{cell}})}$ and every fiber is represented on the common domain ${H^2_{\mathrm{per}}(\Omega_{\mathrm{cell}})}$ . This proposition is invoked, not proved, in this lecture. The finite character-projector calculation above proves the algebraic mechanism the core uses; direct-integral theory supplies its infinite-crystal formulation [Kuc16, arXiv v4 evidence copy, manuscript §§4.1-4.3, pp. 20-25, Eqs. (4.4)-(4.17), numbering collated by work identity].

Bloch states and periodic factors

Fix the Hilbert space ${\mathcal H}$ , the convention ${(\hat T_{\mathbf R}\psi)(\mathbf r)=\psi(\mathbf r-\mathbf R)}$ , and a crystal-momentum class ${[\mathbf k]\in\mathbb R^d/\Lambda^*}$ . A Bloch state in that class is a nonzero vector ${|\psi_{[\mathbf k]}\rangle\in\mathcal H}$ satisfying $\hat T_{\mathbf R}|\psi_{[\mathbf k]}\rangle =\mathrm e^{-\mathrm i\mathbf k\cdot\mathbf R} |\psi_{[\mathbf k]}\rangle \quad\text{for every }\mathbf R\in\Lambda.$ If it is also an eigenvector of ${\hat H}$ , we write ${|\psi_{n\mathbf k}\rangle}$ ; the index ${n}$ identifies an eigenvalue slot inside the fixed-character fiber. Replacing ${\mathbf k}$ by ${\mathbf k+\mathbf G}$ changes the representative and the corresponding cell-periodic factor, but not the crystal-momentum class or the full state [Zak67, APS full-text manifestation, printed p. 1385, Eqs. (2)-(3)].

The nonzero condition excludes the zero vector, which satisfies every linear eigenvalue equation but carries no symmetry label. The definition also separates three data that are often conflated: the class ${[\mathbf k]}$ , a representative ${\mathbf k}$ , and a particular vector within the fiber.

Same state, different factorization. Suppose a fixed Bloch wave has already been written as ${\psi_{\mathbf k}(\mathbf r)= \mathrm e^{\mathrm i\mathbf k\cdot\mathbf r}u_{\mathbf k}(\mathbf r)}$ . For ${\mathbf G\in\Lambda^*}$ , define the periodic function

$u'_{\mathbf k+\mathbf G}(\mathbf r) =\mathrm e^{-\mathrm i\mathbf G\cdot\mathbf r}u_{\mathbf k}(\mathbf r).$

Then

$\mathrm e^{\mathrm i(\mathbf k+\mathbf G)\cdot\mathbf r} u'_{\mathbf k+\mathbf G}(\mathbf r) =\mathrm e^{\mathrm i\mathbf k\cdot\mathbf r}u_{\mathbf k}(\mathbf r).$

The representative and the periodic factor have both changed, but the full Hilbert-space vector has not. Its component vector may nevertheless look different if the basis inside the Bloch block is reindexed.

This is the first decision point in the opening crossing problem. Translation symmetry has justified a common block label and a cell-periodic eigenproblem. It has not supplied an energy, a bandwidth, or an off-diagonal matrix element. Keep the folded pair ${|+\pi/a\rangle,|-\pi/a\rangle}$ in view: the derivation has said where these components can meet, not whether they can mix.

Close the worked factorization before the cell problem. For the same one-dimensional convention, start from ${\psi_k(x-a)=\mathrm e^{-\mathrm ika}\psi_k(x)}$ . Define ${u_k(x)=\mathrm e^{-\mathrm ikx}\psi_k(x)}$ and complete both phase factors in ${u_k(x+a)}$ before cancelling them. Then reconstruct the backward implication from a periodic ${u_k}$ to the translation eigenvalue. Mark the first step at which an incorrect sign would become detectable. Compare that step with the instructor before proceeding.

The cell-periodic eigenproblem

Let ${\Omega_{\mathrm{cell}}\subset\mathbb R^d}$ be one primitive cell and let ${\mathcal H_{\mathrm{per}}=L^2_{\mathrm{per}} (\Omega_{\mathrm{cell}},\mathbb C)}$ denote square-integrable cell-periodic functions. For each eigenvalue label ${n}$ , ${\psi_{n\mathbf k}:\Omega\to\mathbb C}$ is the position-space Bloch wave function, while ${u_{n\mathbf k}:\Omega_{\mathrm{cell}}\to\mathbb C}$ is its cell-periodic factor and belongs to ${\mathcal H_{\mathrm{per}}}$ . Both objects consumed here are functions with the displayed domains. This lecture does not require a distribution or a section, so those mathematical kinds are deferred to the lectures that first use them.

The normalization convention deserves one explicit line because full-crystal and cell normalizations are both common. Since the phase factor has unit magnitude,

$\int_{\Omega}{|\psi_{n\mathbf k}(\mathbf r)|}^2\,\mathrm d^d r =N_{\mathrm c} \int_{\Omega_{\mathrm{cell}}} {|u_{n\mathbf k}(\mathbf r)|}^2\,\mathrm d^d r.$

Here ${N_{\mathrm c}}$ is the number of primitive cells. We normalize the full state to one, so the cell integral equals ${1/N_{\mathrm c}}$ . A code that instead normalizes the periodic factor to one cell must include an overall ${1/\sqrt{N_{\mathrm c}}}$ in the full Bloch state. Neither convention changes matrix elements when used consistently.

On a ring of six cells, ${L=6a}$ , isolate the translation sector with ${k=4\pi/(3a)}$ from the normalized trial function $f(x)=\frac{1+\mathrm e^{\mathrm i kx}}{\sqrt{2L}}.$ Normalize the projected state and express it using a wave-number representative in ${[-\pi/a,\pi/a)}$ and its periodic factor. Check both the translation eigenvalue and the ring/cell normalization.

The two plane waves are orthogonal on this ring. The finite-character projector in Theorem 3.1 weights each translation to retain the chosen phase: $(\hat P_k f)(x)=\frac16\sum_{r=0}^{5}\mathrm e^{\mathrm i kra}f(x-ra) =\frac{1}{\sqrt{2L}}\left[ \frac16\sum_{r=0}^{5}\mathrm e^{\mathrm i4\pi r/3} +\mathrm e^{\mathrm i kx}\right] =\frac{\mathrm e^{\mathrm i kx}}{\sqrt{2L}}.$ The geometric sum vanishes, while the selected harmonic’s phase cancels term by term. Its norm squared is ${1/2}$ , so normalization gives ${\psi(x)=\mathrm e^{\mathrm i kx}/\sqrt L}$ . Choose the representative ${k'=k-2\pi/a=-2\pi/(3a)}$ in the first zone. Its periodic factor is $u_{k'}(x)=\mathrm e^{-\mathrm i k'x}\psi(x) =\frac{\mathrm e^{\mathrm i2\pi x/a}}{\sqrt L},\qquad \psi(x)=\mathrm e^{\mathrm i k'x}u_{k'}(x).$ Check ${u_{k'}(x+a)=u_{k'}(x)}$ , and ${\mathrm e^{-\mathrm i k'a}=\mathrm e^{-\mathrm i ka}}$ : the representative change preserves the translation eigenvalue. Normalization gives ${\int_0^L|\psi|^2\,\mathrm dx=1}$ and ${\int_0^a|u_{k'}|^2\,\mathrm dx=1/6}$ , agreeing with the stated cell convention. The wave numbers have inverse-length units and their translation phases are dimensionless. This constructs a state in a Bloch sector; for a nonzero cosine potential it need not be an energy eigenstate. The cell Hamiltonian must still determine which superpositions diagonalize that sector.

We can now make the earlier distinction between crystal and mechanical momentum operational. Expand the cell-periodic factor in reciprocal-lattice plane waves,

$u_{n\mathbf k}(\mathbf r) =\frac{1}{\sqrt{|\Omega|}} \sum_{\mathbf G\in\Lambda^*} c_{n\mathbf k}(\mathbf G) \mathrm e^{\mathrm i\mathbf G\mathbin{\cdot}\mathbf r},$

which gives

$\psi_{n\mathbf k}(\mathbf r) =\frac{1}{\sqrt{|\Omega|}} \sum_{\mathbf G\in\Lambda^*} c_{n\mathbf k}(\mathbf G) \mathrm e^{\mathrm i(\mathbf k+\mathbf G) \mathbin{\cdot}\mathbf r}.$

For a normalized state in the finite periodic sample, ${\sum_{\mathbf G}{|c_{n\mathbf k}(\mathbf G)|}^2=1}$ . A measurement of canonical momentum can return one of the values ${\hbar(\mathbf k+\mathbf G)}$ , with weights set by the coefficients. In particular,

${\langle\hat{\mathbf p}\rangle}_{n\mathbf k} =\hbar\mathbf k +\hbar\sum_{\mathbf G} {|c_{n\mathbf k}(\mathbf G)|}^2\mathbf G,$

which need not equal ${\hbar\mathbf k}$ . Relabeling the same character by ${\mathbf k+\mathbf G_0}$ merely reindexes the coefficients; it does not change the momentum distribution. Crystal momentum organizes symmetry sectors, whereas mechanical momentum describes outcomes of the momentum operator. This difference matters whenever a probe couples to plane-wave or orbital content rather than to band energy alone.

A normalized Bloch state has nonzero coefficients at two different reciprocal harmonics. A classmate concludes that its definite translation character makes it a momentum eigenstate. Explain which property of discrete translations permits the two components to share a character, and which measurement distinguishes them. State the condition under which this particular state would have a single momentum outcome. Give the physical distinction before using a formula; discuss the response with the instructor.

Let ${\mathcal D_{\mathrm{per}}=H^2_{\mathrm{per}} (\Omega_{\mathrm{cell}})\subset\mathcal H_{\mathrm{per}}}$ . Every fiber operator below acts on this same periodic domain; the calculation therefore changes the operator parameter ${\mathbf k}$ , not the boundary condition. For any ${f\in\mathcal D_{\mathrm{per}}}$ , the essential product-rule step is

$\mathrm e^{-\mathrm i\mathbf k\mathbin{\cdot}\mathbf r} \bigl(-\mathrm i\hbar\partial_j\bigr) \left[ \mathrm e^{\mathrm i\mathbf k\mathbin{\cdot}\mathbf r}f(\mathbf r) \right] =\left(-\mathrm i\hbar\partial_j+\hbar k_j\right)f(\mathbf r), \qquad j\in\{1,\ldots,d\}.$

The derivative acts once on the plane-wave factor and once on the periodic factor. Squaring this transformed momentum and leaving the periodic potential unchanged produces the self-adjoint operator ${\hat H(\mathbf k):\mathcal D_{\mathrm{per}}\to \mathcal H_{\mathrm{per}}}$ . The finite sample uses only its allowed momentum mesh. Continuous momentum labels describe the bulk family, not additional states of that finite sample. To identify this expression as a cell Hamiltonian, first restrict the finite-sample Hamiltonian to its invariant translation sector ${\mathcal H_{\mathbf k}=\operatorname{Ran}P_{\mathbf k}}$ . Define the unitary identification from the periodic cell space to this sector by ${(U_{\mathbf k}u)(\mathbf r)=N_{\mathrm c}^{-1/2} \mathrm e^{\mathrm i\mathbf k\cdot\mathbf r}u(\mathbf r)}$ , extending the cell function periodically over the sample. Here the cell and sample use their usual integral norms. The constant ensures unitarity and cancels in the conjugation; the resulting eigenfunctions can be rescaled to the full-state normalization used above. Removing the Bloch phase after this restriction gives ${\hat H(\mathbf k)=U_{\mathbf k}^{-1} (\hat H|_{\mathcal H_{\mathbf k}})U_{\mathbf k}}$ . No additional projection is needed on a state already in this sector, because the periodic Hamiltonian preserves it. In the infinite crystal, the analogous operators are fibers of the Bloch direct integral, not restrictions to nonzero fixed-momentum subspaces of the full square-integrable space. In either case the cell operator acts on the common periodic domain:

(3.3)

$\hat H(\mathbf k) =U_{\mathbf k}^{-1} \bigl(\hat H|_{\mathcal H_{\mathbf k}}\bigr)U_{\mathbf k} =\frac{{(\hat{\mathbf p}+\hbar\mathbf k)}^2}{2m} +V(\hat{\mathbf r}),$

$\hat H(\mathbf k)|u_{n\mathbf k}\rangle =E_n(\mathbf k)|u_{n\mathbf k}\rangle.$

[Kuc16, arXiv v4 evidence copy, Remark 4.7, manuscript pp. 24-25, Eqs. (4.14)-(4.16), numbering collated by work identity]

For the bounded real periodic potential assumed here, the first-order term in ${\hat H(\mathbf k)}$ has relative bound zero with respect to the periodic Laplacian. In reciprocal-lattice Fourier components this is the estimate

$\lVert\hat{\mathbf p}\psi\rVert_2 \leq \varepsilon\lVert\hat{\mathbf p}^{2}\psi\rVert_2 +C_\varepsilon\lVert\psi\rVert_2, \qquad \varepsilon>0,$

with a finite ${C_\varepsilon}$ . Together with the bounded-potential estimate in Proposition 3.1, this gives an elementary route to self-adjointness of every fixed- ${\mathbf k}$ fiber on the common domain ${\mathcal D_{\mathrm{per}}}$ . Rellich compactness on the torus then makes the embedding ${H^2_{\mathrm{per}}(\Omega_{\mathrm{cell}}) \hookrightarrow L^2_{\mathrm{per}}(\Omega_{\mathrm{cell}})}$ compact; composing that embedding with the elliptic inverse gives compact resolvent. The spectrum is therefore a discrete multiset of real eigenvalues of finite multiplicity, with no finite accumulation point [Kuc16, arXiv v4 evidence copy, Remark 4.7, manuscript pp. 24-25, Eqs. (4.14)-(4.15), and §5, p. 26, Lemma 5.2 with Eq. (5.2), numbering collated by work identity; Tes14a, Second edition, printed pp. 157-159, §6.1, definition (6.1), Lemmas 6.2-6.3, and Theorem 6.4, official errata p. 5; Hun09, Printed pp. 101-102, §4.8 and Theorem 4.23, printed p. 104, §4.10 and Theorem 4.25].

There is also a shorter route under a strictly stronger hypothesis. If ${V\in C^\infty(\mathbb R^d/\Lambda;\mathbb R)}$ , then the whole fiber ${\hat H(\mathbf k)}$ is a symmetric classical elliptic differential operator of order two on the compact boundaryless torus ${\mathbb R^d/\Lambda}$ . The compact-manifold elliptic spectral theorem then packages self-adjointness, compact resolvent, and completeness of the eigenfunctions into one result [Mel22, Printed p. 114, §10, Theorem 7.2 and proof]. That one-theorem route is not a direct proof for the standing assumption ${V\in L^\infty}$ : multiplication by a merely bounded potential need not be a classical order-zero pseudodifferential operator. For bounded ${V}$ , the bounded-perturbation and Rellich route in the preceding paragraph is the controlling argument. The core uses only the spectral conclusion to enumerate ${E_n(\mathbf k)}$ ; proofs of the relative-boundedness, Rellich-compactness, and elliptic spectral theorems lie outside the lecture.

The crystal-momentum-resolved eigenproblem

The cell operator has supplied a spectral problem, while its parameter labels a translation sector. We can now separate what an experiment reads from what the operator family predicts, and then state the boundary of the one-particle description.

What diffraction and spectroscopy determine

For silicon, the operator ${\hat H(\mathbf k)}$ would act on the reciprocal-class plane-wave subspace containing a large basis resolving the periodic potential and orbital content inside a primitive cell. Solving it produces several functions ${E_n(\mathbf k)}$ , not one universal parabola. Translation symmetry guarantees that the calculation may be performed sector by sector and that reciprocal representatives describe the same sector. It does not determine the numerical dispersion, the indirect band gap, or the photoemission intensity. Those outputs depend on the microscopic Hamiltonian and, for intensity, on the probe matrix element.

This separates the roles of the two experimental coordinates introduced at the start. Diffraction determines which displacements ${\mathbf G}$ belong to the reciprocal lattice and hence which labels are equivalent. Photoemission can then place removal-energy features against a chosen representative ${\mathbf k}$ [Hen13, IUCr publisher HTML section 1, diffraction-geometry and structure-solution paragraphs; DHS03, arXiv v1 evidence copy, manuscript PDF pp. 3-7, §II.A-C, Eqs. (1)-(12), numbering collated by work identity]. Feature visibility also depends on surface sensitivity, occupation, lifetime broadening, and dipole matrix elements. The Bloch eigenproblem is therefore the kinematic backbone of a band measurement, not a complete theory of the measurement.

Before reading Figure 3.3, predict which quantity an elastic detector records directly and which reciprocal-lattice statement is inferred from the repeated spots. In particular, decide whether the pattern can by itself supply a band energy.

Figure 3.3. A single-crystal diffractometer connects measured outgoing angles to a computed reciprocal-lattice detector pattern without identifying spot positions as band energies. [BB13, Royal Society printed pp. 428-430 and 433-436, apparatus, reflections, and Bragg-law equation; Hen13, IUCr publisher HTML section 1, paragraph jointly describing the reciprocal lattice, Ewald sphere, Laue conditions, and Bragg law]

Panel (a) identifies the apparatus record: incident setting, outgoing angle, and elastic wave-vector difference. Panel (b) enlarges the sample region. Panel (c) is an illustrative data product, not measured data from a particular material. Repeated bright positions support the inference ${\mathbf q=\mathbf G}$ under the declared elastic model; they do not measure ${E_n(\mathbf k)}$ . Diffraction therefore fixes the translation geometry that makes the Bloch eigenproblem well posed, while spectroscopy is a separate measurement layer.

Reading an actual diffraction record

First connect the sample to the signal. In Figure 3.4, which direction is normal to the marked crystal face, and what quantity becomes the vertical axis of the angular record? Reuse Figure 3.3, panels (a,b), for the elastic incident/outgoing geometry; its modern area detector is not the historical readout device.

Figure 3.4. Rock-salt structure and the historical signal chain. Panel (a) is a schematic conventional cell, not an atomic image inferred from one trace; panel (b) explains how reflected X-rays become an angular ionization record. [BB13, printed pp. 429-430, ionization-current readout and angular sweep; p. 437, alternating species and layers]

The dark-blue and pale-orange spheres distinguish the two FCC sublattices; each ion has six unlike nearest neighbours in the extended crystal. Sphere radii are not ionic radii. The highlighted (100) face has normal [100]; this face label is not an assignment of the first observed reflection to an allowed (100) reciprocal-lattice vector. No absolute lattice scale is inferred here. The Braggs recorded ionization current while locating the reflected maximum for each crystal setting. The incidence angle is ${\theta}$ , while the detector lies near ${2\theta}$ . Panel (b) shows this signal mapping without inventing a historical chamber design. Both panels are schematic, not to scale.

Why does the signal concentrate at particular angles? In Figure 3.5, compare waves scattered at ${P}$ and ${Q}$ , one plane deeper. Identify both extra path segments before reading the peak condition.

Figure 3.5. Extra path length selects the angles of coherent reinforcement. Panel (a) is a schematic scattering-plane section; panel (b) is an ideal single-wavelength probe comparison, not a simulation of the measured rock-salt record. [BB13, printed pp. 435-436, angular selectivity, corrected sine ratios and Bragg equation]

The dashed lines through ${P}$ are perpendicular to the incident and outgoing rays and provide equal-phase comparison fronts. Relative to scattering at ${P}$ , the deeper path contains ${CQ+QD=2d\sin\theta}$ . Its phase delay is ${\varphi=2\pi(2d\sin\theta)/\lambda}$ . When ${2d\sin\theta=n\lambda}$ , successive planes contribute with the same phase: their amplitudes add before the detector measures intensity. The planes represent equivalent scattering layers, not literal mirrors. The ray construction also applies to elastic neutron scattering. This ideal equal-site comparison is separate from the rock-salt sample and historical X-ray apparatus in Figure 3.4 and Figure 3.3.

Panel (b) isolates this mechanism with ${N=8}$ equal-amplitude planes and ${\lambda/d=0.8}$ , using ${I/(N^2I_1)=N^{-2}\left|\sum_{j=0}^{N-1}\mathrm e^{\mathrm i j\varphi}\right|^2}$ , where ${I_1}$ is the intensity from one plane. At a matching angle the normalized intensity is one; between matching angles the phase shifts suppress the sum, without requiring exact cancellation at every angle. Increasing the number of coherent planes narrows the peaks. Atomic scattering strengths, the arrangement within the unit cell, absorption and the instrument alter real peak heights; the atomic basis can even cancel an otherwise geometrically allowed reflection. This expression is the geometric factor and the default neutron-mode curve, using a constant coherent amplitude normalized to its forward value. The HTML X-ray mode multiplies it by the squared neutral-silicon atomic form-factor ratio to its forward value. The implementation uses a reference length of three angstroms to convert the displayed reduced wavelength and spacing to physical units; its coefficient provenance is retained with the figure model. Switching probe preserves spacing, wavelength and selected angle. The modes compare relative angular dependence, not absolute count rates, and neither is a rock-salt or diamond-silicon crystal calculation. The basis, absorption, instrument response and background remain omitted. The geometric matching angles are shared, while the X-ray weighting changes relative peak heights. The interactive scan uses detector angle twice the incidence angle; the historical trace below uses incidence angle.

The detector pattern above is computed. Now ask what the observed angular record contributes. Figure 3.6 shows Bragg and Bragg’s reflected ionization signal from the (100) face of rock-salt. The source contains several X-ray wavelengths: the labelled ${B_1,B_2,B_3}$ belong to one spectral family, whereas the other peaks must not be counted as intervening orders of that same wavelength. Can one family test the integer-order condition? The sample and signal providers are Figure 3.4, panels (a,b), and the elastic geometry in Figure 3.3, panels (a,b).

Figure 3.6. The same spectral family recurs at progressively weaker reflections. Continuous line: digitized experimental trace for rock-salt (100), with the drawn baseline and B1 height used only to set an arbitrary vertical scale. [BB13, fig. 3(I), printed p. 431]

The horizontal axis is the incidence angle ${\theta}$ , not the detector angle ${2\theta}$ . Original acquisition samples and experimental uncertainties are unavailable; recovery accuracy is approximately ${0.2^\circ}$ horizontally and 0.02 on the relative vertical scale. The drawing is not the source of the corrected numerical angles below. Adapted plotting of one historical figure; acknowledgement to W. H. Bragg, W. L. Bragg and the Royal Society under its historical-content reuse policy.

For parallel planes separated by ${d}$ , the path difference is ${2d\sin\theta}$ ; constructive interference at a fixed wavelength requires ${2d\sin\theta_n=n\lambda}$ . The authors separately report corrected detector angles on p. 435. These are the numerical inputs for the test, not more precise readings of the drawing [BB13, printed pp. 435-436, corrected B-line angles and diffraction interpretation].

Reflection	Reported ${2\theta}$	Incidence ${\theta}$	${\sin\theta/\sin\theta_1}$
${B_1}$	${23.1^\circ}$	${11.55^\circ}$	1.000
${B_2}$	${47.3^\circ}$	${23.65^\circ}$	2.004
${B_3}$	${73.3^\circ}$	${36.65^\circ}$	2.981

For example, ${\sin(23.65^\circ)/\sin(11.55^\circ)\simeq2.004}$ , so the second peak is consistent with twice the path difference, not twice the angle. The three ratios approximately follow ${1:2:3}$ ; the displayed digits expose the finite discrepancy and do not assert experimental precision at the last digit. The weakest third reflection is retained rather than replaced by an ideal peak. This is evidence for reciprocal periodicity at a fixed spectral line. It still supplies neither a band energy nor, without an independent wavelength or spacing, an absolute lattice scale. We can now return to the operator family with its geometric labels experimentally grounded.

The operator family over crystal momentum

Expanding the square makes the role of the label more transparent:

$\hat H(\mathbf k) =\hat H(\mathbf 0) +\frac{\hbar}{m}\mathbf k\mathbin{\cdot}\hat{\mathbf p} +\frac{\hbar^2{|\mathbf k|}^2}{2m}\hat I.$

The three terms have different meanings. The first is the periodic cell operator at the origin of the chosen reciprocal-space coordinates. The term linear in ${\mathbf k}$ changes the relative energetic weight of periodic functions with different momentum content. The final term is a scalar shift within the fixed- ${\mathbf k}$ block. Thus ${\mathbf k}$ is a parameter of a family of operators, not an operator acting on the periodic factor. Solving a band problem means diagonalizing this family and following its eigenvalues and eigenspaces as the parameter changes.

This observation also prevents a tempting mistake. Replacing ${\hat{\mathbf p}}$ by ${\hbar\mathbf k}$ would discard the momentum carried by the cell-periodic factor and turn the operator problem into a number before it has been solved. The correct transformation shifts the operator: ${\hat{\mathbf p}\mapsto\hat{\mathbf p}+\hbar\mathbf k}$ . It does not erase ${\hat{\mathbf p}}$ .

The operator family is reciprocal-covariant rather than literally identical at ${\mathbf k}$ and ${\mathbf k+\mathbf G}$ :

$\hat H(\mathbf k+\mathbf G) =\mathrm e^{-\mathrm i\mathbf G\mathbin{\cdot}\hat{\mathbf r}} \hat H(\mathbf k) \mathrm e^{\mathrm i\mathbf G\mathbin{\cdot}\hat{\mathbf r}}.$

Therefore ${\sigma[\hat H(\mathbf k+\mathbf G)] =\sigma[\hat H(\mathbf k)]}$ , including multiplicities, although a chosen eigenvector need not be represented by the same components at the two points.

Bands as spectral families

The discrete-spectrum conclusion in Proposition 3.3 lets us state exactly what the word “band” means in this lecture without making its functional-analysis proof part of the core calculation.

Let ${\mathcal B=\mathbb R^d/\Lambda^*}$ denote the Brillouin torus. For each ${[\mathbf k]\in\mathcal B}$ , list the eigenvalues of the fiber operator ${\hat H(\mathbf k)}$ in nondecreasing order and count multiplicity: $E_1(\mathbf k)\leq E_2(\mathbf k)\leq\cdots, \qquad E_n(\mathbf k)\in\sigma[\hat H(\mathbf k)].$ The ${n}$ th energy band is the energy-ordered function ${E_n:\mathcal B\to\mathbb R}$ . At a degeneracy, adjacent functions take the same value; the energy ordering alone does not select a unique eigenvector or preserve an orbital identity through the degeneracy. When a later discussion instead follows a locally continuous orbital branch, that alternative labeling convention must be stated explicitly [Kuc16, arXiv v4 evidence copy, manuscript pp. 26-27, Definitions 5.3 and 5.7, numbering collated by work identity].

This convention makes ${E_n}$ a spectral value, not a state. The eigenspace may be multidimensional, and a chosen eigenvector may fail to extend as one globally single-valued smooth function even when the energy is continuous.

For a declared one-particle Hamiltonian and declared internal degrees of freedom, the band structure is the spectral-family map $\mathcal E:\mathcal B\longrightarrow \{\text{multisets in }\mathbb R\}, \qquad \mathcal E([\mathbf k])=\sigma[\hat H(\mathbf k)],$ where ${\sigma[\hat H(\mathbf k)]}$ denotes the spectrum of the cell operator ${\hat H(\mathbf k)}$ , here recorded with eigenvalue multiplicities retained [Kuc16, arXiv v4 evidence copy, manuscript pp. 25-27, Definition 5.1 and Theorem 5.5, numbering collated by work identity]. It is not a plot along one selected path in ${\mathcal B}$ , not the density of states obtained by aggregating over ${\mathbf k}$ , and not a photoemission intensity, which also contains occupation, lifetime, surface, and matrix-element information.

The many-body boundary

The one-particle clause now makes the promised interacting boundary precise [Osh00a, arXiv v1 evidence copy, manuscript pp. 1-3, Eqs. (1)-(7), comparing the total-translation step with the separately introduced quasiparticle step; numbering collated by work identity]. Let ${N\geq2}$ , and let ${\mathcal H_N}$ be the appropriate symmetric or antisymmetric subspace of ${L^2(\Omega^N,\mathbb C)}$ . An ${N}$ -particle coordinate wave function is a map ${\Psi:\Omega^N\to\mathbb C}$ in ${\mathcal H_N}$ . Simultaneous translation of every coordinate is the unitary operator ${\hat T_{\mathbf R}^{(N)}:\mathcal H_N\to\mathcal H_N}$ defined by

$(\hat T_{\mathbf R}^{(N)}\Psi) (\mathbf r_1,\ldots,\mathbf r_N) =\Psi(\mathbf r_1-\mathbf R,\ldots,\mathbf r_N-\mathbf R).$

Let ${W:\mathbb R^d\to\mathbb R}$ be a real pair-potential function that is periodic under the finite sample’s boundary-identification lattice, so it is a well-defined function of separation on the sample difference torus. For identical particles also require ${W(-\mathbf s)=W(\mathbf s)}$ . Sample periodicity is not a requirement of primitive-cell periodicity. Using raw coordinate differences inside the sample, without accounting for periodic boundaries, can change the pair energy when both particles are translated together and their coordinates are wrapped back into the sample. This spuriously breaks simultaneous-translation symmetry. A consistent minimum-image prescription accounts for periodic boundaries; summing all periodic images is not required merely to preserve this symmetry.

Now consider the self-adjoint operator ${\hat H_N:\mathcal D(\hat H_N)\subset\mathcal H_N\to\mathcal H_N}$ , on a domain invariant under simultaneous translations,

$\hat H_N =\sum_{i=1}^{N} \left[ \frac{\hat{\mathbf p}_i^{\,2}}{2m}+V(\hat{\mathbf r}_i) \right] +\sum_{i<j}W(\hat{\mathbf r}_i-\hat{\mathbf r}_j).$

The one-body potential is unchanged because ${V(\mathbf r_i+\mathbf R)=V(\mathbf r_i)}$ ; every pair separation is unchanged because ${(\mathbf r_i+\mathbf R)-(\mathbf r_j+\mathbf R)=\mathbf r_i-\mathbf r_j}$ . Consequently,

$[\hat H_N,\hat T_{\mathbf R}^{(N)}]=0, \qquad \hat T_{\mathbf R}^{(N)}\Psi_{\mathbf K} =\mathrm e^{-\mathrm i\mathbf K\cdot\mathbf R}\Psi_{\mathbf K}.$

Thus ${[\mathbf K]\in\mathbb R^d/\Lambda^*}$ is an exact total crystal- momentum label [Osh00a, arXiv v1 evidence copy, manuscript pp. 1-2, Eqs. (1)-(3), numbering collated by work identity]. To compare with translating only one coordinate, work first on the ambient labeled-coordinate space ${L^2(\Omega^N,\mathbb C)}$ , before restricting to symmetric or antisymmetric states. Such a translation generally changes pair separations and hence the interaction expression. It also need not preserve the exchange-symmetric physical subspace, so it is not in general an operator ${\mathcal H_N\to\mathcal H_N}$ whose commutator with the restricted ${\hat H_N}$ could be asserted. Simultaneous translation does preserve that subspace. The interaction can mix individual one-particle labels inside one conserved total sector.

Classify three claims as exact, conditional, or generally false: (i) total crystal momentum is conserved modulo ${\Lambda^*}$ ; (ii) every electron retains its individual ${\mathbf k}$ ; (iii) the one-particle functions ${E_n(\mathbf k)}$ defined above give the exact many-body excitation spectrum. For each classification, name the translation operation or spectral object being tested. Compare the three reasons with the instructor before continuing; the available tools are the coordinate-space Hamiltonian and its symmetry.

Electron addition and removal in an interacting system are described later by the momentum-resolved one-particle spectral distribution. Only under additional conditions does a narrow feature admit the controlled quasiparticle-pole interpretation. Broad continua or interaction-generated features can coexist with total crystal momentum without forming sharp one-particle-like branches. “Band” must therefore be read together with the mathematical object and approximation that it names.

The hat is retained because ${\hat H(\mathbf k)}$ acts on functions. After choosing an ${M}$ -element orthonormal basis ${\mathcal B(\mathbf k)}$ for a projected fiber, write its numerical matrix as

${\bigl[\hat H(\mathbf k)\bigr]}_{\mathcal B(\mathbf k)} =:H(\mathbf k)\in\operatorname{Herm}(M).$

Here ${H:\mathcal B\to\operatorname{Herm}(M)}$ is an ordinary matrix-valued function, without a hat, and ${\operatorname{Herm}(M)}$ is the set of ${M\times M}$ Hermitian complex matrices. A basis change alters the entries of ${H(\mathbf k)}$ but not the underlying operator or the spectral multiset ${\mathcal E([\mathbf k])}$ .

Before returning to the one-particle cell problem, read the symmetry reduction as a matrix. If its rows and columns are grouped by translation character, which entries must vanish, and which entries still require a dynamical model?

Figure 3.7. A Hamiltonian matrix grouped into translation-character sectors has many diagonal blocks and exact zero blocks between different characters. One diagonal block is enlarged into a symbolic Hermitian matrix with internal couplings. Each character corresponds to a crystal-momentum class; for interacting systems this is total crystal momentum.

The Hamiltonian in a symmetry-adapted basis. Each block ${H_{\chi_j}}$ represents ${\hat H}$ restricted to ${\mathcal H_{\chi_j}}$ , with ${\chi_j(\mathbf R)=\mathrm e^{-\mathrm i\mathbf k_j\cdot\mathbf R}}$ . Different characters, equivalently inequivalent crystal-momentum classes, cannot mix. States inside one sector can still couple. The displayed three-state block is symbolic, not a universal block dimension; drawn block sizes carry no dimension information. The finite-periodic decomposition also applies to interacting systems, with total crystal momentum as the label.

The projector proof used only the translation representation and its commutation with the Hamiltonian to obtain ${\hat P_{\chi_i}\hat H\hat P_{\chi_j}=0}$ for ${i\ne j}$ . It did not use a single-particle wavefunction for that step. A one-dimensional character means that translation multiplies every vector in its sector by the same phase, not that the sector contains only one state. Block diagonalization organizes the eigenproblem; diagonalizing each ${H_{\chi_j}}$ remains a separate task. For interacting particles the translations act on the whole system, as above.

It is useful to see what the block construction has and has not simplified. The original eigenproblem lives over the entire crystal. For each allowed ${\mathbf k}$ , ${\hat H(\mathbf k)}$ instead acts on functions in one unit cell with periodic boundary conditions. The label has removed couplings between inequivalent translation characters, but it has not made the within-cell problem one dimensional. There may be many orbitals, plane-wave components, or sublattices in the same block. A “band index” ${n}$ enumerates the eigenvalues of that internal problem.

This reduction is also the practical value of Bloch’s theorem. Analytic tight-binding calculations and numerical electronic-structure methods solve the smaller cell problem over a mesh of ${\mathbf k}$ values instead of diagonalizing one crystal-sized matrix. Translation symmetry provides the computational decomposition; microscopic physics still supplies every entry inside a block.

Zone folding and the separate gap-opening criterion

The opening pair can now be put into its common translation block. First keep the free-particle Hamiltonian fixed and fold its branches; then change a single matrix element and observe which part of the spectrum changes.

Folded free branches

For a free electron, the periodic part may be chosen as a plane wave with a reciprocal vector:

$u_{\mathbf k,\mathbf G}(\mathbf r) =\mathrm e^{\mathrm i\mathbf G\mathbin{\cdot}\mathbf r}, \qquad E_{\mathbf G}(\mathbf k) =\frac{\hbar^2{|\mathbf k+\mathbf G|}^2}{2m}.$

Drawing all ${\mathbf G}$ branches in one first zone produces crossings [Bri30, HAL French manifestation, §§2-4, printed pp. 377-384 and 392-396]. This is a folded representation of the free-particle parabola. When a periodic potential is turned on, its Fourier components may couple branches within the same block. We can already predict the logic: a crossing gaps only if the relevant coupling matrix element is nonzero and allowed by the other symmetries. The tight-binding construction in Lecture 4 will make the same logic visible with localized orbitals and fewer matrix entries.

Locate the meeting of two folded free-electron branches and determine whether they represent the same physical plane wave. In one dimension choose the zone ${[ -\pi/a,\pi/a )}$ and write reciprocal vectors as ${G_{\ell}=2\pi\ell/a}$ , with ${\ell\in\mathbb Z}$ . The free branches are

$E_{\ell}(k)=\frac{\hbar^2}{2m} {\left(k+\frac{2\pi\ell}{a}\right)}^2.$ Compare ${\ell=0}$ and ${\ell=-1}$ , including their velocities at the meeting point. No periodic potential is present.

Compare the same-momentum energies, then use the plane-wave momenta and slopes to distinguish degeneracy from identity of states.

Solving ${E_0(k)=E_{-1}(k)}$ on the extended branches gives the boundary endpoint

$k^2={\left(k-\frac{2\pi}{a}\right)}^2 \quad\Longrightarrow\quad k=\frac{\pi}{a}.$

The endpoint ${+\pi/a}$ is represented inside the chosen half-open zone by ${-\pi/a}$ ; the same two waves then have reciprocal labels ${+1}$ and ${0}$ , respectively. This is a change of labels, not a coupling. The two basis states at the identified boundary are the distinct plane waves ${|+\pi/a\rangle}$ and ${|-\pi/a\rangle}$ . Their group velocities are

$v_0\!\left(\frac{\pi}{a}\right) =\frac{1}{\hbar}{\left.\frac{\mathrm dE_0}{\mathrm dk}\right|}_{k=\pi/a} =+\frac{\hbar\pi}{ma}, \qquad v_{-1}\!\left(\frac{\pi}{a}\right) =\frac{1}{\hbar}{\left.\frac{\mathrm dE_{-1}}{\mathrm dk}\right|}_{k=\pi/a} =-\frac{\hbar\pi}{ma}.$

Both energies there equal ${\hbar^2\pi^2/(2ma^2)}$ , which checks the substitution into either parabola. Their opposite velocities distinguish the two states even though their crystal-momentum characters agree. The velocity scale ${\hbar/(ma)}$ has length/time units; the energy scale has energy units. At ${k=0}$ , the two branches instead have energies zero and ${\hbar^2{(2\pi/a)}^2/(2m)}$ , checking that the equality is localized to the zone edge. Nothing in this relabeling supplies an off-diagonal coupling; that is the additional input in the next calculation.

A coupling opens the avoided crossing

Thus the folded labels meet at equal energy while retaining opposite velocities. Now isolate the missing dynamical move. In the ordered basis ${\mathcal B(k)=(|k\rangle,|k-2\pi/a\rangle)}$ , consider an operator ${\hat H_g(k)}$ whose matrix representation is

${\bigl[\hat H_g(k)\bigr]}_{\mathcal B(k)} {=} \begin{pmatrix} E_0(k) & g\\ g^* & E_{-1}(k) \end{pmatrix}, \qquad g\in\mathbb C.$

The characteristic equation is

$0=\det\!\left( {\bigl[\hat H_g(k)\bigr]}_{\mathcal B(k)}-E I_2 \right) =\bigl[E_0(k)-E\bigr]\bigl[E_{-1}(k)-E\bigr]-{|g|}^2.$

Completing the square in ${E}$ , rather than quoting a two-level formula, gives

$\left[E-\frac{E_0(k)+E_{-1}(k)}2\right]^2 =\left[\frac{E_0(k)-E_{-1}(k)}2\right]^2+|g|^2,$

(3.4)

$E_{\pm}(k) =\frac{E_0(k)+E_{-1}(k)}{2} \pm \sqrt{ {\left[\frac{E_0(k)-E_{-1}(k)}{2}\right]}^2+{|g|}^2 }.$

At ${k=\pi/a}$ , both diagonal entries equal ${E_{\mathrm{edge}}=\hbar^2\pi^2/(2ma^2)}$ , so

$E_{\pm}\!\left(\frac{\pi}{a}\right) =E_{\mathrm{edge}}\pm|g|, \qquad \Delta_{\mathrm{edge}}=2|g|.$

The relabeling in Figure 3.1 supplied only the bare energies. With ${\kappa=ka/\pi}$ they are ${E_0/E_{\mathrm R}=\kappa^2}$ and ${E_{-1}/E_{\mathrm R}={(\kappa-2)}^2}$ . Changing the off-diagonal entry to ${|g|/E_{\mathrm R}=0.2}$ , while keeping those diagonal entries fixed, instead gives ${E_-/E_{\mathrm R}=0.8}$ and ${E_+/E_{\mathrm R}=1.2}$ at the edge, with separation ${0.4E_{\mathrm R}}$ . The later explorer and its separate static fallback display this coupled default. The opening figure does not: its horizontal arrow adds no matrix element.

Two checks distinguish an eigenvalue calculation from a change of labels. The trace is unchanged: ${E_++E_-=E_0+E_{-1}}$ . As ${g\to0}$ , the unordered pair ${\{E_+,E_-\}}$ becomes ${\{E_0,E_{-1}\}}$ , with a degeneracy when the diagonal entries agree. Every term under the square root has units of energy squared. These eigenvalues are exact for the declared two-state matrix; using that matrix as a truncation of a microscopic crystal requires control of the omitted states. The edge splitting ${2|g|}$ is local to this pair. Other momenta or bands may still contribute states in that interval, so it does not by itself establish a global insulating gap.

This is the error-localization checkpoint: if folding alone changes either energy, one has changed the Hamiltonian while claiming only to change its labels. A real periodic potential can also shift the diagonal entries, but those shifts are additional microscopic information, not consequences of the fold.

Suppose two displayed branches share a representative in a reduced Brillouin zone. That coincidence establishes only that the corresponding components can be represented in the same translation block. A local avoided-crossing splitting requires the microscopic Hamiltonian in that block and a nonzero symmetry-allowed coupling at the pair’s meeting [KP31, Royal Society printed §1, pp. 500-503, Eq. (6), Figs. 2-3, summary p. 513]. A global spectral gap additionally requires excluding states from other momenta and bands in the same energy interval. If the coupling vanishes at a meeting, that meeting is not avoided. If another symmetry forces the coupling to vanish, it forbids level repulsion wherever the diagonal energies meet; whether the meeting persists and where it occurs still depend on the perturbed diagonal dispersions.

Use the same two-state construction with ${E_0=2E_{\mathrm R}}$ , ${E_{-1}=5E_{\mathrm R}}$ , and ${g=\mathrm iE_{\mathrm R}/2}$ . Complete ${E_++E_-=\underline{\qquad}}$ and ${{[E-7E_{\mathrm R}/2]}^2=\underline{\qquad}}$ , keeping units explicit. Which unordered pair must be recovered when the coupling is turned off? Compare the completed-square and zero-coupling checks with the instructor.

A proposed argument reads: “The two free plane waves have the same reduced zone-boundary label. Therefore folding assigns their off-diagonal matrix entry the energy ${E_{\mathrm R}}$ , so their separation becomes ${2E_{\mathrm R}}$ , even though the potential is unchanged.” Identify the first invalid step, write the matrix consistent with an unchanged free Hamiltonian, and give a fast eigenvalue check. In your repair, separate the information supplied by translation labels, by the free dispersion, and by an actual coupling. Compare the repaired reasoning before the cell-change test.

Complete the two preceding checks before varying the model. The dashed curves keep the free energies ${E_0/E_{\mathrm R}=\kappa^2}$ and ${E_{-1}/E_{\mathrm R}={(\kappa-2)}^2}$ fixed, where ${\kappa=ka/\pi}$ . Change ${|g|/E_{\mathrm R}}$ and compare them with Equation (3.4). At the edge the splitting is ${2|g|}$ ; away from the edge the pointwise separation also contains the diagonal mismatch. Turning the coupling off recovers the same unordered pair of bare energies, and the trace never changes. This is an exact two-state calculation, not a microscopic prediction of a crystal’s global insulating gap.

The same reasoning transfers immediately to a rectangular lattice. If ${\mathbf a_1=a_x\mathbf e_x}$ and ${\mathbf a_2=a_y\mathbf e_y}$ , then ${\mathbf b_1=(2\pi/a_x)\mathbf e_x}$ and ${\mathbf b_2=(2\pi/a_y)\mathbf e_y}$ . A point with ${k_x=1.4\pi/a_x}$ can be folded by subtracting ${\mathbf b_1}$ , while its ${k_y}$ coordinate is untouched. The phase check factorizes into the two primitive directions. Nothing in the argument depended on the square lattice except the convenient shape of the displayed zone.

Changing the representation cell preserves state count

Take a one-dimensional ring with an even number ${N}$ of primitive cells of length ${a}$ . Keep the physical length ${L=Na}$ , periodic boundary condition and Hamiltonian fixed, but choose representation cells of length ${a'=2a}$ . For one original band, determine the allowed wave-number spacing, the number of reduced-zone labels, and the total state count. Does this change of representation alone couple the folded sectors?

Count the boundary-allowed momenta before pairing their reduced-zone labels. A one-orbital-per-cell projected model has ${N}$ states in its entire one-particle space. A continuum cell generally has infinitely many bands, and each original band contributes its own pair of folded branches. If we merely choose a doubled representation cell ${a'=2a}$ , its reciprocal generator becomes

$G'=\frac{2\pi}{a'}=\frac{\pi}{a},$

and a half-open reduced zone may be chosen as ${[ -\pi/(2a),\pi/(2a) )}$ . Each new-zone representative ${k'}$ collects two old-zone representatives,

$k' \quad\text{and}\quad k'+G' \pmod{\frac{2\pi}{a}}.$

The ring contains ${N/2}$ doubled cells. Its boundary condition gives ${\exp[-\mathrm i k'(2a)(N/2)]=1}$ , hence the spacing remains ${\Delta k'=2\pi/L}$ . There are ${N/2}$ such labels in a zone of width ${G'=\pi/a}$ . At the finite-index level the old labels ${k_m=2\pi m/L}$ pair as

$(m,m+N/2),\qquad m=0,\ldots,N/2-1, \qquad k_{m+N/2}-k_m=\pi/a=G'.$

Reduce either member into the declared half-open new zone to obtain ${k'}$ . The pairs exhaust the original ${N}$ labels exactly once. Each new block therefore displays two branches per original band, and the count is

$\underbrace{\frac{N}{2}}_{\text{new crystal-momentum labels}} \times \underbrace{2}_{\text{folded branches per label}} =N,$

exactly the original one-particle state count for one orbital per primitive cell [ABCS13a, arXiv v1 evidence copy, manuscript pp. 1-3, Eqs. (1)-(7), numbering collated by work identity]. No operator matrix element has changed. The reduced zone and doubled branch count are representational; a local avoided-crossing splitting would still require the nonzero coupling in Equation (3.4), subject to the symmetry and global-gap qualifications in Criterion 3.1.

The unchanged primitive translation supplies an independent check. The paired states have ${\hat T_a}$ eigenvalues ${t_1=\mathrm e^{-\mathrm i k'a}}$ and ${t_2=\mathrm e^{-\mathrm i(k'+G')a}=-t_1}$ . Since ${[\hat H,\hat T_a]=0}$ , their Hamiltonian matrix element obeys $\langle1|\hat H|2\rangle =\langle1|\hat T_a^\dagger\hat H\hat T_a|2\rangle =t_1^*t_2\langle1|\hat H|2\rangle =-\langle1|\hat H|2\rangle=0.$ The larger display block still contains two uncoupled primitive sectors. A physical period-doubling modulation may remove this extra ${a}$ -translation symmetry; choosing a larger cell on paper does not. This count supplies the baseline for deciding whether the next supercell choice merely changes representation or changes the physical Hamiltonian.

For ${N=10}$ , complete the index pairs ${(0,\underline{\quad}),(1,\underline{\quad}),\ldots, (4,\underline{\quad})}$ when the representation cell is doubled. Give the new-zone width, the allowed-label spacing, and the product of labels and branches per original band. Use the retained primitive translation to check whether folding creates a coupling. Compare the count and symmetry check with the instructor before choosing a different supercell.

A twelve-cell ring originally has spacing ${a}$ and a generic potential of primitive period ${a}$ . Compare two changes: (A) describe the unchanged Hamiltonian using cells of length ${3a}$ ; (B) additionally apply ${\delta V(x)=2w\cos[2\pi x/(3a)]}$ , with real nonzero ${w}$ . Choose a useful translation-sector organization for each case. Account for the allowed labels and branches per original band, identify which primitive translation symmetry remains available, and decide whether coincident folded labels merely coexist or may now couple. Does either description alone prove a nonzero global band gap? Defend the choice with a state-count check and the limit ${w\to0}$ . Present the argument before the instructor comparison.

Scope of the construction

Close the equations and reconstruct five claims.

Why is a translation eigenvalue a phase rather than necessarily one?
Why do ${\mathbf k}$ and ${\mathbf k+\mathbf G}$ label the same translation sector even though their plane-wave momenta differ?
Which step turns a crystal-sized eigenproblem into cell-sized problems?
What information about a crossing is still missing after that step?
If the real-space unit cell is doubled as a representation choice, which state count is preserved even though the displayed Brillouin zone shrinks?

Give one connected argument before the instructor comparison. Identify the earliest unsupported link if a conclusion cannot be recovered; return to that calculation rather than substituting the appearance of a plotted curve.

The unresolved crossing. Translation symmetry has now done real work. It has supplied the conserved label ${\mathbf k}$ , identified equivalent representatives, and block-diagonalized the eigenproblem into independent ${\mathbf k}$ sectors. But suppose a fixed block contains two basis states. Translation symmetry alone does not tell us their diagonal energies, their off-diagonal matrix element, or whether another symmetry forces that element to vanish. We need a microscopic model of the block.

Return now to silicon. Its reciprocal lattice fixes which crystal-momentum representatives are equivalent, and Bloch’s theorem allows one-electron energy eigenstates to be chosen in Bloch form. An arbitrary superposition of degenerate energy eigenstates need not carry a single translation character. Neither fact tells us the size or location of its band gap. That requires the matrix elements of an appropriate microscopic Hamiltonian. Likewise, a photoemission band map can locate a dispersing feature, but its absence at one momentum does not prove that the eigenstate is absent: occupation, lifetime, surface projection, and transition matrix elements must be separated. Bloch labeling tells us where states live in symmetry-label space; the Hamiltonian and probe tell us what occurs there and what becomes visible.

This is the larger course distinction established by the lecture. Translation symmetry first partitions states into crystal-momentum sectors. For the one-particle Hamiltonian used here, the spectra inside those sectors are the energy bands. If a translation-invariant interaction is restored, simultaneous translation still supplies total crystal momentum, but the interaction couples different individual electron labels and the exact excitation spectrum is not obtained by listing the one-particle functions ${E_n(\mathbf k)}$ . Later occupation and response theory determines which one-particle states matter; the many-body spectral theory of Lectures 13–14 determines whether an interacting system exhibits narrow quasiparticle branches, broad spectral weight, or both. The hierarchy is therefore symmetry label, declared spectral object, microscopic dynamics, and finally probe—not “crystal momentum implies bands.”

Before treating a band plot as evidence, identify the translation sector, the chosen representative of ${[\mathbf k]}$ , and the spectral subspace whose energy is drawn. A reciprocal relabeling may alter the picture without changing that subspace or spectrum; a gap claim additionally requires a changed Hamiltonian matrix element and, experimentally, a declared probe model. This is the operational distinction that the folding calculation was designed to establish.

Problems

1. Translation characters and reciprocal representatives

A one-dimensional crystal contains ${N}$ cells of spacing ${a}$ with periodic boundary conditions. Let ${\hat T_a}$ translate a state by one cell.

Use ${\hat T_a^N=\hat I}$ and unitarity to find every possible translation eigenvalue. Express the answer as allowed wave numbers ${k_m}$ .
Show directly that ${k_m}$ and ${k_m+2\pi\ell/a}$ , ${\ell\in\mathbb Z}$ , define the same translation character on every lattice vector. Explain why they are different representatives of one label rather than different translation sectors.
For ${\hat H=\hat p^{\,2}/(2m)+V(x)}$ , with ${V(x+a)=V(x)}$ , insert ${\psi_k(x)=\mathrm e^{\mathrm i kx}u_k(x)}$ into the stationary Schrödinger equation and derive the cell operator ${\hat H(k)}$ acting on periodic functions ${u_k:[0,a]\to\mathbb C}$ .
In the free limit, write the folded branches ${E_\ell(k)}$ in the first Brillouin zone and locate the crossing between ${\ell=0}$ and ${\ell=-1}$ in the continuous Bloch family. State which parity of ${N}$ allows the finite ring to sample that exact crossing momentum. State precisely what translation symmetry has determined and what additional information is required before claiming that the crossing opens into a gap.

2. Momentum content of one Bloch state

In one dimension, let a finite ring have length ${L=Na}$ , let ${G=2\pi/a}$ , and choose an allowed ${k}$ . With the common plane-wave normalization made explicit, set

$\psi_k(x)=\frac{1}{\sqrt L}\mathrm e^{\mathrm i kx}u_k(x), \qquad u_k(x)=\alpha+\beta\mathrm e^{\mathrm iGx}, \qquad {|\alpha|}^2+{|\beta|}^2=1.$

Use the convention ${(\hat T_R\psi)(x)=\psi(x-R)}$ .

Verify that ${u_k}$ is cell periodic and find the translation eigenvalue of ${\psi_k}$ for every lattice vector ${R=ja}$ .
List the possible canonical-momentum measurement results and their probabilities. Calculate ${\langle\hat p\rangle}$ .
Relabel the same translation sector by ${k'=k+G}$ . Construct a periodic factor ${u'_{k'}}$ that leaves the full wave function unchanged and show how the two Fourier coefficients are reindexed.
For ${\alpha=\beta=1/\sqrt2}$ , explain why crystal momentum labels a translation sector but does not equal the canonical-momentum expectation.

3. Diagnose a folded crossing and a gap

At the one-dimensional zone boundary ${k=G/2}$ , the free-electron plane waves ${|k\rangle}$ and ${|k-G\rangle}$ are degenerate with energy ${E_0}$ . Suppose a weak real periodic potential has Fourier matrix element ${\langle k|\hat V|k-G\rangle=V_G}$ , where a basis-phase choice makes ${V_G\in\mathbb R}$ . Subtract the mean component ${V_0}$ from the energy zero; thus the diagonal energy in this problem remains ${E_0}$ .

Write the ${2\times2}$ Hamiltonian in this degenerate subspace and find its eigenvalues and normalized eigenvectors.
State the gap and its limit as ${V_G\to0}$ .
Repeat the eigenvalue conclusion when the off-diagonal element is ${|V_G|\mathrm e^{\mathrm i\phi}}$ . Which observable is independent of ${\phi}$ , and which eigenvector data change?
Separate three statements: what zone folding did, what translation symmetry allowed, and what microscopic information actually opened the gap.

(advanced): Translation Bands for Lattice Vibrations and Other Degrees of Freedom

This appendix is optional and deletion safe: no later core lecture requires it. Its purpose is to separate the group-theoretic origin of a band label from the physical meaning of the eigenvalue. The core lectures solved a stationary Schrödinger problem,

$\hat H(\mathbf k)|u_{n\mathbf k}\rangle =E_n(\mathbf k)|u_{n\mathbf k}\rangle.$

Translation symmetry did not know that the spectral value would be an energy. It supplied the character ${\mathrm e^{-\mathrm i\mathbf k\cdot\mathbf R}}$ and the reduction to a fixed- ${\mathbf k}$ internal problem. A periodic vibration uses the same characters but produces squared frequencies and polarization vectors instead.

Consider a one-dimensional diatomic chain. Cell positions are ${R_\ell=\ell a}$ , with ${\ell\in\mathbb Z_N}$ , and each cell contains masses ${M_{\mathrm A}>0}$ and ${M_{\mathrm B}>0}$ . Identical springs of stiffness ${K>0}$ join ${\mathrm A_\ell}$ to ${\mathrm B_\ell}$ and ${\mathrm B_\ell}$ to ${\mathrm A_{\ell+1}}$ . The longitudinal displacements ${u_{\ell\mathrm A},u_{\ell\mathrm B}:\mathbb R\to\mathbb R}$ are ordinary real-valued functions of time. They are classical coordinates here, not wave functions and not operators.

The equations of motion are

$M_{\mathrm A}\ddot u_{\ell\mathrm A} =-K(2u_{\ell\mathrm A}-u_{\ell\mathrm B}-u_{\ell-1,\mathrm B}),$

$M_{\mathrm B}\ddot u_{\ell\mathrm B} =-K(2u_{\ell\mathrm B}-u_{\ell\mathrm A}-u_{\ell+1,\mathrm A}).$

The lattice is periodic, so ask for a displacement pattern with a definite translation character:

$u_{\ell\alpha}(t) =U_\alpha(k)\,\mathrm e^{\mathrm i(kR_\ell-\omega t)}, \qquad \alpha\in\{\mathrm A,\mathrm B\}.$

The physical displacement is the real part. The complex amplitude is a calculation device that turns translation into multiplication by a phase. Substitution gives

$\omega^2 \begin{pmatrix}M_{\mathrm A}&0\\0&M_{\mathrm B}\end{pmatrix} \begin{pmatrix}U_{\mathrm A}\\U_{\mathrm B}\end{pmatrix} =K \begin{pmatrix} 2&-(1+\mathrm e^{-\mathrm i ka})\\ -(1+\mathrm e^{\mathrm i ka})&2 \end{pmatrix} \begin{pmatrix}U_{\mathrm A}\\U_{\mathrm B}\end{pmatrix}.$

This generalized eigenproblem is already a band problem, but the natural inner product is mass weighted [ABCS13a, arXiv v1 evidence copy, Appendix A, manuscript pp. 10-11, Eqs. (A1)-(A8), numbering collated by work identity]. Define ${\mathbf x={(\sqrt{M_{\mathrm A}}U_{\mathrm A}, \sqrt{M_{\mathrm B}}U_{\mathrm B})}^{\mathsf T}\in\mathbb C^2}$ . Then

(3.5)

$D(k)\mathbf x_\lambda(k) =\omega_\lambda^2(k)\mathbf x_\lambda(k),$

$D(k)= \begin{pmatrix} 2K/M_{\mathrm A}&-K(1+\mathrm e^{-\mathrm i ka})/\sqrt{M_{\mathrm A}M_{\mathrm B}}\\ -K(1+\mathrm e^{\mathrm i ka})/\sqrt{M_{\mathrm A}M_{\mathrm B}}&2K/M_{\mathrm B} \end{pmatrix}.$

Here ${\mathrm{BZ}_{1\mathrm D}=[-\pi/a,\pi/a)}$ is the chosen half-open one-dimensional zone, and ${D:\mathrm{BZ}_{1\mathrm D}\to\operatorname{Herm}(2)}$ is an ordinary matrix-valued function. Its eigenvectors are polarization coefficient vectors; after normalization they can be chosen locally as sections of the mode eigenspaces, but no section or gauge construction is needed for the present calculation [ABCS13a, arXiv v1 evidence copy, Appendix A, manuscript pp. 10-11, Eqs. (A1)-(A8), numbering collated by work identity]. Solving the characteristic equation gives

$0=\det[D(k)-\omega^2 I_2] =\left(\frac{2K}{M_{\mathrm A}}-\omega^2\right) \left(\frac{2K}{M_{\mathrm B}}-\omega^2\right) -\frac{4K^2\cos^2(ka/2)}{M_{\mathrm A}M_{\mathrm B}}.$

Expanding the product is a useful factor check:

$\omega^4-2K\left(\frac1{M_{\mathrm A}}+ \frac1{M_{\mathrm B}}\right)\omega^2 +\frac{4K^2\sin^2(ka/2)}{M_{\mathrm A}M_{\mathrm B}}=0.$

The constant term vanishes at ${k=0}$ , so one root must be the acoustic zero mode before the quadratic formula is used. Applying that formula gives

$\omega_\pm^2(k) =K\left(\frac1{M_{\mathrm A}}+\frac1{M_{\mathrm B}}\right) \pm K\sqrt{ {\left(\frac1{M_{\mathrm A}}+\frac1{M_{\mathrm B}}\right)}^2 -\frac{4\sin^2(ka/2)}{M_{\mathrm A}M_{\mathrm B}}}.$

Before interpreting the two curves, check the point at which a wrong normalization is easiest to hide. At ${k=0}$ , the lower branch satisfies ${\omega_-(0)=0}$ . Its physical displacement has ${U_{\mathrm A}=U_{\mathrm B}}$ : translating the entire chain costs no spring energy. The upper branch instead has

$\omega_+^2(0)=2K\left(\frac1{M_{\mathrm A}}+\frac1{M_{\mathrm B}}\right), \qquad M_{\mathrm A}U_{\mathrm A}+M_{\mathrm B}U_{\mathrm B}=0,$

so the two atoms move against one another while the cell center of mass stays fixed. These are the acoustic and optical zone-center patterns. Their limiting motion determines the names.

At the zone edge ${k=\pi/a}$ , the off-diagonal entries vanish and the mass-weighted eigenvectors may be chosen as

$\mathbf x_{\mathrm A}=\begin{pmatrix}1\\0\end{pmatrix}, \quad \omega_{\mathrm A}^2=\frac{2K}{M_{\mathrm A}}, \qquad \mathbf x_{\mathrm B}=\begin{pmatrix}0\\1\end{pmatrix}, \quad \omega_{\mathrm B}^2=\frac{2K}{M_{\mathrm B}}.$

Returning to physical displacements with ${U_\alpha=x_\alpha/\sqrt{M_\alpha}}$ shows that each edge mode occupies one sublattice. At ${k=0}$ , the corresponding normalized physical directions are proportional to ${{(1,1)}^{\mathsf T}}$ and ${{(1,-M_{\mathrm A}/M_{\mathrm B})}^{\mathsf T}}$ . The two eigenvectors distinguish rigid translation from internal cell motion and complete the frequency-based classification.

Before viewing the calculation, predict the relative-motion indicator ${P=\operatorname{Re}(U_{\mathrm A}^*U_{\mathrm B})/ (|U_{\mathrm A}||U_{\mathrm B}|)}$ at ${k=0}$ for both branches. Then identify the momentum at which one sublattice may be stationary and explain why ${P}$ is undefined there rather than indicating either in-phase or out-of-phase motion. An open marker at zero in the plot denotes this undefined boundary by convention; it is not a value of ${P}$ .

The computed comparison in Figure 3.8 fixes ${M_{\mathrm A}/M_{\mathrm B}=2}$ , measures frequency in units of ${\sqrt{K/M_{\mathrm B}}}$ , and keeps the solid/dashed line distinction so the branches remain distinguishable without color.

Figure 3.8. The diatomic-chain acoustic and optical dispersions are paired with a relative-motion diagnostic computed from the physical displacement eigenvectors. [ABCS13a, arXiv v1 evidence copy, Appendix A, manuscript pp. 10-11, Eqs. (A1)-(A8), numbering collated by work identity]

The figure uses both eigenvalues and eigenvectors. The acoustic branch reaches zero at the zone center and has ${P=+1}$ , whereas the optical branch has ${P=-1}$ . At the zone edge the off-diagonal dynamical-matrix entry vanishes; each eigenvector may live on one sublattice, so the normalized product defining ${P}$ has a zero denominator. The open marker placed at zero marks that diagnostic boundary by a plotting convention, not a numerical value of ${P}$ . It does not claim a ninety-degree phase lag.

The comparison with the electronic problem is now precise.

The same translation characters label both problems modulo ${2\pi/a}$ .
The electronic block is a Hamiltonian operator or its coefficient matrix; its spectral value is energy. The vibrational block is a dynamical matrix; its spectral value is ${\omega^2}$ .
Electronic eigenvectors carry orbital amplitudes and use the quantum inner product. Vibrational eigenvectors carry mass-weighted polarizations.
A rigid translation requires an acoustic zero mode when no external pinning breaks continuous displacement symmetry. There is no analogous requirement that an electronic band have zero energy at ${k=0}$ .
Quantizing the normal modes later turns each positive-frequency mode into a harmonic oscillator. That step supplies phonon occupations; it is not part of the classical eigenproblem above.

Magnon and photonic bands obey the same broad separation. Translation can organize either problem into ${\mathbf k}$ -labeled blocks, but the dynamical variables, metric, stability condition, and observable attached to the spectral value are problem specific. A familiar-looking band plot is therefore not permission to import electronic filling or electrical Hall response.

Set ${M_{\mathrm A}=M_{\mathrm B}=M}$ . The two eigenvalues reduce, for ${|k|\leq \pi/a}$ , to $\omega_-^2(k)=\frac{4K}{M}\sin^2\!\frac{ka}{4}, \qquad \omega_+^2(k)=\frac{4K}{M}\cos^2\!\frac{ka}{4}.$ A monatomic chain with spacing ${d=a/2}$ instead has one dispersion, ${\omega^2(q)=4(K/M)\sin^2(qd/2)}$ , over the larger Brillouin zone ${|q|\leq 2\pi/a}$ . Its segment ${q=k}$ gives ${\omega_-}$ , while the translated segment ${q=k+2\pi/a}$ gives ${\omega_+}$ . The two-branch picture is therefore the same spectrum represented in a doubled cell.

At ${|k|=\pi/a}$ the branches meet at ${\omega^2=2K/M}$ ; no gap opens. A gap would require a physical distinction between the two sites, such as unequal masses or unequal spring constants. This is the practical diagnostic: changing the cell can fold labels and move crossings, but only a changed dynamical matrix can create a new restoring scale or an avoided crossing.

The following selected readings extend the representation question while the reference list generated after this appendix covers both core and optional material.

For a projection-based account of hidden primitive-cell Bloch character, see [ABCS13a, arXiv v1 evidence copy, manuscript pp. 1-3, Eqs. (1)-(7), numbering collated by work identity]. The 2013 erratum corrects omitted attribution to Boykin and leaves the scientific conclusions unchanged.

To continue from finite translation projectors to supercell unfolding weights, see [Far15, p. arXiv:1504.04937v3, §§I-II, pp. 1-3, Eqs. (1)-(18)]. The useful comparison is between preserving the full spectrum and displaying the primitive-cell character carried by each state.

For an effective-band representation of supercell calculations, see [KBL10a, arXiv v1 evidence copy, manuscript pp. 1-3, Eqs. (1)-(5), numbering collated by work identity]. Read the spectral weights alongside the doubled-cell count: an unfolding weight supplies information beyond an energy curve drawn in a reduced zone.

Optional support: periodizing a central pair potential

This construction explains the boundary assumption in the many-body Hamiltonian. It is optional: the main argument already states the symmetry it requires.

Start with a relative displacement, not a position in the box. For two particles in free space, a central pair interaction has the form $w(\mathbf s)=v(|\mathbf s|),\qquad \mathbf s=\mathbf r_i-\mathbf r_j.$ It is already a function of displacement: a common translation cancels from the difference. Central means that its value depends only on distance, not direction. Let ${\Lambda_{\mathrm s}}$ be the lattice of translations that identify opposite boundaries of the finite sample. It is generally a sublattice of the primitive crystal lattice. On this sample, ${\mathbf s}$ and ${\mathbf s+\mathbf L}$ , with ${\mathbf L\in\Lambda_{\mathrm s}}$ , describe the same relative position. Our task is to assign them the same pair energy.

Sum periodic images to obtain a sample-periodic interaction. For an absolutely convergent image sum, define $W_{\mathrm s}(\mathbf s)=\sum_{\mathbf L\in\Lambda_{\mathrm s}} w(\mathbf s+\mathbf L).$ A nonsingular example is ${w(\mathbf s)=\varepsilon \exp[-|\mathbf s|^2/(2\ell^2)]}$ , where ${\varepsilon}$ is an energy and ${\ell}$ is a range. On a ring of circumference ${L}$ , $W_L(s)=\varepsilon\sum_{n\in\mathbb Z} \exp[-{(s+nL)}^2/(2\ell^2)].$ Each term is the interaction with one periodic image; the result still depends only on relative displacement. This is a pair kernel, not the total energy including a particle’s own images. Bare Coulomb image sums require a separate neutrality and electrostatic summation prescription; the convergent formula must not be applied to them without those qualifications.

Reindexing proves periodicity; it is not an approximation. For any sample vector ${\mathbf L_0\in\Lambda_{\mathrm s}}$ , set ${\mathbf L'=\mathbf L+\mathbf L_0}$ . This permutes the same image lattice: $W_{\mathrm s}(\mathbf s+\mathbf L_0) =\sum_{\mathbf L}w(\mathbf s+\mathbf L_0+\mathbf L) =\sum_{\mathbf L'}w(\mathbf s+\mathbf L')=W_{\mathrm s}(\mathbf s).$ After a common translation and separate coordinate wrapping, the separation changes only by a sample vector, so the pair energy is unchanged. Separately, centrality gives ${w(-\mathbf s)=w(\mathbf s)}$ ; replacing ${\mathbf L}$ by ${-\mathbf L}$ in the sum gives ${W_{\mathrm s}(-\mathbf s)=W_{\mathrm s}(\mathbf s)}$ . Thus boundary consistency and particle exchange follow from different steps.

The image geometry selects the surviving rotations and reflections. Let ${Q}$ be orthogonal and preserve the sample lattice: ${Q\Lambda_{\mathrm s}=\Lambda_{\mathrm s}}$ . Since lengths are unchanged, $W_{\mathrm s}(Q\mathbf s) =\sum_{\mathbf L}v(|Q\mathbf s+\mathbf L|) =\sum_{\mathbf L}v(|\mathbf s+Q^{-1}\mathbf L|) =W_{\mathrm s}(\mathbf s).$ The last equality reindexes the images. A square sample therefore retains its quarter-turn symmetry; a nonsquare rectangular sample generally does not. Although the original interaction was radial, its periodized version is generally not radial. It respects crystal point-group operations only when the chosen finite boundary lattice is also compatible with them. For the full Hamiltonian, the one-particle potential and operator domain must preserve the same operation too.

Minimum image is another periodic model, not a raw within-box difference. For a central potential, define the shortest periodic distance and its energy by $d_{\mathrm s}(\mathbf s)=\min_{\mathbf L\in\Lambda_{\mathrm s}} |\mathbf s+\mathbf L|,\qquad W_{\mathrm{MI}}(\mathbf s)=v(d_{\mathrm s}(\mathbf s)).$ On a ring, positions ${0.1L}$ and ${0.9L}$ have shortest distance ${0.2L}$ . Translating both by ${0.2L}$ and wrapping gives ${0.3L}$ and ${0.1L}$ : the raw distance changes from ${0.8L}$ to ${0.2L}$ , but the shortest distance does not. Minimum image preserves periodicity, exchange and compatible point-group symmetry. It generally differs from the full image sum. They agree for a potential cut off at a radius smaller than half the shortest nonzero sample lattice vector, since at most one image can contribute. Without a smooth cutoff, switching the nearest image can also introduce force discontinuities.

Do not confuse sample-periodic pair interactions with the crystal field. For a simple convergent model of the one-particle crystal potential, place central site potentials at basis positions ${\boldsymbol\tau_\alpha}$ in every primitive cell of lattice ${\Lambda}$ : $V(\mathbf r)=\sum_{\mathbf R\in\Lambda}\sum_\alpha v_\alpha(|\mathbf r-\mathbf R-\boldsymbol\tau_\alpha|).$ Reindexing ${\mathbf R}$ proves primitive-lattice periodicity of ${V}$ . A crystal rotation or reflection is a symmetry only if it maps the complete site pattern onto itself with the same species potentials. This is a model, not a claim that every crystal potential is a sum of central atomic terms. Return to the many-body Hamiltonian: ${V}$ is tied to the fixed crystal sites; ${W_{\mathrm s}}$ depends on particle separation and uses the sample boundary lattice. These are different reasons for translation symmetry.

For numerical context, the GROMACS periodic-boundary documentation distinguishes minimum-image short-range interactions from lattice-sum methods for long-range electrostatics.

References

[ABCS13a] Allen, P. B., Berlijn, T., Casavant, D. A., and Soler, J. M. (2013). “Recovering hidden bloch character: Unfolding electrons, phonons, and slabs.” Physical Review B 87 (8), 085322. https://doi.org/10.1103/PhysRevB.87.085322.

[Blo29] Bloch, F. (1929). “Über die quantenmechanik der elektronen in kristallgittern.” Zeitschrift für Physik 52 (7–8), 555–600. https://doi.org/10.1007/BF01339455.

[BSW36] Bouckaert, L. P., Smoluchowski, R., and Wigner, E. (1936). “Theory of brillouin zones and symmetry properties of wave functions in crystals.” Physical Review 50 (1), 58–67. https://doi.org/10.1103/PhysRev.50.58.

[BB13] Bragg, W. H., and Bragg, W. L. (1913). “The reflection of x-rays by crystals.” Proceedings of the Royal Society of London. Series A 88 (605), 428–438. https://doi.org/10.1098/rspa.1913.0040.

[Bri30] Brillouin, L. (1930). “Les électrons libres dans les métaux et le rôle des réflexions de bragg.” Journal de Physique et le Radium 1 (11), 377–400. https://doi.org/10.1051/jphysrad:01930001011037700.

[DHS03] Damascelli, A., Hussain, Z., and Shen, Z.-X. (2003). “Angle-resolved photoemission studies of the cuprate superconductors.” Reviews of Modern Physics 75 (2), 473–541. https://doi.org/10.1103/RevModPhys.75.473.

[Far15] Farjam, M. (2015). Projection operator approach to unfolding supercell band structures. https://arxiv.org/abs/1504.04937.

[Hen13] Hendrickson, W. A. (2013). “Evolution of diffraction methods for solving crystal structures.” Acta Crystallographica Section A 69 (1), 51–59. https://doi.org/10.1107/S0108767312050453.

[Hun09] Hunter, J. K. (2009). Notes on partial differential equations: Chapter 4, elliptic PDEs. University of California, Davis. https://www.math.ucdavis.edu/~hunter/m218a_09/ch4.pdf.

[KP31] Kronig, R. de L., and Penney, W. G. (1931). “Quantum mechanics of electrons in crystal lattices.” Proceedings of the Royal Society of London. Series A 130 (814), 499–513. https://doi.org/10.1098/rspa.1931.0019.

[KBL10a] Ku, W., Berlijn, T., and Lee, C.-C. (2010). “Unfolding first-principles band structures.” Physical Review Letters 104 (21), 216401. https://doi.org/10.1103/PhysRevLett.104.216401.

[Kuc16] Kuchment, P. (2016). “An overview of periodic elliptic operators.” Bulletin of the American Mathematical Society 53 (3), 343–414. https://doi.org/10.1090/bull/1528.

[Mel22] Melrose, R. B. (2022). Lectures on microlocal analysis: 18.157, spring 2022. Massachusetts Institute of Technology. https://math.mit.edu/~rbm/18.157Sp22/18.157Sp22%2B.pdf.

[Osh00a] Oshikawa, M. (2000). “Topological approach to luttinger’s theorem and the fermi surface of a kondo lattice.” Physical Review Letters 84 (15), 3370–3373. https://doi.org/10.1103/PhysRevLett.84.3370.

[Tes14a] Teschl, G. (2014). Mathematical methods in quantum mechanics: With applications to schrödinger operators. 2 ed., American Mathematical Society, Providence, Rhode Island. https://radon.mat.univie.ac.at/~gerald/ftp/book-schroe/schroe2.pdf.

[Zak67] Zak, J. (1967). “Finite translations in solid-state physics.” Physical Review Letters 19 (24), 1385–1387. https://doi.org/10.1103/PhysRevLett.19.1385.

Lecture 3: Translation Symmetry and the Bloch Eigenproblem

3.1 Translation symmetry on a finite periodic crystal

3.1.1 Translation operators and common symmetry sectors

3.1.2 Characters on a finite periodic lattice

3.1.3 From generator phases to reciprocal classes

3.2 Reciprocal equivalence and crystal-momentum classes

3.3 From translation characters to Bloch form

3.3.1 Finite character projectors

3.3.2 Bloch states and periodic factors

3.3.3 The cell-periodic eigenproblem

3.4 The crystal-momentum-resolved eigenproblem

3.4.1 What diffraction and spectroscopy determine

3.4.2 Reading an actual diffraction record

3.4.3 The operator family over crystal momentum

3.4.4 Bands as spectral families

3.4.5 The many-body boundary

3.5 Zone folding and the separate gap-opening criterion

3.5.1 Folded free branches

3.5.2 A coupling opens the avoided crossing

3.5.3 Changing the representation cell preserves state count

3.5.4 Scope of the construction

3.6 Problems