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In the plane wave basis of a ring, $V(x)=-\lambda \cos (2\pi x/L)$ has matrix elements $V_{mn}=-(\lambda /2)({\delta }_{m,n+1}+{\delta }_{m,n-1})$. Near $\alpha =1/2$, degenerate perturbation theory on $n=0,1$ givesup to corrections from remote states. Thus the avoided crossing gap is $|\lambda |$ to first order when $|\lambda |\ll E_L$.

For $|\lambda |\gg E_L$, a parabolic approximation about a potential minimum instead gives a local quantum harmonic oscillator spacing $\sqrt{2|\lambda |E_L}$. A local gauge transformation removes the flux from this approximation. Actual ring levels retain exponentially small flux dependence through quantum tunnelling, while nonquadratic corrections to their mean energies are algebraic in $\sqrt{E_L/|\lambda |}$.

An eigenspace of dimension greater than one. In degenerate perturbation theory, first-order energy corrections are the eigenvalues of the perturbing linear operator projected onto the whole unperturbed eigenspace.

Two distinguishable particles with equal one-particle spectra have degenerate ordered orbital products $|r,s⟩$ and $|s,r⟩$ for $r\ne s$. Suppose both individual position means vanish, as in a centred parity-symmetric well. The perturbation $\alpha x_1{x}_2$ then has zero diagonal matrix elements, but its restriction to this two-dimensional subspace isDegenerate perturbation theory gives normalized symmetric and antisymmetric eigenfunctions $(|r,s⟩\pm |s,r⟩)/\sqrt{2}$ with shifts $\pm \alpha |x_{rs}{|}^2$. Thus vanishing diagonal expectations in a chosen degenerate basis do not imply vanishing first-order energy shifts. Distinguishability permits both exchange symmetries; it does not forbid their energy splitting.

For an infinite square well of width $a$, the first two orbitals have $|x_{12}{|}^2=256a^2/(81{\pi }^4)$ by the position matrix elements in a symmetric infinite square well. If the particles are identical spin one-half fermions, the spatially symmetric and antisymmetric combinations must instead be paired with the singlet state and triplet states, respectively. A spin-independent perturbation leaves each allowed spin multiplicity intact.

Let $X$ be a unitary operator with $X^2=I$, and let $P$ be the orthogonal projection operator onto an unperturbed energy eigenspace. Suppose $X$ acts as the same sign $\epsilon =\pm 1$ throughout that eigenspace and $XVX=-V$. For any $u,v$ in the eigenspace, preservation of the inner product gives $⟨u,Vv⟩=⟨Xu,XVv⟩=-⟨\epsilon u,V\epsilon v⟩=-⟨u,Vv⟩$. Hence every projected matrix element vanishes, so $PVP=0$ and degenerate perturbation theory gives zero first-order energy corrections. A zero diagonal alone would be insufficient. For equal spins, the exchange parity of two coupled identical spins is constant on each total-spin sector, while their spin-projection difference changes sign under exchange.

When a ground state exists, its ground-state degeneracy is the dimension of the ground-state subspace:Here $I$ is the identity operator, and the kernel of a linear map is taken within the operator domain of the Hamiltonian operator $H$. The count concerns linearly independent energy eigenstates, rather than every normalized quantum superposition or every choice of global phase. Even a single energy eigenspace contains infinitely many representing vectors; its dimension measures independent freedom. Ground-state degeneracy is the lowest-energy instance of degenerate energy levels.

For an exact finite model with energy unit $\epsilon >0$, let the Hamiltonian operator be the diagonal matrixIts ground-state energy is zero and $g_0=2$: the first two coordinate unit vectors $e_1,e_2$ span the ground-state subspace. Every normalized $a{e}_1+b{e}_2$, with $|a{|}^2+|b{|}^2=1$, is a ground state, including $(e_1+e_2)/\sqrt{2}$. Its energy measurement is certain to return zero, even though a measurement distinguishing $e_1$ from $e_2$ need not have a definite outcome. Under the unchanged Hamiltonian operator, this quantum superposition is a stationary state because both components acquire the same global phase.

Ground-state degeneracy matters when an interaction selects or splits previously equal energy eigenvalues. In the finite model, adding $\epsilon \mathrm{diag}(\delta ,-\delta ,0,0)$ with $0<\delta <2$ gives the formerly lowest pair the distinct energies $\pm \epsilon \delta$, leaving a unique ground state. This direct calculation illustrates why degenerate perturbation theory first examines the perturbation inside the whole ground-state subspace.

A classical continuous-symmetry model can instead have a manifold of minimizing configurations, such as the sphere of orientations of a Néel state in an isotropic Heisenberg antiferromagnet. That geometric description does not by itself count the dimension of a finite quantum ground-state subspace.

Consider a fixed proton and one electron, omitting spin, fine structure and the Lamb shift. Let $e>0$ be the elementary charge, $a_0$ the Bohr radius in this fixed-nucleus model, and $\mathbf{E}=\mathcal{E}\hat{\mathbf{z}}$. Because the electron's electric dipole moment is $\mathbf{d}=-e\mathbf{r}$, its electric-dipole interaction isThe first excited hydrogen eigenspace has a common unperturbed energy eigenvalue $E_2$. Write $|s⟩=|2,0,0⟩$ and $|p_m⟩=|2,1,m⟩$ in the hydrogen energy eigenbasis. The electric-dipole selection rules for hydrogen leave only the $s,p_0$ coupling within this eigenspace. Each diagonal matrix element vanishes by parity, but that does not make the entire perturbation vanish. Degenerate perturbation theory requires its eigenvalues in the whole eigenspace. With the usual real wavefunction phases the off-diagonal position matrix element is $⟨s|z|p_0⟩=-3a_0$.^[ref]

The factor three can be checked directly. Set $\rho =r/a_0$ and use normalized radial wavefunctions and spherical harmonics:The angular integral is $\int Y_{00}^{\ast }\cos \theta Y_{10}d\Omega =1/\sqrt{3}$. Including $z=r\cos \theta$ and the radial volume factor $r^2$ then givesReversing the quantum phase of $|p_0⟩$ reverses this sign, with no change to the predicted energy eigenvalues.

Thus the nontrivial two-dimensional block and its determinant equation areConsequently, for the displayed sign convention,The two remaining energy eigenstates $|p_{+1}⟩,|p_{-1}⟩$ have zero first-order shifts. The slopes correspond to opposite electric dipole moments, $⟨d_z⟩=+3e{a}_0$ and $-3e{a}_0$. The electric field selects combinations of opposite parity that already have these dipole expectations at leading order. This is why applying first-order nondegenerate perturbation theory separately to $2s$ and $2p_0$, and obtaining two zero diagonal entries, misses the splitting.

The approximation requires $e{a}_0|\mathcal{E}|$ small compared with the separation to other principal quantum numbers. For real hydrogen atoms, it also requires the relevant electric-dipole interaction to dominate the small splittings omitted here, including the Lamb shift. At sufficiently small electric fields a pre-existing nonzero $2s$–$2p$ separation must be retained before taking the limit. A simple two-state illustration with unperturbed separation $\delta >0$ has shifts relative to its meanFor $3e{a}_0|\mathcal{E}|\ll \delta$, the lower branch moves by $-(3e{a}_0\mathcal{E}{)}^2/\delta +O({\mathcal{E}}^4)$ relative to its zero-field value. At $\delta =0$ the splitting is linear. This interpolation explains the importance of the hydrogen energy-level degeneracy; it is not a complete model of the fine structure multiplet.

The nearly-free electron model treats a weak periodic potential by degenerate perturbation theory. At a Bragg crossing, a Fourier coefficient $V_n$ couples the two degenerate plane waves and opens an energy gap of width $2|V_n|$.

The Stark effect changes atomic energy levels in a static electric field. For $\mathbf{E}=\mathcal{E}\hat{\mathbf{z}}$, the electric-dipole interaction is $H^{\prime }=-d_z\mathcal{E}=e\mathcal{E}z$ for an electron, with $e>0$. A nondegenerate state of definite parity has $⟨z⟩=0$, so its leading quadratic Stark effect isThe ground state shift is negative, conventionally $-\alpha {\mathcal{E}}^2/2$ with positive electric polarizability.^[ref]

Degenerate opposite-parity states require degenerate perturbation theory. In ideal Coulomb hydrogen, the $2s,2p_0$ subspace has an off-diagonal matrix element of magnitude $3e{a}_{\mathrm{B}}\mathcal{E}$, where $a_{\mathrm{B}}$ is the Bohr radius. Its mixed states shift byThis Linear Stark effect in the first excited hydrogen eigenspace assumes fine-structure and Lamb splittings are negligible compared with the mixing. In an isolated parity eigenstate with such degeneracies resolved, sufficiently weak fields restore quadratic behaviour. Strong fields require further mixing and can ionize the atom.^[ref]

For a bound energy eigenstate with a nondegenerate energy eigenvalue of an even one-dimensional potential, the wavefunction can be chosen with definite parity. A uniform electric field $\mathcal{E}$ gives the electric-dipole interaction $H^{\prime }=-q\mathcal{E}x$, which is odd under reflection. Hencesince $x|{\psi }_n(x){|}^2$ is odd. Second-order nondegenerate perturbation theory then gives the leading allowed Stark effect shift:Only opposite-parity intermediate states have nonzero position matrix elements. More generally, reflection maps $H(\mathcal{E})$ to $H(-\mathcal{E})$, so an analytic nondegenerate energy level is an even function of the field. This explains the absence of all odd orders and the usual quadratic Stark effect.

A level separation changes by the difference of the two quadratic coefficients. Symmetry alone does not guarantee a nonzero differential shift: in a displaced quantum harmonic oscillator every level moves by the same $-q^2{\mathcal{E}}^2/(2m{\omega }^2)$, leaving every spacing unchanged. Degenerate states would instead require degenerate perturbation theory and can split linearly; the nondegeneracy of one-dimensional bound states applies to ordinary isolated well levels.

For distinguishable particles, the ordered spatial products are independent states, and neither singlet states nor triplet states are excluded from any orbital occupation. The unperturbed energy eigenvalues areFor $0<J\ll \epsilon$, the three lowest distinct energy levels arewhere $M=-1,0,1$. The first two levels do have zero first-order correction, since their spatial expectation value factorizes as $⟨1|x|1{⟩}^2=0$.

The last instruction in the original PDF is false for the third level. Although every diagonal product-state matrix element vanishes, the two orbital states have degenerate energy levels. Degenerate perturbation theory requires diagonalizing the perturbation within their span. For each $M$, its matrix in the ordered basis $(u_1{u}_2,u_2{u}_1)$ isThe eigenfunctions are ${\Phi }_{12}^{\pm }{\chi }_{1M}$, withThus the sixfold unperturbed level splits into two triplets. Its average shift is zero, and the diagonal expectation in an individual ordered product state is zero, but neither fact makes all its energy eigenvalue corrections zero. This is the degenerate orbital exchange splitting from a product perturbation; distinguishability allows both combinations rather than removing their coupling.

For negative $J$, the lowest two rows exchange order and the third unperturbed level is instead the twofold singlet orbital pair at $5\epsilon +3J/4$. It splits into ${\Phi }_{12}^{\pm }{\chi }_{00}$ with the same shifts $\pm \alpha D$, each nondegenerate. At $J=0$, the three lowest distinct unperturbed energies are $2\epsilon ,5\epsilon ,8\epsilon$, with multiplicities $4,8,4$. The first and third have zero first-order shift; the middle splits by $\pm \alpha D$, each fourfold. The printed zero-shift assertion therefore fails also in the zero-exchange limit.

In the normalized plane wave basis, the only nonzero potential matrix elements areThus a useful small parameter is $|\lambda |/E_L\ll 1$, which suppresses admixture of well-separated energy levels. First-order nondegenerate perturbation theory further requires $|\lambda |/2\ll |E_n-E_{n\pm 1}|$ for the coupled states. This latter condition fails near a half-integer $\alpha$ even for very small $\lambda$; there one must use degenerate perturbation theory in the nearly degenerate pair. Smallness is measured against the gaps to the other states, not the vanishing gap inside this pair.

Near $\alpha =1/2$, keep $n=0,1$. The resulting Hamiltonian operator matrix and its eigenvalues areIn particular,For positive $\lambda$, the lower quantum state at the crossing is the symmetric combination of $n=0,1$, and the upper state is the antisymmetric combination. A negative $\lambda$ exchanges these combinations but leaves the ordered energy eigenvalues unchanged.

This cosine-potential gap in a quantum ring appears in the middle panel above: each lowest-branch crossing opens into an avoided crossing of gap $|\lambda |$ to first order. Away from these crossings, the diagonal matrix elements vanish, so the leading shifts are second order. At integer $\alpha$, the two first-excited plane waves differ by two units of $n$ and therefore have no first-order coupling; their smaller splitting begins at second order. The numerical plot includes that splitting.

Write $k=e^2/(4\pi {ϵ}_0)$ and retain the nuclear Coulomb potential. In spherical coordinates, the total potential energy iswhen $\beta =-\alpha /2$; $P_2(u)=(3u^2-1)/2$ is a Legendre polynomial. Treat the added term using quadratic anisotropic perturbation of hydrogen and degenerate perturbation theory in the $n=2$ hydrogen energy eigenbasis. It preserves parity and $m$, so it cannot mix $2s$ with $2p$, or distinct $2p$ magnetic sublevels.

For a hydrogen atom, the hydrogen radial second moment givesThe angular expectation values of ${\cos }^2\theta$ are $1/3$ for an $s$ orbital, $3/5$ for $p_0$, and $1/5$ for $p_{\pm 1}$, obtained by integrating the squared spherical harmonics. Therefore the five lowest spatial quantum states, counting statistical weight, have first-order energy levelsEach spatial quantum state has two electron spin states when spin-orbit coupling and the magnetic field are omitted. Thus these are five spatial states but only three distinct first-order energies; five distinct energies would be an incorrect reading. The accidental $2s,2p_{\pm 1}$ equality can be lifted at higher order. Quantitatively the added shifts must also be small compared with neighboring principal-shell gaps; $\alpha \ll k/a_0^3$ is an asymptotic condition, not a guarantee for arbitrary large $n$.

For general $\beta$, the $1s$ expectation value isIt does not cancel every order. At $\beta =-2\alpha$, $V_{\mathrm{ext}}=-2\alpha r^2{P}_2(\cos \theta )$ couples $1s$ to $d$ orbitals. For example its angular matrix element with $Y_{20}$ is $1/\sqrt{5}$, and the $3d$ radial matrix element is nonzero. The second-order nondegenerate perturbation theory sum has negative denominators, so this is a concrete nonzero higher-order contribution.

There is also a global limitation: for either negative $\beta$ above, $V(0,0,z)=-k/|z|+\beta z^2\to -\infty$. A localized wavepacket translated far along $z$ has fixed kinetic energy but arbitrarily negative potential energy, proving there is no true lowest energy level. The results describe local perturbative resonance poles, or a potential completed to a confining form far from the atomic nucleus. No globally exact ground-state cancellation exists for the potential as printed.

Let $H^{(0)}|n⟩=E_n^{(0)}|n⟩$ with a nondegenerate reference eigenvalue, and choose intermediate normalization $⟨n|n^{(1)}⟩=0$. The first-order equation isProjection on $|j⟩$, $j\ne n$, gives $⟨j|n^{(1)}⟩=H_{jn}^{(1)}/(E_n^{(0)}-E_j^{(0)})$. The second-order projected equation gives $E_n^{(2)}=⟨n|H^{(1)}|n^{(1)}⟩$, henceThe sum runs over a complete orthonormal reference basis except the reference state; continuum states require integration if present. Degenerate states with zero denominators require degenerate perturbation theory first. These are expansion coefficients in $E_n=E_n^{(0)}+\lambda E_n^{(1)}+{\lambda }^2{E}_n^{(2)}+\cdots$.

Write $|\uparrow ⟩,|\downarrow ⟩$ for the two $S_3$ eigenstates, of eigenvalues $\pm \hslash /2$. The spin-spin interaction can be evaluated usingFor $|\beta |\gg |\alpha |$, take $H_0=\beta (S_3^{(1)}-S_3^{(2)})/\hslash$ and $V=\alpha {\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}/{\hslash }^2$. The unperturbed energy eigenvalues are $\beta ,-\beta ,0,0$ on $|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\uparrow \uparrow ⟩,|\downarrow \downarrow ⟩$, respectively. For an isolated normalized energy eigenstate $|n⟩$, first-order nondegenerate perturbation theory gives $\Delta E_n^{(1)}=⟨n|V|n⟩$. The two antiparallel spins have expectation $-\alpha /4$, soFor the zero-energy degenerate eigenspace, degenerate perturbation theory requires diagonalizing the restriction of $V$. On the ordered basis $(|\uparrow \uparrow ⟩,|\downarrow \downarrow ⟩)$ it is $(\alpha /4)I$: the ladder operators cannot take one parallel state to the other. Indeed neither parallel state couples to anything else under $H$, so the remaining two levels are exactly $\alpha /4$, retaining their twofold quantum degeneracy.

For $|\alpha |\gg |\beta |$, take $H_0=\alpha {\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}/{\hslash }^2$ and $V=\beta D$, where $D=(S_3^{(1)}-S_3^{(2)})/\hslash$. The spin-one-half triplet state basis and spin-one-half singlet state areThe identity ${\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}=({\mathbf{J}}^2-3{\hslash }^2/2)/2$ gives unperturbed energy eigenvalues $E_t^{(0)}=\alpha /4$ and $E_s^{(0)}=-3\alpha /4$. Direct action givesConsequently the restriction of $V$ to the entire triplet degenerate eigenspace is zero, so no first-order splitting is missed. The singlet expectation is also zero. In second-order nondegenerate perturbation theory, $\Delta E_n^{(2)}={\sum }_{r\ne n}|⟨r|V|n⟩{|}^2/(E_n^{(0)}-E_r^{(0)})$. Within the triplet, the corresponding second-order effective matrix is ${\beta }^2|t_0⟩⟨t_0|/\alpha$, already diagonal in this basis. Thus, through second order,The last two expressions have higher corrections beginning at $O({\beta }^4/|\alpha {|}^3)$, as the exact calculation in part (d) also verifies. For positive $\alpha$, coupling to the lower singlet raises the triplet's $M=0$ level and lowers the singlet; the signs are precisely the energy denominators above.

Let $X$ exchange the two equal-spin factors, so $X^{†}=X=X^{-1}$. For the unperturbed Hamiltonian operator, addition of angular momentum givesFor nonzero $\alpha$, different $J$ have distinct energy eigenvalues, while each eigenspace has the $2J+1$ values of $M$. The stated symmetry of the Clebsch-Gordan coefficients gives the exchange parity of two coupled identical spins:However, $D=(S_3^{(1)}-S_3^{(2)})/\hslash$ changes sign: $XDX=-D$. For any two basis vectors in the same $J$ eigenspace, insertion of $X^2=I$ therefore givesThus the entire projected perturbation $P_J\beta D{P}_J$ is zero, including its off-diagonal matrix elements. Degenerate perturbation theory now proves that every first-order-in-$\beta$ energy correction vanishes. This is first-order cancellation for an exchange-odd perturbation; checking only the individual diagonal expectations would not by itself settle a degenerate problem.

Normalize the plane waves in a periodic crystal of area or volume $\mathcal{V}$. The Fourier coefficients of the periodic potential obeyTranslation of the integral by any lattice vector $\mathbf{R}$ leaves it invariant but multiplies it by $e^{-i(\mathbf{k}-{\mathbf{k}}^{\prime })\cdot \mathbf{R}}$. A nonzero result therefore requires $\mathbf{k}-{\mathbf{k}}^{\prime }$ to be a reciprocal lattice vector.

Near an isolated Bragg scattering degeneracy, ordinary nondegenerate perturbation theory fails because the free energies coincide. Retain the resonant plane waves $|\mathbf{k}⟩$, $|\mathbf{k}+\mathbf{G}⟩$ and apply degenerate perturbation theory. Absorb the constant $V_0$ into the energy origin. Since a real potential has $V_{-\mathbf{G}}=V_{\mathbf{G}}^{\ast }$, the projected Schrödinger equation has coefficient matrixA nonzero eigenstate requiresOff-resonant plane waves give higher-order energy shifts omitted in this leading nearly-free electron model.

Expand a perturbed eigenvector and energy as $\psi ={\psi }_0+\lambda {\psi }_1+O({\lambda }^2)$ and $E(\lambda )=E+\lambda \delta +O({\lambda }^2)$. At first order, $(H-E){\psi }_1+(V-\delta ){\psi }_0=0$. Projecting onto the $E$ eigenspace eliminates the first term and gives $P_{E}V{P}_E{\psi }_0=\delta {\psi }_0$. Thus degenerate perturbation theory diagonalizes the restriction of $V$ to that eigenspace; in the nondegenerate case this reduces to its expectation value.

For the specified two-dimensional eigenspace, the restricted matrix is $\left(\begin{array}{cc}\alpha & \beta \\ \beta & \alpha \end{array}\right)$. Its eigenvectors are $(|1⟩\pm |2⟩)/\sqrt{2}$ and its eigenvalues $\alpha \pm \beta$. Hence the perturbed energies are $E+\lambda (\alpha \pm \beta )+O({\lambda }^2)$.

The square-box unperturbed energies are $E_{pq}={\pi }^2{\hslash }^2(p^2+q^2)/(2m{a}^2)$. Every diagonal matrix element of $xy/a^2$ is $1/4$, because each coordinate has mean $a/2$. The ground state is $(p,q)=(1,1)$, giving $E_{11}+\lambda /4$. The first excited eigenspace is spanned by $(1,2)$ and $(2,1)$. Its off-diagonal entry factors into the square ofThus that entry is $256/(81{\pi }^4)=(4/(3\pi ){)}^4$, and the three levels areThe neglected energy corrections are $O({\lambda }^{2}m{a}^2/{\hslash }^2)$ in the given weak-field regime.

For potential period $a$, a Bloch state has the form ${\psi }_k(x)=e^{ikx}{u}_k(x)$, with $u_k(x+a)=u_k(x)$, or equivalently ${\psi }_k(x+a)=e^{ika}{\psi }_k(x)$. Quasimomentum is defined modulo the reciprocal-lattice vector $2\pi /a$. A Brillouin zone is one fundamental interval for that identification, usually $[-\pi /a,\pi /a)$. Here $a=2\pi$, so it is $[-1/2,1/2)$.

For small $V_0>0$, the first free-particle degeneracy occurs at the zone edge $k=1/2$, between plane waves $e^{ix/2}$ and $e^{-ix/2}$ of energy $1/4$. In this two-dimensional subspace, the diagonal matrix elements of the potential vanish. Its Fourier coefficient at wave number one is $-V_0/2$; the $\cos 2x$ term cannot connect these two waves at first order. Thus degenerate perturbation theory gives the matrixwith energies $1/4\pm V_0/2+O(V_0^2)$. Near the edge the two uncoupled energies have opposite slopes, and diagonalization gives the avoided crossing, so the lowest band gap is

For large $V_0$, expand about each deepest minimum $x=2\pi j$:The local quantum harmonic oscillator $-d^2/d{x}^2+(5V_0/2)x^2$ has levels $(2n+1)\sqrt{5V_0/2}$. Its localization width is $O(V_0^{-1/4})$, so the quartic term changes these low levels by $O(1)$. Tunnelling between successive deep wells gives exponentially narrow energy bands. The secondary minima at $x=(2j+1)\pi$ have potential zero, and their levels lie far above the first two deep-well levels, whose energies are $-2V_0+O(\sqrt{V_0})$. Therefore the gap between the first and second bands is the first oscillator spacing to leading order:The $O(1)$ anharmonic shifts and exponentially small bandwidths do not change this leading estimate.

Bloch theorem states that energy eigenfunctions in a Bravais lattice potential may be chosen asThe wavevector is defined modulo a reciprocal lattice vector, so one can restrict it to the first Brillouin zone. This follows because lattice translations commute with the Hamiltonian and have simultaneous eigenvalues $e^{i\mathbf{k}\cdot \mathbf{l}}$.

In one dimension write $V(x)={\sum }_G{V}_G{e}^{iGx}$ with $G=2\pi j/a$ and expand a Bloch state as ${\psi }_k={\sum }_G{c}_G{e}^{i(k+G)x}$. The Schrödinger equation becomesFor zero potential, the parabolas $E_G^0(k)={\hslash }^2(k+G{)}^2/(2m)$ fold into the Brillouin zone. A weak potential shifts a nondegenerate level by $V_0$ at first order, with second-order correction ${\sum }_{G\ne 0}|V_G{|}^2/(E_0^0-E_G^0)$ where the denominators stay away from zero. This expansion fails near crossings, where degenerate perturbation theory is required.

At the first zone boundary, put $G=2\pi /a$ and $k=G/2+q$. The nearly degenerate waves with momenta $\hslash k$ and $\hslash (k-G)$ are coupled by $V_G$. Retaining their two-dimensional subspace givesDiagonalizing producesAt $q=0$ there is a gap $\boxed{\Delta E=2|V_G|}$: the two standing-wave combinations sample the periodic potential differently. Similar Bragg degeneracies occur at $k=j\pi /a$, with gaps controlled to first order by the corresponding Fourier components. If that component vanishes, a gap need not open at first order. Between the gaps the continuous energy eigenvalues form energy bands, periodically represented in the reduced-zone scheme. For the first zone edge, the curvature of the upper and lower branches gives a minimum and a maximum respectively for a sufficiently weak nonzero potential, so the avoided crossing is a genuine local forbidden energy interval.

At zero temperature electrons occupy states up to the Fermi energy. The Fermi surface is the set $E_n(\mathbf{k})=E_F$ separating occupied from unoccupied states within partially filled bands (Fermi points in one dimension). A partially filled band supplies arbitrarily nearby empty states, allowing occupations to respond to a weak electric field and conduct. Completely filled bands give no net current from such a redistribution; if the next empty band is separated by a positive band gap, the system is a band insulator. Band overlap can instead yield a metal even at an integer band filling. At finite temperature an insulator can conduct through thermally excited electrons and holes.

Put $E_1=E_2=E$. Within the degenerate eigenspace spanned by $|1⟩,|2⟩$, the perturbation isDegenerate perturbation theory diagonalizes this matrix. Its normalized eigenstates $(|1⟩\pm |2⟩)/\sqrt{2}$ have first-order shifts $\pm \epsilon V_0$. The coupling to $|3⟩$ has order ${\epsilon }^2$, so it cannot affect an energy through order $\epsilon$. Therefore