Solution
= Solution
\b[Why <degenerate perturbation theory> is needed.] If several unperturbed states have equal or nearly equal <energy>, a perturbation can mix them at leading order. In that subspace, diagonalize the perturbation matrix rather than using nondegenerate formulas with small <energy> denominators. The resulting eigenvectors identify the combinations selected by the coupling, and the <eigenvalues> give the leading splitting. Other remote states can then be included perturbatively.
For the <linear combination of atomic orbitals> description of <molecular hydrogen>, equivalent atomic $1s$ states on the two nuclei supply the nearly degenerate starting functions. In an orthonormal two-state model, equal diagonal <energy> $E_a$ and real hopping $t$ give $E_a\pm t$ and symmetric/antisymmetric combinations. For actual overlapping atomic functions with overlap $S$ and <Hamiltonian> coupling $h$, one instead solves a <generalized eigenvalue problem>:
$$
E_b=\frac{E_a+h}{1+S},\qquad E_{ab}=\frac{E_a-h}{1-S},\qquad \psi_{b,ab}=\frac{\phi_A\pm\phi_B}{\sqrt{2(1\pm S)}}.
$$
The bonding state has enhanced density between the nuclei and the antibonding state a separating node. Two opposite-spin <Electrons> occupying the <bonding molecular orbital> describe the simplest closed-shell hydrogen-molecule bond. Electron-electron interaction and nuclear repulsion are needed to calculate a realistic total <binding energy>; the two-state splitting alone is not the full molecular <energy>.
In the <nearly-free electron model>, the analogous degenerate states are plane waves $|\mathbf k\rangle$ and $|\mathbf k-\mathbf G\rangle$ at a <Bragg scattering>. A <Fourier coefficient> $U_{\mathbf G}$ of the <periodic potential> couples them. At exact degeneracy $E_B$, diagonalizing their two-by-two matrix gives
$$
\boxed{E_\pm=E_B+U_0\pm|U_{\mathbf G}|,\qquad E_{\rm gap}=2|U_{\mathbf G}|.}
$$
The associated standing waves place electronic density differently relative to the ionic potential. These are the same gap-opening combinations illustrated in Q1(b). The labels of the two starting <basis> states survive as mixed orbital or plane-wave character, rather than independent crossing eigenstates.
\b[Origin of broad and narrow bands.] Spatially extended s/p-derived orbitals have substantial overlap and therefore large hopping amplitudes and broad bands. More localized d orbitals overlap less strongly and produce narrower bands. A multiorbital tight-binding calculation assigns the on-site orbital energies and hopping matrices, forms Bloch sums, and diagonalizes the resulting matrix at each <wavevector>. Different hopping ranges and orbital <symmetry> determine the bandwidths and allowed interband coupling. The two-band description retains one broad combination and one nearly flat d-derived combination; a real transition <metal> contains several d orbitals and more complicated dispersions.
\b[Hybridized <eigenvalues> and eigenstates.] Let $|\psi_{\mathbf k}\rangle=a|\phi_{\mathbf k}\rangle+b|\chi_{\mathbf k}\rangle$, in an <orthonormal basis>. <Hermitian operator> fixes the reverse coupling to be $V^*$. Projection onto the two components gives
$$
\begin{pmatrix}E_1(\mathbf k)&V\\V^*&E_d\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=E\begin{pmatrix}a\\b\end{pmatrix},\qquad (E_1-E)(E_d-E)-|V|^2=0.
$$
Solving this quadratic yields the <band hybridization> dispersion
$$
\boxed{E_\pm(\mathbf k)=\frac{E_1(\mathbf k)+E_d}{2}\pm\sqrt{\left[\frac{E_1(\mathbf k)-E_d}{2}\right]^2+|V|^2}.}
$$
For nonzero $V$, one <eigenvector> representation is $(a,b)\propto(V,E-E_1)$, followed by normalization. Thus the <wavefunctions> exchange broad-band and d-band character through the <avoided crossing>; the levels are equal mixtures at the uncoupled crossing.
At $V=0$, the parabola crosses the horizontal level at $k=\pm k_0$, with $k_0=\sqrt{2mE_d}/\hbar$. For small nonzero $|V|$, the curves avoid crossing with minimum direct separation $2|V|$. The lower branch follows the parabola below the crossing and flattens toward $E_d$ above it; the upper does the opposite. The lower branch satisfies $E_-<E_d$ everywhere and tends to $E_d$ from below as $k\to\infty$, while the upper branch always lies above $E_d$. This is the perfectly flat-band continuum approximation, with zone-boundary intersections excluded as requested.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2013-hybridized-bands.png]
{title=Uncoupled and weakly hybridized broad/flat bands, with the shifted Fermi radius and small lower-band slope}
\b[Fixed particle number and the larger Fermi sphere.] Denote the uncoupled broad-band Fermi radius by $k_0$ to distinguish it from the new one $K$. With two <spin> states, the broad-band <number density> at $V=0$ is $k_0^3/(3\pi^2)$. The stipulated equal population in the flat sector makes the total density twice that value. At zero <temperature> and any nonzero hybridization, all occupied states fit into the monotonic lower branch below $E_d$, while the upper branch is empty. Therefore
$$
\frac{K^3}{3\pi^2}=2\frac{k_0^3}{3\pi^2},\qquad \boxed{K=2^{1/3}k_0.}
$$
This is the stated factor increase. It uses fixed total particle number, equal <spin> counting, an exactly flat unhybridized band and enough available states below any Brillouin-zone restriction. In generic pairs of dispersive bands both hybridized pockets can remain occupied; then <Fermi-volume conservation in two hybridized electron bands> fixes a sum of volumes rather than this one radius. The exactly <flat band> is degenerate at $V=0$, so its initial occupation must be specified, as it is here.
\b[Fermi-velocity suppression.] Set $\Delta(k)=E_1(k)-E_d$. The exact derivative of the lower branch is
$$
\frac{dE_-}{dk}=\frac{dE_1}{dk}\frac12\left[1-\frac{\Delta(k)}{\sqrt{\Delta(k)^2+4|V|^2}}\right].
$$
At $k=K$, $E_1(K)=2^{2/3}E_d$ and $\Delta(K)=(2^{2/3}-1)E_d>0$. For $|V|\ll\Delta(K)$, the square bracket divided by two is $|V|^2/\Delta(K)^2+O(|V|^4/\Delta(K)^4)$. Since <group velocity> is $\hbar^{-1}dE/dk$ and the broad-band reference <velocity> is $v_F^{(1)}=\hbar k_0/m$,
$$
\frac{v_F}{v_F^{(1)}}=\frac{K}{k_0}\frac{|V|^2}{\Delta(K)^2}+O\left(\frac{|V|^4}{E_d^4}\right),
$$
so the <flat-band hybridization Fermi-velocity suppression> is
$$
\boxed{\frac{v_F}{v_F^{(1)}}\simeq\frac{2^{1/3}}{(2^{2/3}-1)^2}\frac{|V|^2}{E_d^2},\qquad A\simeq3.65.}
$$
The reference <velocity> is that of the original broad-band Fermi sphere, not the broad-band slope at the larger $K$; omitting the radius change loses the factor $2^{1/3}$.
The low-temperature qualification becomes more stringent as $V\to0$: the new <chemical potential> approaches $E_d$ with $E_d-\mu\simeq |V|^2/\Delta(K)$, and the relevant thermal <energy> must be small compared with this hybridization scale. At fixed nonzero <temperature> the singular flat-band redistribution is not automatically resolved for arbitrarily small coupling. A small but finite narrow-band dispersion can also change the occupation topology.
\b[<Density of states> and magnetism.] For an isotropic three-dimensional branch with <spin> degeneracy two,
$$
\frac{g(E_F)}{\mathcal V}=\frac{K^2}{\pi^2|dE_-/dk|_{K}}=\frac{K^2}{\pi^2\hbar v_F}.
$$
A small <Fermi velocity> therefore produces a large <density of states>. The localized, partially filled d-derived bands in <iron>, <cobalt> and <nickel> place many states close to the <Fermi level>; broad s/p-derived <conduction bands> in <copper>, <silver> and <gold> coexist with filled d bands lying below it. This explains the substantially larger Fermi-level <density of states> of the former group within the band picture. An exact factor of ten requires the actual band parameters and is not fixed by the two-band toy model.
A high <density of states> lowers the kinetic-energy cost of <spin polarization>. A repulsive <exchange interaction> can then satisfy the <Stoner criterion> and stabilize <ferromagnetism>. With a <single-spin density of states> $g_s$ and an on-site interaction normalization, the criterion is $Ug_s(E_F)>1$; with spin-summed $g=2g_s$, it is $Ug(E_F)>2$. The large d-state contribution favors this instability in Fe, Co and Ni. The noble <metals> do not satisfy the corresponding instability condition in the same ordinary state; this does not imply that they have no weak magnetic response whatsoever.