= Solution
\b[Two atomic orbitals.] In the <linear combination of atomic orbitals> approximation, use $|\psi\rangle=c_A|A\rangle+c_B|B\rangle$, where the two basis functions are localized <atomic orbitals>. Projecting $\widehat H|\psi\rangle=E|\psi\rangle$ onto each orbital gives the <generalized eigenvalue problem>
$$
\boxed{\begin{pmatrix}H_{AA}&H_{AB}\\H_{BA}&H_{BB}\end{pmatrix}
\begin{pmatrix}c_A\\c_B\end{pmatrix}
=E\begin{pmatrix}1&S\\S^*&1\end{pmatrix}\begin{pmatrix}c_A\\c_B\end{pmatrix},\qquad
\det(H-ES)=0.}
$$
Here the scalar overlap in the matrix is $S=\langle A|B\rangle$, while $H_{ab}=\langle a|\widehat H|b\rangle$. Normalization requires $c^\dagger S c=1$, with $S$ now denoting the full overlap matrix. For equivalent real orbitals with diagonal energy $E_a$, real hopping $t$, and real overlap $s$, the eigenstates and energies are
$$
|\psi_\pm\rangle=\frac{|A\rangle\pm|B\rangle}{\sqrt{2(1\pm s)}},\qquad
E_\pm=\frac{E_a\pm t}{1\pm s}.
$$
For the usual negative hopping convention, the symmetric combination is the lower <bonding orbital> and the antisymmetric combination is the higher <antibonding orbital>. In an orthonormal approximation, $s=0$ and the result reduces to a two-by-two <Hermitian matrix> eigenproblem.
\b[Hybridized bands.] At each $\mathbf k$, take the two Bloch basis states to be orthonormal, as implicit in the stated band model, and write $\epsilon_k=\hbar^2k^2/(2m)$. The <Hamiltonian> is
$$
H(\mathbf k)=\begin{pmatrix}\epsilon_k&V\\V^*&E_d\end{pmatrix}.
$$
For $|\psi_{\mathbf k}\rangle=a_{\mathbf k}|\phi_{\mathbf k}\rangle+b_{\mathbf k}|\chi_{\mathbf k}\rangle$, the <eigenvalue> condition is
$$
(\epsilon_k-E)a+Vb=0,\qquad V^*a+(E_d-E)b=0,
$$
so $(E-\epsilon_k)(E-E_d)-|V|^2=0$. Thus <band hybridization> gives
$$
\boxed{E_\pm(k)=\frac{\epsilon_k+E_d}{2}\pm\sqrt{\frac{(\epsilon_k-E_d)^2}{4}+|V|^2}.}
$$
For nonzero $V$, a convenient explicit normalized pair of <eigenstates> is
$$
\boxed{|\psi_{\mathbf k,\pm}\rangle=
\frac{V|\phi_{\mathbf k}\rangle+(E_\pm-\epsilon_k)|\chi_{\mathbf k}\rangle}
{\sqrt{|V|^2+(E_\pm-\epsilon_k)^2}}.}
$$
A phase convention can make $V$ real, but the eigenvalues depend only on its modulus. These are exact within the truncated two-band model, and approximate as a description of a material.
\b[Dispersions and limits.] For $V=0$, the parabola crosses the <flat band> at
$$
k_0=\sqrt{2mE_d}/\hbar.
$$
For $V\ne0$ the crossing is avoided, with a direct <hybridization gap> $2|V|$ at $k_0$. The lower and upper branches exchange orbital character through this <avoided crossing>. At the origin their energies are
$$
E_\pm(0)=\frac{E_d\pm\sqrt{E_d^2+4|V|^2}}2,
$$
so for $|V|\ll E_d$ the small-$k$ expansions are
$$
E_-(k)=-\frac{|V|^2}{E_d}+\left(1-\frac{|V|^2}{E_d^2}\right)\epsilon_k+O(|V|^4/E_d^3,|V|^2\epsilon_k^2/E_d^3),
$$
$$
E_+(k)=E_d+\frac{|V|^2}{E_d}+\frac{|V|^2}{E_d^2}\epsilon_k+O(|V|^4/E_d^3,|V|^2\epsilon_k^2/E_d^3).
$$
At large $k$, where $\epsilon_k-E_d\gg|V|$,
$$
\boxed{E_-(k)=E_d-\frac{|V|^2}{\epsilon_k-E_d}+O\!\left(\frac{|V|^4}{(\epsilon_k-E_d)^3}\right),\qquad
E_+(k)=\epsilon_k+\frac{|V|^2}{\epsilon_k-E_d}+O\!\left(\frac{|V|^4}{(\epsilon_k-E_d)^3}\right).}
$$
The lower branch approaches $E_d$ from below and becomes flat; the upper branch approaches the free-electron parabola from above. At zero coupling the unsorted orbital bands remain exactly flat and parabolic. Sorting them by energy instead gives $E_-=\min(\epsilon_k,E_d)$ and $E_+=\max(\epsilon_k,E_d)$, which cross at $k_0$ without mixing.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2018-b3-hybridized-bands.png]
{title=Avoided crossing and Fermi-volume redistribution in a parabolic-flat two-band model}
{description=Original model curves for zero coupling and $|V|/E_d=0.12$. Wavevectors are in units of $k_0=\sqrt{2mE_d}/\hbar$. The lower-band Fermi radius $2^{1/3}k_0$ and its chemical potential follow fixed total density under the stated equal-population, zero-temperature assumption.}
\b[Fermi wavevector and velocity.] A partially occupied perfectly <flat band> has chemical potential $E_d$. Thus before mixing, the occupied broad band has radius $k_0$ defined above. For spin degeneracy two, its particle density is $n_1=k_0^3/(3\pi^2)$; equal population of the flat sector gives total density $n=2n_1$.
For any nonzero mixing, $E_-(k)<E_d$ at finite $k$, while $E_+(k)>E_d$. Moreover $E_-(k)$ increases monotonically. At zero temperature and this fixed density, a single occupied sphere in the lower branch therefore accommodates all particles, and the upper branch is empty. <Fermi-volume conservation in two hybridized electron bands> gives
$$
\frac{k_F^3}{3\pi^2}=\frac{2k_0^3}{3\pi^2},\qquad
\boxed{k_F=2^{1/3}k_0,\qquad \Delta k_F=(2^{1/3}-1)k_0\simeq0.260k_0.}
$$
The new radius is about $26\%$ larger than the old broad-band radius. The strictly flat uncoupled band has no uniquely determined Fermi radius of its own; its prescribed population is what enters the count.
The radial <group velocity> is $v_-(k)=\hbar^{-1}dE_-/dk$. Differentiating the exact dispersion gives
$$
\frac{dE_-}{d\epsilon_k}=\frac12\left[1-\frac{\epsilon_k-E_d}{\sqrt{(\epsilon_k-E_d)^2+4|V|^2}}\right].
$$
At $k_F$, put $\delta=2^{2/3}-1$, so $\epsilon_{k_F}-E_d=\delta E_d$. Relative to the original broad-band <Fermi velocity> $v_F^{(1)}=\hbar k_0/m$,
$$
\frac{v_F}{v_F^{(1)}}=\frac{2^{1/3}}2\left[1-\frac{\delta}{\sqrt{\delta^2+4|V|^2/E_d^2}}\right].
$$
Expanding for $|V|\ll E_d$ gives the <flat-band hybridization Fermi-velocity suppression>
$$
\boxed{\frac{v_F}{v_F^{(1)}}=\frac{2^{1/3}}{(2^{2/3}-1)^2}\frac{|V|^2}{E_d^2}+O\!\left(\frac{|V|^4}{E_d^4}\right).}
$$
The coefficient is about $3.65$. The enlarged occupied volume coexists with a small velocity because the new <Fermi surface> lies on the almost-flat part of the lower branch. This result assumes fixed particle density, sufficient available states before any <Brillouin zone> boundary, and temperature low enough to resolve energy differences of order $|V|^2/E_d$. The limit $V\to0$ at a fixed nonzero temperature need not give this sharp redistribution. Giving the narrow band a finite dispersion can also leave both pockets occupied, so the result is specific to the ideal model.
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