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Use the stated local axes: $z$ is radial along O–Ru, $x$ is tangential within the Ru–O plane, and $y$ is perpendicular to that plane. The oxygen's three $s{p}^2$ hybrids make the in-plane sigma framework. Its remaining $p_y$ orbital is perpendicular to the plane and can overlap with one $d_{yz}$ orbital on each Ru.

Choose the local orbital phases so their O-facing lobes agree. Of the three metal orbitals, onlyhas the appropriate combination to interact with oxygen $p_y$. The other two independent combinations have coefficients summing to zero and are nonbonding with respect to the central oxygen. Mixing $D$ with $p_y$ gives one lower pi-bonding orbital and one upper pi-antibonding orbital. Each Ru also retains its $d_{xy}$ and $d_{xz}$ orbitals as nonbonding with respect to this central-oxo pi interaction. Thus the complete low-energy diagram contains one bonding level, eight metal-nonbonding levels, and one antibonding level. The eight consist of two $d_{yz}$ combinations, three $d_{xy}$ and three $d_{xz}$ orbitals. They need not be exactly degenerate once other ligands are considered.

For the neutral formula as printed, oxide contributes $-2$ and six acetates contribute $-6$. The Ru oxidation numbers therefore sum to $+8$, giving sixteen metal d electrons overall, formally two Ru(III) and one Ru(II), possibly delocalized. Including the oxygen pi pair gives eighteen electrons, filling the bonding orbital and all eight nonbonding orbitals, leaving pi* empty. If the often encountered singly charged cation were intended instead, there would be seventeen electrons and one singly occupied nonbonding level. This charge distinction changes occupancy, not the orbital connectivity.

Two atomic orbitals. In the linear combination of atomic orbitals approximation, use $|\psi ⟩=c_A|A⟩+c_B|B⟩$, where the two basis functions are localized atomic orbitals. Projecting $\hat{H}|\psi ⟩=E|\psi ⟩$ onto each orbital gives the generalized eigenvalue problemHere the scalar overlap in the matrix is $S=⟨A|B⟩$, while $H_{ab}=⟨a|\hat{H}|b⟩$. Normalization requires $c^{†}Sc=1$, 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 areFor 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.

Hybridized bands. At each $\mathbf{k}$, take the two Bloch basis states to be orthonormal, as implicit in the stated band model, and write ${ϵ}_k={\hslash }^2{k}^2/(2m)$. The Hamiltonian isFor $|{\psi }_{\mathbf{k}}⟩=a_{\mathbf{k}}|{\varphi }_{\mathbf{k}}⟩+b_{\mathbf{k}}|{\chi }_{\mathbf{k}}⟩$, the eigenvalue condition isso $(E-{ϵ}_k)(E-E_d)-|V{|}^2=0$. Thus band hybridization givesFor nonzero $V$, a convenient explicit normalized pair of eigenstates isA 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.

Dispersions and limits. For $V=0$, the parabola crosses the flat band atFor $V\ne 0$ 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 areso for $|V|\ll E_d$ the small-$k$ expansions areAt large $k$, where ${ϵ}_k-E_d\gg |V|$,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 $E_{-}=\min ({ϵ}_k,E_d)$ and $E_{+}=\max ({ϵ}_k,E_d)$, which cross at $k_0$ without mixing.

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 givesThe 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 $v_{-}(k)={\hslash }^{-1}d{E}_{-}/dk$. Differentiating the exact dispersion givesAt $k_F$, put $\delta =2^{2/3}-1$, so ${ϵ}_{k_F}-E_d=\delta E_d$. Relative to the original broad-band Fermi velocity $v_F^{(1)}=\hslash k_0/m$,Expanding for $|V|\ll E_d$ gives the flat-band hybridization Fermi-velocity suppressionThe 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 $V\to 0$ 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.