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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 $2|V_n|$.

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

The potential has period $2\pi$, so the first Brillouin zone is $-1/2\le k<1/2$. At its boundary the free-particle statesare degenerate, with energyThe relevant Fourier coefficient is supplied by $V_0\cos x$:The term $-V_0\cos 2x$ changes wavevector by two and has no matrix element within this degenerate pair. Thus degenerate perturbation theory gives the matrixand the two band-edge energiesThe nearly-free electron model therefore predicts the lowest band gap

Let $G_n=2\pi n/a$. The free states $|k⟩$ and $|k-G_n⟩$ are degenerate at the Bragg point $k=G_n/2=n\pi /a$. Sincedegenerate perturbation theory restricts the Hamiltonian near this crossing toWriting $k=n\pi /a+\kappa$ and diagonalizing gives the nearly-free electron dispersion near a one-dimensional band gapThus the two continuous bands avoid crossing, and at $\kappa =0$ their separation isThis is the nearly-free electron model. The dispersion relation is the relation $E=E(k)$ between energy and Bloch wavevector.

For the specified potential,Henceand all other nonconstant Fourier coefficients vanish. The gaps are thereforeTheir centre energies are respectivelyThe extended-zone sketch consists of the free-particle parabolas shifted upward by $V_0$, with avoided crossings of these two widths at the listed Bragg points; elsewhere they meet to this order because the corresponding Fourier coefficients vanish.

The degenerate $E_0=9/2$ subspace has ordered basisUsing the ladder actions,Thus degenerate perturbation theory asks us to diagonalizeIts normalized eigenvectors and first-order energies areThere is no order-$\lambda$ admixture from other unperturbed levels.

In fact, $H^{\prime }$ preserves $n+2m$, so this two-dimensional eigenspace of $H_0$ is invariant. Directly,Thereforefor every real $\lambda$, with no omitted higher-order terms. This is exact diagonalization within a degenerate subspace, specifically the resonant two-to-one oscillator coupling.

For $H_0=E_2{B}_2^{†}{B}_2$, the eigenspaces of energies $0$ and $E_2$ are each two-dimensional because $n_1$ is free. Degenerate perturbation theory requires diagonalizing $E_1{B}_1^{†}{B}_1$ within each space. It is already diagonal in the product basis, with eigenvalues $0,E_1$. Therefore the first-order eigenvectors remainand their energies are $0,E_1,E_2,E_2+E_1$, respectively. These are the exact answers, so all higher corrections vanish.

Within each degenerate unperturbed eigenspace of fixed $n_2$, the perturbation has matrixIts eigenvectors are $|+{⟩}_1$ and $|-{⟩}_1$, with first-order shifts $+g$ and $-g$. Thus degenerate perturbation theory gives exactly the states and energies found in part (i); again, all higher corrections vanish because the perturbation commutes with the second-oscillator Hamiltonian and is exactly diagonalized within each degenerate block.