= Solution
\b[<mass> as <band curvature>.] In a periodic potential, a weakly forced <Electron> is usefully described by a <wave packet> confined to one <energy band>. Its <group velocity> is
$$
v_i=\frac1\hbar\frac{\partial\varepsilon_n}{\partial k_i},\qquad \hbar\dot k_j=F_j.
$$
Differentiating the <velocity> gives
$$
\dot v_i=\sum_j(m^{*-1})_{ij}F_j,\qquad \boxed{(m^{*-1})_{ij}=\frac1{\hbar^2}\frac{\partial^2\varepsilon_n}{\partial k_i\partial k_j}.}
$$
This <effective-mass tensor> describes the band response, not a change of the <Electron>'s bare <rest mass>. Along a principal direction, a steep upward curvature gives a small positive <mass> and a flatter curvature a larger one. Near a band maximum the <Electron> curvature and <effective mass> are negative; the corresponding missing-electron description uses positive <electron hole> masses. Anisotropy produces distinct principal masses, and strong nonparabolicity makes them <momentum> dependent. The single-band semiclassical description assumes fields weak and slowly varying enough to avoid substantial interband transitions.
\b[<Lattice> <symmetry> and <Bloch's theorem>.] Use the active <spatial translation operator> $T_{\mathbf a}\psi(\mathbf r)=\psi(\mathbf r-\mathbf a)$. A lattice-periodic <scalar> potential satisfies $V(\mathbf r+\mathbf a)=V(\mathbf r)$, while the kinetic term is translation invariant. Therefore $[H,T_{\mathbf a}]=0$. The <lattice> translations also commute with one another, so they can be diagonalized simultaneously with $H$, including within a degenerate <energy> subspace.
For a common <eigenstate>, write $T_{\mathbf a}\psi=t(\mathbf a)\psi$. <Unitarity> gives $|t(\mathbf a)|=1$, and the group law gives $t(\mathbf a+\mathbf b)=t(\mathbf a)t(\mathbf b)$. On primitive Bravais vectors $\mathbf a_j$, choose $t(\mathbf a_j)=e^{-i\mathbf k\cdot\mathbf a_j}$. Every <lattice> vector is an integer combination of these generators, hence
$$
T_{\mathbf a}\psi_{n\mathbf k}=e^{-i\mathbf k\cdot\mathbf a}\psi_{n\mathbf k},\qquad \psi_{n\mathbf k}(\mathbf r+\mathbf a)=e^{i\mathbf k\cdot\mathbf a}\psi_{n\mathbf k}(\mathbf r).
$$
The vector $\mathbf k$ is defined modulo a <reciprocal lattice> vector. The band index labels the different <energy> eigenstates with the same translation character. Define $u_{n\mathbf k}(\mathbf r)=e^{-i\mathbf k\cdot\mathbf r}\psi_{n\mathbf k}(\mathbf r)$. The phase factors cancel under a <lattice translation>, proving <Bloch's theorem> in its usual form:
$$
\boxed{\psi_{n\mathbf k}(\mathbf r)=e^{i\mathbf k\cdot\mathbf r}u_{n\mathbf k}(\mathbf r),\qquad u_{n\mathbf k}(\mathbf r+\mathbf a)=u_{n\mathbf k}(\mathbf r).}
$$
This is the <Translation-character proof of Bloch theorem>; the <commutator> guarantees a <basis> of such states, even when an arbitrary <superposition> of different translation characters is not itself a translation <eigenstate>.
\b[Periodic <Hamiltonian>.] Acting once with <momentum> gives $\mathbf p(e^{i\mathbf k\cdot\mathbf r}u)=e^{i\mathbf k\cdot\mathbf r}(\mathbf p+\hbar\mathbf k)u$. Applying it again and cancelling the common plane-wave factor yields
$$
\boxed{H(\mathbf k)u_{n\mathbf k}=\left[\frac{(\mathbf p+\hbar\mathbf k)^2}{2m}+V(\mathbf r)\right]u_{n\mathbf k}=\varepsilon_n(\mathbf k)u_{n\mathbf k}.}
$$
The <Bloch Hamiltonian> acts on lattice-periodic functions in a <unit cell>. Expanding its square separates a zero-momentum <Hamiltonian>, a term linear in $\mathbf k$ and an identity term quadratic in $\mathbf k$.
\b[Expansion about the zone centre.] Let $|n\rangle=u_{n0}$, $E_n=\varepsilon_n(0)$, and normalize all these states using a unit-cell <inner product>. For the real <scalar> <Hamiltonian> in the question, a nondegenerate zone-centre eigenfunction can be chosen real. Then
$$
\langle n|\mathbf p|n\rangle=-\frac{i\hbar}{2}\int_{\rm cell}\nabla(u_{n0}^2)\,d^3r=0,
$$
because opposite unit-cell faces cancel. Equivalently, <time-reversal symmetry> forbids a nonzero <electric current> in this nondegenerate spinless state. This explains why no linear <energy> term appears; it should not simply be omitted without a <symmetry> or extremum argument.
The first-order <energy> shift is therefore $\hbar^2k^2/(2m)$. In the second-order sum the quadratic term is proportional to the identity and has no off-diagonal <matrix elements>. Only the linear <momentum> vertex contributes. <Second-order nondegenerate perturbation theory> gives the <momentum matrix-element expansion of band energy>
$$
\boxed{\varepsilon_n(\mathbf k)=E_n+\frac{\hbar^2k^2}{2m}+\frac{\hbar^2}{m^2}\sum_{l\ne n}\frac{|\langle n|\mathbf k\cdot\mathbf p|l\rangle|^2}{E_n-E_l}+O(|\mathbf k|^4).}
$$
<Hermitian operator> makes the two reciprocal <matrix elements> <complex conjugates>. All other periodic eigenstates at $\mathbf k=0$ enter the sum. The remainder is even because <time-reversal symmetry> gives $\varepsilon_n(\mathbf k)=\varepsilon_n(-\mathbf k)$ near this isolated state. At a degeneracy the denominators signal the need for <degenerate perturbation theory> instead.
\b[<mass> ratio and the printed formula.] Write $p^i_{nl}=\langle n|p_i|l\rangle$. Differentiating the quadratic <energy> yields
$$
\boxed{(m^{*-1})_{ij}=\frac{\delta_{ij}}m+\frac2{m^2}\sum_{l\ne n}\frac{\operatorname{Re}(p^i_{nl}p^j_{ln})}{E_n-E_l}.}
$$
For a specified unit direction $\mathbf e$, the resulting <scalar> <mass> ratio is
$$
\boxed{\frac m{m^*_{\mathbf e}}=1+\frac2m\sum_{l\ne n}\frac{|\langle n|\mathbf e\cdot\mathbf p|l\rangle|^2}{E_n-E_l}.}
$$
In an isotropic three-dimensional band, averaging equivalent directions gives
$$
\boxed{\frac m{m^*}=1+\frac{2}{3m}\sum_{l\ne n}\frac{|\langle n|\mathbf p|l\rangle|^2}{E_n-E_l}.}
$$
Coupling to lower bands contributes positively to inverse <mass>; coupling to higher bands contributes negatively. Their signs and <matrix elements> determine the actual curvature.
The original PDF prints $m^*=\hbar^2[\nabla_{\mathbf k}^2\varepsilon_n]^{-1}$. \b[Taken literally as a three-dimensional <Laplacian>, this misses a factor of three for an isotropic <scalar> <mass>.] The correct identities are
$$
\frac{\nabla_{\mathbf k}^2\varepsilon_n}{\hbar^2}=\operatorname{tr}(m^{*-1}),\qquad m^*=\frac{3\hbar^2}{\nabla_{\mathbf k}^2\varepsilon_n}\quad\text{in three isotropic dimensions}.
$$
The printed expression would return $m/3$ even for the free dispersion $\hbar^2k^2/(2m)$. It is correct in one dimension, or if its <second derivative> is interpreted directionally. For completeness, its literal three-dimensional trace ratio would instead be $3+(2/m)\sum_{l\ne n}|\langle n|\mathbf p|l\rangle|^2/(E_n-E_l)$; that is the trace of the dimensionless inverse-mass <tensor>, not the usual <scalar> $m/m^*$.
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