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Use the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and the free real scalar field Lagrangian density $\mathcal{L}=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -\frac{1}{2}{m}^2{\varphi }^2$. Its conjugate canonical momentum is $\pi =\dot{\varphi }$. Spatial translation symmetry and Noether's theorem give the canonical stress-energy tensorVanishing boundary flux then makes the four-momentum of a free real scalar field constant. In particular the contravariant spatial momentum isThe minus sign is from ${\partial }^i=-{\partial }_i$; it is essential for the following mode calculation.

In the quantized theory, insert the mode expansions into this integral, using a Hermitian symmetric ordering or normal ordering. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. The spatial integral is $(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}\pm \mathbf{k})$, leavingEach last term integrates to zero: replace $\mathbf{p}$ by $-\mathbf{p}$ and commute the two annihilation operators or the two creation operators. The mixed terms reorder by the canonical commutation relation. Their vacuum contribution is proportional to the integral of the odd vector $\mathbf{p}$ and vanishes with an inversion-symmetric regulator; normal ordering removes it directly. Thus the normal-ordered free scalar four-momentum has spatial partThese manipulations can be made first with box modes and a symmetric momentum cutoff, then used on finite-particle wave packets.

The commutator derivation identity givesFor $U(\mathbf{y})=e^{-i\mathbf{P}\cdot \mathbf{y}}$, the commutator expansion for exponential conjugation therefore gives the translation of a scalar creation operatorThe vacuum has zero momentum, so $U|0⟩=|0⟩$. Hence the one-particle momentum eigenstate obeys $U|\mathbf{q}⟩=e^{-i\mathbf{q}\cdot \mathbf{y}}|\mathbf{q}⟩$: translation preserves its momentum and multiplies it by the associated phase.

The annihilation operator receives the opposite phase. Substituting both phases into the field expansion givesThis is the specified conjugation convention for the spatial translation operator. On states, the active translation moves a wave packet by $\mathbf{y}$, giving a position wavefunction $\psi (\mathbf{x}-\mathbf{y})$. Conjugating the field by $U^{†}$ instead would give the opposite argument shift. Infinitesimally the displayed result is $[P_i,\varphi ]=i{\partial }_i\varphi$.

The single-particle Hamiltonian $H_0=-{\hslash }^2{\partial }_x^2/(2m)+V(x)$ has a periodic potential $V(x+d)=V(x)$. Define the unitary spatial translation operator $T_{d}f(x)=f(x+d)$. Its derivatives commute with translation, and the potential's periodicity gives $H_0{T}_d=T_d{H}_0$.

Each energy eigenspace is therefore preserved by $T_d$, so within it one may choose simultaneous energy and translation eigenstates. On a large periodic interval of $M$ cells, $T_d^M=1$ and its eigenvalues are phases $\lambda =e^{ikd}$, with $k=2\pi j/(Md)$; the infinite-volume limit gives continuous quasimomentum modulo $2\pi /d$. For a simultaneous state in band $j$, this provesThen $u_{jk}(x)=e^{-ikx}{\psi }_{jk}(x)$ obeys $u_{jk}(x+d)=u_{jk}(x)$, which constructs the usual Bloch state rather than assuming its form. The label $j$ distinguishes energy bands at the same crystal momentum $\hslash k$. In a degenerate energy eigenspace an arbitrary superposition need not itself have one translation phase; the result refers to the simultaneous Bloch states.

The spatial translation operator is $U(\alpha )=e^{-i\alpha \hat{p}/\hslash }$. From $[\hat{x},\hat{p}]=i\hslash$ and the preceding central-commutator result,Thus $\hat{x}U|x⟩=U\hat{x}|x⟩+\alpha U|x⟩=(x+\alpha )U|x⟩$, consistent with $U|x⟩=|x+\alpha ⟩$. Since $⟨x|U=⟨x-\alpha |$, the translated wavefunction is $\boxed{\psi (x-\alpha )}$.

For the quantum harmonic oscillator, the creation and annihilation operators give $\hat{p}=i\sqrt{m\hslash \omega /2}(a^{†}-a)$. Put $s=\alpha \sqrt{m\omega /(2\hslash )}$. The central commutator of $s{a}^{†}$ and $-sa$ is $s^2$, soAs $a|0⟩=0$ and $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$,Taking position wavefunctions provesThe coefficient squares sum to $e^{-s^2}{\sum }_n{s}^{2n}/n!=1$, checking normalization of this coherent state.

Similarly $[\mathbf{P},a_{\mathbf{p}}]=-\mathbf{p},a_{\mathbf{p}}$. Transforming both terms of the field expansion givesTherefore the spatial translation operator generated by $\mathbf{P}$ translates the field argument. Equivalently, its action on states translates a wavefunction in the opposite argument convention.

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 isDifferentiating the velocity givesThis 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.

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, henceThe 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: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.

Periodic Hamiltonian. Acting once with momentum gives $\mathbf{p}(e^{i\mathbf{k}\cdot \mathbf{r}}u)=e^{i\mathbf{k}\cdot \mathbf{r}}(\mathbf{p}+\hslash \mathbf{k})u$. Applying it again and cancelling the common plane-wave factor yieldsThe 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 $\mathbf{k}$.

Expansion about the zone centre. Let $|n⟩=u_{n0}$, $E_n={\epsilon }_n(0)$, 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. Thenbecause 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 ${\hslash }^2{k}^2/(2m)$. 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 energyHermitian 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 ${\epsilon }_n(\mathbf{k})={\epsilon }_n(-\mathbf{k})$ near this isolated state. At a degeneracy the denominators signal the need for degenerate perturbation theory instead.

mass ratio and the printed formula. Write $p_{nl}^i=⟨n|p_i|l⟩$. Differentiating the quadratic energy yieldsFor a specified unit direction $\mathbf{e}$, the resulting scalar mass ratio isIn an isotropic three-dimensional band, averaging equivalent directions givesCoupling 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 $m^{\ast }={\hslash }^2[{\nabla }_{\mathbf{k}}^2{\epsilon }_n{]}^{-1}$. Taken literally as a three-dimensional Laplacian, this misses a factor of three for an isotropic scalar mass. The correct identities areThe printed expression would return $m/3$ even for the free dispersion ${\hslash }^2{k}^2/(2m)$. 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 $3+(2/m){\sum }_{l\ne n}|⟨n|\mathbf{p}|l⟩{|}^2/(E_n-E_l)$; that is the trace of the dimensionless inverse-mass tensor, not the usual scalar $m/m^{\ast }$.

Define the signed dimensionless displacementThe spatial translation operator is $T(x_0)=e^{-ip{x}_0/\hslash }$, and the phases of the new energy eigenstates are fixed by $|n^{\prime }⟩=T(x_0)|n⟩$. Their expansion coefficients are thereforeSince $p=i\sqrt{m\hslash \omega /2}(a^{†}-a)$,Set $A=-\lambda a^{†}$ and $B=\lambda a$. Their commutator is $[A,B]={\lambda }^{2}I$, which commutes with both operators. Thus the stated special case of the Baker--Campbell--Hausdorff formula is applicable and givesBecause $a|0⟩=0$,Taking the overlaps with $⟨n|$ therefore provesThis is the coherent state after shifting an oscillator equilibrium. Its normalization follows from $e^{-{\lambda }^2}{\sum }_n{\lambda }^{2n}/n!=1$.

There is a sign qualification in the printed expansion: $(m\omega x_0^2/(2\hslash ){)}^{n/2}=|\lambda {|}^n$ agrees with $(-\lambda {)}^n$ when $x_0\le 0$, equivalently $\gamma \ge 0$. For a general signed displacement $x_0>0$, the odd coefficients have the opposite sign in the specified translated basis. The squared displacement alone fixes the occupation probabilities, not these coherent relative phases. The two-exponential identity also requires the central-commutator condition used above; it is not valid in that simple form for arbitrary operators.

The canonical commutation relation gives $[P^{\mu },a_p^{†}]=p^{\mu }{a}_p^{†}$ for a free real scalar field. Differentiate the exponential conjugation: its derivative is a scalar multiple of the same operator, yielding the displayed phase. The annihilation operator receives the opposite phase, so at equal time $e^{-i\mathbf{P}\cdot \mathbf{y}}\varphi (\mathbf{x})e^{i\mathbf{P}\cdot \mathbf{y}}=\varphi (\mathbf{x}+\mathbf{y})$. The spatial translation operator translates a state's position wavefunction in the opposite argument direction.