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A density operator describes both a quantum state prepared in one pure state and a statistical preparation whose members occupy different pure states. If normalized vectors $|{\psi }_i⟩$ occur with probabilities $p_i$, defineThe resulting operator is Hermitian, positive semidefinite and has trace one. Positivity follows because $⟨v|\rho |v⟩={\sum }_i{p}_i|⟨{\psi }_i|v⟩{|}^2\ge 0$. Conversely, the spectral theorem gives such a representation of any finite-dimensional operator with these properties, using its eigenvectors and eigenvalues. The matrix representing $\rho$ depends on the chosen basis; the operator and its predictions do not.

One operator gives every measurement probability. For a projective measurement with orthogonal projections ${\Pi }_a$ and outcomes $a$,For example, the first formula follows by averaging $⟨{\psi }_i|{\Pi }_a|{\psi }_i⟩$ over the preparation. The trace expression is basis independent. A general measurement replaces ${\Pi }_a$ by positive measurement effects that sum to the identity, without changing the probability formula. Thus $\rho$ specifies the statistics accessible to measurements on that system.

The distinction between pure and mixed states is especially transparent in the eigenvalues. A pure state has $\rho =|\psi ⟩⟨\psi |$, ${\rho }^2=\rho$, and $\mathrm{Tr}{\rho }^2=1$. A mixed state has at least two positive eigenvalues and $\mathrm{Tr}{\rho }^2<1$. For a qubit,pure states lie on the Bloch sphere; mixed states lie inside the Bloch ball. The maximally mixed qubit $\rho =I/2$ can be prepared by choosing equally between the two $z$ spin eigenstates, or equally between the two $x$ spin eigenstates. These ensembles are different preparation procedures but give the same density operator. No measurement on the resulting qubit alone can identify which ensemble was used.

quantum coherence must be distinguished from merely having two populated levels. In the $z$ basis, the state $|+x⟩$ hasBoth give equal probabilities for the two $z$ outcomes. A measurement along $x$ gives the positive result with probability one for the first state and one half for the second. The off-diagonal entries encode the relative phase information responsible for this difference. Whether a matrix entry is off diagonal is basis dependent; purity is not.

For an isolated system, each component evolves with the same unitary time-evolution operator. Therefore $\rho (t)=U(t)\rho (0)U^{†}(t)$ andUnitary evolution preserves the eigenvalues and purity of $\rho$. A density operator commuting with a time-independent Hamiltonian operator is stationary. A thermal equilibrium example is the canonical ensemble ${\rho }_{\beta }=e^{-\beta H}/Z$, where $Z=\mathrm{Tr}(e^{-\beta H})$ and $\beta =1/(k_{\mathrm{B}}T)$. expectation values then reproduce quantum statistical mechanics without assigning the system a definite energy eigenstate on every run.

A mixed state can also arise without classical randomness in the joint preparation. For a bipartite quantum system, the reduced density operator ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ gives every measurement probability accessible to subsystem $A$. For the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, the joint state is pure, whereas ${\rho }_A=I/2$. Its local mixedness reflects entanglement with the inaccessible system; the partial trace discards correlations that matter for joint measurements. More generally, interaction with an environment can suppress off-diagonal entries in a selected basis, producing decoherence while the combined system still evolves unitarily.

Finally, state evolution conditional on a measurement outcome differs from isolated evolution. An ideal projective measurement yielding $a$ updates the state by the Lüders rule $\rho \mapsto {\Pi }_a\rho {\Pi }_a/P(a)$. Discarding the outcome gives ${\sum }_a{\Pi }_a\rho {\Pi }_a$, which generally removes coherence between different eigenspaces. The Von Neumann entropy $-\mathrm{Tr}(\rho \ln \rho )$ quantifies mixedness: it vanishes for pure states and reaches $\ln d$ for $I/d$. Multiplying by $k_{\mathrm{B}}$ gives thermodynamic entropy units. These uses make the density operator a common language for preparation uncertainty, measurement, thermal states and entangled subsystems, while retaining their distinct physical origins.

orbital angular momentum generates spatial rotations and is $\mathbf{L}=\mathbf{r}\times \mathbf{p}$. With $\mathbf{p}=-i\hslash \nabla$, its components satisfyThus $L^2$ and one component, conventionally $L_z$, have simultaneous eigenstates, while all three components cannot generally have sharp values. On ordinary single-valued spatial wavefunctions,The spherical harmonics form an orthonormal angular basis. The integer restriction is specific to orbital wavefunctions; half-integer spin is an intrinsic degree of freedom, not the angular dependence of a scalar spatial wavefunction.

The angular momentum ladder operators obeyThe ladder ends at $m=\pm \ell$, producing $2\ell +1$ states. A state with specified $m$ has a sharp projection $\hslash m$, but the angular momentum magnitude is $\hslash \sqrt{\ell (\ell +1)}$, not $\hslash \ell$. No classical picture of a vector with simultaneously fixed Cartesian components reproduces all these measurement properties.

In a central potential, rotational symmetry gives $[H,L_i]=0$. separation of variables yields $\psi (\mathbf{r})=R_{\ell }(r)Y_{\ell m}(\theta ,\varphi )$. The radial Schrödinger equation includes a centrifugal potential ${\hslash }^2\ell (\ell +1)/(2m{r}^2)$, which suppresses small-radius amplitude for nonzero $\ell$. The energy is independent of $m$ when rotational symmetry is intact, but need not be independent of $\ell$. A magnetic field can break that degeneracy. Spatial inversion gives $Y_{\ell m}(-\hat{\mathbf{r}})=(-1{)}^{\ell }{Y}_{\ell m}(\hat{\mathbf{r}})$, connecting orbital angular momentum to parity and to the even/odd rotational restriction for the hydrogen spin isomers.

For the nonrelativistic Coulomb Hamiltonian with reduced electron-proton mass $\mu$, separation in Spherical coordinate system gives ${\psi }_{n\ell m}(r,\theta ,\varphi )=R_{n\ell }(r)Y_{\ell m}(\theta ,\varphi )$. The quantum numbers obey $n\ge 1$, $0\le \ell \le n-1$ and $-\ell \le m\le \ell$. These are joint eigenstates of energy, $L^2$ and $L_z$, with the latter two eigenvalues $\ell (\ell +1){\hslash }^2$ and $m\hslash$.

The Coulomb bound-state energies and characteristic length areFor example, ${\psi }_{100}=(\pi a_{\mu }^3{)}^{-1/2}{e}^{-r/a_{\mu }}$ is spherically symmetric and has no radial node. In general the radial part contains an exponential times a Generalized Laguerre polynomial, with $n-\ell -1$ radial nodes. The spherical harmonic specifies angular structure and azimuthal phase; real linear combinations of degenerate $m$ states give familiar directional orbitals.

The leading Coulomb energy depends on $n$ alone, giving $n^2$ spatial states for each principal shell, or $2n^2$ including electron spin. Fine structure, Lamb shift and Hyperfine structure lift parts of this ideal degeneracy. An atomic orbital is a probability amplitude, not a classical planetary path.^[ref]

Hamiltonian and commutators. Use the canonical commutation relation $[\hat{x},\hat{p}]=i\hslash I$ and define $c_0=\sqrt{m\omega /(2\hslash )}$. Expanding the product of the given creation and annihilation operators,Therefore the quantum harmonic oscillator Hamiltonian operator isSimilarly,Using $[AB,C]=A[B,C]+[A,C]B$ gives

Lowering a normalized energy eigenstate. The commutator impliesFor $n>0$, nondegeneracy of this one-dimensional energy level means $\hat{a}|{\varphi }_n⟩=c_n|{\varphi }_{n-1}⟩$. Its norm is fixed by the number operator:Choosing successive eigenstate phases in the conventional way makes $c_n=\sqrt{n}$, soFor $n=0$, the same norm is zero, hence $\hat{a}|{\varphi }_0⟩=0$. A nonzero lowered ground state would have energy below the lowest allowed level, and positivity of ${\hat{a}}^{†}\hat{a}$ forbids that possibility.

Coherent-state eigenvalue. Applying the annihilation operator term by term, with the $n=0$ contribution zero, givesThusThe coherent state is normalized because $e^{-|\alpha {|}^2}{\sum }_n|\alpha {|}^{2n}/n!=1$. Its annihilation operator is not Hermitian, so the eigenvalue $\alpha$ may be complex.

Time evolution. Under the time-independent Hamiltonian operator, each energy component acquires its own phase:Since $|\alpha e^{-i\omega t}|=|\alpha |$, the normalization prefactor is unchanged. HenceThe state remains a coherent state, with its complex parameter rotating at the oscillator's angular frequency; the displayed extra factor is a global phase from the zero-point energy.

Operators, quantization and rotations. Angular momentum is the generator of spatial rotations. In quantum mechanics, its components are Hermitian operators on their appropriate domains and obeyThese angular momentum commutation relations prevent all three components from having simultaneous sharp values. For example, the Robertson uncertainty principle gives $\Delta J_x\Delta J_y\ge \hslash |⟨J_z⟩|/2$. One conventionally diagonalizes ${\hat{J}}^2$ and a selected component ${\hat{J}}_z$:The allowed labels are $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and $m=-j,-j+1,\dots ,j$. An angular-momentum multiplet has $2j+1$ states. The magnitude $\hslash \sqrt{j(j+1)}$ differs from the largest possible component $\hslash j$; the remaining components fluctuate even in a state with maximal $m$.

The angular momentum lowering operator and its raising counterpart, ${\hat{J}}_{\pm }={\hat{J}}_x\pm i{\hat{J}}_y$, satisfyThe ladder terminates at $m=\pm j$. A rotation through angle $\theta$ about unit vector $\hat{\mathbf{n}}$ acts through the unitary operator $\exp [-i\theta \hat{\mathbf{n}}\cdot \hat{\mathbf{J}}/\hslash ]$. Thus quantized components and the continuous rotation of a state are compatible.

Orbital angular momentum. The orbital contribution is associated with motion in space:For a spinless spatial wavefunction, the simultaneous angular eigenfunctions are the spherical harmonics $Y_{\ell m}(\theta ,\varphi )$, with $\ell =0,1,2,\dots$ and $m=-\ell ,\dots ,\ell$. Single-valued ordinary scalar spatial wavefunctions give integer orbital labels. In a central potential energy, $[\hat{H},{\hat{L}}^2]=[\hat{H},{\hat{L}}_z]=0$, so stationary states may be labelled by these angular quantum numbers. The angular kinetic contribution appears in the radial equation as $\ell (\ell +1){\hslash }^2/(2m{r}^2)$, the centrifugal potential. Noncentral interactions need not conserve these orbital labels.

Intrinsic spin and addition. Spin angular momentum is an intrinsic internal degree of freedom with the same commutation algebra. It is not the angular momentum of a small classical charged sphere literally rotating about its centre. For an electron, $s=1/2$ and $m_s=\pm 1/2$, so ${\hat{S}}_i=(\hslash /2){\sigma }_i$ in terms of the Pauli matrices. Its squared spin is $3{\hslash }^2/4$. Other particles can have integer or half-integer spin, including spin zero.

For one particle with orbital and spin degrees of freedom, $\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$; the two component sets commute with each other. Addition of angular momentum permits $j=|\ell -s|,|\ell -s|+1,\dots ,\ell +s$. The coupled states $|j,m_j⟩$ are linear combinations of the uncoupled $|\ell ,m_{\ell }⟩|s,m_s⟩$ with $m_j=m_{\ell }+m_s$, determined by Clebsch-Gordan coefficients. A rotationally invariant spin-orbit interaction can conserve $J^2,J_z$ while mixing the separate component labels.

Angular momentum also carries a magnetic moment. For an electron's orbital motion and intrinsic spin,The minus signs reflect negative electric charge. With the Bohr magneton ${\mu }_B=e\hslash /(2m_e)$, a spin projection $m_s$ has ${\mu }_z=-g_s{\mu }_B{m}_s$. A magnetic field can therefore distinguish angular-momentum states through their interaction energy $-\mathbf{\mu }\cdot \mathbf{B}$.

Stern–Gerlach measurement. In the Stern-Gerlach experiment, atoms emerge from a heated source, pass through collimating apertures, traverse a strongly inhomogeneous magnetic field, and strike a position-sensitive screen. The beam is neutral, so its separation is interpreted through the atomic magnetic moment rather than the ordinary charged-particle Lorentz force. Near a chosen quantization direction, the interaction givesA uniform field can exert a torque but would not separate the beam by this translational force; the gradient is essential. Different allowed magnetic-moment components become correlated with different trajectories. A classical ensemble with continuously varying orientations would yield a continuous spread, whereas discrete quantum components yield separated channels.

In the modern electronic description of ground-state silver, filled shells cancel their electronic angular momenta and the unpaired $5s$ electron has $L=0$, $S=1/2$, $J=1/2$. The effective electronic measurement therefore has two components, approximately ${\mu }_z=\pm {\mu }_B$, rather than an orbital $\ell =1/2$, which is not allowed. For positive field gradient the electronic $m_J=+1/2$ component has negative ${\mu }_z$ and deflects downward in this coordinate convention. Nuclear spin and hyperfine structure are neglected in this simplified two-channel account. Historically the original splitting preceded the electron-spin interpretation, so the modern explanation should not be confused with the original investigators' interpretation.

More generally a prepared spin state $a|+z⟩+b|-z⟩$ gives the two outcomes with probabilities $|a{|}^2,|b{|}^2$. Selecting one channel prepares a state for further measurements. An ideal second analyser along the same axis then gives a definite result, whereas one along a perpendicular axis gives two equally likely outcomes for a spin-one-half eigenstate. This connects the apparatus directly to incompatible spin components and the Born rule. Angular momentum is quantized through its operator algebra; the Stern–Gerlach gradient turns that discrete internal information into spatially distinguishable outcomes.

For removal of a half-oscillator wall, the sudden approximation keeps the state unchanged during the change of potential. Its probability in the new full-oscillator ground state ${\Psi }_0={\pi }^{-1/4}{\ell }^{-1/2}{e}^{-x^2/(2{\ell }^2)}$ is the squared overlap:where ${\int }_0^{\infty }x{e}^{-x^2/{\ell }^2}dx={\ell }^2/2$. ThereforeThe integral runs only over the initially occupied half-line. The initial state is not the odd full-line eigenstate: it is zero, rather than negative, on the left. Indeed its overlap with that final first excited state is $1/\sqrt{2}$, giving probability $1/2$, and the remaining probability lies in higher states. Subsequent evolution under the new fixed Hamiltonian operator changes the quantum phases of its energy components without changing these energy-measurement probabilities.

Put $\Omega =\lambda B$, including its sign. The initial vector is the $+\hslash /2$ eigenstate of $S_z$; it is not an energy eigenstate of $H=\Omega S_x$. In terms of the spin eigenstates found above, $|+z⟩=(|+x⟩+|-x⟩)/\sqrt{2}$. Each energy eigenstate acquires its own unitary time evolution phase:Equivalently, $e^{-iHt/\hslash }=I\cos (\Omega t/2)-i{\sigma }_x\sin (\Omega t/2)$ follows from ${\sigma }_x^2=I$. The Born rule givesThe expectation value of the spin angular momentum is $(\hslash /2)(0,-\sin \Omega t,\cos \Omega t)$, a rotation around the $x$ axis. Its direction follows the specified sign of the Hamiltonian; changing the magnetic coupling sign reverses the precession.

The commutators implyThus any nonzero resulting vector is the adjacent energy eigenstate. Since $a^{†}a|n⟩=n|n⟩$ and $a{a}^{†}=a^{†}a+1$, their squared norms areChoosing consistent phases for the eigenstates yieldsThe vanishing last expression terminates the lowering sequence at the ground state, rather than introducing a negative-energy oscillator state.

The total angular momentum operator is $\mathbf{J}=\mathbf{L}+\mathbf{S}$. Since orbital and spin operators act on different factors of the tensor product, they commute, and expanding $J^2$ givesA coupled angular momentum eigenstate $|l,s;j,m_j⟩$ is a simultaneous eigenstate of these three squared operators, with eigenvalues ${\hslash }^{2}j(j+1)$, ${\hslash }^{2}l(l+1)$ and ${\hslash }^{2}s(s+1)$. Thus the angular factor in the spin-orbit interaction is the numberThe other factor, $r^{-1}dV/dr$, is radial. It commutes with the angular operators, and a central-potential energy eigenstate separates into a radial wavefunction and an angular spinor. Taking its radial expectation value therefore givesThis spin-orbit energy shift in a central potential is independent of $m_j$, so rotational symmetry preserves the $2j+1$ magnetic degeneracy within a fixed $j$ multiplet. It is a first-order fine structure correction; the given coefficient already includes the relativistic factor needed for the electron's spin-orbit coupling.

For either number operator $\hat{N}={\hat{a}}^{†}\hat{a}$, positivity follows fromHence an eigenvalue $\nu$ cannot be negative. The commutatorsshow that $\hat{a}|\nu ⟩$, if nonzero, is an eigenstate with eigenvalue $\nu -1$. For a normalized eigenstate, $\Vert \hat{a}|\nu ⟩{\Vert }^2=\nu$.

If $\nu$ were not an integer, successive nonzero lowerings would eventually reach an eigenstate with $0<{\nu }^{\prime }<1$. Its next lowering would have nonzero norm but a negative eigenvalue, contradicting positivity. The ladder must therefore terminate at $\nu =0$, with $\hat{a}|0⟩=0$. Conversely, successive applications of the creation operator never vanish, becauseThe normalizable bottom state exists explicitly: solving $\hat{a}{\psi }_0=0$ gives ${\psi }_0(x)\propto \exp [-m\omega x^2/(2\hslash )]$. Thus the number operator has exactly the non-negative integer eigenvalues.

Since the two coordinate number operators commute, use the product eigenstates $|n_x,n_y⟩$. Their energy eigenvalues areAt fixed $N=n_x+n_y$, the allowed pairs are $(0,N),(1,N-1),\dots ,(N,0)$. Thus the energy level $E_N=(N+1)\hslash \omega$ has quantum degeneracy $N+1$, and the ground-state energy is $\hslash \omega$.

The number operator commutators established above imply that ${\hat{a}}_x^{†}|n⟩$ is proportional to $|n+1⟩$ and ${\hat{a}}_x|n⟩$ to $|n-1⟩$. Their norms determine the magnitudes of the proportionality constants:Choose the phase of the normalized ground eigenstate once, then define recursivelyThis removes the otherwise arbitrary relative phases of the eigenstates. The Hermitian adjoint relation then fixes the lowering coefficient with the same positive sign, givingAt $n=0$, the second expression means the zero vector, ${\hat{a}}_x|0⟩=0$; it does not introduce a physical state $|-1⟩$.

The specified energy requires $n_x+n_y=2$, so the relevant energy eigenspace has the orthonormal basisApplying the creation and annihilation operators givesThus its matrix in this ordered basis isWriting $q=\hslash \lambda$, the characteristic polynomial gives $\lambda ({\lambda }^2-4)=0$. The possible angular momentum eigenvalues are thereforeFor completeness, normalized simultaneous energy and angular momentum eigenstates areEach value occurs once in this energy eigenspace. Equivalently, circular ladder operators of a two-dimensional oscillator give $\hat{H}=\hslash \omega ({\hat{N}}_{+}+{\hat{N}}_{-}+1)$ and ${\hat{L}}_z=\hslash ({\hat{N}}_{+}-{\hat{N}}_{-})$. With $N_{+}+N_{-}=2$, the difference can only be $2,0,-2$.

Solving the time-independent Schrödinger equation with $\psi \propto e^{ik\varphi }$ gives $E={\hslash }^2{k}^2/(2m{R}^2)$. The periodic boundary condition requires $e^{2\pi ik}=1$, hence $k=n\in \mathbb{Z}$. With the angular integration measure $d\varphi$, wavefunction normalization givesso a convenient orthonormal eigenbasis isThe $n=0$ ground state is nondegenerate. For each $|n|>0$, $u_n$ and $u_{-n}$ span a two-dimensional energy eigenspace; real sine and cosine combinations are an equally valid basis. These opposite-circulation eigenstates have angular momentum $n\hslash$ and $-n\hslash$.

If instead the wavefunction is normalized with respect to physical arc length $ds=Rd\varphi$, its expression is $e^{in\varphi }/\sqrt{2\pi R}$. The two conventions describe the same probability, since $|u_n{|}^{2}d\varphi =|u_n/\sqrt{R}{|}^{2}ds$. The remaining parts use $d\varphi$.

Diagonalization and conventions. The spin-dependent part is $\lambda (-k_y{\sigma }_x+k_x{\sigma }_y)$. For a Hermitian Hamiltonian $\lambda$ is real; the supplied lower-band spinor uses the convention $\lambda >0$, which will be adopted below. The squared spin matrix is ${\lambda }^2(k_x^2+k_y^2)I$. Hence the two ordered energies areIn the $k_z=0$ plane, let $k=\sqrt{k_x^2+k_y^2}$ and $q_R=2m\lambda /{\hslash }^2>0$. ThenIf $\lambda <0$, use $q_R=2m|\lambda |/{\hslash }^2$ to retain the energy ordering, and interchange the spin directions assigned below. This is an isotropic Rashba effect; $q_R$ has units of inverse length, whereas $\lambda$ has units of energy times length.

Band minima. Define $k_R=q_R/2$ and $E_R={\hslash }^2{q}_R^2/(8m)$. Completing the square givesAlong $k_y=k_z=0$, the radial coordinate is $|k_x|$, so the lower branch has two minima:In the full plane these are points on a circle of minima $k=k_R$. Both branches meet at $k=0$, $E=0$. Their ordered energy curves contain $|k_x|$: the lower curve descends from the crossing on either side, while the upper rises. Treating them as globally labeled smooth spin branches without switching labels across $k_x=0$ would obscure the lower-band minima.

Occupied regions and Fermi contours. At zero temperature each spin-split eigenstate with energy below $E_F$ is occupied. PutFor $-E_R<E_F<0$, $0<D<k_R$. The lower-band roots are $k_R-D$ and $k_R+D$, while the upper band is entirely empty. Thus the central disk $k<k_R-D$ is unoccupied, the annulus $k_R-D<k<k_R+D$ is singly occupied, and the exterior is unoccupied. Both Fermi circles belong to the lower band; there is no doubly occupied region.

At $E_F=0$, the lower band is occupied for $0<k<2k_R=q_R$, giving a singly occupied disk with outer Fermi circle at $q_R$. The two bands also meet the Fermi energy at the origin. That isolated degenerate point is not a finite doubly occupied disk, and occupation exactly at the Fermi energy is a boundary convention of zero area. Outside the outer circle the states are empty.

For $E_F>0$, $D>k_R$. The lower branch has outer radius $k_{-}=D+k_R$, while the upper branch has inner radius $k_{+}=D-k_R$. The disk $k<k_{+}$ is doubly occupied, with one state in each band at every momentum. The annulus $k_{+}<k<k_{-}$ is singly occupied, in the lower band only, and the exterior is unoccupied. These are the Fermi contours of an isotropic Rashba band; double occupancy here refers to two orthogonal spinor states at the same momentum, not two electrons occupying one identical quantum state.

Inferring the splitting. Above the crossing, the two positive radii differ by $2k_R=q_R$:The minimum radius is therefore $k_R=0.05{\mathring{\mathrm{A}}}^{-1}$. It is the difference of the measured radii that is required for $E_F>0$; below the crossing the two lower-band radii instead add to $q_R$.

Spinors and spin direction. Write $a=(k_y+i{k}_x)/k$, which has $|a|=1$ for $k>0$. An orthogonal normalized partner to the supplied lower-band state isIndeed $u_{-}^{†}{u}_{+}=(-|a{|}^2+1)/2=0$. Direct multiplication by the spin matrix gives eigenvalues $-\lambda k$ and $+\lambda k$ for $u_{-}$ and $u_{+}$, respectively, so orthogonality has selected the other band rather than merely another arbitrary spin basis.

For a normalized spinor $(\alpha ,\beta {)}^T$, the Pauli expectation values are $(2\mathrm{Re}({\alpha }^{\ast }\beta ),2\mathrm{Im}({\alpha }^{\ast }\beta ),|\alpha {|}^2-|\beta {|}^2)$. Consequently the Rashba spin texture isThe spin lies in the plane and is perpendicular to momentum. On the outer circle $k_1$, which belongs to $E_{-}$, it points clockwise as viewed from $+z$; at $\mathbf{k}=(k_1,0)$ it points along $-y$. On the inner circle $k_2$, belonging to $E_{+}$, it points counterclockwise; at $\mathbf{k}=(k_2,0)$ it points along $+y$. Opposite momenta have opposite spins, consistent with time-reversal symmetry. At the origin the eigenstates are degenerate and no unique spin orientation is defined.

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.