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For observables $A,B$, nonzero $[A,B]$ obstructs a complete common eigenbasis. In a finite-dimensional system, commuting Hermitian operators can be simultaneously diagonalized; quantum degeneracy may require choosing an appropriate basis inside the degenerate eigenspace. Noncommuting observables can still share particular eigenvectors, so incompatibility is not a claim that every individual state must have nonzero spreads in both quantities.

The Robertson uncertainty relation follows by applying Cauchy-Schwarz inequality to $(A-⟨A⟩)|\psi ⟩$ and $(B-⟨B⟩)|\psi ⟩$:In particular $[x,p_x]=i\hslash$ gives $\Delta x\Delta p_x\ge \hslash /2$. These are preparation spreads, not necessarily measurement errors. A zero commutator expectation can make this bound zero even when the linear operators do not commute.

Measurement order can matter: preparing $|+z⟩$, measuring $S_x$, then measuring $S_z$ gives both z outcomes with probability $1/2$ when the intermediate result is ignored. A direct z measurement would give $+\hslash /2$ certainly. Finally the Heisenberg equation of motion $dA/dt=(i/\hslash )[H,A]+\partial A/\partial t$ relates commutation to conservation: an observable with no explicit time dependence and $[H,A]=0$ is conserved. For unbounded self-adjoint operators these statements require the appropriate common domains; a formal commutator calculation alone need not justify simultaneous spectral measurements.

If $\hat{A}|a⟩=\lambda |a⟩$, the Hermitian operator property makes $\lambda$ real. Applying the operator identity to this eigenvector gives ${\lambda }^3=1$. Its only real root is $\lambda =1$; the other cube roots of unity are nonreal. In an orthonormal eigenbasis, $\hat{A}={\sum }_a|a⟩⟨a|=\boxed{\hat{I}}$. Equivalently, the spectral theorem places the entire spectrum of the self-adjoint observable at one, giving the same conclusion. The quantum-mechanical setting assumes the usual well-defined observable and operator domains.

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.

A Hermitian operator satisfies the adjoint symmetry conditionfor vectors in its specified domain. For a finite matrix this means $A=A^{†}$. For an unbounded differential operator, boundary conditions and domains matter: symmetry on a chosen domain is not alone sufficient for self-adjointness. Physical observables require the corresponding self-adjoint operator and its spectral measure. This distinction prevents purely formal integration by parts from being mistaken for a complete operator definition.

Real eigenvalues. If $A|a⟩=a|a⟩$ with nonzero norm, Hermiticity impliesso $a=a^{\ast }$. Thus eigenvalues representing measurement outcomes are real.

Orthogonality. For eigenvectors $|a⟩,|b⟩$,Hence $(a-b)⟨b|a⟩=0$, and distinct eigenvalues have orthogonal eigenvectors. Within a degenerate eigenspace, Gram-Schmidt orthogonalization can choose an orthonormal basis. In finite dimensions the spectral theorem gives a complete orthonormal eigenbasis andwhere $P_a$ projects onto each eigenspace. In infinite dimensions a continuous spectrum may require a spectral integral rather than a normalizable eigenvector for every possible outcome; a position operator on the full line is a familiar example.

Measurements and probabilities. In the discrete case the Born rule assigns probability $⟨\psi |P_a|\psi ⟩$ to the outcome $a$. For a nondegenerate eigenbasis, $|\psi ⟩={\sum }_a{c}_a|a⟩$ gives probabilities $|c_a{|}^2$, withThe variance vanishes precisely when the state lies in an eigenspace with a single sharp outcome. A repeatable ideal projective measurement prepares the corresponding eigenspace; superpositions of different eigenvalues do not have a preassigned sharp value of that observable in this description.

Compatibility and dynamics. Commuting Hermitian matrices admit simultaneous diagonalization, permitting a common set of sharp labels. For example $L^2,L_z$ commute, whereas $L_x,L_y$ do not. In the differential-operator setting the corresponding spectral compatibility requires appropriate domain conditions. Noncommuting observables obey the Robertson uncertainty principleThe canonical commutation relation then gives $\Delta x\Delta p\ge \hslash /2$.

The Hamiltonian operator represents energy. A self-adjoint $H$ generates unitary time evolution, $U(t)=e^{-iHt/\hslash }$, preserving norm and total probability; its eigenvalues determine stationary-state phases and transition energies. Hermiticity supplies real outcomes and orthogonal eigenspaces, while self-adjointness and the spectral theorem make these measurement and evolution rules mathematically complete.

Why degenerate perturbation theory is needed. If several unperturbed states have equal or nearly equal energy, a perturbation can mix them at leading order. In that subspace, diagonalize the perturbation matrix rather than using nondegenerate formulas with small energy denominators. The resulting eigenvectors identify the combinations selected by the coupling, and the eigenvalues give the leading splitting. Other remote states can then be included perturbatively.

For the linear combination of atomic orbitals description of molecular hydrogen, equivalent atomic $1s$ states on the two nuclei supply the nearly degenerate starting functions. In an orthonormal two-state model, equal diagonal energy $E_a$ and real hopping $t$ give $E_a\pm t$ and symmetric/antisymmetric combinations. For actual overlapping atomic functions with overlap $S$ and Hamiltonian coupling $h$, one instead solves a generalized eigenvalue problem:The bonding state has enhanced density between the nuclei and the antibonding state a separating node. Two opposite-spin electrons occupying the bonding molecular orbital describe the simplest closed-shell hydrogen-molecule bond. Electron-electron interaction and nuclear repulsion are needed to calculate a realistic total binding energy; the two-state splitting alone is not the full molecular energy.

In the nearly-free electron model, the analogous degenerate states are plane waves $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$ at a Bragg scattering. A Fourier coefficient $U_{\mathbf{G}}$ of the periodic potential couples them. At exact degeneracy $E_B$, diagonalizing their two-by-two matrix givesThe associated standing waves place electronic density differently relative to the ionic potential. These are the same gap-opening combinations illustrated in Q1(b). The labels of the two starting basis states survive as mixed orbital or plane-wave character, rather than independent crossing eigenstates.

Origin of broad and narrow bands. Spatially extended s/p-derived orbitals have substantial overlap and therefore large hopping amplitudes and broad bands. More localized d orbitals overlap less strongly and produce narrower bands. A multiorbital tight-binding calculation assigns the on-site orbital energies and hopping matrices, forms Bloch sums, and diagonalizes the resulting matrix at each wavevector. Different hopping ranges and orbital symmetry determine the bandwidths and allowed interband coupling. The two-band description retains one broad combination and one nearly flat d-derived combination; a real transition metal contains several d orbitals and more complicated dispersions.

Hybridized eigenvalues and eigenstates. Let $|{\psi }_{\mathbf{k}}⟩=a|{\varphi }_{\mathbf{k}}⟩+b|{\chi }_{\mathbf{k}}⟩$, in an orthonormal basis. Hermitian operator fixes the reverse coupling to be $V^{\ast }$. Projection onto the two components givesSolving this quadratic yields the band hybridization dispersionFor nonzero $V$, one eigenvector representation is $(a,b)\propto (V,E-E_1)$, followed by normalization. Thus the wavefunctions exchange broad-band and d-band character through the avoided crossing; the levels are equal mixtures at the uncoupled crossing.

At $V=0$, the parabola crosses the horizontal level at $k=\pm k_0$, with $k_0=\sqrt{2m{E}_d}/\hslash$. For small nonzero $|V|$, the curves avoid crossing with minimum direct separation $2|V|$. The lower branch follows the parabola below the crossing and flattens toward $E_d$ above it; the upper does the opposite. The lower branch satisfies $E_{-}<E_d$ everywhere and tends to $E_d$ from below as $k\to \infty$, while the upper branch always lies above $E_d$. This is the perfectly flat-band continuum approximation, with zone-boundary intersections excluded as requested.

Fixed particle number and the larger Fermi sphere. Denote the uncoupled broad-band Fermi radius by $k_0$ to distinguish it from the new one $K$. With two spin states, the broad-band number density at $V=0$ is $k_0^3/(3{\pi }^2)$. The stipulated equal population in the flat sector makes the total density twice that value. At zero temperature and any nonzero hybridization, all occupied states fit into the monotonic lower branch below $E_d$, while the upper branch is empty. ThereforeThis is the stated factor increase. It uses fixed total particle number, equal spin counting, an exactly flat unhybridized band and enough available states below any Brillouin-zone restriction. In generic pairs of dispersive bands both hybridized pockets can remain occupied; then Fermi-volume conservation in two hybridized electron bands fixes a sum of volumes rather than this one radius. The exactly flat band is degenerate at $V=0$, so its initial occupation must be specified, as it is here.

Fermi-velocity suppression. Set $\Delta (k)=E_1(k)-E_d$. The exact derivative of the lower branch isAt $k=K$, $E_1(K)=2^{2/3}{E}_d$ and $\Delta (K)=(2^{2/3}-1)E_d>0$. For $|V|\ll \Delta (K)$, the square bracket divided by two is $|V{|}^2/\Delta (K{)}^2+O(|V{|}^4/\Delta (K{)}^4)$. Since group velocity is ${\hslash }^{-1}dE/dk$ and the broad-band reference velocity is $v_F^{(1)}=\hslash k_0/m$,so the flat-band hybridization Fermi-velocity suppression isThe reference velocity is that of the original broad-band Fermi sphere, not the broad-band slope at the larger $K$; omitting the radius change loses the factor $2^{1/3}$.

The low-temperature qualification becomes more stringent as $V\to 0$: the new chemical potential approaches $E_d$ with $E_d-\mu \simeq |V{|}^2/\Delta (K)$, and the relevant thermal energy must be small compared with this hybridization scale. At fixed nonzero temperature the singular flat-band redistribution is not automatically resolved for arbitrarily small coupling. A small but finite narrow-band dispersion can also change the occupation topology.

Density of states and magnetism. For an isotropic three-dimensional branch with spin degeneracy two,A small Fermi velocity therefore produces a large density of states. The localized, partially filled d-derived bands in iron, cobalt and nickel place many states close to the Fermi level; broad s/p-derived conduction bands in copper, silver and gold coexist with filled d bands lying below it. This explains the substantially larger Fermi-level density of states of the former group within the band picture. An exact factor of ten requires the actual band parameters and is not fixed by the two-band toy model.

A high density of states lowers the kinetic-energy cost of spin polarization. A repulsive exchange interaction can then satisfy the Stoner criterion and stabilize ferromagnetism. With a single-spin density of states $g_s$ and an on-site interaction normalization, the criterion is $U{g}_s(E_F)>1$; with spin-summed $g=2g_s$, it is $Ug(E_F)>2$. The large d-state contribution favors this instability in Fe, Co and Ni. The noble metals do not satisfy the corresponding instability condition in the same ordinary state; this does not imply that they have no weak magnetic response whatsoever.

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 $m/m^{\ast }$.

For a Hermitian operator $Q$, the operator exponential $C_{\lambda }=e^{i\lambda Q}$ is a unitary operator, with $C_{\lambda }^{†}=e^{-i\lambda Q}$. Differentiation and $[Q,C_{\lambda }]=0$ giveWhen $[Q,A]=cA$, this is the first-order ordinary differential equation $A^{\prime }(\lambda )=icA(\lambda )$ with $A(0)=A$. Multiplication by $e^{-ic\lambda }$ makes its derivative zero, provingEquivalently, the commutator expansion for exponential conjugation has successive nested commutators $c^{r}A$, whose exponential series sums to this expression.

Take parity symmetry in quantum field theory to preserve free time evolution, so the conjugate initial data reflect along with the field. Substitute the mode expansion of a free field into the defining parity operator relation and change the integration variable $\mathbf{p}\mapsto -\mathbf{p}$. Since $E_{-\mathbf{p}}=E_{\mathbf{p}}$, coefficient comparison, or comparison of both the field and its conjugate initial data, gives the intrinsic-even conventionHere parity symmetry in quantum field theory reverses spatial momentum and preserves time. The creation operator formulas follow by taking Hermitian adjoints. The given field relation has fixed the intrinsic phase to $+1$.

For one real scalar field, let $R$ act on one-particle wave functions by $(Rf)(\mathbf{p})=f(-\mathbf{p})$. It is a Hermitian operator with $R^2=1$, so $P_{-}=(1-R)/2$ is the orthogonal projector onto odd wave functions. Its second quantization is the odd-mode number operatorThe full momentum space is integrated in both expressions; the factors prevent double counting the pairs $\{\mathbf{p},-\mathbf{p}\}$. A suitable parity operator isFor verification, $[N_{\mathrm{odd}},a_{\mathbf{p}}]=-(a_{\mathbf{p}}-a_{-\mathbf{p}})/2$. Thus the even combination $a_{\mathbf{p}}+a_{-\mathbf{p}}$ is unchanged, and the odd combination $a_{\mathbf{p}}-a_{-\mathbf{p}}$ changes sign under conjugation by $e^{i\pi N_{\mathrm{odd}}}$, exactly reproducing $a_{\mathbf{p}}\mapsto a_{-\mathbf{p}}$. The same holds for creation operators. The odd-mode number operator has integer eigenvalues, so ${\mathcal{P}}^2=1$, and $\mathcal{P}|0⟩=|0⟩$. Multiplying by an overall constant phase would not change its conjugation action. This is parity as odd-mode number parity.

Define the constant Dirac gamma matrix combinations ${\alpha }^k={\gamma }^0{\gamma }^k$ and $\beta ={\gamma }^0$. The Hamiltonian operator isEvaluate its commutator with $n(\mathbf{x})$ using the bilocal result in a. The mass term cancels immediately when $\mathbf{z}=\mathbf{y}$, because its two Dirac delta distributions coincide. For the derivative term first keep the two points distinct:The terms in which the derivative acts on ${\psi }_c$ cancel against each other; the derivative of the second Dirac delta distribution remains. ThusThe last equality is integration by parts, or the defining derivative rule for a distribution. Consequently the literal conserved current whose divergence equals $C$ isBecause a commutator of two Hermitian operators is anti-Hermitian, this $\mathbf{J}$ includes $-i$. The physical Hermitian Dirac current is instead $j^k=\bar{\psi }{\gamma }^k\psi$, so $C=-i\nabla \cdot \mathbf{j}$. The Heisenberg equation of motion now givesThis establishes the continuity equation directly from the required commutator, without assuming the Dirac equation or invoking the Noether theorem as a substitute for the computation.

A quantum state ensemble of normalized states $|{\psi }_i⟩$, with probabilities $p_i\ge 0$ and ${\sum }_i{p}_i=1$, has density operatorIt is a positive semidefinite Hermitian operator with trace one, and the Born rule assigns a measurement outcome with projector $\Pi$ the probability $\mathrm{Tr}(\rho \Pi )$.

A pure state has $\rho =|\psi ⟩⟨\psi |$, hence ${\rho }^2=\rho$ and purity of a density operator $\mathrm{Tr}({\rho }^2)=1$. Its only nonzero eigenvalue is one. A mixed state has at least two nonzero eigenvalues and $\mathrm{Tr}({\rho }^2)<1$. The spectral theorem proves the converse criteria, since the nonnegative eigenvalues sum to one. Different quantum state ensembles may give the same density operator; the mere presence of several terms in a preparation list does not make the state mixed if all positive-weight terms represent the same quantum state ray.

A bounded Hermitian operator $T$ on a Hilbert space is positive, also called a positive semidefinite operator, whenThe value must be real as well as nonnegative. This permits zero values on nonzero vectors: the zero linear operator is positive. For a bounded Hermitian operator, positivity is equivalent to its spectrum being contained in $[0,\infty )$; in finite dimension, this is the condition that all eigenvalues are nonnegative.^[ref]

The Hermitian operator diagram separates three cases. $D=\mathrm{diag}(1,-1)$ is self-adjoint but not positive, because $q_D((0,1{)}^{\mathrm{T}})=-1$. The orthogonal projection $P=\mathrm{diag}(1,0)$ is positive semidefinite, with $q_P(x)=|x_1{|}^2$, but is not strictly positive on every nonzero vector. The complex Hermitian matrix $H=\left(\begin{array}{cc}2& i\\ -i& 2\end{array}\right)$ is positive definite because $q_H(a{e}_1+b{e}_3)=|a{|}^2+3|b{|}^2>0$ whenever $(a,b)\ne (0,0)$. For real symmetric matrices, MIT's lecture 27 relates these tests to the geometry of quadratic forms and their minima.^[ref]

A fundamental source of positive operators is $T^{\ast }T$ for a bounded linear operator $T:H\to K$:For the complex matrix $T=\left(\begin{array}{cc}1& i\\ 2& 1-i\end{array}\right)$ used above,Its eigenvalues are $4-\sqrt{6}$ and $4+\sqrt{6}$, both positive. Positivity organizes the squared error minimized by a least-squares solution: the same $T^{\ast }T$ occurs in the normal equation for a linear inverse problem. The positive square root of an operator gives the converse squared-length description, $q_T(x)=\Vert T^{1/2}x{\Vert }^2$ for every bounded positive operator $T$.

On a real Hilbert space, nonnegativity of $⟨Tx,x⟩$ alone does not imply self-adjointness, so the Hermitian operator hypothesis is part of the definition. On a complex Hilbert space, real-valuedness of all these expressions already forces self-adjointness by the polarization identity.

Strict positivity and a uniform positive lower bound also differ in infinite dimension. On the l2 sequence space, the bounded Hermitian operator $(Tx{)}_n=x_n/n$ satisfiesConsequently no $\gamma >0$ satisfies $q_T(x)\ge \gamma \Vert x{\Vert }^2$ for every $x$. Compare the positive definite symmetric operator and uniformly positive definite symmetric operator conditions.

For a Hermitian operator with spectral decomposition $X={\sum }_i{\lambda }_i{P}_i$, its positive part is $X_{+}={\sum }_i\max ({\lambda }_i,0)P_i=(|X|+X)/2$. It is a positive operator supported on the positive spectral subspace. Together with the negative part of a Hermitian operator it satisfies $X=X_{+}-X_{-}$ and $|X|=X_{+}+X_{-}$.

For a fixed dimensionless Hermitian operator $A$ driven by $H(t)=\Omega (t)A$, all instantaneous generators commute. The exact unitary operator is $U(T)=\exp (-i\theta A)$, so the area, rather than the detailed envelope, fixes the rotation. For $A^2=I$, it is $\cos \theta I-i\sin \theta A$. This convention uses the Quantum Hamiltonian coefficient directly; a convention with $\hslash {\Omega }_{R}A/2$ instead has $\theta =\frac{1}{2}\int {\Omega }_{R}dt$.

Let a Hermitian operator $A$ have dyadic eigenvalues ${\lambda }_j=c_j/2^t$ in $[0,1)$ for known $t$, with controlled access to the unitary operator $U=e^{2\pi iA}$. This interval ensures that different eigenvalues have different eigenphases; exact representability alone would not exclude phase aliasing, as $0$ and $1$ both give phase zero. Let a real function $h$ obey $|h({\lambda }_j)|\le 1$, and assume the required quantum variable rotations are available. Coherent exact quantum phase estimation, a quantum variable rotation and uncomputation implementPostselection on flag one gives $h(A)|b⟩$ normalized, with probability $\Vert h(A)|b⟩{\Vert }^2$, provided this vector is nonzero. Erasing the eigenvalue label by uncomputation is essential to preserve coherence between different eigenvectors. On this dyadic spectrum, $U^{2^t}=I$ gives $U^{-1}=U^{2^t-1}$, so reversing the phase-estimation gates is possible using forward controlled-$U$ calls. The choice $h(\lambda )=\lambda$ multiplies by $A$; a scaled reciprocal gives the different filter used in the HHL algorithm.

If a Fermi interaction contains vector and axial operators with the same time-reversal symmetry component signs, antiunitarity conjugates their coupling constants without exchanging the two operators. After removing a common field phase, invariance requires $g_A/g_V$ real when $g_V\ne 0$. Equivalently $\mathrm{Im}(g_A{g}_V^{\ast })=0$. Fixed intrinsic phases can additionally constrain the common phase; the relative condition is independent of that choice. Adding a Hermitian-conjugate interaction guarantees a Hermitian operator Hamiltonian, but does not guarantee time-reversal symmetry.

A stabilizer code is the common $+1$ eigenspace of a stabilizer group $S$, an abelian subgroup of the Pauli group excluding $-I$. Choose $r$ independent stabilizer generators $g_1,\dots ,g_r$ on $n$ physical qubits. They are commuting Hermitian operators, each squaring to $I$. The stabilizer subspace and its orthogonal projector areThe stabilizer-projector formula gives $\dim \mathcal{C}=2^{n-r}$, so $k=n-r$ logical qubits remain. Independent here means that no nonempty product of the chosen stabilizer generators is $I$. Excluding $-I$ is necessary: a vector cannot simultaneously be fixed by $I$ and $-I$ unless it is zero.

For the three-qubit bit-flip repetition code, take $g_1=Z_1{Z}_2$ and $g_2=Z_2{Z}_3$, with $Z_j$ the Pauli Z gate on site $j$ and identities elsewhere. The stabilizer group is $S=\{I,Z_1{Z}_2,Z_2{Z}_3,Z_1{Z}_3\}$. Both stabilizer generators have eigenvalue $+1$ exactly on the span of $|000⟩$ and $|111⟩$, giving $2^{3-2}=2$ dimensions for one logical qubit. A projective measurement of these checks distinguishes four orthogonal two-dimensional sectors rather than eight individual computational basis states.

A Pauli error that anticommutes with a stabilizer generator changes that check's sign, producing an error syndrome. A Pauli operator commuting with every check can still act on the logical qubits; the centralizer of a stabilizer group describes these possibilities. Thus the number of checks determines the encoded dimension, while the quantum code distance determines protection against arbitrary local errors. The five-qubit error correcting code, for instance, has four stabilizer generators and quantum code distance three; the three-qubit bit-flip repetition code has two checks but quantum code distance one because a single phase flip is unflagged. ^[ref].

The trace norm of a finite-dimensional operator $A$ is the sum of its singular values:For a Hermitian operator $A=A_{+}-A_{-}$, it equals $\mathrm{Tr}{A}_{+}+\mathrm{Tr}{A}_{-}$.

A Hermitian operator satisfies $\Vert X{\Vert }_1={\max }_{-I\le T\le I}\mathrm{Tr}(XT)$. Each diagonal entry of $T$ in an eigenbasis of $X$ lies between minus one and one, giving the upper bound ${\sum }_i|{\lambda }_i|$. The sign operator of $X$ attains it. The substitution $T=2E-I$ converts this formula to optimization over binary POVM effects and proves the Holevo–Helstrom theorem.