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Take the nonnegative scalar Laplace-Beltrami operator $\Delta =-\mathrm{div}\nabla$ on a closed compact Riemannian manifold, and include every eigenvalue with its multiplicity. The spectrum contains substantial global geometric information, but generally does not determine the Riemannian metric up to isometry. Specifying the operator and its domain matters: the spectrum on functions is different information from the spectra of the Hodge Laplacian on differential forms, and a boundary requires fixed boundary conditions.

The Weyl law makes dimension and volume audible. If $N(\lambda )$ counts eigenvalues at most $\lambda$, thenwhere ${\omega }_d$ is the volume of the Euclidean unit ball. Thus the growth exponent recovers $d$, and its leading coefficient recovers $\mathrm{Vol}(M)$. The multiplicity of zero gives the number of connected components: a function in the kernel has $\int |\nabla f{|}^2=0$ and hence is constant on each component.

More detailed invariants come from the heat trace, which is determined directly by the spectrum:The first heat invariants for a closed manifold areHere $R$ is scalar curvature, $\mathrm{Ric}$ the Ricci curvature, and $\mathrm{Rm}$ the Riemann curvature tensor. A divergence term in the local coefficient integrates to zero on a closed manifold. The expansion supplies further integrated curvature expressions, rather than a pointwise curvature map.

On a closed surface, $R=2K$, $|\mathrm{Ric}{|}^2=2K^2$ and $|\mathrm{Rm}{|}^2=4K^2$, so the same coefficients becomeThe Gauss-Bonnet theorem gives $\int KdA=2\pi \chi (M)$; hence the constant term of the heat trace is $\chi (M)/6$. In particular the Euler characteristic, and for a connected orientable surface its genus, are spectrally determined. There is even spectral determination of hyperbolic curvature on a surface: if this spectrum agrees with that of a curvature-$-1$ surface, the coefficients give $\int K=-A$ and $\int K^2=A$, whenceThus $K=-1$ everywhere. This determines local curvature, while leaving room for globally nonisometric hyperbolic surfaces.

Geodesic information is also accessible in important settings. For closed hyperbolic surfaces, the Selberg trace formula makes the scalar spectrum equivalent to the unmarked length spectrum with its appropriate multiplicities. This does not supply a marked length spectrum: it does not identify which group element or intersecting curve has a given length. For general manifolds, one must not simply assert this equivalence; the singularities of the wave trace can be affected by degeneracies and cancellations.

The historic development separates these positive results from the inverse problem. Weyl's 1911 eigenvalue asymptotics established the leading geometric information. The Minakshisundaram-Pleijel heat expansion of 1949 brought local curvature into the short-time analysis. In 1964 Milnor exhibited nonisometric flat tori in dimension sixteen with identical scalar spectra, showing that all eigenvalues together need not recover the metric. Kac's 1966 question about a vibrating drum popularized the corresponding inverse problem for planar domains. Vignéras's 1980 hyperbolic examples and the Sunada theorem of 1985 extended the supply of counterexamples; the latter turns Gassmann equivalence of finite subgroups into a geometric construction. The four-dimensional tetracode length-isospectral lattice construction and the planar examples of Gordon, Webb and Wolpert in 1992 made nonuniqueness explicit in smaller dimensions and in the original drum setting.

Nor is scalar isospectrality a universal topological invariant strong enough to determine the manifold. For example, Tetra and Didi are scalar-isospectral closed orientable flat three-manifolds with first Betti numbers one and zero. Their topology therefore differs. The spectra of all Hodge Laplacians would detect these Betti numbers through the dimensions of their zero eigenspaces, illustrating the extra information obtained by changing the operator. Dimension, volume and many integrated invariants are determined; the full metric and, in general, the topology are not.

The Bloch vector representation of a qubit density matrix is $\rho =\frac{1}{2}(I+r_{x}X+r_{y}Y+r_{z}Z)$, where $r_j=\mathrm{Tr}(\rho {\sigma }_j)$ and $|\mathbf{r}|\le 1$. Pure states have $|\mathbf{r}|=1$, while the origin represents the maximally mixed state $I/2$, not a pure state.

For the two preparations, direct evaluation of the Pauli matrices givesTheir dot product is $-1/2$, so their Bloch vectors meet at angle $\boxed{2\pi /3}$, or $120$ degrees. This is the angle between Bloch vectors; the angle determined by the modulus of the Hilbert-space overlap is different.

Completeness of the POVM requires $E_3=I-E_1-E_2$. In the computational basis, subtraction givesAn overall phase of ${\varphi }_3$ is immaterial. The matrix has eigenvalues $2/3$ and zero, so it is positive; together with $E_1,E_2$ it is a valid trine qubit POVM.

The Born rule gives $\mathrm{Pr}(j\mid {\rho }_i)=\mathrm{Tr}(E_j{\rho }_i)=\frac{2}{3}|⟨{\varphi }_j|{\psi }_i⟩{|}^2$. The conditional probabilities areOutcome one identifies ${\rho }_2$, since ${\varphi }_1$ is orthogonal to ${\psi }_1$; outcome two identifies ${\rho }_1$, since ${\varphi }_2$ is orthogonal to ${\psi }_2$. Outcome three is inconclusive. If Alice's prior probability for ${\rho }_1$ is $q$, the unconditional outcome probabilities are $(1-q)/2$, $q/2$, and $1/2$ respectively. Because outcome three has the same likelihood for both preparations, it leaves the prior odds unchanged. Bob's conclusive identification succeeds with probability one half and is never wrong; he does not identify the state on the inconclusive trials. This is unambiguous quantum state discrimination.

Here a quantum operation is interpreted as a deterministic quantum channel, so it is completely positive and matrix trace preserving. Write its Kraus representation with ${\sum }_k{A}_k^{†}{A}_k=I$ and define the isometry in its Stinespring dilationThen $V^{†}V=I$ and $\Phi (\rho )={\mathrm{Tr}}_E(V\rho V^{†})$. Acting on $B$ gives ${\sigma }_{A{B}^{\prime }E}=(I_A\otimes V){\rho }_{AB}(I_A\otimes V^{†})$. An isometry preserves the nonzero eigenvalues of a density operator. Hence $S(A{B}^{\prime }E)=S(AB)$, $S(B^{\prime }E)=S(B)$, and $S(A)$ is unchanged. It follows that $I(A:B^{\prime }E{)}_{\sigma }=I(A:B{)}_{\rho }$.

Now discard $E$ and apply part 1. This proves data processing for quantum mutual information:The matrix trace-preserving convention matters. Postselection can increase conditional quantum mutual information: take a uniform classical bit $A$, copy it to $B$ with probability $\epsilon$, and otherwise put $B$ in an erasure state independent of $A$. Initially $I(A:B)=\epsilon$ bits. Projecting onto the nonerased sector and conditioning on success gives a perfectly correlated bit pair with quantum mutual information one. That success branch is matrix trace decreasing, and its renormalized action is not a deterministic channel. Retaining the full outcome flag and averaging all branches restores the ordinary channel inequality.

Trace out $Q^{\prime }$ in the final state. With $q(x,y)=p(x)\mathrm{Tr}(E_y{\rho }_x)$, the remaining state is diagonal in the product classical basis:The Von Neumann entropy of a classical diagonal density operator equals the Shannon entropy of its diagonal probabilities, by its spectrum. Therefore $I(A^{\prime }:B^{\prime })=H(X)+H(Y)-H(X,Y)=I(X:Y)$, the mutual information between source symbol and measurement outcome.

For the input, put $\bar{\rho }={\sum }_{x}p(x){\rho }_x$. The marginal on $Q$ is $\bar{\rho }$, the marginal on $A$ has Von Neumann entropy $H(X)$, and the entropy of a classical-quantum state isTo justify the last formula, diagonalize each ${\rho }_x$ with eigenvalues ${\lambda }_{xj}$. The joint block-diagonal state has eigenvalues $p(x){\lambda }_{xj}$; expanding $-{\sum }_{x,j}p(x){\lambda }_{xj}{\log }_2[p(x){\lambda }_{xj}]$ yields exactly the displayed Von Neumann entropy. Thus $I(A:Q)=S(\bar{\rho })-{\sum }_{x}p(x)S({\rho }_x)$, the Holevo quantity of the ensemble. Part (ii) becomes the Holevo boundIt holds for every POVM, and hence also for the accessible information obtained by maximizing the mutual information over all measurements. The only extra Von Neumann entropy identities used were the classical-diagonal and classical–quantum block-spectrum formulas, both justified above; the information inequality itself followed from the requested strong-subadditivity argument.

Principal component analysis replaces the original coordinates by orthogonal linear combinations, ordered so that the first few retain as much variance as possible. It gives a lower-dimensional summary for visualization or compression and reveals the main directions of variation. A direction of large variance need not be the direction most useful for classification.

Write $\mu =\mathbb{E}X$. For a unit vector $u$, the centered score $Y=u^T(X-\mu )$ has varianceMaximizing this Rayleigh quotient with a Lagrange multiplier givesThus a maximizing direction must be an eigenvector of the covariance matrix. To identify the maximum, use the spectral theorem for real symmetric matrices to choose an orthonormal basis of eigenvectors $v_1,\dots ,v_p$, with eigenvalues ${\lambda }_1\ge \cdots \ge {\lambda }_p\ge 0$. Writing $u={\sum }_j{c}_j{v}_j$ givesEquality is attained at $u=v_1$. Consequently the first population principal component isFor each subsequent principal component, maximize the same variance over unit coefficient vectors orthogonal to all the previously chosen ones. Restricting the expansion above to the remaining eigenvectors gives $v_2$, then $v_3$, and so on. ThusThe principal components are uncorrelated; they need not be independent unless additional distributional assumptions, such as joint Gaussianity, hold. Their total variance is $\mathrm{tr}\Sigma ={\sum }_j{\lambda }_j$. Signs of eigenvectors are arbitrary, and a repeated eigenvalue permits any orthonormal basis in its eigenspace. In sample principal component calculations, replace $\mu$ by the sample mean and $\Sigma$ by the sample covariance matrix.

Let $v_2,v_3,v_4$ be the three supplied coefficient vectors, in their displayed order. Multiplication by the sample covariance matrix givesThe four vectors are pairwise orthogonal, with norms $\sqrt{3}$, $\sqrt{102}$, $\sqrt{6}$ and $\sqrt{204}$. Hence $u_j=v_j/\Vert v_j\Vert$ form an orthonormal basis of eigenvectors. The variance of the principal component with unit coefficient vector $u_j$ is its eigenvalue ${\lambda }_j$. Using unnormalized coefficient vectors would instead give variances ${\lambda }_j\Vert v_j{\Vert }^2$ and would not produce the desired proportions.

The total variance isTherefore the explained variance of a principal component isThe first two principal components together account for $2658/3162\simeq 84.06\%$, and the first three account for $2958/3162\simeq 93.55\%$ of the total variance.

For $v_1=(1,1,1,0{)}^T$, direct multiplication by the sample covariance matrix givesIt is therefore an eigenvector, with eigenvalue $2250$. Normalization gives $u_1=v_1/\sqrt{3}$, so the corresponding sample principal component isIts coefficients are proportional to the required vector. The remaining eigenvalues calculated in (ii) are smaller, confirming that this is the first principal component.

Work in units $\hslash =1$ and with the particle mass absorbed into the stated kinetic term. Divide the interval into $N$ pieces of length $\Delta =T/N$. For a suitable self-adjoint Hamiltonian operator, the Trotter product formula gives the regulated limitInsert $1=\int d{q}_j|q_j⟩⟨q_j|$ between neighboring factors. Because $V(\hat{q})$ is diagonal in the position operator basis, the given free-particle propagator yieldsThe exponent is the time-sliced action. Its continuum notation isThe measure is defined by the displayed finite-dimensional limits and their normalization, not by a translation-invariant flat measure on an infinite-dimensional space. The oscillatory limit needs a damping/analytic-continuation prescription. This is the time-sliced configuration-space path integral derived from the operator kernel.

For a quadratic potential energy, write $q=q_c+\eta$, where the classical path satisfies ${\ddot{q}}_c+V^{\prime }(q_c)=0$ and the endpoint conditions, and $\eta (0)=\eta (T)=0$. Expansion is exact:Integration by parts makes the linear term equal to $-[{\int }_0^T({\ddot{q}}_c+V^{\prime }(q_c))\eta dt]$, since the endpoint term vanishes. It is zero by the Euler-Lagrange equation. Also $V^{\prime \prime }$ is constant, independent of the endpoints. Thus classical-path factorization of a quadratic path integral is exact:Only the classical action carries the endpoint dependence.

Let $D={\partial }_t^2+V^{\prime \prime }$ on homogeneous Dirichlet boundary conditions. Expand $\eta$ in its normalized eigenfunctions, with eigenvalues ${\lambda }_n$. The fluctuation integral becomes a product of regulated integrals $\int d{a}_n\exp (-i{\lambda }_n{a}_n^2/2)$. Each is proportional to ${\lambda }_n^{-1/2}$, with a prescribed complex phase. Relative to $D_0={\partial }_t^2$, the normalization is thereforeThe subscript denotes the Dirichlet operator domain, and the square-root branch is continued from the free kernel with its convergence prescription. Equivalently one can use determinants of $-D$ and $-D_0$. For $V^{\prime \prime }={\omega }^2$, the Dirichlet oscillator determinant ratio isThis also checks the free limit. At conjugate times, the fluctuation operator has a zero eigenvalue and the ordinary finite-prefactor expression fails; the kernel is obtained by distributional continuation. Away from these times the classical boundary problem is unique and the stated factorization applies directly.

For the infinite-time oscillator with an external source, use vacuum boundary conditions and the Feynman i-epsilon prescription. For a smooth source of compact support, defineThe imaginary term damps the real-field integral. Integration by parts gives $S_{ϵ}[q]=\frac{1}{2}q{A}_{ϵ}q+Jq$, with the pairings understood as time integrals. Completing the square givesTranslate the integration variable. The source-independent Gaussian functional integral is $N$, soOn Fourier modes $e^{i\omega t}$, $A_{ϵ}$ has eigenvalue ${\omega }^2-m^2+iϵ$, henceFor $m>0$, closing the contour above for $t>0$ and below for $t<0$ gives $G(t)=-i{e}^{-im|t|}/(2m)$. The physical time-ordered oscillator quantum field theory propagator is $iG$, not $G$ itself. The shifted stationary path is $q_c=-GJ$ and obeys ${\ddot{q}}_c+m^2{q}_c=J$ with the same vacuum prescription. A retarded inverse would define a different boundary problem. Dividing by the zero-source integral removes $N$ and gives the vacuum source functional of a quantum oscillator. At zero mass an additional infrared prescription is needed for the inverse on constant modes.

For $X\in L$, define the Adjoint representation of a Lie algebra by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity gives, for every $Z$,Thus $\mathrm{ad}$ preserves brackets and is a Lie algebra representation. Its Killing form isUsing the representation identity and the cyclic property of the trace,The Killing form is also symmetric by cyclicity.

In a basis $T_a$, write $[T_a,T_b]=c_{ab}{}^d{T}_d$. The lowered structure constants satisfyThe first expression is antisymmetric in $a,b$ and the second in $b,c$. These transpositions generate all permutations, so $c_{abc}$ is totally antisymmetric.

An invariant subspace of $L$ under its Adjoint representation of a Lie algebra is exactly an ideal of a Lie algebra: $[L,I]\subseteq I$. A simple Lie algebra has no nonzero proper ideal, proving the adjoint representation is irreducible.

Put $D_X=d(X)$. The same trace calculation proves invariance of the Trace form of a Lie algebra representation:Because the $D_X$ are anti-Hermitian, $H$ is a real symmetric bilinear form. The Killing form of a simple compact Lie algebra is negative definite. Define a real linear map $T$ by $H(X,Y)=\kappa (TX,Y)$. Invariance of both forms implies that $T$ commutes with every ${\mathrm{ad}}_Z$, while symmetry makes $T$ self-adjoint for the positive inner product $-\kappa$. By the spectral theorem, its real eigenspaces are invariant under the Adjoint representation of a Lie algebra. Irreducibility forces a single eigenvalue, so $T=\mu I$ and $H=\mu \kappa$. This also avoids the extra care needed when applying the complex form of the Schur lemma to a real vector space.

The kernel of $d$ is an ideal of a Lie algebra. It cannot be all of $L$: a trivial representation of dimension greater than one is reducible. Simplicity therefore makes $d$ faithful. For every nonzero $X$,Hence the positive trace index for a compact simple Lie algebra givesin the adapted basis.

For the cubic trace tensor of a Lie algebra representation, the identityexpands toSet $Y=T_d$, $X=T_a$, $Z=T_b$, $W=T_c$. Cyclicity of $B$ converts $B_{\ell bc}$ to $B_{bc\ell }$ and $B_{a\ell c}$ to $B_{ca\ell }$. Thus the invariance identity for a cubic trace tensor is

For the final contraction, it is important to keep the specified index-raising convention. In the adapted basis write $C_{ab}{}^c=c_{ab}{}^c$. Then $c_{abc}=-C_{ab}{}^c$, while raising the last two indices with ${\kappa }^{ab}=-{\delta }^{ab}$ gives $c_a{}^{mn}=-C_{am}{}^n$. Total antisymmetry and the definition of the Killing form implyOnly the antisymmetric part of $B$ in $m,n$ contributes, soConsequently the Killing-normalized contraction of a cubic trace tensor isThe negative sign printed in the last requested identity is incompatible with raising indices by the inverse Killing form. The opposite sign is obtained if one instead defines the lowered structure constants using the positive Euclidean metric, a different convention from the one specified.

For a concrete check, take the two-dimensional representation of $\mathfrak{su}(2)$ with $T_a=-i{\sigma }_a/(2\sqrt{2})$, where the ${\sigma }_a$ are Pauli matrices. ThenTheir contraction is $+{\delta }_{ab}/8$, confirming the positive sign with $\mu =1/4$ and providing a counterexample to the printed sign.

Use the metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and first consider the connected Poincaré group, ${\mathbb{R}}^{1,3}⋊S{O}^{+}(1,3)$. Physical half-integer spins require unitary representations of its double cover ${\mathbb{R}}^{1,3}⋊SL(2,\mathbb{C})$; genuine representations of the group without this cover permit only integer spin. Write $P^{\mu }=d(P^{\mu })$ and $M^{\mu \nu }=d(M^{\mu \nu })$. Strong continuity gives self-adjoint generators, with commutators understood on a common invariant dense domain.

The commuting translation generators have a joint spectral resolution. Lorentz transformations carry the spectral four-momentum $p$ to $\Lambda p$, so an irreducible representation has its momentum spectral measure concentrated on a single Lorentz orbit. For positive-energy particles the orbits of interest are the future mass shells $p^2=m^2>0$, or $p^2=0$ with $p\ne 0$ and $p^0>0$. The zero-momentum orbit and spacelike or negative-energy orbits are outside these two particle cases.

Choose ${ϵ}^{0123}=1$ and define the Pauli-Lubanski pseudovector with the overall sign conventionThis sign makes the spatial rest-frame vector equal to $m\mathbf{J}$. The generators' Hermiticity and the antisymmetric epsilon contraction make $W^{\mu }$ Hermitian: the ordering correction from commuting $P$ through $M$ contracts metric tensors into epsilon and vanishes. The same antisymmetry proves Pauli-Lubanski orthogonality, $W^{\mu }{P}_{\mu }=0$.

The assumed commutators say that $P^{\mu }$ and $W^{\mu }$ transform as four-vectors and that $W$ commutes with all translations. Consequentlycommute with the entire Poincare algebra. For instance the two terms in the Lorentz commutator with a contracted vector square cancel. The Casimir operators therefore have fixed spectral values in a unitary irreducible representation, by the Schur lemma applied to their spectral projections. A second ingredient, the little group, distinguishes the internal states.

Fix a standard momentum $k$ on the orbit. Its little group is the subgroup of Lorentz transformations leaving $k$ fixed. Choose $L(p)$ with $L(p)k=p$. A Lorentz transformation acts on the internal fibre by the Wigner rotationIf $d_K$ is a unitary irreducible representation of this stabilizer, the induced representation acts on wavefunctions asThe measure $d^{3}p/(2p^0)$ on either future mass shell is Lorentz invariant, so this action is unitary. The identityproves the group multiplication law. Conversely, covariance of the translation spectral resolution supplies precisely these fibres and their stabilizer action. A commuting operator acts fibrewise; transitivity relates its fibre values, and irreducibility of $d_K$ makes it scalar. This explains why the construction gives, and classifies, the unitary irreducible representations on each orbit.

For a timelike momentum choose $k=(m,0,0,0)$. A Lorentz transformation fixing it acts only as a rotation of its spatial orthogonal complement, so the massive particle little group is $SO(3)$, or $SU(2)$ in the double cover. PutThe Poincare algebra gives $[J_i,J_j]=i{ϵ}_{ijk}{J}_k$. The unitary irreducible representations of $SU(2)$ are labelled by $j=0,\frac{1}{2},1,\dots$, with ${\mathbf{J}}^2=j(j+1)$ and $J_3$ eigenvalues $-j,-j+1,\dots ,j$. Evaluation of the Pauli-Lubanski pseudovector at $k$ gives $W^0=0$, $\mathbf{W}=m\mathbf{J}$. Thus the massive Pauli-Lubanski Casimir and particle labels areThe full massive induced representation of the Poincare double cover is infinite-dimensional because momentum ranges continuously over the orbit; its finite internal multiplicity is $2j+1$.

For a null momentum choose $k=(E,0,0,E)$ with $E>0$. The infinitesimal stabilizer generators areTheir commutators with momentum, evaluated at $k$, vanish. From the rotation and boost commutators of the Poincare algebra one obtainsThese are the massless little-group generators of $ISO(2)$: a planar rotation and two commuting null rotations, which act like translations in the little-group algebra. Hence the massless particle little group is the Euclidean plane group, with the appropriate double cover for half-integer spin.

The definition of $W$ gives $W^0=\mathbf{P}\cdot \mathbf{J}$ and $\mathbf{W}=P^0\mathbf{J}-\mathbf{P}\times \mathbf{K}$. At the standard null momentum,Suppose the little-group representation has finite dimension, as for particles with finitely many polarizations. The commuting Hermitian $N_1,N_2$ have a finite joint spectrum. Conjugation by $e^{-i\varphi R}$ rotates this spectrum through every planar angle. The only finite set invariant under all these rotations is the origin. Thus both $N_i$ vanish, proving that finite-dimensional unitary representations of the massless little group have trivial translations. Irreducibility then leaves a one-dimensional rotation character, $R=hI$. The massless longitudinal Pauli-Lubanski eigenvalues arefor the double cover, with $h\in \mathbb{Z}$ for a genuine $SO(2)$ character. The number $h$ is helicity. Different helicities have the same two Casimir eigenvalues, so the little group is indispensable for the classification. A connected-group irreducible representation has one helicity; if parity is also implemented, it exchanges $h$ and $-h$ and normally requires their pair.

Unitarity alone also permits nontrivial null-rotation generators. The rotation-invariant operator $N_1^2+N_2^2$ is scalar in an irreducible little-group representation; if it has value $(\rho /E{)}^2>0$, one obtains a continuous-spin representation. A concrete realization on angular wavefunctions isThese operators satisfy the same commutators. Periodic or antiperiodic angular wavefunctions give integer or half-integer rotation modes, and the translation generators connect infinitely many such modes. The invariant labels areThese are the additional infinite-polarization massless unitary irreducible representations. Restricting to finitely many particle polarizations excludes them; they must not be ruled out merely by assuming unitarity. This completes the timelike and null cases of the Wigner classification.

Let $t=(T-T_c)/T_c$, and suppose $r=r_{t}t$ with $r_t>0$ while $u>0$ and $\kappa >0$ remain finite at the transition. Critical exponents describe the leading singular powers: $M\sim (-t{)}^{\beta }$ below the transition, $\chi \sim |t{|}^{-\gamma }$, $\xi \sim |t{|}^{-\nu }$, $C_{V,s}\sim |t{|}^{-\alpha }$, $M(t=0)\sim \mathrm{sgn}(h)|h{|}^{1/\delta }$, and $G(R,t=0)\sim R^{-(D-2+\eta )}$.

The quartic equation of state $h=rM+u{M}^3$ immediately gives $M=\sqrt{-r/u}$ at zero field below the transition, so $\beta =1/2$. Differentiating with respect to the conjugate field gives ${\chi }^{-1}=r+3u{M}^2$, which is $r$ above and $2|r|$ below; therefore $\gamma =1$. At $r=0$, $M=(h/u{)}^{1/3}$ with its real sign, giving $\delta =3$. The quadratic correlation function has $\xi \propto |r{|}^{-1/2}$ and critical Ornstein--Zernike correlation function proportional to $q^{-2}$, giving $\nu =1/2$ and $\eta =0$.

Finally the minimized potential below the transition is $V_s=-r^2/(4u)$, while it is zero above. Its second temperature derivative is a finite step, so the mean-field heat-capacity critical exponent is $\alpha =0$. In particular,These are derived mean-field critical exponents. In the fluctuation-dominated critical region of a phase transition, one instead derives exponents from the scaling eigenvalues of the renormalization-group fixed point, as in the next question. A zero heat-capacity power exponent can also accompany logarithms in other theories, so specifying the power alone does not completely describe that singularity.

Use ${\beta }_T=1/(k_{B}T)$ for inverse temperature, to distinguish it from the order-parameter critical exponent $\beta$. In a translation-invariant pure thermodynamic phase, the connected two-point correlation function isFor finite correlation length, at $|r|\gg \xi$ it decays exponentially, possibly with an algebraic prefactor; ${\xi }^{-1}$ is the asymptotic exponential decay rate. Below an ordered transition the pure phase is selected by taking the thermodynamic limit before $h\to 0^{+}$ or $h\to 0^{-}$. A zero-field symmetric mixture of the two ordered phases has a nondecaying connected term and does not describe the usual pure-phase magnetic susceptibility.

Let $\mathcal{F}=-k_{B}T\log Z$ denote extensive physical free energy and $S={\sum }_r{\sigma }_r$. Since the field contributes $-hS$ to the statistical Hamiltonian, differentiation givesExpanding the variance and using translation invariance proves the correlation-function susceptibility sum rule:The sum includes $r=0$. If the source is instead defined as the dimensionless quantity ${\beta }_{T}h$, the corresponding derivative omits the displayed inverse-temperature factor.

A real-space renormalization group step of length factor $b>1$ uses a local blocking kernel $B_b({\sigma }^{\prime },\sigma )\ge 0$ satisfying ${\sum }_{{\sigma }^{\prime }}{B}_b({\sigma }^{\prime },\sigma )=1$. For example, a deterministic majority-spin rule assigns one coarse Ising spin to each block; tied blocks can be resolved with equal probabilities. Define the blocked statistical Hamiltonian byThe spacing becomes $a^{\prime }=ba$. All generated interactions, including multispin terms and the field-independent constant, must be retained for an exact transformation. Summing over ${\sigma }^{\prime }$ uses normalization to prove $Z(u^{\prime },C^{\prime },N^{\prime })=Z(u,C,N)$. Iteration gives $N_p=b^{-pD}N$ and $Z(u_p,C_p,N_p)=Z(u_0,C_0,N)$.

The coarse probability law is the exact marginal law of these block variables. Indeed, for any coarse observable $A$, its expected value is the microscopic expected value of ${\sum }_{{\sigma }^{\prime }}{B}_b({\sigma }^{\prime },\sigma )A({\sigma }^{\prime })$. A blocking rule that retains the long-wavelength order-parameter mode, with source and observable transformations tracked, therefore retains long-distance response and correlation functions. Their field normalization may change, while their physical length scales and critical singularities do not. Partition-function equality alone would not justify this assertion for an arbitrary kernel that discards all slow variables. Truncating the generated operator family is likewise an approximation.

For the free-energy recursion use dimensionless free energy per site,in the thermodynamic limit. Exact partition-function invariance gives $F(u_0,C_0)=b^{-pD}F(u_p,C_p)$. If one step sends $C$ to $C^{\prime }=b^{D}C+c(u)$, the coefficient $c(u)$ records the identity-operator contribution from eliminated fluctuations and normalization. ConsequentlyRepeated substitution derives the additive free-energy recursion under blocking:In physical energy units, divide this convention by ${\beta }_T$. The additive term is essential for absolute free energies even when it does not affect the leading critical powers.

A renormalization-group fixed point satisfies $R_b(u_{\ast })=u_{\ast }$. Linearize in scaling coordinates $v_i$ so that $v_i^{\prime }=b^{y_i}{v}_i+O(v^2)$. A relevant operator has $y_i>0$, an irrelevant operator has $y_i<0$, and a marginal operator has $y_i=0$ with behaviour decided by nonlinear terms. Thus the eigenvalues of the discrete linear map are $b^{y_i}$; the sign criterion applies to the generator exponents $y_i$, not to the discrete multipliers. The critical surface is the stable manifold on which every relevant scaling field has been tuned away. Irrelevant coordinates flow to the renormalization-group fixed point along it. A repulsive renormalization-group trajectory leaves the renormalization-group fixed point along an unstable relevant direction, ultimately entering a noncritical phase or another scaling regime.