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Extensive and Intensive Quantities

Extensive and intensive are not permanent labels attached to symbols. They are asymptotic claims about a specified quantity along a specified family of systems. Energy can be extensive in one family and superextensive in another. An operator can be a sum over every site even when symmetry forces its expectation value to vanish. A boundary term can be subextensive relative to volume while remaining the leading signal of the physics one wants to measure.

The central diagnostic is

XL=x BL+rL,rL=o(BL),X_L = x\,\mathcal B_L + r_L, \qquad r_L = o(\mathcal B_L),

where BL\mathcal B_L is the chosen bulk-size variable: for example, the number of lattice sites Ns,LN_{s,L} or the continuum volume VLV_L. The coefficient xx is the limiting bulk density when it exists, and rLr_L contains boundary and other subleading contributions.

This page owns the vocabulary and the scaling audit. Thermodynamic Limit owns existence and convergence of the limiting sequence, while Thermodynamic Potentials owns Legendre transforms, natural variables, Euler relations, and response derivatives.

Helpful background. Locality in Many-Body Systems explains support, range, and interaction strength. Asymptotic Analysis reviews OO and oo notation. Use the Statistical Mechanics Checklist if temperature, pressure, or chemical potential are unfamiliar.

A scaling label has no meaning until the family is declared. Before classifying XLX_L, record:

Ledger entryWhat must be specified
bulk sizeBL=Ns,L\mathcal B_L=N_{s,L}, VLV_L, particle number, or another justified variable
geometrydimension, shape, aspect ratio, and which lengths grow
state pointdensity or filling, composition, temperature, fields, and other controls held fixed
interactionscoupling constants, range or decay, and any size-dependent normalization
boundary dataopen, periodic, twisted, fixed, trapped, or interfacial conditions
state or ensembleground state, eigenstate, canonical ensemble, driven state, and so on
quantityoperator, expectation, variance, free energy, response, or estimator

Two uses of the vocabulary must be separated:

  • the leading size dependence of a numerical sequence such as ⟨ML⟩\langle M_L\rangle;
  • the homogeneity class of an operator or thermodynamic state function under replication.

A numerical sequence has a nonzero extensive leading term when

XLBL⟶x≠0,\frac{X_L}{\mathcal B_L} \longrightarrow x\ne0,

equivalently

XL=x BL+o(BL).X_L = x\,\mathcal B_L + o(\mathcal B_L).

The stronger statement includes a limiting density. Merely proving

XL=O(BL)X_L=O(\mathcal B_L)

says only that XLX_L is at most extensive. It permits x=0x=0, oscillation without a limit, or a genuinely subextensive sequence.

By contrast, an extensive operator or state function is degree one in the extensive data a\boldsymbol a:

X(λa)=λX(a)X(\lambda\boldsymbol a) = \lambda X(\boldsymbol a)

at leading bulk order. An intensive thermodynamic variable is degree zero:

Y(λa)=Y(a).Y(\lambda\boldsymbol a) = Y(\boldsymbol a).

These structural classifications survive special state points where the value happens to vanish. A finite-size sequence YL=y+o(1)Y_L=y+o(1) is compatible with an intensive bulk variable, but convergence alone is not sufficient: a correction 1/L1/L also converges and is not thereby a thermodynamic field.

For total-like numerical sequences, compare with the bulk size:

Total-like sequenceDiagnosticTypical example under ordinary bulk assumptions
nonzero extensive leading valueXL/BL→x≠0X_L/\mathcal B_L\to x\ne0particle number at fixed density
subextensive contributionXL/BL→0X_L/\mathcal B_L\to0a regular surface term relative to volume
superextensive value sequence∣XL∣/BL→∞\lvert X_L\rvert/\mathcal B_L\to\inftyunnormalized coherent all-to-all pair energy

Separately, classify replicated objects and state variables by homogeneity:

Replication classDiagnosticTypical example
degree-one extensive objectX(λa)=λX(a)X(\lambda\boldsymbol a)=\lambda X(\boldsymbol a)a local-sum operator or bulk internal-energy function
degree-zero intensive variableY(λa)=Y(a)Y(\lambda\boldsymbol a)=Y(\boldsymbol a)temperature or pressure

These are not competing bins. An O(1)O(1) impurity contribution is subextensive as a total relative to volume, but that does not make it an intensive thermodynamic field. Conversely, temperature and chemical potential are intensive without being obtained by dividing some universal total by volume.

Scaling linear size is also not the same operation as replicating volume. In dd dimensions, L↦λLL\mapsto\lambda L sends a regular volume to λdV\lambda^d V, whereas thermodynamic extensivity concerns scaling all extensive data by a common factor.

A global sum of local densities has the structural form

ML=∑i∈ΛLmi.M_L = \sum_{i\in\Lambda_L} m_i.

When the state is homogeneous and correlations are sufficiently controlled,

⟨ML⟩Ns,L⟶m\frac{\langle M_L\rangle}{N_{s,L}} \longrightarrow m

defines an intensive density. Three distinctions prevent common mistakes:

  1. MLM_L is a global operator even though every mim_i is local.
  2. The operator, its expectation value, its variance, and its standard deviation can have different scaling.
  3. A symmetry-forced value ⟨ML⟩=0\langle M_L\rangle=0 does not turn the sum into an intensive operator.

For example, in a product state of NsN_s independent spins with

⟨Zi⟩=0,⟨Zi2⟩=1,\langle Z_i\rangle=0, \qquad \langle Z_i^2\rangle=1,

the total magnetization M=∑iZiM=\sum_i Z_i obeys

⟨M⟩=0,Var⁡(M)=Ns,ΔM=Ns.\begin{aligned} \langle M\rangle &= 0, & \operatorname{Var}(M) &= N_s, \\ \Delta M &= \sqrt{N_s}. \end{aligned}

For the density m=M/Nsm=M/N_s,

Var⁡(m)=1Ns,Δm=1Ns.\operatorname{Var}(m) = \frac{1}{N_s}, \qquad \Delta m = \frac{1}{\sqrt{N_s}}.

For commuting local densities, write the finite-system connected covariance as Cij(L)=⟨mimj⟩c,LC_{ij}^{(L)}=\langle m_i m_j\rangle_{\mathrm c,L}. A size-uniform summability condition is

sup⁡Lsup⁡i∈ΛL∑j∈ΛL∣Cij(L)∣≤C\sup_L \sup_{i\in\Lambda_L} \sum_{j\in\Lambda_L} \left| C_{ij}^{(L)} \right| \le C

for a constant CC independent of LL. It implies that the variance is at most volume order:

Var⁡(M)=∑i,jCij(L)=O(Ns).\operatorname{Var}(M) = \sum_{i,j} C_{ij}^{(L)} = O(N_s).

If the system has a regular translation-invariant bulk and the bulk-averaged integrated covariance approaches a positive constant, then Var⁡(M)/Ns\operatorname{Var}(M)/N_s approaches that constant. At a critical point, with long-range correlations, or in a constrained collective state, the scaling can change. Fluctuations and Susceptibilities owns those response and critical-scaling qualifications.

“Local” and “intensive” therefore answer different questions. Locality concerns support in space or on a graph. Intensity concerns scaling under enlargement at fixed thermodynamic state.

Extensivity asks how a quantity scales when a system is enlarged. Additivity asks what happens when two macroscopic pieces are combined. Define the composition defect

ΔXAB=X(A∪B)−X(A)−X(B).\Delta X_{AB} = X(A\cup B)-X(A)-X(B).

Then:

  • exact additivity means ΔXAB=0\Delta X_{AB}=0;
  • asymptotic additivity means ΔXAB=o(BA+BB)\Delta X_{AB}=o(\mathcal B_A+\mathcal B_B);
  • nonadditivity means a cross contribution remains of bulk order.

When a coupling itself depends on total system size, the restriction convention is part of this definition. A physical cut of one size-NN system retains the parent couplings in its AA, BB, and cross terms. Rebuilding AA and BB as independently normalized size-NAN_A and size-NBN_B models changes their Hamiltonians and answers a different question.

For short-range interactions between regular regions, coupling the pieces normally creates terms near their common boundary. Those terms are often subextensive, so the bulk quantity is asymptotically additive.

The converse is false: extensive does not imply additive. Consider the Kac-normalized complete-graph Ising Hamiltonian

HN=−JN∑i<jZiZj.H_N = - \frac{J}{N} \sum_{i<j} Z_iZ_j.

In the aligned state,

EN=−JNN(N−1)2=−J2(N−1),E_N = - \frac{J}{N} \frac{N(N-1)}{2} = - \frac{J}{2}(N-1),

so the energy has a nonzero extensive leading term. Split this one parent Hamiltonian into two macroscopic groups of sizes NAN_A and NBN_B:

HN=HA(N)+HB(N)+HAB(N),H_N = H_A^{(N)} + H_B^{(N)} + H_{AB}^{(N)},

where all three terms retain the parent factor 1/N1/N, and

HAB(N)=−JN∑i∈A∑j∈BZiZj.H_{AB}^{(N)} = - \frac{J}{N} \sum_{i\in A} \sum_{j\in B} Z_iZ_j.

For NA=λNN_A=\lambda N and NB=(1−λ)NN_B=(1-\lambda)N, the aligned cross energy is

⟨HAB(N)⟩=−Jλ(1−λ)N,\left\langle H_{AB}^{(N)}\right\rangle = - J\lambda(1-\lambda)N,

which remains bulk order. The two regions cannot be decoupled by discarding a surface correction. If instead one compares the special aligned energies of separately rebuilt HNAH_{N_A} and HNBH_{N_B}, changing 1/N1/N to 1/NA1/N_A and 1/NB1/N_B can cancel this leading cross term; that cancellation reflects a change of Hamiltonian normalization, not additivity of the parent system. The Kac factor repairs the energy scale but does not restore geometric locality or separability of macroscopic regions.

Subextensive does not mean negligible. For a regular dd-dimensional region, a useful conditional expansion is

FL=fbLd+fsLd−1+feLd−2+⋯ .F_L = f_{\mathrm b}L^d + f_{\mathrm s}L^{d-1} + f_{\mathrm e}L^{d-2} + \cdots.

The surface contribution vanishes in FL/LdF_L/L^d, yet it can encode surface tension, edge modes, boundary critical behavior, wetting, or topological information.

A concrete lattice count makes the hierarchy visible. An open L×LL\times L square lattice has

Nbondopen=2L(L−1)=2L2−2LN_{\mathrm{bond}}^{\mathrm{open}} = 2L(L-1) = 2L^2-2L

nearest-neighbor bonds, while its periodic counterpart has

Nbondperiodic=2L2.N_{\mathrm{bond}}^{\mathrm{periodic}} = 2L^2.

The bulk term is O(L2)O(L^2), the missing open-boundary bonds are O(L)O(L), and the correction to the bond density is O(L−1)O(L^{-1}).

The same logic appears in continuum thermodynamics. A grand potential with an interface can take the schematic form

Ω=−PV+γA+⋯ ,\Omega = - PV + \gamma A + \cdots,

where γA\gamma A is subextensive relative to volume but extensive in interface area. Thus Ω=−PV\Omega=-PV is a bulk relation, not an exact identity for every finite, trapped, or interfacial system.

The hierarchy requires a regular family whose boundary-to-volume ratio vanishes. Thin strips, fractal or perforated regions, growing impurity sets, and size-dependent shapes can reorder the terms. The Thermodynamic Limit page owns those geometric convergence conditions.

For a homogeneous equilibrium phase with conventional bulk scaling, internal energy is a first-degree homogeneous function of its extensive natural variables:

U(λS,λV,λNa)=λU(S,V,Na).U(\lambda S,\lambda V,\lambda N_a) = \lambda U(S,V,N_a).

Its conjugate fields are degree zero. For example,

T(λS,λV,λNa)=T(S,V,Na),T(\lambda S,\lambda V,\lambda N_a) = T(S,V,N_a),

and similarly for pressure PP and chemical potentials μa\mu_a.

This ideal homogeneity permits the Euler relation

U=TS−PV+∑aμaNa.U = TS-PV + \sum_a \mu_aN_a.

It is not a microscopic identity valid without qualifications. Surfaces, traps, and finite-size corrections spoil exact homogeneity, while long-range nonadditivity can spoil even its leading bulk form. In ordinary additive matter, phase coexistence can preserve leading first-degree homogeneity while making derivatives nonunique or discontinuous; interfaces then supply subextensive finite-size terms. The derivation and the associated Gibbs–Duhem relation belong to Thermodynamic Potentials.

Each potential must be scaled while its intensive controls are held fixed:

PotentialReplication operation in a conventional one-component bulk phase
U(S,V,N)U(S,V,N)scale SS, VV, and NN together
F(T,V,N)F(T,V,N)scale VV and NN at fixed TT
G(T,P,N)G(T,P,N)scale NN at fixed TT and PP
Ω(T,V,μ)\Omega(T,V,\mu)scale VV at fixed TT and μ\mu

For example,

F(T,V,N)=Vf(T,n),n=NV,F(T,V,N) = Vf(T,n), \qquad n=\frac{N}{V},

in a homogeneous extensive phase. The bulk relations G=μNG=\mu N and Ω=−PV\Omega=-PV require the same assumptions.

The partition function itself is generally not extensive. If

ln⁡ZL=−βf BL+o(BL),\ln Z_L = -\beta f\,\mathcal B_L + o(\mathcal B_L),

then

ZL=exp⁡ ⁣[−βf BL+o(BL)].Z_L = \exp\!\left[ -\beta f\,\mathcal B_L + o(\mathcal B_L) \right].

Thus ln⁡ZL\ln Z_L and the corresponding free energy have a linear leading term, while ZLZ_L scales exponentially.

Units do not determine scaling class. Internal energy and chemical potential both carry energy units, yet the former is ordinarily extensive and the latter intensive.

Long-Range Interactions and Kac Normalization

Section titled “Long-Range Interactions and Kac Normalization”

Suppose a regular dd-dimensional lattice has a coherent, same-sign pair interaction

J(r)∼1rα.J(r) \sim \frac{1}{r^\alpha}.

The number of sites in a shell of radius rr grows as rd−1r^{d-1}, so the coupling sum seen by one site scales schematically as

∑r≲Lrd−1−α∼{constant,α>d,log⁡L,α=d,Ld−α,α<d.\sum_{r\lesssim L} r^{d-1-\alpha} \sim \begin{cases} \text{constant}, & \alpha>d, \\ \log L, & \alpha=d, \\ L^{d-\alpha}, & \alpha<d. \end{cases}

At fixed density, with Ns∼LdN_s\sim L^d, this counting suggests

EL∼{Ns,α>d,Nslog⁡L,α=d,NsLd−α∼Ns 2−α/d,α<d.E_L \sim \begin{cases} N_s, & \alpha>d, \\ N_s\log L, & \alpha=d, \\ N_sL^{d-\alpha} \sim N_s^{\,2-\alpha/d}, & \alpha<d. \end{cases}

This is a diagnostic, not a universal theorem. Alternating signs, screening, charge neutrality, angular structure, correlations, and geometry can change the result. Neutral Coulomb matter is a crucial warning: despite the long-range interaction, appropriate stability and neutrality conditions can support a thermodynamic free-energy density.

A generalized Kac factor divides by the divergent per-site coupling sum so that energy per site remains controlled. As the complete-graph example showed, normalizing the leading energy does not by itself establish locality, additivity, ensemble equivalence, or a conventional thermodynamic limit.

The statement “entropy is extensive” is safe only for equilibrium thermodynamic entropy under appropriate bulk assumptions. Quantum theory contains several inequivalent entropy questions.

For a bipartite density operator,

S(AB)=S(A)+S(B)−I(A:B),S(AB) = S(A)+S(B)-I(A:B),

where I(A:B)≥0I(A:B)\ge0 is the mutual information. Von Neumann entropy is exactly additive for a product state, but correlations reduce S(AB)S(AB) relative to S(A)+S(B)S(A)+S(B).

Further distinctions are essential:

  • a global pure state has zero von Neumann entropy even when the system is macroscopic;
  • equilibrium thermal entropy often has an extensive leading term when correlations are sufficiently well behaved;
  • ground-state entanglement entropy can follow an area law and be subextensive relative to subsystem volume;
  • highly entangled or thermalizing states can have volume-law subsystem entropy;
  • logarithmic and topological terms can be physically decisive despite being subleading.

Entropy in Quantum Statistical Mechanics owns entropy definitions, and Thermal Entropy versus Entanglement Entropy owns their detailed comparison.

Similar care applies beyond entropy. An extensive eigenvalue of a one-body density matrix, a structure-factor peak proportional to volume, and an additive thermodynamic potential are all volume-scaling signals, but they are not the same mathematical property.

Before writing “XX is extensive” or “YY is intensive,” complete this ledger:

QuestionRequired answer
What grows?State BL\mathcal B_L, geometry, dimension, and aspect ratio.
What is fixed?Density or filling, composition, temperature, fields, and coupling normalization.
What object is classified?Distinguish operator, expectation, cumulant, estimator, or state function.
What is the leading law?Give XL/BL→xX_L/\mathcal B_L\to x, another exponent, or a justified bound.
What is the normalization?Write the density or per-site quantity explicitly.
What is subleading?Record surface, edge, corner, impurity, shell, or logarithmic terms.
Is it additive?Estimate the cross contribution between macroscopic pieces.
Which assumptions matter?State locality or decay, stability, ensemble, correlations, and boundary conditions.

A concise defensible claim has the form:

Along the stated family, at fixed controls and normalization, XL=xBL+o(BL)X_L=x\mathcal B_L+o(\mathcal B_L); the listed boundary or correlation terms are subleading, and additivity is a separate tested property.

If the density limit has not been established, report the weaker evidence honestly: for example, “the available sizes are consistent with linear leading growth.”

Ordinary bulk vocabulary may need modification when:

  • a long-range interaction is not summable or remains nonadditive;
  • attractive interactions are unstable against collapse;
  • critical correlations change fluctuation or response exponents;
  • a trap or spatially varying field destroys homogeneous replication;
  • the boundary grows as fast as the nominal bulk;
  • a conserved sector or global constraint couples distant regions;
  • several limits, such as L→∞L\to\infty and a source tending to zero, do not commute;
  • the word entropy refers to entanglement, diagonal, coarse-grained, or another nonthermodynamic quantity.

Do not force every case into “extensive versus intensive.” State the observed exponent or asymptotic form and the family that produced it.

  • A finite value is not a scaling law. Register the family and held-fixed controls, then show a density limit or a weaker asymptotic bound. Write “at most extensive” when only O(BL)O(\mathcal B_L) is known.
  • The classification axes are distinct. State whether the claim concerns a total-like value, a replicated operator or state function, or a degree-zero field. Local, bounded, dimensionless, and numerically small do not mean intensive; a zero expectation does not reclassify a global sum.
  • Examples require assumptions. Do not say every energy or entropy is extensive. Name the stability, correlation, state, ensemble, interaction, and geometry conditions that support the claimed leading behavior.
  • Scaling does not prove separability. Test additivity and locality independently. Kac normalization can control energy per site while leaving a bulk-order cross interaction, nonlocal couplings, and possible ensemble nonequivalence.
  • Normalization and subleading terms are data. State whether heat capacity, susceptibility, or structure factor is total or per volume. Classify ln⁡Z\ln Z or the free energy rather than calling ZZ extensive, and retain surface, edge, impurity, or logarithmic terms when they carry the target physics.

Let

XL=aLd+bLd−1+clog⁡L,d>1.X_L = aL^d + bL^{d-1} + c\log L, \qquad d>1.

Classify each term relative to the volume BL=Ld\mathcal B_L=L^d, find the limiting density, and identify the leading correction to XL/LdX_L/L^d when b≠0b\ne0.

Solution

The aLdaL^d term is extensive, while bLd−1bL^{d-1} and clog⁡Lc\log L are subextensive relative to volume. Dividing by LdL^d gives

XLLd=a+bL+clog⁡LLd.\frac{X_L}{L^d} = a + \frac{b}{L} + \frac{c\log L}{L^d}.

Therefore the density tends to aa. When b≠0b\ne0, the leading correction is b/Lb/L because log⁡L/Ld\log L/L^d decays faster for d>1d>1. The logarithmic term may still encode important universal or topological information.

Exercise 2: Boundary Counting in d Dimensions

Section titled “Exercise 2: Boundary Counting in d Dimensions”

For L>2L>2, count undirected nearest-neighbor bonds on a dd-dimensional hypercubic lattice of side LL with open and periodic boundary conditions. Separate bulk and boundary pieces and compare bonds per site.

Solution

In each of the dd coordinate directions, an open lattice has Ld−1L^{d-1} rows containing L−1L-1 bonds. Therefore

Nbondopen=dLd−1(L−1)=dLd−dLd−1.N_{\mathrm{bond}}^{\mathrm{open}} = dL^{d-1}(L-1) = dL^d-dL^{d-1}.

Periodic closure adds one bond to every row:

Nbondperiodic=dLd.N_{\mathrm{bond}}^{\mathrm{periodic}} = dL^d.

Thus the open system has a bulk term dLddL^d and a boundary deficit −dLd−1-dL^{d-1}. Since the site count is LdL^d, the bond densities are

NbondopenLd=d−dL,NbondperiodicLd=d.\frac{N_{\mathrm{bond}}^{\mathrm{open}}}{L^d} = d-\frac{d}{L}, \qquad \frac{N_{\mathrm{bond}}^{\mathrm{periodic}}}{L^d} = d.

Both boundary choices approach dd bonds per site even though their finite-size energies can differ by a surface term.

Exercise 3: A Normalization-Sensitive Composition Defect

Section titled “Exercise 3: A Normalization-Sensitive Composition Defect”

Consider

Hγ=−JNγ∑i<jZiZj.H_\gamma = - \frac{J}{N^\gamma} \sum_{i<j} Z_iZ_j.

First find the value of γ\gamma for which the aligned-state energy has a nonzero extensive leading term. Now set that value and divide the spins into NA=λNN_A=\lambda N and NB=(1−λ)NN_B=(1-\lambda)N with magnetization densities mAm_A and mBm_B. Evaluate the combined energy using HNH_N, but evaluate each isolated block with the same Kac rule using its own denominator NAN_A or NBN_B. Find the leading composition defect

ΔEAB=EN−ENA−ENB.\Delta E_{AB} = E_N-E_{N_A}-E_{N_B}.

Explain why the result differs from simply retaining the parent normalization in a physical cut.

Solution

There are N(N−1)/2=O(N2)N(N-1)/2=O(N^2) aligned pairs, so

Eγ∼−J2N2−γ.E_\gamma \sim - \frac{J}{2} N^{2-\gamma}.

A nonzero extensive leading term requires 2−γ=12-\gamma=1, hence γ=1\gamma=1. For an Ising configuration with total magnetization M=nmM=nm, the identity

∑i<jZiZj=M2−n2\sum_{i<j} Z_iZ_j = \frac{M^2-n}{2}

gives

En(m)=−J2(nm2−1).E_n(m) = - \frac{J}{2} \left( nm^2-1 \right).

The combined magnetization density is

m=λmA+(1−λ)mB.m = \lambda m_A + (1-\lambda)m_B.

Substitution yields

ΔEAB=JN2λ(1−λ)(mA−mB)2−J2.\begin{aligned} \Delta E_{AB} &= \frac{JN}{2} \lambda(1-\lambda) \left( m_A-m_B \right)^2 \\ &\quad- \frac{J}{2}. \end{aligned}

For unequal block magnetizations, the defect is bulk order even though every separately normalized energy is O(N)O(N). If mA=mBm_A=m_B, the bulk term cancels in this special comparison and only the O(1)O(1) term remains. That cancellation occurs because rebuilding the blocks changes their intrablock coupling denominators. In a physical cut that retains the parent 1/N1/N, the cross interaction itself remains O(N)O(N), as shown in the body. The two conventions answer different questions and must not be mixed.

Let M=∑i=1NXiM=\sum_{i=1}^{N}X_i, where the XiX_i are independent with ⟨Xi⟩=0\langle X_i\rangle=0 and Var⁡(Xi)=σ2\operatorname{Var}(X_i)=\sigma^2. Find the scaling of ⟨M⟩\langle M\rangle, Var⁡(M)\operatorname{Var}(M), ΔM\Delta M, the normalized average m=M/Nm=M/N, Var⁡(m)\operatorname{Var}(m), and Δm\Delta m. Which step needs replacement when connected correlations do not obey a size-uniform absolute-summability bound?

Solution

Independence gives

⟨M⟩=0,Var⁡(M)=∑iVar⁡(Xi)=Nσ2.\begin{aligned} \langle M\rangle &= 0, \\ \operatorname{Var}(M) &= \sum_i \operatorname{Var}(X_i) = N\sigma^2. \end{aligned}

Therefore

ΔM=σN,⟨m⟩=0,Var⁡(m)=σ2N,Δm=σN.\begin{aligned} \Delta M &= \sigma\sqrt N, & \langle m\rangle &= 0, \\ \operatorname{Var}(m) &= \frac{\sigma^2}{N}, & \Delta m &= \frac{\sigma}{\sqrt N}. \end{aligned}

The operator MM remains a degree-one extensive sum despite its zero mean. The normalized mm is an intensive average, while its typical zero-mean fluctuation decays as N−1/2N^{-1/2}. With correlations,

Var⁡(M)=∑i,j⟨XiXj⟩c.\operatorname{Var}(M) = \sum_{i,j} \langle X_iX_j\rangle_{\mathrm c}.

If no size-uniform absolute-summability bound controls the off-diagonal correlations, this independence argument no longer fixes the scaling. The variance may still be O(N)O(N) in a particular state, but it need not be, and the density fluctuations need not decay as N−1/2N^{-1/2}.

At fixed temperature, let a family of independent identical units have

ZN=zN,z>0.Z_N = z^N, \qquad z>0.

Classify ZNZ_N, ln⁡ZN\ln Z_N, the Helmholtz free energy FN=−kBTln⁡ZNF_N=-k_{\mathrm B}T\ln Z_N, and FN/NF_N/N. Then repeat the leading classification if a subexponential factor changes the partition function to

ZN=zNNaZ_N = z^N N^a

with fixed aa.

Solution

The partition function is exponential in NN, not extensive. Its logarithm is

ln⁡ZN=Nln⁡z,\ln Z_N = N\ln z,

so the free energy is

FN=−NkBTln⁡z.F_N = - Nk_{\mathrm B}T\ln z.

Both ln⁡ZN\ln Z_N and FNF_N have degree-one leading scaling, while

FNN=−kBTln⁡z\frac{F_N}{N} = - k_{\mathrm B}T\ln z

is intensive. With ZN=zNNaZ_N=z^N N^a,

ln⁡ZN=Nln⁡z+aln⁡N.\ln Z_N = N\ln z + a\ln N.

The logarithmic term is subextensive, so it changes finite-size corrections without changing the leading linear scaling of ln⁡ZN\ln Z_N or FNF_N. This is why the exponential form should be written with an o(N)o(N) term in the exponent rather than by assuming a ratio asymptotic to one.