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

$$X_L = x\,\mathcal B_L + r_L, \qquad r_L = o(\mathcal B_L),$$

where $\mathcal B_L$ is the chosen bulk-size variable: for example, the number of lattice sites $N_{s,L}$ or the continuum volume $V_L$. The coefficient $x$ is the limiting bulk density when it exists, and $r_L$ contains boundary and other subleading contributions.

The key distinction is between a scaling law and a composition rule: a quantity can grow proportionally to system size even when two macroscopic pieces remain strongly coupled.

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

Scaling Along a Family of Systems

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

Feature of the system	What must be specified
bulk size	$\mathcal B_L=N_{s,L}$, $V_L$, particle number, or another justified variable
geometry	dimension, shape, aspect ratio, and which lengths grow
state point	density or filling, composition, temperature, fields, and other controls held fixed
interactions	coupling constants, range or decay, and any size-dependent normalization
boundary data	open, periodic, twisted, fixed, trapped, or interfacial conditions
state or ensemble	ground state, eigenstate, canonical ensemble, driven state, and so on
quantity	operator, 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 $\langle M_L\rangle$;
• the homogeneity class of an operator or thermodynamic state function under replication.

A numerical sequence has a nonzero extensive leading term when

$$\frac{X_L}{\mathcal B_L} \longrightarrow x\ne0,$$

equivalently

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

The stronger statement includes a limiting density. Merely proving

$$X_L=O(\mathcal B_L)$$

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

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

$$X(\lambda\boldsymbol a) = \lambda X(\boldsymbol a)$$

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

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

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

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

Total-like sequence	Diagnostic	Typical example under ordinary bulk assumptions
nonzero extensive leading value	$X_L/\mathcal B_L\to x\ne0$	particle number at fixed density
subextensive contribution	$X_L/\mathcal B_L\to0$	a regular surface term relative to volume
superextensive value sequence	$\lvert X_L\rvert/\mathcal B_L\to\infty$	unnormalized coherent all-to-all pair energy

Separately, classify replicated objects and state variables by homogeneity:

Replication class	Diagnostic	Typical example
degree-one extensive object	$X(\lambda\boldsymbol a)=\lambda X(\boldsymbol a)$	a local-sum operator or bulk internal-energy function
degree-zero intensive variable	$Y(\lambda\boldsymbol a)=Y(\boldsymbol a)$	temperature or pressure

These are not competing bins. An $O(1)$ 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 $d$ dimensions, $L\mapsto\lambda L$ sends a regular volume to $\lambda^d V$, whereas thermodynamic extensivity concerns scaling all extensive data by a common factor.

Totals, Densities, and Local Observables

A global sum of local densities has the structural form

$$M_L = \sum_{i\in\Lambda_L} m_i.$$

When the state is homogeneous and correlations are sufficiently controlled,

$$\frac{\langle M_L\rangle}{N_{s,L}} \longrightarrow m$$

defines an intensive density. Three distinctions prevent common mistakes:

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

For example, in a product state of $N_s$ independent spins with

$$\langle Z_i\rangle=0, \qquad \langle Z_i^2\rangle=1,$$

the total magnetization $M=\sum_i Z_i$ obeys

$$\begin{aligned} \langle M\rangle &= 0, & \operatorname{Var}(M) &= N_s, \\ \Delta M &= \sqrt{N_s}. \end{aligned}$$

For the density $m=M/N_s$,

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

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

$$\sup_L \sup_{i\in\Lambda_L} \sum_{j\in\Lambda_L} \left| C_{ij}^{(L)} \right| \le C$$

for a constant $C$ independent of $L$. It implies that the variance is at most volume order:

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

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

“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 Is Not Additivity

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

$$\Delta X_{AB} = X(A\cup B)-X(A)-X(B).$$

Then:

• exact additivity means $\Delta X_{AB}=0$;
• asymptotic additivity means $\Delta X_{AB}=o(\mathcal B_A+\mathcal B_B)$;
• 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-$N$ system retains the parent couplings in its $A$, $B$, and cross terms. Rebuilding $A$ and $B$ as independently normalized size-$N_A$ and size-$N_B$ 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

$$H_N = - \frac{J}{N} \sum_{i<j} Z_iZ_j.$$

In the aligned state,

$$E_N = - \frac{J}{N} \frac{N(N-1)}{2} = - \frac{J}{2}(N-1),$$

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

$$H_N = H_A^{(N)} + H_B^{(N)} + H_{AB}^{(N)},$$

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

$$H_{AB}^{(N)} = - \frac{J}{N} \sum_{i\in A} \sum_{j\in B} Z_iZ_j.$$

For $N_A=\lambda N$ and $N_B=(1-\lambda)N$, the aligned cross energy is

$$\left\langle H_{AB}^{(N)}\right\rangle = - J\lambda(1-\lambda)N,$$

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 $H_{N_A}$ and $H_{N_B}$, changing $1/N$ to $1/N_A$ and $1/N_B$ 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.

Boundary and Other Subextensive Terms

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

$$F_L = f_{\mathrm b}L^d + f_{\mathrm s}L^{d-1} + f_{\mathrm e}L^{d-2} + \cdots.$$

The surface contribution vanishes in $F_L/L^d$, yet it can encode surface tension, boundary critical behavior, or wetting. Other subleading contributions, such as those to entanglement entropy, can encode topological information; they are different observables.

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

$$N_{\mathrm{bond}}^{\mathrm{open}} = 2L(L-1) = 2L^2-2L$$

nearest-neighbor bonds, while its periodic counterpart has

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

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

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

$$\Omega = - PV + \gamma A + \cdots,$$

where $\gamma A$ is subextensive relative to volume but extensive in interface area. Thus $\Omega=-PV$ 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. See Thermodynamic Limit for these geometric conditions.

Homogeneity and Thermodynamic Potentials

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

$$U(\lambda S,\lambda V,\lambda N_a) = \lambda U(S,V,N_a).$$

Its conjugate fields are degree zero. For example,

$$T(\lambda S,\lambda V,\lambda N_a) = T(S,V,N_a),$$

and similarly for pressure $P$ and chemical potentials $\mu_a$.

This ideal homogeneity permits the Euler relation

$$U = TS-PV + \sum_a \mu_aN_a.$$

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:

Potential	Replication operation in a conventional one-component bulk phase
$U(S,V,N)$	scale $S$, $V$, and $N$ together
$F(T,V,N)$	scale $V$ and $N$ at fixed $T$
$G(T,P,N)$	scale $N$ at fixed $T$ and $P$
$\Omega(T,V,\mu)$	scale $V$ at fixed $T$ and $\mu$

For example,

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

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

The partition function itself is generally not extensive. If

$$\ln Z_L = -\beta f\,\mathcal B_L + o(\mathcal B_L),$$

then

$$Z_L = \exp\!\left[ -\beta f\,\mathcal B_L + o(\mathcal B_L) \right].$$

Thus $\ln Z_L$ and the corresponding free energy have a linear leading term, while $Z_L$ 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

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

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

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

$$\sum_{r\lesssim L} r^{d-1-\alpha} \sim \begin{cases} \text{constant}, & \alpha>d, \\ \log L, & \alpha=d, \\ L^{d-\alpha}, & \alpha<d. \end{cases}$$

At fixed density, with $N_s\sim L^d$, this counting suggests

$$E_L \sim \begin{cases} N_s, & \alpha>d, \\ N_s\log L, & \alpha=d, \\ N_sL^{d-\alpha} \sim N_s^{\,2-\alpha/d}, & \alpha<d. \end{cases}$$

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.

Quantum Qualifications

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,

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

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

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 defines these entropies, and Thermal Entropy versus Entanglement Entropy compares their scaling.

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.

Exercises

Exercise 1: Bulk and Boundary Terms

Let

$$X_L = aL^d + bL^{d-1} + c\log L, \qquad d>1.$$

Classify each term relative to the volume $\mathcal B_L=L^d$, find the limiting density, and identify the leading correction to $X_L/L^d$ when $b\ne0$.

Solution

For $a\ne0$, the $aL^d$ term has nonzero extensive leading behavior, while $bL^{d-1}$ and $c\log L$ are subextensive relative to volume. Dividing by $L^d$ gives

$$\frac{X_L}{L^d} = a + \frac{b}{L} + \frac{c\log L}{L^d}.$$

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

Exercise 2: Boundary Counting in d Dimensions

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

Solution

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

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

Periodic closure adds one bond to every row:

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

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

$$\frac{N_{\mathrm{bond}}^{\mathrm{open}}}{L^d} = d-\frac{d}{L}, \qquad \frac{N_{\mathrm{bond}}^{\mathrm{periodic}}}{L^d} = d.$$

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

Exercise 3: A Normalization-Sensitive Composition Defect

Consider

$$H_\gamma = - \frac{J}{N^\gamma} \sum_{i<j} Z_iZ_j.$$

First find the value of $\gamma$ for which the aligned-state energy has a nonzero extensive leading term. Now set that value and divide the spins into $N_A=\lambda N$ and $N_B=(1-\lambda)N$ with magnetization densities $m_A$ and $m_B$. Evaluate the combined energy using $H_N$, but evaluate each isolated block with the same Kac rule using its own denominator $N_A$ or $N_B$. Find the leading composition defect

$$\Delta E_{AB} = E_N-E_{N_A}-E_{N_B}.$$

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

Solution

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

$$E_\gamma \sim - \frac{J}{2} N^{2-\gamma}.$$

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

$$\sum_{i<j} Z_iZ_j = \frac{M^2-n}{2}$$

gives

$$E_n(m) = - \frac{J}{2} \left( nm^2-1 \right).$$

The combined magnetization density is

$$m = \lambda m_A + (1-\lambda)m_B.$$

Substitution yields

$$\Delta E_{AB} = \frac{JN}{2} \lambda(1-\lambda) \left( m_A-m_B \right)^2 - \frac{J}{2}.$$

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

Exercise 4: Means and Fluctuations

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

Solution

Independence gives

$$\begin{aligned} \langle M\rangle &= 0, \\ \operatorname{Var}(M) &= \sum_i \operatorname{Var}(X_i) = N\sigma^2. \end{aligned}$$

Therefore

$$\begin{aligned} \Delta M &= \sigma\sqrt N, & \langle m\rangle &= 0, \\ \operatorname{Var}(m) &= \frac{\sigma^2}{N}, & \Delta m &= \frac{\sigma}{\sqrt N}. \end{aligned}$$

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

$$\operatorname{Var}(M) = \sum_{i,j} \langle X_iX_j\rangle_{\mathrm c}.$$

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 $O(N)$ in a particular state, but it need not be, and the density fluctuations need not decay as $N^{-1/2}$.

Exercise 5: Partition-Function Scaling

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

$$Z_N = z^N, \qquad z>0.$$

Classify $Z_N$, $\ln Z_N$, the Helmholtz free energy $F_N=-k_{\mathrm B}T\ln Z_N$, and $F_N/N$. Then repeat the leading classification if a subexponential factor changes the partition function to

$$Z_N = z^N N^a$$

with fixed $a$.

Solution

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

$$\ln Z_N = N\ln z,$$

so the free energy is

$$F_N = - Nk_{\mathrm B}T\ln z.$$

For $z\ne1$, both $\ln Z_N$ and $F_N$ have nonzero degree-one leading scaling, while

$$\frac{F_N}{N} = - k_{\mathrm B}T\ln z$$

is intensive. With $Z_N=z^N N^a$,

$$\ln Z_N = N\ln z + a\ln N.$$

For $z\ne1$, the logarithmic term is subextensive and does not change the leading linear scaling. If $z=1$, the linear coefficient vanishes: the logarithm becomes the leading term, and the free-energy density tends to zero. This is why the exponential form should be written with an $o(N)$ term in the exponent rather than by assuming a ratio asymptotic to one.

References

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• M. Kac, G. E. Uhlenbeck, and P. C. Hemmer, “On the van der Waals Theory of the Vapor-Liquid Equilibrium. I. Discussion of a One-Dimensional Model”, Journal of Mathematical Physics 4, 216–228 (1963).
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