Fluctuations and Susceptibilities

Equilibrium fluctuations and static susceptibilities are two descriptions of how weakly constrained a macroscopic quantity is. A broad equilibrium distribution means that nearby values cost little thermodynamic weight; a source conjugate to that quantity can then shift its mean strongly. For a canonical state with a linear source,

$$H_f = H_0-fB,$$

the isothermal response of an observable $A$ is

$$\chi_T^{AB} = \left. \frac{\partial\langle A\rangle_f} {\partial f} \right|_{f=0}.$$

If $A$ has no explicit source dependence, the exact quantum identity is

$$\chi_T^{AB} = \beta\, \operatorname{Cov}_{\mathrm{KM}}(A,B),$$

where $\operatorname{Cov}_{\mathrm{KM}}$ is the Kubo–Mori covariance. The familiar formula

$$\chi_T^{BB} = \beta\operatorname{Var}(B)$$

is recovered when $B$ commutes with the equilibrium density operator, in particular when $[B,H_0]=0$. Replacing the Kubo–Mori covariance by an ordinary equal-time variance without checking this condition is one of the most important quantum corrections to classical fluctuation intuition.

Three standard special cases are

$$\operatorname{Var}(E) = k_{\mathrm B}T^2C_V,$$ $$\operatorname{Var}(N) = k_{\mathrm B}T\,Vn^2\kappa_T,$$

and, for a commuting magnetization $M$ coupled through $-BM$,

$$\frac{1}{V} \left( \frac{\partial\langle M\rangle} {\partial B} \right)_T = \frac{\beta}{V} \operatorname{Var}(M).$$

Each formula silently specifies an ensemble, held-fixed variables, source convention, normalization, and equilibration protocol. This page develops those qualifications rather than treating fluctuation formulas as context-free substitutions.

Helpful background. Canonical and grand-canonical states determine which global quantities fluctuate; source derivatives connect them to response.

Equilibrium fluctuations and conjugate sources

Unless stated otherwise, take $T>0$, a convergent equilibrium trace, and source- and temperature-independent observables. A susceptibility compares separately equilibrated states at the same temperature. The linear coupling $-fB$ fixes the sign and units: increasing $f$ lowers the energy of large-$B$ states. Number response uses a grand-canonical trace, whereas the canonical energy identity uses fixed particle number.

For normalized negative-temperature states, the derivative identity $\chi_T=\beta\operatorname{Cov}_{\mathrm{KM}}$ still holds, but positivity of $\chi_T$ and concavity of free energy in a linear field reverse. The oscillator example below requires positive temperature.

What Counts as a Fluctuation?

For a Hermitian observable $A$ in a state $\rho$, define

$$\Delta A = A-\langle A\rangle_\rho\mathbb I,$$ $$\operatorname{Var}_\rho(A) = \langle(\Delta A)^2\rangle_\rho,$$

and

$$\operatorname{Cov}_\rho(A,B) = \frac{1}{2} \langle \Delta A\Delta B + \Delta B\Delta A \rangle_\rho.$$

The symmetrized covariance is real for Hermitian $A$ and $B$. When the observables commute, it reduces to the ordinary classical covariance of their joint outcome distribution.

The word fluctuation is used for several related but distinct objects:

1. Outcome fluctuation: the spread of repeated projective measurements of $A$ in the fixed state $\rho$.
2. Ensemble fluctuation: the spread of quantities such as energy or particle number across the sectors weighted by an equilibrium ensemble.
3. Temporal noise: the time-dependent random signal or correlation spectrum produced by a detector coupled to the system.

The first is always defined by the quantum variance. The second becomes an ordinary distribution over sectors when the fluctuating quantity commutes with the equilibrium state. The third depends on dynamics, operator ordering, detector conventions, and bandwidth. It cannot generally be reconstructed from one equal-time variance.

There is also no decomposition-independent split of a mixed-state variance into “classical ignorance” and “quantum uncertainty.” A density operator has many pure-state decompositions. If a particular preparation ensemble is operationally specified, a law-of-total-variance decomposition may be useful, but it belongs to that preparation, not to $\rho$ alone.

Ensemble Dependence

Whether a global quantity fluctuates is part of the ensemble definition.

Ensemble	Controlled variables	Global quantities fixed by construction	Typical global fluctuations
microcanonical	$E,V,N$	energy shell and particle sector	observables within the shell
canonical	$T,V,N$	particle number	energy and unconstrained observables
grand canonical	$T,V,\mu$	neither $E$ nor $N$	energy, particle number, and their covariance

An idealized microcanonical state has zero variance of the coarse-grained shell label, while a canonical state has nonzero energy variance. A canonical fixed-$N$ state has

$$\operatorname{Var}(N)=0,$$

whereas a grand-canonical state generally does not. These statements do not imply that the corresponding material has zero heat capacity or compressibility. A response coefficient compares nearby equilibrium states as a control variable changes; a global variance describes the distribution inside one specified ensemble.

For ordinary short-range systems, local observables can agree across ensembles in the thermodynamic limit even when global fluctuations remain different. Ensemble Equivalence explains the conditions and failures of that statement.

Source-Dependent Equilibrium States

Let

$$H_f = H_0-fB$$

with $B=B^\dagger$, and define

$$Z_f = \operatorname{Tr}e^{-\beta H_f}, \qquad \rho_f = \frac{e^{-\beta H_f}}{Z_f}.$$

The sign convention matters: positive $f$ energetically favors larger $B$. The first derivative of the free energy

$$F(f) = -\frac{1}{\beta}\ln Z_f$$

is

$$\frac{\partial F}{\partial f} = -\langle B\rangle_f.$$

This identity is exact even if $B$ does not commute with $H_f$. It follows from the Duhamel derivative

$$\frac{\partial e^{-\beta H_f}} {\partial f} = \beta \int_0^1ds\, e^{-(1-s)\beta H_f} B e^{-s\beta H_f}$$

and cyclicity of the trace.

The second derivative probes response:

$$-\frac{\partial^2F}{\partial f^2} = \frac{\partial\langle B\rangle_f} {\partial f} = \chi_T^{BB}(f).$$

Thus at positive temperature the free energy is concave in a linearly coupled field. The susceptibility is nonnegative for a stable diagonal source–detector pair under these conventions.

The Kubo–Mori Covariance

For a full-rank equilibrium state $\rho$, define

$$\operatorname{Cov}_{\mathrm{KM},\rho}(A,B) = \int_0^1ds\, \operatorname{Tr} \left( \rho^s\Delta A\, \rho^{1-s}\Delta B \right).$$

The definition extends to rank-deficient states by a suitable limit on the support when that limit exists. For Hermitian observables it is real and symmetric, and

$$\operatorname{Cov}_{\mathrm{KM}}(A,A) \geq 0.$$

In an eigenbasis of $\rho$,

$$\rho = \sum_n p_n \lvert n\rangle\langle n\rvert,$$

one obtains

$$\operatorname{Cov}_{\mathrm{KM}}(A,A) = \sum_{n,m} L(p_n,p_m) \left| \langle n\lvert\Delta A\rvert m\rangle \right|^2,$$

where the logarithmic mean is

$$L(x,y) = \begin{cases} \dfrac{x-y}{\ln x-\ln y},&x\neq y,\\ x,&x=y. \end{cases}$$

Every coefficient is nonnegative. If $[A,\rho]=0$, only matrix elements inside equal-$p$ blocks survive and

$$\operatorname{Cov}_{\mathrm{KM}}(A,A) = \operatorname{Var}(A).$$

For $\rho=e^{-\beta H}/Z$, the same object has the imaginary-time form

$$\beta\operatorname{Cov}_{\mathrm{KM}}(A,B) = \int_0^\beta d\lambda\, \left\langle \Delta A(-i\hbar\lambda) \Delta B(0) \right\rangle,$$

with

$$A(-i\hbar\lambda) = e^{\lambda H}Ae^{-\lambda H}.$$

The isothermal source response is therefore

$$\chi_T^{AB} = \beta\operatorname{Cov}_{\mathrm{KM}}(A,B)$$

when $A$ has no explicit $f$ dependence. The imaginary-time integral, not an arbitrary equal-time ordering, is the generally correct quantum static correlator.

The logarithmic mean obeys $L(x,y)\leq(x+y)/2$. Applying this to the spectral sum gives

$$0\leq\operatorname{Cov}_{\mathrm{KM}}(A,A)\leq\operatorname{Var}(A), \qquad 0\leq\chi_T^{AA}\leq\beta\operatorname{Var}(A).$$

For a full-rank state, equality in the upper bound holds exactly when $[A,\rho]=0$. Equal-time quantum fluctuations can therefore overestimate the isothermal response. At zero temperature, take the limit of $\beta\operatorname{Cov}_{\mathrm{KM}}$ as a whole; inserting a rank-one state before multiplying by divergent $\beta$ can lose a finite susceptibility.

Explicit Source Dependence

Suppose the Hamiltonian is a nonlinear function $H(f)$ and define the generalized displacement

$$B_f = -\frac{\partial H(f)}{\partial f}.$$

Then

$$\frac{\partial F}{\partial f} = -\langle B_f\rangle_f$$

and

$$\frac{\partial\langle B_f\rangle_f} {\partial f} = \left\langle \frac{\partial B_f}{\partial f} \right\rangle_f + \beta \operatorname{Cov}_{\mathrm{KM},f}(B_f,B_f).$$

The first term is an explicit-source or contact contribution. For a strictly linear coupling, $B_f=B$ and it vanishes. For electromagnetic response, elastic coordinates, and source-dependent effective operators, omitting it can give the wrong static limit.

The same structure holds for a detector $A(f)$ distinct from the coupled operator:

$$\frac{\partial\langle A(f)\rangle_f} {\partial f} = \left\langle \frac{\partial A}{\partial f} \right\rangle_f + \beta \operatorname{Cov}_{\mathrm{KM},f}(A,B_f).$$

Classical Tilting Intuition

When $X$ commutes with the equilibrium state, the source simply tilts its probability distribution:

$$P_f(X) = \frac{ P_0(X)e^{\beta fX} }{ \left\langle e^{\beta fX}\right\rangle_0 }.$$

Differentiating at $f=0$ gives

$$\left. \frac{\partial\langle X\rangle_f} {\partial f} \right|_{f=0} = \beta\operatorname{Var}_0(X).$$

For a Gaussian distribution,

$$P_0(X) \propto \exp \left[ -\frac{(X-\overline X)^2}{2\sigma_X^2} \right],$$

the tilted mean is

$$\langle X\rangle_f = \overline X + \beta\sigma_X^2f.$$

A broader distribution therefore produces a steeper linear response.

For a commuting observable with a linear conjugate source, equilibrium width and isothermal response carry the same information. The broad distribution has the larger variance and therefore the larger slope $\chi_T=\beta\operatorname{Var}(X)$. Noncommuting quantum observables require the Kubo–Mori covariance instead.

This picture is local. Far from $f=0$, higher cumulants control nonlinear response. Near phase coexistence, $P_0(X)$ can be bimodal rather than Gaussian, and one variance no longer describes its shape.

Energy Fluctuations and Heat Capacity

In the canonical ensemble,

$$\rho_\beta = \frac{e^{-\beta H}}{Z}, \qquad Z = \operatorname{Tr}e^{-\beta H},$$

with a temperature-independent Hamiltonian. Because $H$ commutes with $\rho_\beta$,

$$U = \langle H\rangle = -\frac{\partial\ln Z}{\partial\beta}$$

and

$$\frac{\partial U}{\partial\beta} = -\operatorname{Var}(H).$$

Using

$$\frac{d\beta}{dT} = -\frac{1}{k_{\mathrm B}T^2},$$

the constant-volume heat capacity is

$$\begin{aligned} C_V &= \left( \frac{\partial U}{\partial T} \right)_{V,N} \\ &= k_{\mathrm B}\beta^2 \operatorname{Var}(H). \end{aligned}$$

Equivalently,

$$\operatorname{Var}(E) = k_{\mathrm B}T^2C_V.$$

This is exact for the canonical ensemble under the assumptions above. It immediately gives

$$C_V\geq0$$

for a finite temperature-independent Hamiltonian with a convergent partition function. It does not prove nonnegative heat capacity in every ensemble. Microcanonical long-range and nonadditive systems require separate stability analysis.

If $C_V=c_VV$ is extensive, then

$$\operatorname{Var}(E) = k_{\mathrm B}T^2c_VV,$$

so the standard deviation grows as $V^{1/2}$ while a nonzero bulk mean energy grows as $V$. Relative fluctuations then scale as $V^{-1/2}$. The statement should not be expressed as $\Delta E/U$ when the chosen zero makes $U$ vanish or changes sign; the scaling of the intensive energy $e=E/V$ is unambiguous:

$$\operatorname{Var}(e) = \frac{k_{\mathrm B}T^2c_V}{V}.$$

The Canonical Ensemble develops the full energy distribution, Schottky anomaly, and oscillator examples. Here the formula serves as one member of a unified response dictionary.

Number Fluctuations and Compressibility

For a grand-canonical state,

$$\rho_{\mathrm G} = \frac{ e^{-\beta(H-\mu N)} }{\Xi},$$

ordinary equilibrium requires the charge to be conserved,

$$[H,N]=0.$$

Then

$$\overline N = \frac{1}{\beta} \frac{\partial\ln\Xi}{\partial\mu}$$

and

$$\left( \frac{\partial\overline N}{\partial\mu} \right)_{T,V} = \beta\operatorname{Var}(N).$$

For a homogeneous one-component system with density

$$n = \frac{\overline N}{V},$$

the isothermal compressibility is

$$\kappa_T = \frac{1}{n^2} \left( \frac{\partial n}{\partial\mu} \right)_T.$$

Therefore

$$\kappa_T = \frac{ \beta\operatorname{Var}(N) }{ Vn^2 }$$

or

$$\operatorname{Var}(N) = k_{\mathrm B}T\,Vn^2\kappa_T.$$

This relation is sometimes called the compressibility sum rule. It requires consistent definitions: some authors call

$$\kappa_n = \left( \frac{\partial n}{\partial\mu} \right)_T$$

the density susceptibility, while $\kappa_T=\kappa_n/n^2$ is the thermodynamic compressibility. The factors of $V$, $n$, and $n^2$ are not optional conventions.

Why fixed particle number is not zero compressibility

In a canonical ensemble,

$$\operatorname{Var}_{N}(N)=0$$

by construction. A finite compressibility can still be obtained from the curvature of the fixed-$N$ Helmholtz free energy, from the equation of state, or from an appropriate long-wavelength density response. The grand-canonical global-variance formula and the canonical free-energy derivative are alternative equilibrium representations under conditions of ensemble equivalence; they are not identities inside the same finite ensemble.

At exactly zero wavevector, the density operator is total $N$. If $N$ is conserved, its isolated finite-frequency retarded self-response vanishes. The thermodynamic derivative instead compares re-equilibrated states with different $\mu$. This is an order-of-limits issue, not a contradiction.

Strict zero temperature

For a finite system at $T=0$, the grand-canonical ground state usually lies in one definite-$N$ sector except at a level crossing. Then $\operatorname{Var}(N)=0$ and $\overline N(\mu)$ is stepwise constant. A smooth finite compressibility emerges only after an appropriate thermodynamic and zero-temperature limit. The expression $\beta\operatorname{Var}(N)$ must therefore be interpreted as a limit, not as the product of two separately evaluated quantities $\infty$ and $0$.

Magnetization Fluctuations and Susceptibility

For one field component, choose

$$H(B) = H_0-BM.$$

The total isothermal susceptibility is

$$\chi_{M,T} = \left( \frac{\partial\langle M\rangle} {\partial B} \right)_T.$$

If $[M,H_0]=0$, then

$$\chi_{M,T} = \beta\operatorname{Var}(M).$$

For a bulk susceptibility, define

$$\chi_T = \frac{1}{V}\chi_{M,T} = \frac{\beta}{V} \operatorname{Var}(M).$$

Depending on electromagnetic unit conventions, additional factors such as $\mu_0$ may be included in the definition of susceptibility or field. The Hamiltonian coupling determines the convention safely.

Independent moments and the Curie law

Consider $N_s$ independent two-state magnetic moments $m s_i$, with $s_i=\pm1$:

$$M = m\sum_{i=1}^{N_s}s_i, \qquad H(B) = -BM.$$

The one-site partition function is

$$Z_1 = 2\cosh(\beta mB),$$

and

$$\langle M\rangle = N_sm\tanh(\beta mB).$$

Thus

$$\chi_{M,T} = N_s\beta m^2 \operatorname{sech}^2(\beta mB).$$

The independent-site variance is

$$\operatorname{Var}(M) = N_sm^2 \operatorname{sech}^2(\beta mB),$$

which verifies $\chi_{M,T}=\beta\operatorname{Var}(M)$. At zero field,

$$\chi_T = \frac{n_sm^2}{k_{\mathrm B}T},$$

where $n_s=N_s/V$. This is the Curie law for independent moments.

Interactions can enhance, suppress, or singularly reorganize these fluctuations. A Curie–Weiss fit is not by itself evidence for independent spins, and a susceptibility peak in a finite sample is not by itself proof of a phase transition.

The Kondo Model Preview gives a canonical impurity example: a high-temperature Curie moment crosses over to a finite low-temperature susceptibility when an antiferromagnetic conduction bath screens the spin.

The Noncommuting Quantum Correction

The ordinary variance formula can fail dramatically when the source-coupled observable does not commute with the unperturbed Hamiltonian.

Consider a harmonic oscillator under a static force:

$$H_f = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2 - fx.$$

Completing the square gives

$$H_f = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 \left( x- \frac{f}{m\omega^2} \right)^2 - \frac{f^2}{2m\omega^2}.$$

Therefore

$$\langle x\rangle_f = \frac{f}{m\omega^2}, \qquad \chi_T^{xx} = \frac{1}{m\omega^2}.$$

At $f=0$, however,

$$\operatorname{Var}(x) = \frac{\hbar}{2m\omega} \coth \left( \frac{\beta\hbar\omega}{2} \right).$$

Hence

$$\beta\operatorname{Var}(x) = \frac{\beta\hbar}{2m\omega} \coth \left( \frac{\beta\hbar\omega}{2} \right)$$

is not $1/(m\omega^2)$ except in the high-temperature limit. At low temperature it even grows proportional to $\beta$ because the equal-time variance retains zero-point fluctuations.

The Kubo–Mori expression gives the correct answer. With

$$\overline n = \frac{1}{e^{\beta\hbar\omega}-1},$$

the imaginary-time correlator is

$$\langle x(-i\hbar\lambda)x(0)\rangle = \frac{\hbar}{2m\omega} \bigl[ (\overline n+1)e^{-\lambda\hbar\omega} + \overline n e^{\lambda\hbar\omega} \bigr].$$

Integrating gives

$$\int_0^\beta d\lambda\, \langle x(-i\hbar\lambda)x(0)\rangle = \frac{1}{m\omega^2}.$$

The static displacement susceptibility is controlled by the imaginary-time average. Equal-time zero-point uncertainty is physically real, but it is not interchangeable with a re-equilibrated force response.

Several Sources and Cross Susceptibilities

For linear sources

$$H(\mathbf f) = H_0- \sum_b f_bB_b,$$

define

$$\chi_{ab} = \frac{\partial\langle B_a\rangle} {\partial f_b}.$$

The equilibrium response matrix is

$$\chi_{ab} = \beta \operatorname{Cov}_{\mathrm{KM}}(B_a,B_b).$$

For real fields and Hermitian observables,

$$\chi_{ab} = \chi_{ba},$$

provided both derivatives refer to the same equilibrium potential and held-fixed variables. This is a static Maxwell reciprocity relation, not a generic statement about nonequilibrium retarded cross response.

For any real coefficients $c_a$, let

$$C = \sum_a c_aB_a.$$

Then

$$\sum_{a,b}c_a\chi_{ab}c_b = \beta \operatorname{Cov}_{\mathrm{KM}}(C,C) \geq 0.$$

The susceptibility matrix is therefore positive semidefinite. A zero eigenvalue can signal a redundant source direction, an exact constraint, or an operator combination that is constant on the support of the state.

When the source observables commute with the equilibrium state, the response reduces to ordinary covariance. Its positivity yields the covariance inequality

$$\operatorname{Cov}(A,B)^2 \leq \operatorname{Var}(A) \operatorname{Var}(B).$$

This inequality bounds cross responses by diagonal susceptibilities.

Energy–Number Cross Fluctuations

In grand-canonical equilibrium with $[H,N]=0$,

$$\Xi = \operatorname{Tr} e^{-\beta(H-\mu N)}.$$

At fixed $T$,

$$\left( \frac{\partial U}{\partial\mu} \right)_{T,V} = \beta\operatorname{Cov}(H,N).$$

At fixed $\mu$,

$$\left( \frac{\partial\overline N}{\partial\beta} \right)_{\mu,V} = -\operatorname{Cov} \left( N,H-\mu N \right).$$

The grand-Hamiltonian variance is

$$\operatorname{Var}(H-\mu N) = {}\operatorname{Var}(H) + \mu^2\operatorname{Var}(N) - 2\mu\operatorname{Cov}(H,N).$$

Consequently, a temperature derivative at fixed $\mu$ does not isolate $\operatorname{Var}(H)$. The derivative path through thermodynamic parameter space must be stated.

One useful path adjusts $\mu(T)$ to hold the mean number fixed. For a temperature- and chemical-potential-independent $H$ conserving $N$, with convergent derivatives and $\operatorname{Var}(N)>0$, eliminating $d\mu/dT$ from $d\overline N/dT=0$ gives

$$C_{\overline N}^{\mathrm{gc}} \equiv\left.\frac{dU}{dT}\right|_{\overline N,V} =k_{\mathrm B}\beta^2 \left[ \operatorname{Var}(H) -\frac{\operatorname{Cov}(H,N)^2}{\operatorname{Var}(N)} \right]\geq0.$$

The bracket is the variance of the residual $\Delta H-a\Delta N$ minimized over $a$, with $a=\operatorname{Cov}(H,N)/\operatorname{Var}(N)$. It is therefore nonnegative. This is a response along a family of grand-canonical states; it is not generally the canonical fixed-$N$ heat capacity of a finite system. The first exercise derives it and compares the two in a solvable model. If $\operatorname{Var}(N)=0$, solve the constraint separately rather than dividing by zero.

Local Correlations and Bulk Response

Let an extensive observable be a sum of local densities,

$$X = \sum_i x_i.$$

For commuting equal-time variables,

$$\operatorname{Var}(X) = \sum_{i,j} \langle\Delta x_i\Delta x_j\rangle.$$

In a homogeneous continuum state observed in a finite region,

$$\frac{\operatorname{Var}(X)}{V} = \int d^dr\,w_V(\mathbf r)C_{xx}(\mathbf r),$$

where

$$C_{xx}(\mathbf r) = \langle \Delta x(\mathbf r) \Delta x(\mathbf 0) \rangle.$$

The window factor $w_V(\mathbf r)=|V\cap(V-\mathbf r)|/V$ accounts for pairs that fit inside the observed region. For a regular bulk limit with integrable correlations it approaches one at fixed separation. A periodic homogeneous system has the corresponding exact integral over its periodic domain. Thus, in the bulk limit, a commuting susceptibility per volume obeys

$$\frac{\chi_T^{XX}}{V} = \beta \int d^dr\, C_{xx}(\mathbf r).$$

This identity explains why long-ranged correlations enhance susceptibilities. If $C_{xx}$ is absolutely integrable, the variance is at most extensive. It is proportional to $V$ only when the limiting integral is nonzero; anticorrelations or an exact constraint can cancel it. If a correlation length diverges or correlations decay too slowly, the integral can grow with system size.

For particle density,

$$\operatorname{Var}(N) = \int_Vd^dr \int_Vd^dr'\, \langle \Delta n(\mathbf r) \Delta n(\mathbf r') \rangle.$$

With the common static-structure-factor normalization

$$S(\mathbf 0) = \frac{\operatorname{Var}(N)}{\overline N},$$

the finite-temperature grand-canonical compressibility relation becomes

$$S(\mathbf 0) = n k_{\mathrm B}T\kappa_T.$$

Finite windows, exact global conservation, boundaries, and nonuniform density modify the naive $\mathbf q=0$ identification. Correlation Functions Overview develops connected correlators and spatial ordering conventions.

Thermodynamic-Limit Scaling

Away from criticality, an additive system with finite correlation length typically has

$$\langle X\rangle \sim V, \qquad \operatorname{Var}(X) \sim V.$$

Therefore

$$\Delta X \sim V^{1/2}$$

and, when the mean density is nonzero,

$$\frac{\Delta X}{|\langle X\rangle|} \sim V^{-1/2}.$$

Macroscopic relative fluctuations vanish even though the absolute fluctuation grows. This concentration is one reason deterministic thermodynamics emerges from statistical mechanics.

For the intensive variable $x=X/V$,

$$\operatorname{Var}(x) \sim V^{-1}.$$

If its distribution has a large-deviation form

$$P(x) \asymp e^{-\beta Vg(x)},$$

and $g$ has a unique minimum at $x_\star$ with strictly positive curvature $g''(x_\star)>0$ and regular local prefactors, then

$$g(x) \simeq g(x_\star) + \frac{1}{2}g''(x_\star) (x-x_\star)^2.$$

The Gaussian width and response per volume are

$$\operatorname{Var}(x) \simeq \frac{1}{\beta Vg''(x_\star)},$$ $$\frac{1}{V} \frac{\partial\langle X\rangle}{\partial f} \simeq \frac{1}{g''(x_\star)}.$$

Small curvature means both broad fluctuations and large response.

Criticality and Phase Coexistence

A susceptibility can diverge only after an appropriate thermodynamic limit. For every finite system with a finite-dimensional Hilbert space, the partition function is analytic at finite temperature and source, and equilibrium susceptibilities are finite. Numerical or experimental finite systems display rounded peaks.

Near a continuous transition, a growing correlation length can make

$$\int d^dr\,C_{xx}(\mathbf r)$$

diverge. Then $\operatorname{Var}(X)$ grows faster than $V$, ordinary central-limit scaling fails, and the susceptibility acquires finite-size scaling governed by critical exponents.

At first-order coexistence, the order-parameter distribution may be bimodal. If the two peaks differ by an extensive amount of order $V$, then the between-peak contribution can scale as

$$\operatorname{Var}(X) \sim V^2.$$

The susceptibility per volume can then grow as $V$ at coexistence. This is phase switching, not a single broad Gaussian phase.

Finite-Temperature Phase Transitions explains thermal criticality and two-phase scaling.

Long-range interactions, nonadditive systems, constraints, and disorder can alter these scaling statements. The Thermodynamic Limit page states the order of limits and additivity assumptions in detail.

Static Thermodynamic Versus Retarded Response

The isothermal susceptibility compares equilibrium states:

$$\chi_T^{BB} = \left. \frac{\partial\langle B\rangle_f} {\partial f} \right|_{f=0}.$$

The retarded susceptibility instead asks how an isolated system evolves after a time-dependent perturbation:

$$\chi_{BB}^R(t) = \frac{i}{\hbar} \theta(t) \langle[B(t),B(0)]\rangle.$$

If $[B,H_0]=0$, then

$$B(t)=B$$

and

$$\chi_{BB}^R(t)=0.$$

The thermodynamic response can nevertheless be

$$\chi_T^{BB} = \beta\operatorname{Var}(B) \neq 0.$$

There is no paradox. The equilibrium protocol allows a bath to reweight sectors with different $B$. Isolated unitary evolution preserves their populations. Equilibration, bath coupling, system size, wavevector, frequency, and switching rate determine which static limit is measured.

For spatial response, the limits

$$\lim_{\mathbf q\to0} \lim_{\omega\to0} \chi(\mathbf q,\omega)$$

and

$$\lim_{\omega\to0} \lim_{\mathbf q\to0} \chi(\mathbf q,\omega)$$

need not agree. The Kubo formula develops these dynamical limits; susceptibilities compares channel normalizations and measured responses.

From static response to noise spectra

The dynamical fluctuation–dissipation theorem relates frequency-resolved fluctuations to absorptive response. Integrating a symmetrized noise spectrum gives an equal-time variance; integrating the imaginary-time correlation gives the isothermal susceptibility. The oscillator demonstrates why those results differ.

Exactly zero-frequency fluctuations in conserved or degenerate sectors require separate care: their commutator response can vanish even though their re-equilibrated variance contributes to static response. The sum rules distinguish this contribution from the inverse-frequency moment of the dissipative spectrum.

Estimating a response from measured fluctuations

Independent equilibrium samples directly estimate a variance. Correlated samples need an autocorrelation correction, and detector noise adds its own variance. Slow drift or metastable switching can broaden a time trace without measuring the intended stationary equilibrium distribution. For a measured subregion, account for its window and exchange with the remainder of a globally fixed-number sample.

When possible, compare the inferred fluctuation response with an independently measured slope under a weak calibrated source. A discrepancy can reveal incomplete equilibration, detector error, a missing source term, or an inappropriate ensemble. It need not challenge the underlying statistical identity.

Exercises

Fixed mean number is not fixed particle number

Derive the displayed formula for $C_{\overline N}^{\mathrm{gc}}$ by adjusting $\mu(T)$ to keep $\overline N$ constant. Then take two spinless fermionic modes with energies $0$ and $\Delta>0$, so $H=\Delta n_2$ and $N=n_1+n_2$. Find $\mu(T)$ for $\overline N=1$, and compare this heat capacity with the exact canonical result in the sector $N=1$. Use $q=e^{-\beta\Delta/2}$.

Solution

At fixed $\mu$, the energy–number derivative identities give

$$\begin{aligned} \left(\frac{\partial U}{\partial T}\right)_\mu &=\frac{\operatorname{Var}(H)-\mu\operatorname{Cov}(H,N)}{k_{\mathrm B}T^2},\\ \left(\frac{\partial\overline N}{\partial T}\right)_\mu &=\frac{\operatorname{Cov}(H,N)-\mu\operatorname{Var}(N)}{k_{\mathrm B}T^2}. \end{aligned}$$

Combine the second identity with $\partial_\mu\overline N=\beta\operatorname{Var}(N)$ to obtain

$$\left.\frac{d\mu}{dT}\right|_{\overline N} =\frac1T\left[\mu-\frac{\operatorname{Cov}(H,N)}{\operatorname{Var}(N)}\right].$$

Substituting into $dU/dT=(\partial_TU)_\mu+(\partial_\mu U)_T\,d\mu/dT$, with $\partial_\mu U=\beta\operatorname{Cov}(H,N)$, proves the formula.

For the two modes, $\mu=\Delta/2$ gives $\langle n_1\rangle=1/(1+q)$ and $\langle n_2\rangle=q/(1+q)$ at every $T>0$. The mean is exactly one, but independent mode occupations still allow

$$P(N=0)=P(N=2)=\frac{q}{(1+q)^2},\qquad \operatorname{Var}(N)=\frac{2q}{(1+q)^2}>0.$$

The grand-canonical energy and its constrained derivative are

$$U_{\mathrm{gc}}=\frac{\Delta q}{1+q},\qquad \frac{C_{\overline N=1}^{\mathrm{gc}}}{k_{\mathrm B}} =\frac{(\beta\Delta)^2q}{2(1+q)^2}.$$

By contrast, the canonical sector contains only $|10\rangle$ and $|01\rangle$, with $Z_1=1+q^2$. Consequently

$$U_{\mathrm c}=\frac{\Delta q^2}{1+q^2},\qquad \frac{C_{N=1}^{\mathrm c}}{k_{\mathrm B}} =\frac{(\beta\Delta)^2q^2}{(1+q^2)^2}.$$

Both preparations have mean number one, but the probabilities of the zero- and two-particle sectors distinguish them. The different thermal responses are a finite-system ensemble effect, not an interaction effect.

Positivity of a susceptibility matrix

Let Hermitian observables $A$ and $B$, each commuting with the equilibrium state, couple to sources $f_A$ and $f_B$. Use positivity of

$$\operatorname{Var}(uA+vB)$$

for all real $u,v$ to prove

$$\operatorname{Cov}(A,B)^2 \leq \operatorname{Var}(A) \operatorname{Var}(B).$$

Interpret the result as a constraint on cross susceptibility.

Solution

Expanding gives

$$\operatorname{Var}(uA+vB) = u^2\operatorname{Var}(A) + 2uv\operatorname{Cov}(A,B) + v^2\operatorname{Var}(B).$$

This quadratic form must be nonnegative for every $(u,v)$, so its covariance matrix is positive semidefinite. Its determinant must satisfy

$$\operatorname{Var}(A) \operatorname{Var}(B) - \operatorname{Cov}(A,B)^2 \geq 0.$$

For these linear sources commuting with the equilibrium state,

$$\chi_{AA}=\beta\operatorname{Var}(A), \quad \chi_{BB}=\beta\operatorname{Var}(B), \quad \chi_{AB}=\beta\operatorname{Cov}(A,B).$$

Hence

$$|\chi_{AB}| \leq \sqrt{\chi_{AA}\chi_{BB}}.$$

A cross response cannot exceed the geometric mean of the two diagonal responses under these assumptions.

Fixed particle number

A finite canonical system has $\operatorname{Var}(N)=0$. Does the fluctuation formula imply $\kappa_T=0$? Give two ways to determine a nonzero compressibility without changing the canonical trace space.

Solution

No. The formula

$$\kappa_T = \frac{\beta\operatorname{Var}(N)}{Vn^2}$$

is a grand-canonical global-fluctuation identity. In a canonical trace, $N$ is fixed and its variance vanishes by definition.

A canonical compressibility can instead be computed from the volume curvature of the Helmholtz free energy,

$$P = -\left( \frac{\partial F}{\partial V} \right)_{T,N}, \qquad \kappa_T = -\frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_{T,N},$$

or from a long-wavelength density response with the order of limits specified. A subvolume can also exchange particles with the rest of a globally fixed-$N$ sample and therefore have nonzero number fluctuations.

A conserved quantity and two static protocols

Suppose $[B,H_0]=0$. Show that the isolated retarded self-response vanishes while the isothermal equilibrium response can be nonzero. What physical operation distinguishes the protocols?

Solution

Conservation gives

$$B(t) = e^{iH_0t/\hbar}Be^{-iH_0t/\hbar} = B.$$

Therefore

$$\chi_{BB}^R(t) = \frac{i}{\hbar} \theta(t) \langle[B(t),B]\rangle = 0.$$

For re-equilibrated states of $H_f=H_0-fB$,

$$\chi_T^{BB} = \left. \frac{\partial\langle B\rangle_f} {\partial f} \right|_{f=0} = \beta\operatorname{Var}(B),$$

which can be positive. The equilibrium protocol permits exchange with an environment or preparation procedure that changes the statistical weights of different $B$ sectors. Isolated unitary evolution cannot change those populations.

An explicit contact term

Set $k_{\mathrm B}=1$ and consider $H(f)=-f\sigma_z+cf^2I/2$, with $c>0$. Find the response of $B_f=-\partial_fH$ at $f=0$. Can it be negative at positive temperature?

Solution

The scalar term cancels from the normalized density operator, so $\langle\sigma_z\rangle=\tanh(\beta f)$. But the generalized displacement is $B_f=\sigma_z-cfI$, hence

$$\partial_f\langle B_f\rangle\big|_0=\beta-c.$$

The covariance contributes $\beta$ and the explicit derivative contributes $-c$. The result is negative for $c>\beta$. Positive semidefiniteness of the covariance response applies to fixed observables coupled linearly; this nonlinear Hamiltonian also has a contact term.

References

• R. Balian, From Microphysics to Macrophysics, Volume I, Springer (1991), chapters 4–7.
• H. B. Callen and T. A. Welton, “Irreversibility and Generalized Noise”, Physical Review 83, 34–40 (1951).
• H. B. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed., Wiley (1985), chapters 7, 15, and 19.
• M. Kardar, Statistical Physics of Particles, Cambridge University Press (2007), chapters 1, 4, and 6.
• R. Kubo, “Statistical-Mechanical Theory of Irreversible Processes. I. General Theory and Simple Applications to Magnetic and Conduction Problems”, Journal of the Physical Society of Japan 12, 570–586 (1957).
• R. Kubo, “The Fluctuation-Dissipation Theorem”, Reports on Progress in Physics 29, 255–284 (1966).
• L. D. Landau and E. M. Lifshitz, Statistical Physics, Part 1, 3rd ed., Butterworth–Heinemann (1980), sections 19–24 and 112–117.
• R. K. Pathria and P. D. Beale, Statistical Mechanics, 4th ed., Elsevier (2021), chapters 3–5 and 11–13.