Fluctuations and response functions

A counter that will not sit still

Imagine twenty independent units, each either unexcited or excited. A counter reports how many are excited. At 300 K its equilibrium average is five, but individual readings need not be five. Can that unforced wandering tell us how the average would change if we made excitation slightly easier?

Yes, under specific equilibrium conditions. In our assigned model the count variance is 3.75 and the standard deviation is about 1.936 units. The same variance gives the initial slope of the mean count against a properly defined dimensionless bias. We will first see why, then ask what a real time record adds.

The units are labelled and noninteracting, with equal degeneracy for the two alternatives. An excited unit has energy Δ; an unexcited one has energy zero. Choose the fixed gap Δ = k_B T₀ ln 3, with T₀ = 300 K. This is an illustrative model, not a fitted measurement of a particular material.

Choose a reading path

On a first reading, follow the count distribution, the small bias, the relaxing step and the finite observation record. Read variance as a mean squared spread, a correlation as memory between two readings, and a response as a change caused by a controlled perturbation. The worked examples use the same assigned sample throughout.

The return path derives general covariance identities, examines spatial correlations and finite reservoirs, connects causality with frequency response, and gives a quantum counterexample. Partial derivatives and integrals help with that path; the physical meanings are introduced locally before the formulas.

The ensembles and partition-functions topics develop the probability weights used here. This lesson asks what their fluctuations reveal about a response and what extra assumptions are needed for dynamics. Phase transitions, critical scaling and detailed transport equations have their own permanent topics.

Define the spread before using it

For a random observable A, write its equilibrium average as ⟨A⟩ and its deviation as δA = A − ⟨A⟩. The variance is Var(A) = ⟨(δA)²⟩ = ⟨A²⟩ − ⟨A⟩². Its square root is the standard deviation. Variance has the square of A’s units; standard deviation has A’s units.

A broad distribution is not automatically a bad instrument. The sample can fluctuate physically even with an ideal detector. Instrument noise, uncertain calibration and uncertainty in an estimated mean are different contributions that require their own measurement model.

For two observables define Cov(A,B) = ⟨δA δB⟩. It can be positive, negative or zero. Zero covariance alone does not establish statistical independence. For a sum, every pair matters: Var(ΣᵢAᵢ) = Σᵢⱼ Cov(Aᵢ,Aⱼ). Independence is one sufficient reason for the off-diagonal terms to vanish.

Count independent binary units

At zero bias and temperature T, the excitation probability is p = 1/[1 + exp(Δ/(k_B T))]. At T₀ our chosen gap gives p = 1/4. Since n² = n, one unit has variance p − p² = p(1 − p) = 3/16.

Let K = Σᵢnᵢ be the excited count. Independence gives ⟨K⟩ = Np and Var(K) = Np(1 − p). The probability of exactly k excitations is P(K = k) = C(N,k) pᵏ(1 − p)ᴺ⁻ᵏ for integers k from zero to N. C(N,k) counts which labelled units are excited; it is not an additional excitation degeneracy.

For twenty units, ⟨K⟩ = 5 and Var(K) = 3.75. The standard deviation 1.936 is not a hard interval containing every reading. The exact distribution has finite support 0 through 20 and is not exactly Gaussian. A continuous bell-shaped approximation is unnecessary for this small sample.

The relative standard deviation is √[(1 − p)/(Np)]. Increasing N from 20 to 80 at the same p quadruples the mean but only doubles the standard deviation: the relative spread halves. Later we identify the correlation and ensemble conditions behind this familiar size scaling.

Tilt the weights and measure a slope

Apply a bias energy f so that the energy of an excited unit is Δ − f. At the reference temperature, p(g) = 1/[1 + exp(ln 3 − g)]. Positive g makes excitation cheaper, so the mean count rises. The slope at g = 0 is d⟨K⟩/dg = Np(1 − p) = Var(K).

For our sample that slope is 3.75 counts per unit g. A small positive change δg = 0.050 therefore predicts a mean-count increase of 3.75 × 0.050 = 0.1875. Evaluating the full curve instead gives an increase of about 0.189833. The tangent is a local approximation, not a second exact probability law.

In physical energy units, ∂⟨K⟩/∂f = Var(K)/(k_B T₀), with units J⁻¹ because a count is dimensionless. At a different temperature T, our fixed definition g = f/(k_B T₀) gives ∂⟨K⟩/∂g = (T₀/T) Var(K). Omitting T₀/T would confuse two different choices of horizontal axis.

The spread does not exert a new force. Both the spread and the susceptibility describe how the same equilibrium weights change when an energy preference is varied. A fluctuation measurement predicts this derivative only after the coupling, temperature and ensemble are specified.

Energy fluctuations and heat capacity

At zero bias, the sample energy is E = ΔK. Thus Var(E) = Δ²Np(1 − p). Changing the bath temperature changes p while Δ stays fixed. Differentiating the mean energy gives C = ∂⟨E⟩/∂T = Nk_B[Δ/(k_B T)]²p(1 − p).

Combining those results gives Var(E) = k_B T²C. This is a canonical-equilibrium identity at fixed volume, particle number and external parameters for a temperature-independent Hamiltonian. It is not a statement that an exactly fixed-energy microcanonical system has the same total-energy variance.

For a canonical system the variance is nonnegative, so this particular heat capacity is nonnegative. That does not license replacing other ensembles, finite reservoirs or constrained response functions by the canonical formula. When energy levels or effective model parameters explicitly depend on temperature, their derivative terms must be retained.

Return path: particle-number fluctuations and compressibility

A sample exchanging particles with a reservoir has a different fluctuating quantity. In grand-canonical equilibrium at fixed T and V, Var(N) = k_B T (∂⟨N⟩/∂μ), where μ is chemical potential. The derivative changes the particle preference while the temperature and volume remain fixed.

For a homogeneous single-component bulk system, the number density is ⟨N⟩/V. Define κ_T = −(1/V)(∂V/∂P) at fixed T and particle number. The equilibrium relation dP = (⟨N⟩/V) dμ at fixed T then gives Var(N)/⟨N⟩ = (⟨N⟩/V) k_B T κ_T. The right-hand side is dimensionless: pressure times compressibility is dimensionless.

An ideal classical gas has κ_T = V/(⟨N⟩ k_B T), giving Var(N) = ⟨N⟩ in this grand-canonical bulk setting. A whole closed vessel with exactly fixed N has zero total-number variance instead. A small open observation volume inside it can still fluctuate, with finite-volume and conservation corrections. Always identify the observed region and what can cross its boundary.

Equilibrium does not supply a clock

The excitation probability tells us how often each state occurs in equilibrium. It does not tell us how quickly a unit switches. To introduce time, assign a continuous-time two-state Markov process: k₀₁ = p_eq(g)/τ and k₁₀ = [1 − p_eq(g)]/τ, with τ = 0.020 s at the reference temperature.

Here a rate has units s⁻¹, and the next transition depends only on the present state in this model. Their ratio k₀₁/k₁₀ = exp(g − ln 3) gives the equilibrium probability ratio. Detailed balance means the two opposite equilibrium probability currents are equal. Multiplying both rates by the same factor changes the clock while leaving that ratio and the equilibrium distribution unchanged.

At zero bias, k₀₁ = 12.5 s⁻¹ and k₁₀ = 37.5 s⁻¹. The mean probability obeys dp/dt = [p_eq(g) − p]/τ. This rate law is an assigned reversible kinetic model, not a consequence of the partition function alone and not a universal law for every two-level physical system.

Correlation measures memory

In a stationary equilibrium record define C(t) = ⟨δn(t)δn(0)⟩. Stationary means that shifting both observation times together does not change the statistics. At t = 0, C(0) is the variance, 3/16. At later times C measures how much information about the initial occupation remains in this two-time average.

Solving the assigned two-state process gives C(t) = (3/16) exp(−|t|/τ). After one τ the correlation has fallen to 1/e of its initial value; after five τ it is about 0.006738 of that value. The smooth curve is an ensemble correlation, not the jagged trajectory of a particular unit.

Twenty independent units have count correlation C_K(t) = 20C(t). If the units interact, cross-correlations must be added instead. A faster detector can resolve more points on a trace without making those points statistically independent.

A step separates memory from response

Start in zero-bias equilibrium and switch g to 0.050 at time zero. In the assigned rate law, the exact mean probability is p(t) = p_eq(0.050) + [p₀ − p_eq(0.050)] exp(−t/τ). It moves toward the new equilibrium; it does not jump immediately because the prescribed transition rates are finite.

Linear response keeps only the first derivative in the bias: δp(t) ≈ p₀(1 − p₀) δg [1 − exp(−t/τ)]. At t = τ, the exact increase is about 0.00599988 and the linear prediction is about 0.00592613. Both use the same relaxation time in this deliberately chosen kinetic model, but their final amplitudes differ.

More generally write δ⟨n(t)⟩ = ∫ R(t − t′) f(t′) dt′. The response kernel R is zero for negative delays: a perturbation cannot change an earlier reading. For this model, R(t) = β₀C(0) exp(−t/τ)/τ for t ≥ 0, where β₀ = 1/(k_B T₀). Its units are J⁻¹ s⁻¹.

For positive delays R(t) = −β₀ dC(t)/dt. That is the time-domain fluctuation–dissipation relation for this equilibrium model and its specified conjugate perturbation. An arbitrary fitted decay curve does not establish that relation for a driven or ageing sample.

Name the spectrum convention

A power spectral density distributes a stationary fluctuation’s variance over frequency. Here use angular frequency ω in rad/s and a two-sided spectrum S(ω) = ∫ from −∞ to ∞ C(t) exp(iωt) dt. The inverse convention gives C(0) = ∫ S(ω) dω/(2π). Since occupation is dimensionless, S has units seconds.

For the assigned process, S(ω) = 2C(0)τ/[1 + (ωτ)²]. Its zero-frequency value is 0.00750 s. The curve is even in ω, and integrating over both negative and positive frequencies gives the variance 3/16. A figure showing only ω ≥ 0 must not be silently integrated as if it were the whole two-sided spectrum.

With the same exp(iωt) transform, define χ(ω) = ∫₀∞ R(t) exp(iωt) dt. A real drive written Re[F exp(−iωt)] gives response Re[χ(ω)F exp(−iωt)]. Here χ = β₀C(0)/(1 − iωτ). Its real part is in-phase response; its positive imaginary part at positive ω describes a lag with this convention.

The classical relation is S(ω) = 2k_B T₀ χ″(ω)/ω, with χ″ denoting the imaginary part, not a second derivative. At ω = 0 use the continuous limit rather than divide zero by zero. Frequency in cycles/s, one-sided spectra, other Fourier signs and quantum operator orderings require corresponding factors and definitions.

Dissipation is a cycle average

Take a weak physical bias f(t) = F cos(ωt). With H = H₀ − fn, the external work rate associated with changing the control is −⟨n⟩ df/dt. In the linear periodic state δ⟨n⟩ = F[χ′ cos(ωt) + χ″ sin(ωt)]. Multiplying and averaging over a cycle leaves ⟨P⟩ = F²ωχ″/2 because the mean of sin² is 1/2 and that of sin cos is zero.

For a passive equilibrium system this average is nonnegative. At ωτ = 1 the assigned model has a 45° lag and |χ| = χ₀/√2. A dimensionless bias amplitude 0.020 predicts an occupation amplitude about 0.00265165 and absorbed power about 3.88308 × 10⁻²⁴ W per unit.

Those amplitude and power values are linear-response and leading quadratic-power predictions, not an exact finite-amplitude solution of the nonlinear time-dependent rate law. Nor is the maximum of χ″ automatically a resonance: our simple relaxing unit has no oscillatory restoring dynamics. The factor 1/2 belongs to the stated peak cosine amplitude F, not an RMS-amplitude convention.

Time averaging has a memory cost

A finite-time average is n̄ = (1/t_obs)∫₀ᵗ_obs n(t) dt. Its variance is a double integral of the covariance, not simply the single-reading variance divided by the number of digitized points. For a real stationary scalar record it becomes Var(n̄) = (2/t_obs²)∫₀ᵗ_obs (t_obs − t) C(t) dt.

For the exponential model, let u = t_obs/τ. Then Var(n̄)/C(0) = 2[u − 1 + exp(−u)]/u², with continuous limit one at u = 0. At long times the ratio approaches 2τ/t_obs. The factor two reflects the full area under the normalized covariance on both time sides.

A two-second observation has u = 100, variance ratio about 0.0198 and standard deviation of the averaged occupation about 0.0609303. Its variance equals that of about 50.505 independent equilibrium readings, not every sample from a fast digitizer. This noninteger effective count is a variance equivalence, not a claim that precisely that many independent events occurred.

The formula assumes equilibrium at the start and the stated continuous ideal observation. Drift, finite detector exposure, filtering, missed transitions and measurement noise change the recorded covariance and must enter the estimator’s model. Repeating a long measurement is not a substitute for checking stationarity.

Return path: differentiate the entire average

Let a classical equilibrium state have probability proportional to exp[−βH(z;f)] with β = 1/(k_B T) and a field-independent state-space measure. Differentiating the numerator and normalization of ⟨A⟩ at fixed temperature gives ∂f⟨A⟩ = ⟨∂fA⟩ − β Cov(A,∂fH). Both terms matter if the observable explicitly changes with the control.

For H = H₀ − fB and field-independent A and B, this reduces to ∂f⟨A⟩ = β Cov(A,B). When A = B the covariance is a variance, but a cross-response can have either sign. If the measure or allowed state space depends on f, their derivatives are additional contributions; do not use the short formula unchanged.

The sign can be checked physically at positive temperature. Increasing f favours larger B, so the conjugate slope is β Var(B) ≥ 0; it is strictly positive where Var(B) > 0. If the Hamiltonian is instead H₀ + fB, the covariance contribution has the opposite sign. The variable held fixed is part of the definition of a susceptibility.

Return path: size, correlations and constraints

For an extensive sum with a mean proportional to N, relative fluctuations scale as 1/√N when the summed connected covariances grow only proportionally to N. Independence suffices, but sufficiently short-range, summable correlations can also give that scaling. Large N alone does not prove it.

Near criticality a growing correlation length can make many constituents fluctuate together and change finite-size scaling. Long-range interactions and constraints require their own analysis. A system can have locally similar mean observables in two ensembles while their global conserved-quantity variances differ.

The equilibrium static structure factor expresses spatial density covariance in wave-number space; in the open homogeneous bulk limit its zero-wave-number value connects to compressibility. A closed fixed-N sample removes its exact total-number mode. Taking the bulk, small-wave-number and fixed-constraint limits in different orders can therefore answer different questions. The critical-phenomena topic develops those limits in detail.

Return path: a reservoir can be finite

For a weakly coupled isolated sample plus bath, the energy probability is proportional to exp{[S_s(E_s) + S_b(E_total − E_s)]/k_B}. Expand the combined entropy around a stable interior equilibrium. Its first derivative vanishes because the temperatures agree.

For positive finite heat capacities C_s and C_b, the second derivative is −(1/T²)(1/C_s + 1/C_b). A quadratic expansion therefore gives the Gaussian approximation Var(E_s) ≈ k_B T² C_s C_b/(C_s + C_b). The infinite-bath limit recovers k_B T²C_s; a finite bath suppresses the fluctuation because it changes temperature when it supplies energy.

If C_b = 4C_s, this approximation predicts 0.8 times the infinite-bath variance and about 0.894427 times its standard deviation. It is not an exact law for a tiny discrete system, a boundary maximum or a broad non-Gaussian distribution. Strong coupling also changes the additive-energy construction. The ensembles lesson’s finite-reservoir example shows why exact state counting can be preferable.

Return path: causality links the frequency parts

A causal response with suitable convergence is analytic in the upper half of the complex-frequency plane for our exp(iωt) convention. Our relaxation susceptibility has its pole at ω = −i/τ, in the lower half-plane, consistently with a decaying response rather than an anticipation of the drive.

For a real response with the required high-frequency decay, the Kramers–Kronig relation is χ′(ω) = (1/π) PV∫ χ″(ω′)/(ω′ − ω) dω′ over the whole real frequency axis. PV means the symmetric principal-value treatment of the apparent singularity. In particular χ′(0) = (2/π)∫₀∞ χ″(ω)/ω dω under those conditions.

Inserting χ″ = χ₀ωτ/[1 + (ωτ)²] returns χ₀. Together with the classical spectrum relation this recovers C(0) = k_B T₀χ₀. Instantaneous or nondecaying response contributions require appropriate subtractions or separate terms. A finite measured frequency band does not supply all missing tails simply because a causal relation exists.

Return path: quantum variance is not enough

The classical covariance shortcut cannot be applied blindly to noncommuting quantum observables. Consider H = −Γσₓ − fσ_z with an energy scale Γ > 0, where σₓ and σ_z are Pauli matrices with eigenvalues ±1. The measured observable is A = σ_z. Its unperturbed Hamiltonian does not commute with A.

The two energies are ±r, with r = √(Γ² + f²), so Z = 2 cosh(βr) and ⟨A⟩ = (f/r)tanh(βr). Here β = 1/(k_B T), and tanh x = [exp(x) − exp(−x)]/[exp(x) + exp(−x)]. At f = 0, ⟨A⟩ = 0 and A² = 1, so its equal-time variance is one. Yet ∂f⟨A⟩ = tanh(βΓ)/Γ, not generally β.

The ratio of the exact susceptibility to β is tanh x/x with x = βΓ. At x = 1 it is about 0.761594; at x approaching zero it tends to one. The discrepancy is not experimental error. The perturbation changes quantum eigenstates as well as their populations, so ordinary equal-time variance omits information about operator response.

Quantum fluctuation–dissipation relations use specified operator ordering or a symmetrized spectrum and contain the corresponding ℏω/(k_B T) factor. Here ℏ = h/(2π), where h is Planck’s constant. The classical low-frequency limit requires ℏ|ω| ≪ k_B T in addition to the relevant equilibrium conditions. This static counterexample illustrates a failed shortcut; it is not itself a derivation of a quantum noise spectrum or of a kinetic relaxation time.

Return path: stationary is not equilibrium

A steady driven state can have time-translation-invariant statistics while carrying probability currents and continually dissipating energy. Stationarity permits a spectrum; it does not by itself grant the equilibrium fluctuation–dissipation relation.

Outside equilibrium, the response can depend on how a perturbation changes the dynamics, not just on the unperturbed stationary distribution or a measured covariance. Active matter, externally pumped devices and ageing systems require models of those changes. Calling a ratio an effective temperature does not prove that it is a single thermodynamic temperature for every observable and frequency.

Even in equilibrium, an observed signal may include detector noise unrelated to the sample’s conjugate response. A comparison of noise and susceptibility must use the same observable, coupling, units, baseline conditions, frequency convention and measurement transfer function. A discrepancy is a prompt to check those conditions, not immediate proof of exotic physics.

Use a fluctuation as a physical question

First specify the system boundary, ensemble and observable. Decide which spread is intrinsic to the sample and which part belongs to measurement. Then name the field that couples to the observable and all fixed controls. A correctly matched covariance can predict a static derivative without applying a large perturbation.

For time-dependent inference, add the kinetic assumptions and examine a stationary correlation record. State whether frequency means radians or cycles per second and whether a spectrum is one-sided, two-sided, classical or quantum ordered. Compare the response in the linear regime, and keep finite-duration uncertainty separate from instantaneous variance.

Fluctuations can therefore reveal heat capacity, compressibility, susceptibility and relaxation, but not through a universal noise-to-response slogan. Their usefulness comes from the physical conditions that make the connection precise.

Worked examples

Try it yourself

1. Must the twenty-unit counter read five whenever it is in equilibrium?

   Show the answer and reasoning

   No; five is an ensemble average, not every counter reading.

   No. Five is the ensemble mean; a particular integer reading fluctuates. The variance 3.75 describes its spread, not a forbidden deviation threshold.

2. At 300 K, what happens to the relative count spread when N changes from 20 to 80?

   Show the answer and reasoning

   It halves to about 0.193649 under the stated independence assumption.

   It halves, from about 0.387298 to 0.193649. At fixed p, the mean grows fourfold and the standard deviation grows twofold, under the stated independence assumption.

3. If the count variance is 3.75, what are the units and value of its energy variance?

   Show the answer and reasoning

   Var(E)=3.75Δ² with units J²; its standard deviation is Δ√3.75 in J.

   Var(E) = Δ² × 3.75, with units J². Standard deviation of energy is Δ√3.75 with units J; replacing Δ² by Δ would have the wrong units.

4. For the coupling H = H₀ − fK, does a small positive f raise or lower the mean count at fixed T?

   Show the answer and reasoning

   It raises the conjugate mean in this classical fixed-measure equilibrium model.

   It raises it. The conjugate derivative is βVar(K) ≥ 0 in this classical fixed-measure equilibrium model. The sign would reverse for an energy term +fK.

5. Find d⟨K⟩/dg at 150 K for twenty units when g retains the fixed 300 K reference scale.

   Show the answer and reasoning

   d⟨K⟩/dg=3.6.

   The excitation probability is 0.1 and Var(K) = 1.8. Multiply by T₀/T = 2 to obtain 3.6. A derivative equal simply to the variance uses a different dimensionless field definition.

6. Multiply both zero-bias transition rates by ten. What changes?

   Show the answer and reasoning

   The equilibrium probabilities stay the same; the relaxation time becomes 0.002 s.

   Their ratio and the equilibrium occupation and variance stay unchanged. Their sum grows tenfold, so the relaxation time becomes 0.002 s and the spectrum changes width. Equilibrium weights alone do not fix dynamics.

7. What fraction of the initial correlation remains after 0.020 s?

   Show the answer and reasoning

   C(τ)/C(0)=e⁻¹≈0.367879.

   One relaxation time has elapsed, so C(t)/C(0) = exp(−1) ≈ 0.367879. This is a two-time ensemble covariance, not the fraction of units forbidden from switching.

8. Why are the exact and linear step changes different even though they share τ here?

   Show the answer and reasoning

   They use different final-amplitude expressions, not different assigned clocks.

   The exact result uses p_eq(0.050) − p₀, whereas linear response substitutes p₀(1 − p₀) × 0.050. The assigned rates keep their sum fixed; the approximation is in the bias-dependent final preference.

9. Does S = 2k_B Tχ″/ω predict an undefined physical noise at ω = 0?

   Show the answer and reasoning

   No; its continuous limit is S(0)=0.00750 s.

   No. For this model χ″ is proportional to ω near zero. Taking the continuous limit gives S(0) = 2C(0)τ = 0.00750 s for one unit.

10. A calculation integrates the displayed ω ≥ 0 half of the two-sided spectrum with dω/(2π). Why does it find only 3/32?

    Show the answer and reasoning

    It integrated only half the frequency axis without changing the spectral normalization.

    The spectrum is even and that integral covers only half the frequency axis. The complete two-sided integral gives 3/16. A one-sided spectral definition must include its corresponding factor.

11. At ωτ = 1, are χ′ and χ″ equal, and what does that mean for a weak cosine drive?

    Show the answer and reasoning

    Yes: χ′=χ″=χ₀/2, lag 45°, and leading mean power F²ωχ₀/4.

    Both equal χ₀/2. The linear response lags by 45°, and a peak field amplitude F absorbs mean power F²ωχ₀/4. This power is the leading quadratic prediction, not an exact arbitrary-amplitude law.

12. If the same two-second ideal record is sampled a hundred times more densely, does its continuous-time mean variance necessarily fall by one hundred?

    Show the answer and reasoning

    No. More correlated samples of the same ideal record need not reduce the continuous-average variance.

    No. Closely spaced points remain correlated. The ideal continuous-average variance is set by the covariance integral, about 0.0198 times a single-reading variance here. Detector bandwidth and discrete sampling need their own model.

13. A whole vessel has exactly fixed particle number. Must its total-number variance equal (⟨N⟩/V) k_B T κ_T times its mean?

    Show the answer and reasoning

    No. An exactly fixed total particle number has zero total-number variance.

    No. That fluctuation identity concerns the appropriate open grand-canonical bulk setting. The closed vessel’s total-number variance is zero; a subvolume can still fluctuate with conservation and finite-volume corrections.

14. Why is ∂f⟨A⟩ not always just a covariance term?

    Show the answer and reasoning

    The explicit derivative of the observable and any field-dependent measure/domain must also be retained.

    If A itself depends on f there is an explicit ⟨∂fA⟩ contribution. A field-dependent Hamiltonian, state-space measure or allowed domain also has to be differentiated consistently. The short formula states conditions, not just notation.

15. Does a large N alone guarantee relative fluctuations proportional to 1/√N?

    Show the answer and reasoning

    No; the connected covariance sum and mean need the appropriate size scaling.

    No. The connected covariance sum must have the required extensive scaling and the mean must scale appropriately. Long correlations, criticality, long-range interactions or constraints can change the result.

16. Under the stated Gaussian finite-bath assumptions, what variance ratio follows from C_b = C_s?

    Show the answer and reasoning

    Half the infinite-bath variance and 1/√2 of its standard deviation, within the stated approximation.

    C_sC_b/(C_s + C_b) = C_s/2, giving half the infinite-bath variance and 1/√2 of its standard deviation. Do not interpret the Gaussian approximation as exact for an arbitrary small discrete sample.

17. For the quantum counterexample at large βΓ, what is the approximate static susceptibility and why does βVar(A) fail?

    Show the answer and reasoning

    The exact susceptibility approaches 1/Γ, not β.

    tanh(βΓ) approaches one, so χ approaches 1/Γ, while βVar(A) = β. The noncommuting perturbation changes eigenstates; ordinary equal-time variance does not supply the full quantum static response.

18. A driven device has stationary noise. Is that enough to infer its response with the equilibrium FDT?

    Show the answer and reasoning

    No. Stationarity alone does not establish the equilibrium response relation.

    No. Stationarity permits a time-translation-invariant spectrum but does not establish detailed balance or equilibrium. The chosen perturbation, driven currents and detector transfer function can matter.

What to carry forward

• A fluctuation is a deviation from a specified average. Its physical spread, an instrument’s noise and uncertainty in an estimated mean are not the same thing.
• Classical equilibrium covariances generate static responses only with the specified coupling, ensemble, measure and fixed controls.
• Equilibrium weights do not fix a relaxation time. A dynamical model is additional physical input.
• Causality, Fourier sign, spectral sidedness and units are part of a time-dependent fluctuation–response statement.
• Correlated readings impose a memory cost on averaging; denser recording does not automatically create independent observations.
• Finite baths, conservation laws, critical correlations, non-equilibrium states and noncommuting quantum observables require their appropriate models rather than an unqualified variance shortcut.