Partition Functions and Thermodynamic Response

A source-dependent partition function generates equilibrium thermodynamics. Its first derivatives give expectation values, its second derivatives give isothermal response, and Legendre transforms exchange controlled sources for mean extensive variables. In a classical Gibbs ensemble the response is a connected covariance. In a quantum ensemble it generally involves an imaginary-time integral, even for a time-independent source. Every derivative below keeps the stated temperature, volume and other sources fixed.

Required background. Statistical Ensembles and Field Configurations defines the normalized measures used here. Helpful background. Current Sources and Generating Functionals gives the QFT source and Ward-identity framework.

One generating function, several thermodynamic potentials

Keep a finite volume $V$ and an ultraviolet regulator, and assume that the trace and derivatives below exist in the chosen source domain. For inverse temperature $\beta>0$ , chemical potentials $\mu_a$ , and a real source $h$ coupled linearly to a Hermitian extensive observable $M$ , define

\mathcal K=H-\sum_a\mu_aQ_a-hM, \qquad \Xi=\operatorname{Tr}e^{-\beta\mathcal K}, \qquad \rho=\Xi^{-1}e^{-\beta\mathcal K}.

The operators and couplings in $\mathcal K$ are independent of temperature and of the source being differentiated, apart from the displayed linear dependence. Chemical potentials describe conserved charges compatible with the physical Hamiltonian at the chosen external fields; more general linear Gibbs sources obey the same derivative identities without automatically having that chemical-potential interpretation. Replacing the trace by a phase-space integral gives the classical formulas.

The Massieu function $\Psi=\log\Xi$ is dimensionless and the grand potential is

\Omega=-T\log\Xi.

In a homogeneous extensive thermodynamic limit with negligible boundary work, $\Omega=-pV$ . Finite-volume pressure is instead defined by the volume derivative with its boundary conditions fixed. Differentiation at fixed $\beta,V$ , and all other sources gives

\langle Q_a\rangle =\frac1\beta\frac{\partial\Psi}{\partial\mu_a}, \qquad \langle M\rangle =\frac1\beta\frac{\partial\Psi}{\partial h}, \qquad \langle\mathcal K\rangle =-\frac{\partial\Psi}{\partial\beta}.

The last derivative is not $\langle H\rangle$ when $\beta$ varies while $\mu_a$ and $h$ are held fixed. Equivalently one may use dimensionless sources $\alpha_a=\beta\mu_a$ ; translating between the two parameterizations changes which derivative is taken.

For a canonical partition function $Z(\beta,V)$ ,

F=-T\log Z, \qquad S=-\left(\frac{\partial F}{\partial T}\right)_V, \qquad E=F+TS, \qquad p=-\left(\frac{\partial F}{\partial V}\right)_T.

These relations assume that the regulator and boundary work are held consistently while differentiating. A lattice calculation that changes the lattice spacing with temperature needs an explicit line of constant physics; differentiating bare parameters as though they were fixed physical couplings does not yield the thermodynamic energy.

Connected cumulants and susceptibility matrices

One cannot differentiate a noncommuting operator exponential as an ordinary scalar exponential. At fixed $\beta$ , its first variation is

\partial_h e^{-\beta\mathcal K} =\int_0^\beta\!\mathrm d\tau\, e^{-(\beta-\tau)\mathcal K}M e^{-\tau\mathcal K}.

Cyclicity of the trace makes the first derivative $\partial_h\Psi=\beta\langle M\rangle$ . Differentiating a normalized expectation also differentiates $\Xi^{-1}$ ; this subtracts the disconnected product. Define the imaginary-time evolution and integrated connected correlation by

\begin{aligned} A(\tau)&=e^{\tau\mathcal K}Ae^{-\tau\mathcal K}, &\delta A&=A-\langle A\rangle,\\ \mathcal B(A,B)&=\int_0^\beta\!\mathrm d\tau\, \langle\delta A(\tau)\delta B(0)\rangle. \end{aligned}

This is the integrated Kubo–Mori correlation; no extra factor of $\beta$ is included in its definition. For $n_a=\langle Q_a\rangle/V$ the responses are

\begin{aligned} \partial_h^2\Psi&=\beta\,\mathcal B(M,M),\\ \chi_{ab} &=\left(\frac{\partial n_a}{\partial\mu_b}\right)_{T,V,h,\mu_{c\ne b}} =\frac{1}{V}\mathcal B(Q_a,Q_b). \end{aligned}

These are isothermal equilibrium derivatives: the state is allowed to re-equilibrate at the same temperature. They need not equal the zero-frequency retarded response of an isolated system with conserved, nonrelaxing components. The operator-exponential expansion and this distinction appear in Kubo 1957, § 3, pp. 576–577, Eqs. (3.15)–(3.21).

For a classical Gibbs measure, or for a quantum observable commuting with the full $\mathcal K$ , the integrand is independent of $\tau$ . In that case

\partial_h^2\Psi=\beta^2\langle(\delta M)^2\rangle, \qquad \chi_{ab}=\frac{\beta}{V}\langle\delta Q_a\delta Q_b\rangle

for the corresponding commuting observables. Commutation of the charges with one another alone is not enough: their commutation with the actual Gibbs generator matters. Higher source derivatives likewise generate connected imaginary-time integrals in the quantum case, rather than unrestricted equal-time moments.

The response matrix is positive semidefinite for Hermitian linear sources in a positive Gibbs state, including the noncommuting case. To see this, take a real linear combination $A=\sum_a u_aQ_a$ and diagonalize $\mathcal K$ , with eigenvalues $\epsilon_i$ and probabilities $p_i=e^{-\beta\epsilon_i}/\Xi$ . Then

\mathcal B(A,A) =\sum_{i,j} w_{ij}\lvert(\delta A)_{ij}\rvert^2, \qquad w_{ij}=\begin{cases} \dfrac{p_i-p_j}{\epsilon_j-\epsilon_i},&\epsilon_i\ne\epsilon_j,\\[3pt] \beta p_i,&\epsilon_i=\epsilon_j. \end{cases}

Every weight is nonnegative, since the Gibbs probability decreases with energy. Thus $u_a\chi_{ab}u_b\ge0$ ; symmetry follows also from the commuting derivatives of $\Psi$ . A negative eigenvalue cannot describe this exact finite-volume equilibrium Hessian. It can arise from an unstable approximate branch, a different constraint or an incorrectly assembled response.

In the canonical ensemble, the heat capacity at fixed volume is a variance because the energy commutes with its Gibbs generator:

C_V=\left(\frac{\partial E}{\partial T}\right)_V =\frac{\langle(\Delta H)^2\rangle}{T^2},

provided $H$ has no explicit temperature dependence. Effective Hamiltonians with temperature-dependent matched coefficients require derivative terms from those coefficients.

Legendre transforms and convexity

The source $h$ and mean $m=\langle M\rangle/V$ are conjugate. In a differentiable finite-volume system one may define

g(h)=-\frac{T}{V}\log\Xi(h), \qquad f(m)=g(h)+hm, \qquad m=-\frac{\partial g}{\partial h}.

Then $\partial f/\partial m=h$ and

\frac{\partial m}{\partial h} =-\left(\frac{\partial^2 g}{\partial h^2}\right) =\frac{1}{V}\mathcal B(M,M)\ge0.

Where this susceptibility is strictly positive the local inverse exists and $f''(m)=1/\chi$ . A zero response requires a constrained or generalized inverse. At phase coexistence the infinite-volume function may not be differentiable. The correct transform is Legendre–Fenchel, and the inverse map can become set-valued. A metastable nonconvex effective potential can remain useful as a coarse-grained approximation, but it is not the exact convex thermodynamic Legendre transform.

A noncommuting two-level check

Let $H=\Delta\sigma_z/2$ , $M=\sigma_x$ and $\Delta>0$ , with no chemical potential. The source-dependent energies are $\pm R$ , where $R=\sqrt{\Delta^2/4+h^2}$ . Direct diagonalization gives

\Xi(h)=2\cosh(\beta R), \qquad \langle M\rangle_h=\frac{h}{R}\tanh(\beta R), \qquad \left.\partial_h\langle M\rangle_h\right|_{h=0} =\frac{2}{\Delta}\tanh\frac{\beta\Delta}{2}.

At zero source $\langle M\rangle=0$ and $M^2=1$ , so the naive equal-time expression would instead give $\beta$ . The two agree to leading order at high temperature, $\beta\Delta\ll1$ , but not generally. The two off-diagonal terms in the spectral sum for $\mathcal B(M,M)$ give exactly $2\tanh(\beta\Delta/2)/\Delta$ . This checks both the imaginary-time integration and its normalization.

Free scalar consistency check

For independent bosonic modes of frequencies $\omega_{\mathbf k}$ ,

\log Z=-\sum_{\mathbf k}\log\left(1-e^{-\beta\omega_{\mathbf k}}\right)

after subtracting the temperature-independent vacuum energy, so that $Z$ here contains only thermal excitations. Their energy is

E=\sum_{\mathbf k}\omega_{\mathbf k}n_B(\omega_{\mathbf k}), \qquad n_B(\omega)=\frac1{e^{\beta\omega}-1}.

Differentiating once more gives a positive heat capacity mode by mode. The same answer follows from the variance of occupation number, $\langle(\Delta n)^2\rangle=n_B(1+n_B)$ , independently checking the sign and the use of $\log Z$ rather than $Z$ as the cumulant generator.

Common pitfalls

Unstated held-fixed data. $\partial_\beta$ at fixed $\mu$ differs from $\partial_\beta$ at fixed $\beta\mu$ . Declare the source coordinates before differentiating.

Intensive–extensive mismatch. A charge variance is extensive away from criticality. Divide by $V$ before comparing it with an intensive susceptibility.

Contact terms omitted. A source can enter both the state and the operator or action explicitly. Second derivatives then contain contact or diamagnetic terms; the connected two-point function alone is not always the complete response.

The shared convention and limit table records which finite-volume definitions and limit statements must remain separate. Spatial correlator representations of the susceptibility continue on Correlations, Susceptibilities, and Correlation Lengths.

The schematic below organizes the relationships used on this page. Inspect it with this question in mind: How do ensembles, partition functions, response derivatives, Legendre transforms, and phase probabilities connect?

The finite-volume partition function generates response, Legendre transforms change controlled variables, and large-deviation rate functions encode phase weights only after the thermodynamic-limit order is declared. Solid connections show the primary relation; dashed outlines or arrows mark qualifications and failure boundaries. The diagram is schematic and not to scale.

The surrounding discussion supplies the relevant equations and checks in text form; the figure is a navigational summary.

Exercise

Show that the grand-canonical number susceptibility of one bosonic mode is $\partial_\mu\langle n\rangle=\beta n(1+n)$ for $\mu<\omega$ .

Solution

With $n=[e^{\beta(\omega-\mu)}-1]^{-1}$ , differentiation gives $\partial_\mu n=\beta e^{\beta(\omega-\mu)}/[e^{\beta(\omega-\mu)}-1]^2=\beta n(1+n)$ . The divergence as $\mu\to\omega^-$ signals loss of stability of the uncondensed single-mode description.

References

Kubo, Ryogo. “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, no. 6 (1957): 570–586. doi:10.1143/JPSJ.12.570; Open PDF.