Thermal States and Partition Functions

Finite-spectrum thermodynamics converts energy levels into normalized Boltzmann weights. Stable arithmetic is straightforward, but incomplete spectra and omitted symmetry multiplicities can bias results while producing smooth temperature curves. This page treats numerical evaluation; thermal density operators explains the statistical interpretation.

Helpful background. Exact diagonalization supplies energies and eigenvectors.

Shifted Boltzmann weights

Assume a finite Hilbert space and positive temperature, with inverse temperature $\beta=1/(k_BT)>0$. For complete energies $E_n$, choose $E_0=\min_nE_n$ and compute

$$w_n=e^{-\beta(E_n-E_0)},\quad Z'= \sum_n w_n,\quad p_n=w_n/Z'.$$

Then

$$\log Z=-\beta E_0+\log Z',\qquad F=E_0-\beta^{-1}\log Z'.$$

The shift prevents overflow at low temperature without changing probabilities. A degeneracy $g_n$ multiplies the weight if each distinct energy appears only once. If eigenvalues are listed with multiplicity, multiplying by $g_n$ again double-counts states.

For an observable, $\langle A\rangle_\beta=\sum_n p_n\langle n|A|n\rangle$. Build the full density matrix only when needed. Noncommuting observables still use their energy-basis diagonal elements for equilibrium expectations.

Derivatives and fluctuations

For a temperature-independent Hamiltonian,

$$U=\sum_np_nE_n,\qquad C_V=k_B\beta^2\left(\langle H^2\rangle-U^2\right).$$

These relations follow from $\partial_\beta p_n=-(E_n-U)p_n$, hence $\partial_\beta U=-\operatorname{Var}(H)$ and $\partial_T\beta=-k_B\beta^2$. A Hamiltonian fitted separately at each temperature has additional derivative terms and does not obey this fluctuation formula without modification.

For stable evaluation, retain the shifted energies $\varepsilon_n=E_n-E_0$, compute $u'=\sum_np_n\varepsilon_n$, and use $U=E_0+u'$ and $\operatorname{Var}(H)=\sum_np_n(\varepsilon_n-u')^2$. This avoids both subtracting large second moments and subtracting a rounded large $U$ from every energy. It cannot recover splittings already lost when the input energies were rounded. Entropy is $S=-k_B\sum_np_n\log p_n$, with $0\log0=0$.

A susceptibility is not always $\beta$ times the equal-time variance: that formula requires the coupled observable to commute with the Hamiltonian, or an appropriate classical limit. For general quantum response use the imaginary-time correlation or spectral Kubo expression in the many-body treatment.

Two-level benchmark

For levels $0,\Delta$ with $\Delta>0$,

$$Z=1+e^{-\beta\Delta},\quad U=\frac{\Delta}{1+e^{\beta\Delta}},\quad C_V=k_B(\beta\Delta)^2 \frac{e^{-\beta\Delta}}{(1+e^{-\beta\Delta})^2}.$$

At low temperature $U$ and $C_V$ vanish; at high temperature $U\to\Delta/2$ and $S\to k_B\log2$. Shifting both levels by a constant changes $U$ and $F$ by that constant but leaves $C_V$, probabilities, and entropy unchanged. This is an excellent numerical invariance check.

At extremely low temperature do not compute $e^{+\beta\Delta}$ in the population formula; use the negative-exponent form $U=\Delta e^{-\beta\Delta}/(1+e^{-\beta\Delta})$.

Incomplete spectra and sectors

If omitted states all lie above $E_{\rm cut}$ and their number is at most $D_{\rm miss}$,

$$Z'_{\rm miss}\leq D_{\rm miss} e^{-\beta(E_{\rm cut}-E_0)}.$$

A gap to the first omitted state alone is insufficient without controlling the multiplicity. Write the retained partition sum as $R$ and the displayed upper bound as $B$. Then the omitted thermal probability obeys

$$q=\frac{Z'_{\rm miss}}{R+Z'_{\rm miss}} \leq\frac{B}{R+B}.$$

The complete state decomposes as $\rho=(1-q)\rho_{\rm ret}+q\rho_{\rm miss}$ on orthogonal spectral subspaces. Its trace distance from the normalized retained state is exactly $q$. Therefore

$$|\langle A\rangle-\langle A\rangle_{\rm ret}| \leq 2\|A\|q,\qquad 0\leq F_{\rm ret}-F \leq\beta^{-1}\log(1+B/R).$$

Evaluate the last logarithm with a log1p operation when $B/R$ is small. These bounds assume that the retained eigenpairs and the omitted-state threshold are themselves reliable; they do not include eigensolver or energy-estimation errors.

For retained levels $0,1$ at $\beta=2$, and at most two omitted states above energy $4$, $R=1+e^{-2}$ and $B=2e^{-8}$. The bound is $q<5.91\times10^{-4}$, giving an expectation error below $1.182\times10^{-3}$ for $\|A\|\leq1$. This makes the temperature range of a truncated calculation quantitatively assessable.

Sum over all physical symmetry sectors required by the ensemble. A canonical fixed-number calculation and a grand-canonical sum with weights $e^{-\beta(E-\mu N)}$ answer different questions. At high temperature, $\operatorname{Tr}\rho=1$ will hold even for an incorrectly omitted sector; check $Z(\beta=0)=D$ instead.

For larger spaces, finite-temperature Lanczos and stochastic trace methods introduce additional Krylov and sampling errors. They should be benchmarked against complete small-system spectra.

Exercise

A ground state is $g$-fold degenerate and separated by a gap from all other states. What is the zero-temperature entropy of the canonical Gibbs limit?

Solution

The Gibbs state approaches the uniform mixture on the ground-state subspace, so $S\to k_B\log g$. Choosing one ground-state eigenvector instead produces a pure state and zero entropy; that is a different limiting preparation. Numerical splitting smaller than the physical resolution can spuriously select one member if the temperature is taken too low first.

References

• Jaklič, J., and Prelovšek, P., “Lanczos method for the calculation of finite-temperature quantities in correlated systems,” Physical Review B 49, 5065–5068 (1994), doi:10.1103/PhysRevB.49.5065.
• Weinberg, P., and Bukov, M., “QuSpin. Part I: spin chains,” SciPost Physics 2, 003 (2017), arXiv:1610.03042 — finite-spectrum calculations.