Spectral Representation

A thermal spectral representation rewrites an equilibrium two-point function as a transform of exact transition energies and thermally weighted matrix elements.

For the propagator convention below, the following identity describes fermionic channels and the dynamical part of bosonic channels. An exact bosonic zero-energy fluctuation must be added separately at zero Matsubara frequency:

$$\mathcal G_{AB}(i\zeta_\ell) = \int_{-\infty}^{\infty} dE\, \frac{ \rho_{AB}^{(\eta)}(E) }{ i\zeta_\ell-E }.$$

Here $\eta=+1$ for an even, bosonic channel and $\eta=-1$ for an odd, fermionic channel. The same spectral density defines an analytic function

$$\mathcal G_{AB}(z) = \int_{-\infty}^{\infty} dE\, \frac{ \rho_{AB}^{(\eta)}(E) }{ z-E },$$

whose imaginary-axis samples are Matsubara values and whose upper real-axis boundary is retarded:

$$\mathcal G_{AB}^{\mathrm R}(E) = \lim_{\epsilon\to0^+} \mathcal G_{AB}(E+i\epsilon).$$

This compact bridge is exact under its stated assumptions. Its content is not the symbolic replacement $i\zeta_\ell\mapsto E+i0^+$ by itself. The real work lies in defining the channel, thermal weights, operator order, overall sign, static terms, and analytic class correctly.

Helpful background. Thermal Green functions defines ordering and endpoint jumps; Matsubara frequencies fixes the grids and units. This page derives the exact transform; analytic continuation treats inference from finite noisy data.

Thermal spectra and their assumptions

The derivation assumes:

1. a positive-temperature equilibrium state generated by one self-adjoint operator $\mathcal K$;
2. a trace or thermodynamic-limit KMS state for which the stated correlators exist;
3. time-translation invariance;
4. operators with definite fermion parity;
5. a declared Fourier normalization;
6. enough spectral integrability, or suitable subtractions, for each transform used.

No weak-coupling, quasiparticle, perturbative, Gaussian, or thermodynamic-limit assumption is required. A finite interacting system has an exact spectral representation just as a free system does. Its spectral measure is generally a sum of delta functions rather than a smooth curve.

Operator, spectral, and transform conventions

Thermal generator

Use

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

For a grand-canonical state with $[H,\hat N]=0$,

$$\mathcal K = H-\mu\hat N.$$

Let

$$\mathcal K|n\rangle = K_n|n\rangle, \qquad p_n = \frac{ e^{-\beta K_n} }{ \mathcal Z }.$$

Imaginary-time and real-time operators evolve with the same generator:

$$A(\tau) = e^{\tau\mathcal K/\hbar} A e^{-\tau\mathcal K/\hbar},$$

and

$$A(t) = e^{i\mathcal Kt/\hbar} A e^{-i\mathcal Kt/\hbar}.$$

If real time instead uses $H$, number-changing transitions acquire an explicit chemical-work shift, derived below. For number-changing fields the Gibbs trace here runs over the full Fock space. Fixed-$N$ canonical traces obey the simple cyclic KMS argument only for operators that preserve the traced sector.

Thermal interval

Keep $\tau$ in time units:

$$L_\tau := \beta\hbar, \qquad 0 \leq \tau \lt L_\tau.$$

The inverse temperature $\beta$ has inverse-energy units. Matsubara variables on this page have energy units.

Graded channel

Let $A$ and $B$ have the same definite parity, and define

$$[A,B]_\eta := AB-\eta BA,$$

with

$$\eta = \begin{cases} +1, & \text{even or bosonic channel},\\ -1, & \text{odd or fermionic channel}. \end{cases}$$

Thus $[A,B]_{+1}$ is a commutator and $[A,B]_{-1}$ is an anticommutator.

The derivation does not assign statistics from the microscopic particles alone. A density, spin, current, or pair bilinear made from fermions is an even operator and therefore uses a bosonic thermal boundary condition.

Matrix elements and transfer energy

Write

$$A_{nm} := \langle n|A|m\rangle, \qquad B_{mn} := \langle m|B|n\rangle,$$

and define the energy transferred to the state by the insertion:

$$E_{mn} := K_m-K_n.$$

Positive $E_{mn}$ raises the eigenvalue of $\mathcal K$. In a number-changing channel this is an energy measured relative to chemical work.

Matsubara transform

For a periodic or antiperiodic function,

$$\mathcal G(i\zeta_\ell) = \frac1\hbar \int_0^{L_\tau} d\tau\, e^{i\zeta_\ell\tau/\hbar} \mathcal G(\tau),$$

with inverse

$$\mathcal G(\tau) = \frac1\beta \sum_\ell e^{-i\zeta_\ell\tau/\hbar} \mathcal G(i\zeta_\ell).$$

The grid obeys

$$e^{i\beta\zeta_\ell} = \eta.$$

Explicitly,

$$\zeta_\ell = \begin{cases} \Omega_\ell=2\pi\ell/\beta, & \eta=+1,\\ \nu_\ell=(2\ell+1)\pi/\beta, & \eta=-1. \end{cases}$$

Two overall-sign conventions

This page first uses the propagator convention

$$\mathcal G_{AB}(\tau) = - \left\langle \mathcal T_\tau A(\tau)B(0) \right\rangle_\beta.$$

It pairs naturally with

$$\mathcal G_{AB}^{\mathrm R}(t) = - \frac{i}{\hbar} \theta(t) \left\langle [A(t),B(0)]_\eta \right\rangle_\beta.$$

For an ordinary bosonic observable correlator, many authors instead define

$$C_{AB}(\tau) = + \left\langle \mathcal T_\tau \delta A(\tau)\delta B(0) \right\rangle_\beta.$$

With the source convention used in Retarded and Advanced Response, its susceptibility begins with $+i\theta(t)/\hbar$. The two conventions contain the same transition data but differ by an overall sign and denominator orientation. They are reconciled explicitly below.

Ordered Lehmann Sum

For

$$0 \lt \tau \lt \beta\hbar,$$

imaginary-time ordering leaves $A(\tau)$ to the left. Insert a complete energy eigenbasis:

$$\begin{aligned} \mathcal G_{AB}(\tau) ={}& - \sum_{n,m} p_n \langle n|A(\tau)|m\rangle B_{mn} \\ ={}& - \sum_{n,m} p_n e^{-E_{mn}\tau/\hbar} A_{nm}B_{mn}. \end{aligned}$$

This is the thermal Lehmann sum in imaginary time. Each term contains three logically separate ingredients:

• $E_{mn}$ fixes the exponential scale;
• $A_{nm}B_{mn}$ fixes visibility in the chosen channel;
• $p_n$ fixes how often the initial state occurs in the Gibbs ensemble.

Interactions change all three through the exact many-body eigenstates. The representation itself remains valid.

Ordered Transition Spectra

Define the forward ordered transition spectrum

$$\mathcal S_{AB}^{>}(E) :={} \sum_{n,m} p_n A_{nm}B_{mn} \times \delta \left( E-E_{mn} \right).$$

Then

$$\mathcal G_{AB}(\tau) = - \int_{-\infty}^{\infty} dE\, e^{-E\tau/\hbar} \mathcal S_{AB}^{>}(E).$$

The reversed operator order can be expressed on the same energy support:

$$\mathcal S_{AB}^{<}(E) :={} \sum_{n,m} p_m A_{nm}B_{mn} \times \delta \left( E-E_{mn} \right).$$

On the support of the delta function,

$$\frac{p_m}{p_n} = e^{-\beta E}.$$

Therefore

$$\mathcal S_{AB}^{<}(E) = e^{-\beta E} \mathcal S_{AB}^{>}(E).$$

This is the frequency-domain KMS balance needed for the present derivation. The full fluctuation–dissipation dictionary, including symmetrized spectra and Bose factors, remains in its dedicated page.

For $B=A^\dagger$,

$$\mathcal S_{AA^\dagger}^{>}(E) \geq 0$$

as a distribution. This positivity belongs to the ordered transition spectrum. A commutator spectral density need not be nonnegative.

Graded Spectral Density

Define

$$\rho_{AB}^{(\eta)}(E) := \mathcal S_{AB}^{>}(E) - \eta \mathcal S_{AB}^{<}(E).$$

Using KMS balance,

$$\rho_{AB}^{(\eta)}(E) = \left( 1-\eta e^{-\beta E} \right) \mathcal S_{AB}^{>}(E).$$

Equivalently, the exact Lehmann form is

$$\rho_{AB}^{(\eta)}(E) ={} \sum_{n,m} \left( p_n-\eta p_m \right) A_{nm}B_{mn} \times \delta \left( E-K_m+K_n \right).$$

The word spectral density is incomplete unless $\eta$, the operator pair, energy variable, generator, and normalization have been declared.

Fermionic weight

For $\eta=-1$,

$$p_n-\eta p_m = p_n+p_m.$$

Every conjugate-channel coefficient is nonnegative. A normal fermionic single-particle spectral matrix is therefore positive semidefinite, and canonical anticommutation fixes its zeroth moment.

The sum $p_n+p_m$ is not itself a Fermi–Dirac function. In an interacting many-body problem it combines exact Gibbs probabilities from adjacent particle-number sectors. The familiar Fermi factor emerges when the ordered spectra are reconstructed from the full spectral function.

Bosonic weight

For $\eta=+1$,

$$p_n-\eta p_m = p_n-p_m = p_n \left( 1-e^{-\beta E} \right).$$

For $B=A^\dagger$ in a positive-temperature Gibbs state, the sign of this coefficient follows the sign of $E$. Consequently,

$$E\, \rho_{AA^\dagger}^{(+)}(E) \geq 0$$

in a diagonal channel, apart from distributional qualifications at $E=0$. Negative-energy bosonic commutator weight is normally negative. Applying fermionic positivity to it is an error.

Temperature dependence

The exact transition energies and matrix elements of a fixed $\mathcal K$ do not depend on temperature, but their ensemble weights do. Thus a thermal spectral density can change with temperature even before self-consistent parameters, thermal expansion, or a temperature-dependent effective Hamiltonian are introduced.

Special quadratic examples can hide this dependence. For one free fermion or boson mode, Gibbs factors cancel in the graded spectral density, leaving a unit delta function. That cancellation is a benchmark, not a universal theorem.

Imaginary-Time Spectral Kernel

Away from a bosonic zero-energy singularity,

$$\mathcal S_{AB}^{>}(E) = \frac{ \rho_{AB}^{(\eta)}(E) }{ 1-\eta e^{-\beta E} }.$$

The imaginary-time representation becomes

$$\mathcal G_{AB}(\tau) = - \int_{-\infty}^{\infty} dE\, K_\eta(\tau,E) \rho_{AB}^{(\eta)}(E),$$

where

$$K_\eta(\tau,E) := \frac{ e^{-E\tau/\hbar} }{ 1-\eta e^{-\beta E} }, \qquad 0\lt\tau\lt\beta\hbar.$$

This kernel is the direct bridge from real-energy spectral weight to Euclidean data.

Fermionic kernel

For $\eta=-1$,

$$K_{\mathrm F}(\tau,E) = \frac{ e^{-E\tau/\hbar} }{ 1+e^{-\beta E} }.$$

Using

$$f(E) = \frac1{ e^{\beta E}+1 },$$

one may write

$$K_{\mathrm F}(\tau,E) = \left[ 1-f(E) \right] e^{-E\tau/\hbar}.$$

Positive-energy addition weight is emphasized near $\tau=0^+$; negative-energy removal weight becomes visible near $\tau=\beta\hbar^-$. Both branches are needed for a balanced reconstruction.

Bosonic kernel

For $\eta=+1$,

$$K_{\mathrm B}(\tau,E) = \frac{ e^{-E\tau/\hbar} }{ 1-e^{-\beta E} }.$$

The denominator vanishes linearly at $E=0$. For regular dynamic weight, the numerator of the bosonic spectral density also vanishes linearly, so their ratio can remain finite. An exact zero-energy delta contribution is different: multiplication by $1-e^{-\beta E}$ removes it from the commutator spectrum. That static information must be retained separately.

Why Euclidean data are smooth

The kernel integrates spectral information over all energies. Fine real-axis structures can therefore produce very similar functions of $\tau$, especially over a finite noisy grid. Exact forward transformation is stable; inversion is not.

This is an information statement, not a claim that imaginary-time data are unphysical. Euclidean correlators are exact equilibrium observables of the thermal formalism, and their moments, endpoints, symmetries, and parameter dependence can be highly constraining.

Matsubara Representation

The discrete frequency formula follows directly from the kernel. First exclude the simultaneous bosonic case $E=0$, $\Omega_\ell=0$: the elementary integral there is $\beta$, although the quotient below has the indeterminate form $0/0$. Its contribution is the static term derived later. Begin with

$$\mathcal G_{AB}(i\zeta_\ell) ={} - \int_{-\infty}^{\infty} dE\, \mathcal S_{AB}^{>}(E) \times \frac1\hbar \int_0^{\beta\hbar} d\tau\, e^{(i\zeta_\ell-E)\tau/\hbar}.$$

The elementary integral is

$$\frac1\hbar \int_0^{\beta\hbar} d\tau\, e^{(i\zeta_\ell-E)\tau/\hbar} = \frac{ e^{\beta(i\zeta_\ell-E)}-1 }{ i\zeta_\ell-E }.$$

The thermal boundary condition gives

$$e^{i\beta\zeta_\ell} = \eta.$$

Therefore

$$- \left[ e^{\beta(i\zeta_\ell-E)}-1 \right] = 1-\eta e^{-\beta E},$$

and hence

$$\mathcal G_{AB}(i\zeta_\ell) = \int_{-\infty}^{\infty} dE\, \frac{ \rho_{AB}^{(\eta)}(E) }{ i\zeta_\ell-E }.$$

The statistics appear twice and consistently:

• in the thermal factor $p_n-\eta p_m$ inside $\rho^{(\eta)}$;
• in the allowed discrete values of $\zeta_\ell$.

Changing one without the other destroys the derivation.

The Common Analytic Function

For complex $z$ away from the real spectral support, define

$$\mathcal G_{AB}(z) = \int_{-\infty}^{\infty} dE'\, \frac{ \rho_{AB}^{(\eta)}(E') }{ z-E' }.$$

Under the stated integrability conditions, this Cauchy transform is analytic off its real spectral support. Its dynamical Matsubara values are

$$\mathcal G_{AB}(i\zeta_\ell).$$

Approaching the real axis from above gives

$$\mathcal G_{AB}^{\mathrm R}(E) = \mathcal G_{AB}(E+i0^+),$$

while approaching from below gives

$$\mathcal G_{AB}^{\mathrm A}(E) = \mathcal G_{AB}(E-i0^+).$$

The statement

$$i\zeta_\ell \longrightarrow E+i0^+$$

means: identify the analytic function fixed by the physics, then evaluate its retarded boundary value. It does not mean that a finite list of samples can be converted by textual substitution.

The matrix element $B_{mn}$ takes $|n\rangle$ to $|m\rangle$, while $A_{nm}$ returns it. The transition energy, operator product, and Gibbs populations determine the spectral weight. A Cauchy transform produces the nonstatic Matsubara values and the retarded boundary. An exact bosonic static term remains separate at zero Matsubara frequency; the commutator measure does not determine it.

Retarded Boundary Value

For the propagator convention,

$$\mathcal G_{AB}^{\mathrm R}(t) = - \frac{i}{\hbar} \theta(t) \left\langle [A(t),B(0)]_\eta \right\rangle_\beta.$$

Fourier transformation in energy gives

$$\mathcal G_{AB}^{\mathrm R}(E) = \int_{-\infty}^{\infty} dE'\, \frac{ \rho_{AB}^{(\eta)}(E') }{ E-E'+i0^+ }.$$

Similarly,

$$\mathcal G_{AB}^{\mathrm A}(E) = \int_{-\infty}^{\infty} dE'\, \frac{ \rho_{AB}^{(\eta)}(E') }{ E-E'-i0^+ }.$$

Using

$$\frac1{x+i0^+} = \operatorname{PV} \frac1x - i\pi\delta(x),$$

the discontinuity is

$$\mathcal G^{\mathrm R}(E) - \mathcal G^{\mathrm A}(E) = -2\pi i\, \rho^{(\eta)}(E).$$

Thus

$$\rho^{(\eta)}(E) = \frac{i}{2\pi} \left[ \mathcal G^{\mathrm R}(E) - \mathcal G^{\mathrm A}(E) \right].$$

For a scalar conjugate channel with the usual reality properties,

$$\rho^{(\eta)}(E) = - \frac1\pi \operatorname{Im} \mathcal G^{\mathrm R}(E).$$

The infinitesimal $i0^+$ selects causal support. It is not a linewidth. Replacing it by finite $\gamma$ is a separate physical, limiting, or numerical assumption.

Propagator and observable-response signs

Two sign conventions recur often enough to deserve a direct dictionary.

Feature	Propagator convention	Observable-response convention
imaginary-time object	$\mathcal G=-\langle\mathcal T_\tau AB\rangle$	$C=+\langle\mathcal T_\tau\delta A\delta B\rangle$
real-time object	$\mathcal G^{\mathrm R}=-(i/\hbar)\theta\langle[A,B]_\eta\rangle$	$\chi^{\mathrm R}=+(i/\hbar)\theta\langle[\delta A,\delta B]\rangle$
Matsubara kernel	$\rho/(i\zeta-E)$	$\rho/(E-i\Omega)$
retarded kernel	$\rho/(E_{\rm ext}-E+i0^+)$	$\rho/(E-E_{\rm ext}-i0^+)$
common use	single-particle propagation	response to a classical source

For a bosonic observable channel with no static term,

$$C_{AB}(i\Omega_\ell) = \int_{-\infty}^{\infty} dE\, \frac{ \rho_{AB}^{(+)}(E) }{ E-i\Omega_\ell }.$$

For the source coupling $H^{\prime}=-fB$,

$$\chi_{AB}^{\mathrm R}(E_{\rm ext}) = \int_{-\infty}^{\infty} dE\, \frac{ \rho_{AB}^{(+)}(E) }{ E-E_{\rm ext}-i0^+ }.$$

These expressions are the same analytic function in the observable-response convention. Since $C=-\mathcal G$ for the same bosonic operator pair, there is no physical contradiction.

A formula copied between the two columns without its defining sign will appear to disagree by a minus sign even when both sources are internally correct.

Bosonic Static Term

The most important qualification to the compact bosonic formula occurs at exactly zero transfer energy.

For connected bosonic observables, define the degenerate-transition weight

$$\begin{aligned} \Gamma_{AB} :={}& \sum_{\substack{n,m\\K_m=K_n}} p_n (\delta A)_{nm} (\delta B)_{mn}. \end{aligned}$$

These terms contribute a constant to the positive-sign Euclidean correlator:

$$C_{AB}^{\mathrm{static}}(\tau) = \Gamma_{AB}.$$

Its Matsubara transform is

$$C_{AB}^{\mathrm{static}}(i\Omega_\ell) = \beta\Gamma_{AB} \delta_{\ell0}.$$

But the bosonic commutator weight contains

$$p_n-p_m = 0$$

whenever $K_n=K_m$. The commutator spectral density therefore loses this exact static contribution.

Let $\rho_{AB,\mathrm{dyn}}^{(+)}$ denote the contribution from nondegenerate transitions. A complete observable representation is

$$C_{AB}(i\Omega_\ell) ={} \int_{-\infty}^{\infty} dE\, \frac{ \rho_{AB,\mathrm{dyn}}^{(+)}(E) }{ E-i\Omega_\ell } + \beta\Gamma_{AB} \delta_{\ell0},$$

The dynamic measure may have regular or continuum support through $E=0$. Only the exact degenerate contribution $\Gamma_{AB}$ is stored separately.

Connected does not mean dynamic

Subtracting

$$\delta A = A-\langle A\rangle$$

removes the disconnected product $\langle A\rangle\langle B\rangle$. It does not remove fluctuations inside a degenerate subspace or fluctuations of a conserved quantity.

If $Q$ commutes with $\mathcal K$, then

$$C_{QQ}(\tau) = \left\langle (\delta Q)^2 \right\rangle_\beta$$

is constant, while

$$\rho_{QQ}^{(+)}(E) = 0.$$

Consequently,

$$C_{QQ}(i\Omega_\ell) = \beta \operatorname{Var}_\beta(Q) \delta_{\ell0}.$$

The dynamic commutator response and the equilibrium static fluctuation answer different limiting questions. Discarding the Matsubara zero mode would erase the entire correlator in this example.

Why fermions differ

For a fermionic channel,

$$1+e^{-\beta E} \longrightarrow 2$$

as $E\to0$. A zero-energy fermionic transition is not annihilated by the graded anticommutator weight. The special missing-static-term mechanism is therefore bosonic.

High-Frequency Moments

For large $|z|$, formally write

$$\frac1{z-E} = \frac1z \sum_{r=0}^{\infty} \left( \frac{E}{z} \right)^r$$

and interpret the result as an asymptotic moment expansion when the required moments exist:

$$\frac1{z-E} = \frac1z + \frac{E}{z^2} + \frac{E^2}{z^3} + \cdots.$$

Then

$$\mathcal G_{AB}(z) \sim \frac{M_0}{z} + \frac{M_1}{z^2} + \frac{M_2}{z^3} + \cdots,$$

with

$$M_r := \int_{-\infty}^{\infty} dE\, E^r \rho_{AB}^{(\eta)}(E).$$

The first two moments are

$$M_0 = \left\langle [A,B]_\eta \right\rangle_\beta,$$

and

$$M_1 = \left\langle \left[ [A,\mathcal K], B \right]_\eta \right\rangle_\beta.$$

Higher moments contain further commutators with $\mathcal K$. They connect equal-time algebra to large-Matsubara-frequency tails.

For a canonical fermionic orbital,

$$M_0 = \left\langle \{c,c^\dagger\} \right\rangle = 1,$$

so

$$\mathcal G(i\nu_\ell) \sim \frac1{i\nu_\ell}$$

at large $|\nu_\ell|$. A computed tail with the wrong coefficient indicates missing spectral weight, a transform error, or a violated canonical algebra.

The full hierarchy of nested-commutator constraints belongs to Sum Rules.

Worked example: temperature-dependent interaction peaks

Consider two spinless fermionic orbitals with no hopping and

$$\mathcal K=\epsilon_1 n_1+\epsilon_2 n_2+Un_1n_2, \qquad n_j=c_j^\dagger c_j.$$

The four Fock states have energies $0$, $\epsilon_1$, $\epsilon_2$, and $\epsilon_1+\epsilon_2+U$. Adding a fermion to orbital 1 costs $\epsilon_1$ when orbital 2 is empty and $\epsilon_1+U$ when orbital 2 is occupied. Summing the addition and removal thermal weights gives

$$\rho_1^{(-)}(E) =(1-\bar n_2)\delta(E-\epsilon_1) +\bar n_2\delta(E-\epsilon_1-U),$$

where

$$\begin{aligned} \mathcal Z&=1+e^{-\beta\epsilon_1}+e^{-\beta\epsilon_2} +e^{-\beta(\epsilon_1+\epsilon_2+U)},\\ \bar n_2&=\frac{e^{-\beta\epsilon_2} +e^{-\beta(\epsilon_1+\epsilon_2+U)}}{\mathcal Z}. \end{aligned}$$

Each line is nonnegative and their weights sum to one. The resolvent is

$$\mathcal G_1(z)=\frac{1-\bar n_2}{z-\epsilon_1} +\frac{\bar n_2}{z-\epsilon_1-U}.$$

For $\epsilon_1,\epsilon_2,U>0$, cooling to zero temperature empties orbital 2, leaving only the first peak. At infinite temperature $\bar n_2\to1/2$, so the peaks have equal weights. Their energies remain fixed while their thermal weights change. If $U=0$, they coincide and recover the free result $\mathcal G_1(z)=1/(z-\epsilon_1)$ for every temperature.

The first spectral moment is $\epsilon_1+U\bar n_2$, matching the equal-time anticommutator of $[c_1,\mathcal K]=(\epsilon_1+Un_2)c_1$ with $c_1^\dagger$.

Worked Benchmark: One Boson Mode

Let

$$\mathcal K = \epsilon b^\dagger b, \qquad \epsilon\gt0.$$

For $A=b$ and $B=b^\dagger$,

$$\mathcal S^{>}(E) = \left( 1+n_{\mathrm B}(\epsilon) \right) \delta(E-\epsilon),$$

while

$$\mathcal S^{<}(E) = n_{\mathrm B}(\epsilon) \delta(E-\epsilon).$$

Here

$$n_{\mathrm B}(\epsilon) = \frac1{ e^{\beta\epsilon}-1 }.$$

The bosonic commutator spectrum is

$$\rho^{(+)}(E) = \delta(E-\epsilon).$$

Again, the free-mode thermal factors cancel in the graded difference. The leading-minus propagator is

$$\mathcal D(\tau) = - \left( 1+n_{\mathrm B} \right) e^{-\epsilon\tau/\hbar},$$

for $0\lt\tau\lt\beta\hbar$, and

$$\mathcal D(i\Omega_\ell) = \frac1{ i\Omega_\ell-\epsilon }.$$

Its retarded boundary value is

$$\mathcal D^{\mathrm R}(E) = \frac1{ E-\epsilon+i0^+ }.$$

The stability condition $\epsilon\gt0$ is essential. At $\epsilon=0$, the grand-canonical occupation diverges and the one-mode Gibbs trace does not exist without further physics.

Worked Benchmark: Two-Level Observable

Consider

$$\mathcal K = \Delta|e\rangle\langle e|, \qquad \Delta\gt0,$$

with

$$A = |g\rangle\langle e| + |e\rangle\langle g|.$$

The Gibbs probabilities are

$$p_g = \frac1{ 1+e^{-\beta\Delta} }, \qquad p_e = \frac{ e^{-\beta\Delta} }{ 1+e^{-\beta\Delta} }.$$

The ordered spectrum is

$$\mathcal S^{>}_{AA}(E) = p_g\delta(E-\Delta) + p_e\delta(E+\Delta).$$

It obeys detailed balance because the negative-energy line has the smaller Boltzmann weight. The commutator spectral density is

$$\rho_{AA}^{(+)}(E) ={} \left( p_g-p_e \right) \times \left[ \delta(E-\Delta) - \delta(E+\Delta) \right].$$

Since

$$p_g-p_e = \tanh \left( \frac{\beta\Delta}{2} \right),$$

the response weight vanishes at infinite temperature and approaches its ground-state value at low temperature.

Use the positive observable correlator

$$C_{AA}(\tau) = \left\langle \mathcal T_\tau A(\tau)A(0) \right\rangle.$$

For $0\lt\tau\lt\beta\hbar$,

$$C_{AA}(\tau) = p_g e^{-\Delta\tau/\hbar} + p_e e^{+\Delta\tau/\hbar}.$$

Its Matsubara transform is

$$C_{AA}(i\Omega_\ell) = \frac{ 2\Delta \tanh(\beta\Delta/2) }{ \Omega_\ell^2+\Delta^2 }.$$

At zero Matsubara frequency,

$$C_{AA}(i\Omega_0) = \frac{ 2 \tanh(\beta\Delta/2) }{ \Delta }.$$

This is a dynamic, nondegenerate example: no extra $\Gamma$ term is needed.

Chemical Potential and Energy Labels

When both the Gibbs state and time evolution use

$$\mathcal K = H-\mu\hat N,$$

the transfer energy is

$$E_{mn}^{(\mathcal K)} = \left( E_m^{(H)}-E_n^{(H)} \right) - \mu \left( N_m-N_n \right).$$

The KMS ratio is simply

$$\frac{p_m}{p_n} = e^{-\beta E_{mn}^{(\mathcal K)}}.$$

If operators evolve with $H$ instead, let

$$E_{mn}^{(H)} = E_m^{(H)}-E_n^{(H)},$$

and

$$\Delta N_{mn} = N_m-N_n.$$

Then

$$\frac{p_m}{p_n} = \exp \left[ -\beta \left( E_{mn}^{(H)} - \mu\Delta N_{mn} \right) \right].$$

For number-preserving observables, $\Delta N_{mn}=0$ and the two energy labels coincide. For particle addition or removal, the distinction shifts the spectral origin by chemical work. A spectral plot should state which generator defines zero energy.

Finite and Infinite Systems

For a finite closed system with a discrete spectrum,

$$\rho(E) = \sum_j w_j\delta(E-E_j).$$

The exact retarded function has distributional real-axis poles. Interactions can split and redistribute the residues, but they do not by themselves give exact finite-system eigenstates a decay width.

Smooth spectral weight can emerge through:

• a thermodynamic limit with dense levels;
• coupling to multiparticle continua;
• a genuinely open system;
• disorder or ensemble averaging;
• a declared numerical broadening;
• instrumental convolution.

These mechanisms are physically different. The spectral representation accommodates all of them through a measure, but it does not identify their origin automatically.

Numerical Evaluation

Exact diagonalization

For a finite basis:

1. diagonalize the same $\mathcal K$ used in the Gibbs state;
2. compute $E_{mn}$ and the relevant matrix elements;
3. exploit conserved quantum numbers to skip forbidden transitions;
4. accumulate $\mathcal S^>$ before forming graded differences;
5. keep exact zero-energy bosonic weight in a separate accumulator;
6. verify moments before applying any plotting kernel.

This ordering preserves positivity of the ordered spectrum and avoids subtracting nearly equal large numbers too early.

Stable thermal differences

For a bosonic transition,

$$p_n-p_m = p_n \left( 1-e^{-\beta E} \right).$$

When $|\beta E|\ll1$, evaluate

$$1-e^{-\beta E} = - \operatorname{expm1}(-\beta E)$$

numerically. Direct subtraction loses relative precision.

For fermionic weights $p_n+p_m$, use scaled exponentials or a log-sum-exp strategy when the spectrum spans many thermal decades.

Kernel evaluation

The bosonic factor

$$\frac{ \rho(E) }{ 1-e^{-\beta E} }$$

should be evaluated from a known regular limit near $E=0$, not by dividing two independently noisy arrays. An exact $\delta(E)$ term requires separate symbolic or discrete bookkeeping.

Broadening

Replacing

$$\delta(E-E_j)$$

with a Lorentzian or Gaussian may help visualization. Record the kernel, width, normalization, and whether moments are preserved. Never feed a broadened plot back into an exact sum rule without accounting for the tails and finite integration window.

Continuation boundary

On a regulated spectral domain with the static term separated, forward evaluation of $\mathcal G(\tau)$ or $\mathcal G(i\zeta_\ell)$ is a direct integration problem. Stable thermal-factor evaluation still matters near zero energy or across a large dynamic range. Recovering $\rho(E)$ from finitely many noisy values is an ill-posed inverse problem. Positivity, moments, symmetry, and causality constrain that inversion but do not manufacture resolution absent from the data. Analytic Continuation develops covariance whitening, singular-value diagnostics, regularization families, and tests of which spectral features the data resolve.

Interpreting a thermal spectrum

An excitation contributes only if the operator pair has a nonzero matrix element. A missing line can therefore reflect a selection rule, while a temperature-dependent weight can reflect changing initial populations without a changing Hamiltonian.

The energy in $e^{-E\tau/\hbar}$ is a transition energy, not a decay rate in laboratory time. Similarly, the infinitesimal retarded prescription fixes causal support; it does not provide an intrinsic linewidth. Static thermal fluctuations and dynamical commutator response should be compared only after specifying the physical limiting procedure.

Common pitfalls

The signed spectrum depends on the bracket. Fermionic anticommutators give $p_n+p_m$, while bosonic commutators give $p_n-p_m$. Their positivity properties differ even though both enter a Cauchy transform.

A zero divided by zero can hide the whole answer. An exactly conserved fluctuation has no commutator spectral weight but a nonzero static Euclidean correlation. Keep that term before dividing by $1-e^{-\beta E}$.

A few exact moments do not determine a unique line shape. They constrain any continuation, but sharply different real-frequency spectra can still fit finite uncertain imaginary-time data.

Further spectral calculations

• Single-particle Green functions interprets addition/removal weights and self-energies.
• Fluctuation–dissipation theorem relates ordered noise and absorption.
• Sum rules derives higher nested-commutator constraints.
• Resolvent spectral representation develops the spectral theorem for one Hamiltonian.

Exercises

1. Derive the Matsubara denominator

Starting from

$$\mathcal G(\tau) = - \int dE\, e^{-E\tau/\hbar} \mathcal S^>(E),$$

derive the Matsubara representation for a channel satisfying

$$e^{i\beta\zeta_\ell} = \eta.$$

Identify the step at which the graded thermal weight appears. For a bosonic channel, separate any exact zero-energy line before applying the quotient at $\Omega_0=0$.

Solution

Transform over one thermal interval:

$$\mathcal G(i\zeta_\ell) ={} - \int dE\, \mathcal S^>(E) \times \frac1\hbar \int_0^{\beta\hbar} d\tau\, e^{(i\zeta_\ell-E)\tau/\hbar}.$$

The integral is

$$\frac{ e^{\beta(i\zeta_\ell-E)}-1 }{ i\zeta_\ell-E }.$$

Using $e^{i\beta\zeta_\ell}=\eta$,

$$\begin{aligned} \mathcal G(i\zeta_\ell) &= \int dE\, \frac{ \left( 1-\eta e^{-\beta E} \right) \mathcal S^>(E) }{ i\zeta_\ell-E } \\ &= \int dE\, \frac{ \rho^{(\eta)}(E) }{ i\zeta_\ell-E }. \end{aligned}$$

The graded thermal factor enters when the endpoint exponential is replaced by the channel’s periodic or antiperiodic boundary phase. A bosonic ordered line $W_0\delta(E)$ contributes $-\beta W_0\delta_{\ell0}$ to this leading-minus propagator and is not included in the commutator-density quotient.

2. Recover the free fermion branches

For

$$\rho^{(-)}(E) = \delta(E-\xi),$$

derive $\mathcal G(\tau)$ on $0\lt\tau\lt\beta\hbar$, its $0^+$ and $\beta\hbar^-$ limits, and $\mathcal G(i\nu_\ell)$.

Solution

The fermionic kernel gives

$$\mathcal G(\tau) = - \frac{ e^{-\xi\tau/\hbar} }{ 1+e^{-\beta\xi} }.$$

Since

$$1-f(\xi) = \frac1{ 1+e^{-\beta\xi} },$$

the positive-time branch is

$$\mathcal G(\tau) = - \left[ 1-f(\xi) \right] e^{-\xi\tau/\hbar}.$$

Therefore

$$\mathcal G(0^+) = - \left[ 1-f(\xi) \right],$$

and

$$\mathcal G(\beta\hbar^-) = -f(\xi).$$

Antiperiodicity identifies the latter with $-\mathcal G(0^-)$. Thus

$$\mathcal G(0^-) = f(\xi).$$

The spectral Cauchy transform gives

$$\mathcal G(i\nu_\ell) = \frac1{ i\nu_\ell-\xi }.$$

The contact jump is

$$\mathcal G(0^+) - \mathcal G(0^-) = -1.$$

3. Check two-level detailed balance

For the two-level observable in the worked example, compute the ratio of the integrated line weights $W_-$ at $-\Delta$ and $W_+$ at $+\Delta$, denoted

$$\frac{W_-}{W_+}$$

and show that the commutator spectrum is odd.

Solution

The two ordered weights are

$$W_+ = p_g, \qquad W_- = p_e.$$

Their ratio is

$$\frac{p_e}{p_g} = e^{-\beta\Delta}.$$

This is detailed balance. The commutator density is

$$\rho^{(+)}(E) ={} \left( p_g-p_e \right) \delta(E-\Delta) - \left( p_g-p_e \right) \delta(E+\Delta).$$

Replacing $E$ by $-E$ exchanges the delta functions and contributes a minus sign:

$$\rho^{(+)}(-E) = -\rho^{(+)}(E).$$

4. Diagnose a conserved zero mode

Let $Q$ be Hermitian and satisfy

$$[Q,\mathcal K] = 0.$$

For $\delta Q=Q-\langle Q\rangle$, find the positive Euclidean correlator, its Matsubara transform, and its commutator spectral density.

Solution

Conservation gives

$$\delta Q(\tau) = \delta Q.$$

Therefore

$$C_{QQ}(\tau) = \left\langle (\delta Q)^2 \right\rangle_\beta = \operatorname{Var}_\beta(Q).$$

The transform of a constant is

$$\begin{aligned} C_{QQ}(i\Omega_\ell) &= \frac{ \operatorname{Var}_\beta(Q) }{\hbar} \int_0^{\beta\hbar} d\tau\, e^{i\Omega_\ell\tau/\hbar} \\ &= \beta \operatorname{Var}_\beta(Q) \delta_{\ell0}. \end{aligned}$$

Because the commutator vanishes,

$$\rho_{QQ}^{(+)}(E) = 0.$$

The entire correlator is a static bosonic Matsubara zero mode. It cannot be reconstructed from the commutator spectrum alone.

5. Obtain the first two moments

Use the Lehmann definition of $\rho^{(\eta)}$ to prove

$$M_0 = \langle[A,B]_\eta\rangle_\beta,$$

and

$$M_1 = \left\langle \left[ [A,\mathcal K], B \right]_\eta \right\rangle_\beta.$$

Solution

Integrating the zeroth moment removes the delta function:

$$M_0 = \sum_{n,m} \left( p_n-\eta p_m \right) A_{nm}B_{mn}.$$

The first term is $\langle AB\rangle$. Relabeling $n\leftrightarrow m$ in the second term gives $\langle BA\rangle$. Hence

$$M_0 = \langle AB-\eta BA\rangle = \langle[A,B]_\eta\rangle.$$

For the first moment, use

$$\left( [A,\mathcal K] \right)_{nm} = \left( K_m-K_n \right) A_{nm}.$$

The same relabeling then yields

$$M_1 = \left\langle [A,\mathcal K]B - \eta B[A,\mathcal K] \right\rangle,$$

which is

$$M_1 = \left\langle \left[ [A,\mathcal K], B \right]_\eta \right\rangle.$$

6. Translate the chemical-potential convention

Suppose

$$[H,\hat N]=0$$

and an operator lowers particle number by $q$ (so its conjugate insertion raises it by $q$):

$$[\hat N,A] = -qA.$$

Relate its $H$-evolved and $\mathcal K=H-\mu\hat N$-evolved forms, and state the corresponding shift of spectral energy.

Solution

Since $H$ and $\hat N$ commute,

$$\begin{aligned} A_{\mathcal K}(t) ={}& e^{i(H-\mu\hat N)t/\hbar} A \times e^{-i(H-\mu\hat N)t/\hbar} \\ ={}& e^{iHt/\hbar} e^{-i\mu\hat Nt/\hbar} A \times e^{i\mu\hat Nt/\hbar} e^{-iHt/\hbar}. \end{aligned}$$

The number commutator gives

$$e^{-i\mu\hat Nt/\hbar} A e^{i\mu\hat Nt/\hbar} = e^{iq\mu t/\hbar}A.$$

Therefore

$$A_{\mathcal K}(t) = e^{iq\mu t/\hbar} A_H(t).$$

The energy-domain transform is shifted by

$$E_{\mathcal K} = E_H-q\mu.$$

Equivalently, a transition changing particle number by $\Delta N=q$ has

$$\Delta K = \Delta H-\mu q.$$

7. Prove the channel-dependent sign constraint

Assume $B=A^\dagger$. Show that the fermionic graded spectral density is nonnegative, while a bosonic commutator density satisfies

$$E\rho^{(+)}(E) \geq 0$$

for each diagonal transition in a positive-temperature Gibbs state.

Solution

For a conjugate channel,

$$A_{nm}B_{mn} = |A_{nm}|^2 \geq 0.$$

In a fermionic channel the coefficient is

$$p_n+p_m \geq 0.$$

Every Lehmann line is therefore nonnegative.

In a bosonic channel,

$$p_n-p_m = p_n \left( 1-e^{-\beta E} \right)$$

on a line at $E=K_m-K_n$. The factor $1-e^{-\beta E}$ has the same sign as $E$. Thus

$$E \left( p_n-p_m \right) |A_{nm}|^2 \geq 0$$

term by term. Summing all transitions gives the stated distributional sign constraint. The density itself is negative at negative energy, so ordinary nonnegativity would be the wrong test.

References

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