research39 min studyLesson 741 in Physics

Identify thermal equilibrium through correlations

Derive KMS detailed balance from the Gibbs trace while keeping real-time Fourier conventions and bosonic versus fermionic Green-function signs explicit.

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What you’ll understand

  • Derive the equilibrium relation between greater and lesser correlations.
  • Distinguish a trace identity from signs introduced by fermionic Green-function definitions.
  • Test thermal detailed balance in a finite-level model.

A stationary expectation value does not by itself establish thermal equilibrium. Thermal correlations obey an additional relation between positive and negative energy transfer, fixed by the Gibbs ensemble. The Kubo–Martin–Schwinger condition expresses that relation without requiring weak interactions or a particle picture. It becomes the bridge between imaginary time, spectral functions, and fluctuation–dissipation relations, provided the operator definitions and Fourier signs are kept consistent.

Declare the ensemble and time convention

Use ℏ=kB=c=1\hbar=k_B=c=1, metric (+,−,−,−)(+,-,-,-), and Heisenberg evolution A(t)=eiHtAe−iHtA(t)=e^{iHt}Ae^{-iHt}. Temperature therefore has energy units, and β=1/T\beta=1/T has inverse-energy units. Our real-time Fourier transform uses eiωt−ik⋅xe^{i\omega t-i\mathbf k\cdot\mathbf x}, with inverse factors of two pi. For a neutral Hermitian bosonic operator A, define raw unordered correlations without factors of i. These conventions will also fix the retarded signs later in the module. Chemical potentials are set to zero throughout these elementary relations.

The Gibbs state commutes with H, giving time-translation invariance. Cyclicity of the trace, combined with the Boltzmann weight, relates an operator moved around that trace to evolution by an imaginary time interval. The resulting KMS identity holds in the appropriate analytic strip for well-defined correlations. It is stronger than stationarity and does not assume free fields. With nonzero conserved-charge chemical potentials, charged operators acquire additional charge-dependent factors; substituting a chemical potential by intuition would not preserve the present formulas.

ϱβ=e−βHZ,C>(t)=⟨A(t)A(0)⟩β,C<(t)=⟨A(0)A(t)⟩β,C>(t−iβ)=C<(t)\varrho_\beta=\frac{e^{-\beta H}}{Z},\qquad C^>(t)=\langle A(t)A(0)\rangle_\beta,\quad C^<(t)=\langle A(0)A(t)\rangle_\beta,\qquad C^>(t-i\beta)=C^<(t)
The normalized density operator is denoted varrho to distinguish it from a spectral density. The imaginary shift is a thermal trace identity, not physical time evolution into a complex laboratory clock. The displayed relation concerns the specified raw bosonic products at zero chemical potential.↔ Scroll wide equations horizontally.

Translate KMS into energy-transfer balance

Fourier transforming the imaginary-time shift yields C>(ω)=eβωC<(ω)C^>(\omega)=e^{\beta\omega}C^<(\omega) under our positive-sign forward transform. A spectral decomposition makes the factor transparent: swapping the initial and final energy eigenstates changes their Gibbs probabilities by the exponential of the exchanged energy. For a Hermitian A, the lesser spectrum at positive frequency equals the greater spectrum at negative frequency. Positive-frequency fluctuations can therefore outweigh negative-frequency fluctuations without violating stationarity. The ratio is a thermal population factor, not an independently adjustable emission parameter.

Fermionic signs require separating raw products from signed Green functions. For an annihilation operator, the raw greater product and raw lesser product still satisfy a positive Boltzmann ratio from the ordinary trace. Standard fermionic conventions instead define the greater function with minus i and the lesser with plus i. Their KMS relation consequently contains a minus sign. Fermionic imaginary-time antiperiodicity additionally follows from fermionic time ordering. Attributing every minus sign directly to cyclicity would conflate three different operations and spoil the occupation factors.

C>(ω)=eβωC<(ω),GF>=−iCF>,GF<=+iCF< ⟹ GF>(ω)=−eβωGF<(ω)C^>(\omega)=e^{\beta\omega}C^<(\omega),\qquad G_F^>=-iC_F^>,\quad G_F^<=+iC_F^<\ \Longrightarrow\ G_F^>(\omega)=-e^{\beta\omega}G_F^<(\omega)
For the fermionic statement, CF> = ⟨c(t)c†(0)⟩ and CF< = ⟨c†(0)c(t)⟩ are raw products at zero chemical potential. Their signed Green functions differ in their prefactors. βω is dimensionless; all paired spectra have the same units.↔ Scroll wide equations horizontally.

Use detailed balance as a diagnostic

A finite system diagonal in its energy basis is stationary for any chosen populations. Only special population ratios correspond to a Gibbs temperature. Checking correlations at several transition frequencies can distinguish those cases, whereas fitting one transition can always produce a misleading effective temperature. In a field theory, operator choice and accessible frequency range similarly determine how demanding the diagnostic is. An equilibrium claim should be supported by compatible spectral ratios and conserved quantities rather than a single approximately constant occupation number.

KMS does not say that a thermal state has no fluctuations or no dissipation. It constrains how fluctuations and response are related. Conversely, a driven stationary state may have constant macroscopic properties while violating this thermal relation. Nonequilibrium contour methods can describe such states without imposing KMS. Reserve the thermal identity for an actual equilibrium ensemble or a controlled local-equilibrium approximation, and record which time and length scales justify using the latter. A transient fit to a Boltzmann-looking slope is not a proof of global equilibrium.

Detailed balance in a two-level probe

A two-level Hamiltonian has energies zero and Δ>0\Delta>0, with βΔ=log⁡3\beta\Delta=\log3. The Hermitian operator A exchanges the two states with unit matrix element. Find the Gibbs populations and the ratio of the positive- and negative-frequency weights in C>C^>.

  1. The partition function is Z=1+e−βΔ=1+1/3=4/3Z=1+e^{-\beta\Delta}=1+1/3=4/3. Thus the ground-state probability is 3/43/4 and the excited-state probability is 1/41/4.

  2. Expanding the correlation gives C>(t)=(3/4)e−iΔt+(1/4)eiΔtC^>(t)=(3/4)e^{-i\Delta t}+(1/4)e^{i\Delta t}. Under the declared Fourier transform, these terms sit at positive and negative frequency, respectively.

  3. The common delta-function normalization cancels in the ratio: the weights at +Δ+\Delta and −Δ-\Delta are in the ratio three. This equals eβΔe^{\beta\Delta} and checks the KMS sign convention.

The result

The populations are three quarters and one quarter, with a three-to-one greater-spectrum weight ratio. Nonzero negative-frequency weight represents the finite thermal population of the upper level.

Explore the model

Interactive lab

Interactive lab · probability

Inspect a declared population law

Reset and set success probability to 0.25 and trials to 4. Predict mean count 1 and variance 0.75. Compare after a fresh Reset with p = 0.5 and n = 4, where mean is 2 and variance 1. The lab displays exact independent binomial probabilities, not thermal operator correlations or sampled experiments. It does not derive a temperature from those populations or implement a KMS test.

Inspect a declared population lawThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA population model is an input; thermal detailed balance is an additional relation between transitions.

Check your understanding

For the specified raw unordered fermion products at zero chemical potential, where does the minus sign in the standard signed Green-function KMS relation arise?

Try it yourself

A free fermionic level of energy ϵ>0\epsilon>0 has βϵ=log⁡3\beta\epsilon=\log3. Calculate its occupation and the coefficients of e−iϵte^{-i\epsilon t} in CF>C_F^> and CF<C_F^<. State the corresponding ratio for the signed Green functions defined above.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The empty and occupied probabilities add to one. The greater raw product detects an available empty state, while the lesser detects an occupied state.

Compare with the solution

The occupation is nF=1/(eβϵ+1)=1/4n_F=1/(e^{\beta\epsilon}+1)=1/4. Thus CF>(t)=(3/4)e−iϵtC_F^>(t)=(3/4)e^{-i\epsilon t} and CF<(t)=(1/4)e−iϵtC_F^<(t)=(1/4)e^{-i\epsilon t}. Their coefficient ratio is positive three. Multiplication by the specified prefactors gives GF>=−3GF<G_F^>=-3G_F^<. The minus sign changes the Green-function convention, not the positivity of the occupation. This single-mode example assumes a Gibbs ensemble, zero chemical potential, and a positive level energy.

Take it with you

  • KMS is an equilibrium analytic relation stronger than time-translation invariance.
  • Real-time Fourier signs and Green-function prefactors determine spectral signs.
  • Bosonic and fermionic thermal factors must be derived with the operator definitions held fixed.

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