Skip to content

Correlation Functions

Correlation functions are the time-domain language of quantum noise. They say how fluctuations at one time are related to fluctuations at another time, and they are the quantities that enter weak-coupling master equations before any Fourier transform is taken.

For a bath operator B(t)B(t) in a reference state ρB\rho_B, the basic ordered two-point function is

CBB(t,s)=Tr⁡B ⁣[B(t)B(s)ρB].C_{BB}(t,s) = \operatorname{Tr}_B \!\left[ B(t)B(s)\rho_B \right].

If the bath state is stationary, this depends only on the time difference:

CBB(t,s)=CBB(t−s).C_{BB}(t,s) = C_{BB}(t-s).

Spectra, rates, fluctuation–dissipation relations, and Markov approximations all begin with this object. The frequency-domain version is Noise Spectra.

An instantaneous bath expectation value is usually not the main source of dissipation. If a system couples through

Hint=∑αSα⊗Bα,H_{\mathrm{int}} = \sum_\alpha S_\alpha\otimes B_\alpha,

then the mean bath force

⟨Bα⟩B=Tr⁡B(BαρB)\langle B_\alpha\rangle_B = \operatorname{Tr}_B(B_\alpha\rho_B)

produces a coherent first-order shift. One often removes it by defining centered operators

δBα=Bα−⟨Bα⟩BIB.\delta B_\alpha = B_\alpha-\langle B_\alpha\rangle_B I_B.

The irreversible dynamics appears at second order through correlations such as

Cαβ(τ)=Tr⁡B ⁣[δBα(τ)δBβ(0)ρB].C_{\alpha\beta}(\tau) = \operatorname{Tr}_B \!\left[ \delta B_\alpha(\tau) \delta B_\beta(0) \rho_B \right].

The bath affects the system through how long these correlations persist, what frequencies they contain, and how operator ordering distinguishes absorption from emission.

A bath state is stationary when

[HB,ρB]=0.[H_B,\rho_B]=0.

Then, in the bath Heisenberg picture,

Bα(t)=eiHBt/ℏBαe−iHBt/ℏ.B_\alpha(t) = e^{iH_Bt/\hbar} B_\alpha e^{-iH_Bt/\hbar}.

Using stationarity, the two-time correlation becomes translation invariant:

Cαβ(t,s)=Tr⁡B ⁣[Bα(t)Bβ(s)ρB]=Tr⁡B ⁣[Bα(t−s)Bβ(0)ρB].\begin{aligned} C_{\alpha\beta}(t,s) &= \operatorname{Tr}_B \!\left[ B_\alpha(t)B_\beta(s)\rho_B \right] \\ &= \operatorname{Tr}_B \!\left[ B_\alpha(t-s)B_\beta(0)\rho_B \right]. \end{aligned}

This is why stationary baths can be summarized by functions of one time variable, and why Fourier spectra are useful.

Stationarity is not the same as equilibrium. A driven steady state, squeezed bath, biased electronic lead, or engineered reservoir may be stationary without satisfying thermal detailed balance.

Quantum operator order matters. Define the ordered correlation

Cαβord(t)=⟨Bα(t)Bβ(0)⟩.C_{\alpha\beta}^{\mathrm{ord}}(t) = \langle B_\alpha(t)B_\beta(0)\rangle.

The reversed-order correlation is

Cβαrev(t)=⟨Bβ(0)Bα(t)⟩.C_{\beta\alpha}^{\mathrm{rev}}(t) = \langle B_\beta(0)B_\alpha(t)\rangle.

In general,

Cαβord(t)≠Cβαrev(t).C_{\alpha\beta}^{\mathrm{ord}}(t) \ne C_{\beta\alpha}^{\mathrm{rev}}(t).

This noncommutativity is the time-domain reason quantum spectra need not be even in frequency. It is also why a zero-temperature bath can absorb energy from a system while not thermally exciting it.

For a Hermitian bath operator BB, the relation

CBBord(t)∗=CBBord(−t)C_{BB}^{\mathrm{ord}}(t)^* = C_{BB}^{\mathrm{ord}}(-t)

often holds in stationary states. This ensures that the corresponding full spectrum is real and nonnegative, but not necessarily symmetric in ω\omega.

Two combinations are especially useful. The symmetrized correlation is

CBBsym(t)=12⟨{B(t),B(0)}⟩.C_{BB}^{\mathrm{sym}}(t) = \frac12 \langle \{B(t),B(0)\} \rangle.

The commutator correlation is

CBBcom(t)=⟨[B(t),B(0)]⟩.C_{BB}^{\mathrm{com}}(t) = \langle [B(t),B(0)] \rangle.

The ordered correlation can be reconstructed as

⟨B(t)B(0)⟩=CBBsym(t)+12CBBcom(t).\langle B(t)B(0)\rangle = C_{BB}^{\mathrm{sym}}(t) + \frac12 C_{BB}^{\mathrm{com}}(t).

These two parts play different roles. The symmetrized part is often what appears in classical-looking noise power and detector readouts. The commutator part is tied to response and dissipation. In equilibrium their Fourier transforms are related by the fluctuation–dissipation relation.

For the response relation, see Fluctuation–Dissipation Relation.

The correlation time is the timescale over which C(τ)C(\tau) remains appreciable. There is no single universal definition, but a useful estimate is

τB∼∫0∞dτ ∣C(τ)∣∣C(0)∣\tau_B \sim \frac{ \int_0^\infty d\tau\,|C(\tau)| }{ |C(0)| }

when the integral exists.

The Markov approximation requires the system state to change little over this time. If Γ\Gamma is a characteristic system relaxation or dephasing rate, the schematic condition is

ΓτB≪1.\Gamma\tau_B\ll1.

This condition can fail because correlations decay slowly, because the system evolves quickly, or because the bath has narrow resonances that store memory. Long algebraic tails, strong coupling, finite reservoirs, delay lines, and structured spectra all require caution.

For the approximation itself, see Markov Approximation.

A simple stationary classical-looking model is

C(τ)=σ2e−∣τ∣/τB.C(\tau) = \sigma^2 e^{-|\tau|/\tau_B}.

The two-sided spectrum is Lorentzian:

S(ω)=∫−∞∞dτ eiωτC(τ)=2σ2τB1+ω2τB2.S(\omega) = \int_{-\infty}^{\infty} d\tau\, e^{i\omega\tau}C(\tau) = \frac{2\sigma^2\tau_B} {1+\omega^2\tau_B^2}.

As τB\tau_B becomes small with 2σ2τB2\sigma^2\tau_B held fixed, the correlation approaches a delta function. That is the white-noise or Markov limit. If τB\tau_B is comparable to system timescales, replacing this noise by white noise loses memory and spectral structure.

In a weak-coupling derivation, bath correlations enter one-sided transforms such as

Γαβ(ω)=∫0∞dτ eiωτCαβ(τ).\Gamma_{\alpha\beta}(\omega) = \int_0^\infty d\tau\, e^{i\omega\tau} C_{\alpha\beta}(\tau).

The real parts supply dissipative rates after the appropriate system-frequency decomposition and secular or coarse-graining steps. The imaginary parts contribute Hamiltonian shifts, often called Lamb shifts.

A full two-sided transform is

γαβ(ω)=∫−∞∞dτ eiωτCαβ(τ).\gamma_{\alpha\beta}(\omega) = \int_{-\infty}^{\infty} d\tau\, e^{i\omega\tau} C_{\alpha\beta}(\tau).

The matrix γαβ(ω)\gamma_{\alpha\beta}(\omega) must be positive semidefinite for each ω\omega when it comes from a stationary quantum bath. This positivity is one reason secular Lindblad–GKSL generators have positive rates.

The derivational details live in Redfield Equation and Thermal Master Equations.

For a thermal bath,

ρB=e−βHBZ,\rho_B = \frac{e^{-\beta H_B}}{Z},

correlations satisfy the Kubo–Martin–Schwinger condition. One common form is

⟨A(t)B(0)⟩β=⟨B(0)A(t+iβℏ)⟩β.\langle A(t)B(0)\rangle_\beta = \langle B(0)A(t+i\beta\hbar)\rangle_\beta.

This analytic relation encodes equilibrium. In frequency language it implies detailed balance. For one Hermitian bath operator with the spectrum convention used in Noise Spectra,

SBB(−ω)=e−βℏωSBB(+ω),ω>0.S_{BB}(-\omega) = e^{-\beta\hbar\omega} S_{BB}(+\omega), \qquad \omega>0.

Thus upward transitions are suppressed relative to downward transitions at low temperature. A bath that violates this relation may still be stationary, but it is not an equilibrium thermal bath at temperature T=1/(kBβ)T=1/(k_B\beta).

For the rate-level form, see Detailed Balance.

For a Gaussian bath linearly coupled to the system, two-point correlation functions determine all higher moments by Wick’s theorem. Oscillator baths, input–output vacuum fields, and many weakly perturbed thermal environments are treated this way.

For a non-Gaussian bath, two-point functions are not enough. Higher cumulants can matter:

⟨B(t1)B(t2)B(t3)⟩c,⟨B(t1)B(t2)B(t3)B(t4)⟩c,…\langle B(t_1)B(t_2)B(t_3)\rangle_c, \qquad \langle B(t_1)B(t_2)B(t_3)B(t_4)\rangle_c, \quad \ldots

Examples include telegraph noise, rare switching events, strongly nonlinear detectors, finite spin environments, and shot noise outside a Gaussian approximation. A master equation derived only from two-point functions should not be overinterpreted in such cases.

  • Treating a nonstationary environment as if all correlations depended only on time differences.
  • Using symmetrized correlations in a formula that requires ordered correlations.
  • Reading a short correlation time from a plot without comparing it to the system timescale.
  • Assuming a stationary bath is automatically thermal.
  • Forgetting to subtract nonzero bath means before interpreting second-order terms as noise.
  • Confusing a mode spectral density J(ω)J(\omega) with an ordered noise spectrum S(ω)S(\omega).
  • Assuming two-point functions fully describe a non-Gaussian bath.
  • Ignoring the imaginary part of one-sided transforms, which can produce Hamiltonian shifts.
  • H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press, 2002.
  • C. W. Gardiner and P. Zoller, Quantum Noise, 3rd ed., Springer, 2004.
  • U. Weiss, Quantum Dissipative Systems, 4th ed., World Scientific, 2012.
  • R. Kubo, “Statistical-mechanical theory of irreversible processes. I,” Journal of the Physical Society of Japan 12, 570–586, 1957.
  • D. F. Walls and G. J. Milburn, Quantum Optics, 2nd ed., Springer, 2008.
  • A. Rivas and S. F. Huelga, Open Quantum Systems: An Introduction, Springer, 2012.
  1. Show that if [HB,ρB]=0[H_B,\rho_B]=0, then
Tr⁡B[Bα(t)Bβ(s)ρB]=Tr⁡B[Bα(t−s)Bβ(0)ρB].\operatorname{Tr}_B[B_\alpha(t)B_\beta(s)\rho_B] = \operatorname{Tr}_B[B_\alpha(t-s)B_\beta(0)\rho_B].
Solution

Write U(t)=e−iHBt/ℏU(t)=e^{-iH_Bt/\hbar}, so B(t)=U†(t)BU(t)B(t)=U^\dagger(t)BU(t). Then

Bα(t)Bβ(s)=U†(s)Bα(t−s)Bβ(0)U(s).B_\alpha(t)B_\beta(s) = U^\dagger(s) B_\alpha(t-s)B_\beta(0) U(s).

Therefore

Tr⁡B[Bα(t)Bβ(s)ρB]=Tr⁡B[Bα(t−s)Bβ(0)U(s)ρBU†(s)].\operatorname{Tr}_B[B_\alpha(t)B_\beta(s)\rho_B] = \operatorname{Tr}_B[ B_\alpha(t-s)B_\beta(0) U(s)\rho_BU^\dagger(s) ].

Stationarity gives U(s)ρBU†(s)=ρBU(s)\rho_BU^\dagger(s)=\rho_B, yielding the result.

  1. Let
Csym(t)=12⟨{B(t),B(0)}⟩,Ccom(t)=⟨[B(t),B(0)]⟩.C^{\mathrm{sym}}(t) = \frac12\langle\{B(t),B(0)\}\rangle, \qquad C^{\mathrm{com}}(t) = \langle[B(t),B(0)]\rangle.

Express the ordered correlation ⟨B(t)B(0)⟩\langle B(t)B(0)\rangle in terms of these two quantities.

Solution

By adding and subtracting the reversed order,

⟨B(t)B(0)⟩=12⟨B(t)B(0)+B(0)B(t)⟩+12⟨B(t)B(0)−B(0)B(t)⟩.\langle B(t)B(0)\rangle = \frac12 \langle B(t)B(0)+B(0)B(t)\rangle + \frac12 \langle B(t)B(0)-B(0)B(t)\rangle.

The first term is Csym(t)C^{\mathrm{sym}}(t) and the second term is 12Ccom(t)\frac12 C^{\mathrm{com}}(t). Thus

⟨B(t)B(0)⟩=Csym(t)+12Ccom(t).\langle B(t)B(0)\rangle = C^{\mathrm{sym}}(t) + \frac12 C^{\mathrm{com}}(t).
  1. For
C(τ)=σ2e−∣τ∣/τB,C(\tau)=\sigma^2e^{-|\tau|/\tau_B},

compute the two-sided spectrum.

Solution

Because C(τ)C(\tau) is even,

S(ω)=2σ2∫0∞dτ e−τ/τBcos⁡(ωτ).S(\omega) = 2\sigma^2 \int_0^\infty d\tau\, e^{-\tau/\tau_B} \cos(\omega\tau).

Using

∫0∞dτ e−aτcos⁡(ωτ)=aa2+ω2,a=1τB,\int_0^\infty d\tau\, e^{-a\tau}\cos(\omega\tau) = \frac{a}{a^2+\omega^2}, \qquad a=\frac1{\tau_B},

gives

S(ω)=2σ2τB1+ω2τB2.S(\omega) = \frac{2\sigma^2\tau_B} {1+\omega^2\tau_B^2}.
  1. A stationary bath has a long algebraic correlation tail C(τ)∼τ−1C(\tau)\sim \tau^{-1} over the experimentally relevant window. Why is a Markov approximation suspicious?
Solution

A Markov approximation assumes that bath correlations decay over a short memory time compared with the system evolution. A τ−1\tau^{-1} tail is long lived and does not provide a clean finite correlation time over the relevant window. The system can remain correlated with earlier bath fluctuations, so replacing the delayed state by the present state and extending integrals to infinity may give wrong rates or miss memory effects.