Bath 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 $B(t)$ in a reference state $\rho_B$, the basic ordered two-point function is

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

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

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

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

Fluctuations Around the Mean Bath Force

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

$$H_{\mathrm{int}} = \sum_\alpha S_\alpha\otimes B_\alpha,$$

then the mean bath force

$$\langle B_\alpha\rangle_B = \operatorname{Tr}_B(B_\alpha\rho_B)$$

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

$$\delta B_\alpha = B_\alpha-\langle B_\alpha\rangle_B I_B.$$

The irreversible dynamics appears at second order through correlations such as

$$C_{\alpha\beta}(\tau) = \operatorname{Tr}_B \!\left[ \delta B_\alpha(\tau) \delta B_\beta(0) \rho_B \right].$$

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

Stationary Baths

A bath state is stationary when

$$[H_B,\rho_B]=0.$$

Then, in the bath Heisenberg picture,

$$B_\alpha(t) = e^{iH_Bt/\hbar} B_\alpha e^{-iH_Bt/\hbar}.$$

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

$$\begin{aligned} C_{\alpha\beta}(t,s) &= \operatorname{Tr}_B \!\left[ B_\alpha(t)B_\beta(s)\rho_B \right] \\ &= \operatorname{Tr}_B \!\left[ B_\alpha(t-s)B_\beta(0)\rho_B \right]. \end{aligned}$$

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 nonequilibrium reservoir can have time-translation-invariant correlations without thermal detailed balance. Driven or squeezed fields often require a rotating frame or an enlarged driven model to make that statement; they need not commute with the undriven bare Hamiltonian in the laboratory frame.

Ordered Correlations

Quantum operator order matters. Define the ordered correlation

$$C_{\alpha\beta}^{\mathrm{ord}}(t) = \langle B_\alpha(t)B_\beta(0)\rangle.$$

The reversed-order correlation is

$$C_{\beta\alpha}^{\mathrm{rev}}(t) = \langle B_\beta(0)B_\alpha(t)\rangle.$$

In general,

$$C_{\alpha\beta}^{\mathrm{ord}}(t) \ne C_{\beta\alpha}^{\mathrm{rev}}(t).$$

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 $B$, the relation

$$C_{BB}^{\mathrm{ord}}(t)^* = C_{BB}^{\mathrm{ord}}(-t)$$

holds for every stationary state. It ensures a real spectrum; nonnegativity additionally follows from the positive-type property of physical correlations. In a bath energy basis with stationary populations $p_m$, the spectral representation makes both the sign convention and positivity explicit:

$$S_{BB}(\omega)=2\pi\sum_{m,n}p_m|B_{nm}|^2 \delta\!\left(\omega-\frac{E_n-E_m}{\hbar}\right)\ge0.$$

Positive frequency corresponds to the bath gaining energy. A generic function obeying $C(t)^*=C(-t)$ need not have a nonnegative Fourier transform.

Symmetrized and Commutator Parts

Two combinations are especially useful. The symmetrized correlation is

$$C_{BB}^{\mathrm{sym}}(t) = \frac12 \langle \{B(t),B(0)\} \rangle.$$

The commutator correlation is

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

The ordered correlation can be reconstructed as

$$\langle B(t)B(0)\rangle = C_{BB}^{\mathrm{sym}}(t) + \frac12 C_{BB}^{\mathrm{com}}(t).$$

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.

Correlation Time

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

$$\tau_B \sim \frac{ \int_0^\infty d\tau\,|C(\tau)| }{ |C(0)| }$$

when the integral exists.

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

$$\Gamma\tau_B\ll1.$$

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.

Example: Exponential Correlation

A simple stationary classical-looking model is

$$C(\tau) = \sigma^2 e^{-|\tau|/\tau_B}.$$

The two-sided spectrum is Lorentzian:

$$S(\omega) = \int_{-\infty}^{\infty} d\tau\, e^{i\omega\tau}C(\tau) = \frac{2\sigma^2\tau_B} {1+\omega^2\tau_B^2}.$$

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

Master-Equation Rates

For dimensionless system coupling operators and energy-valued Hermitian bath forces, define the one-sided rate transform

$$\Gamma_{\alpha\beta}(\omega) = \frac1{\hbar^2}\int_0^\infty d\tau\, e^{i\omega\tau} C_{\alpha\beta}(\tau).$$

For several bath forces, use the Hermitian and anti-Hermitian matrix combinations, not elementwise real and imaginary parts: $\gamma_{\alpha\beta}=\Gamma_{\alpha\beta}+\Gamma_{\beta\alpha}^*$ gives the dissipative matrix, while $(\Gamma_{\alpha\beta}-\Gamma_{\beta\alpha}^*)/(2i)$ gives the Lamb-shift coefficients. For one Hermitian force these reduce to $2\operatorname{Re}\Gamma$ and $\operatorname{Im}\Gamma$.

A full two-sided transform is

$$\gamma_{\alpha\beta}(\omega) = \frac1{\hbar^2}\int_{-\infty}^{\infty} d\tau\, e^{i\omega\tau} C_{\alpha\beta}(\tau).$$

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

The factor $1/\hbar^2$ gives these matrices units of inverse time. For non-Hermitian bath operators, the conjugate/dagger placement in the correlation matrix must be adjusted consistently. The derivation is developed in Redfield Equation and Thermal Master Equations.

Thermal Correlations and KMS

For a thermal bath,

$$\rho_B = \frac{e^{-\beta H_B}}{Z},$$

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

$$\langle A(t)B(0)\rangle_\beta = \langle B(0)A(t+i\beta\hbar)\rangle_\beta.$$

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,

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

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 $T=1/(k_B\beta)$.

For the rate-level form, see Detailed Balance.

Gaussian and Non-Gaussian Baths

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:

$$\langle B(t_1)B(t_2)B(t_3)\rangle_c, \qquad \langle B(t_1)B(t_2)B(t_3)B(t_4)\rangle_c, \quad \ldots$$

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.

Exercises

1. Show that if $[H_B,\rho_B]=0$, then

$$\operatorname{Tr}_B[B_\alpha(t)B_\beta(s)\rho_B] = \operatorname{Tr}_B[B_\alpha(t-s)B_\beta(0)\rho_B].$$

Solution

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

$$B_\alpha(t)B_\beta(s) = U^\dagger(s) B_\alpha(t-s)B_\beta(0) U(s).$$

Therefore

$$\operatorname{Tr}_B[B_\alpha(t)B_\beta(s)\rho_B] = \operatorname{Tr}_B[ B_\alpha(t-s)B_\beta(0) U(s)\rho_BU^\dagger(s) ].$$

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

2. Let

$$C^{\mathrm{sym}}(t) = \frac12\langle\{B(t),B(0)\}\rangle, \qquad C^{\mathrm{com}}(t) = \langle[B(t),B(0)]\rangle.$$

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

Solution

By adding and subtracting the reversed order,

$$\langle B(t)B(0)\rangle = \frac12 \langle B(t)B(0)+B(0)B(t)\rangle + \frac12 \langle B(t)B(0)-B(0)B(t)\rangle.$$

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

$$\langle B(t)B(0)\rangle = C^{\mathrm{sym}}(t) + \frac12 C^{\mathrm{com}}(t).$$

3. A bath qubit has $H_B=\hbar\Omega\lvert e\rangle\langle e\rvert$ and $B=b\sigma_x$, with populations $p_g,p_e$. Find $C(t)$ and the two spectral line weights.

Solution

The lowering operator evolves as $e^{-i\Omega t}\lvert g\rangle\langle e\rvert$, so

$$C(t)=b^2[p_ge^{-i\Omega t}+p_ee^{i\Omega t}],\qquad S(\omega)=2\pi b^2[p_g\delta(\omega-\Omega)+p_e\delta(\omega+\Omega)].$$

Both weights are nonnegative. At equilibrium their ratio is $p_e/p_g=e^{-\beta\hbar\Omega}$. In the bath ground state only the positive-frequency absorption line remains.

4. A stationary bath has a long algebraic correlation tail $C(\tau)\sim \tau^{-1}$ 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 $\tau^{-1}$ 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.

References

• 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.
• R. Kubo, “Statistical-mechanical theory of irreversible processes. I,” Journal of the Physical Society of Japan 12, 570–586, 1957.
• A. Rivas and S. F. Huelga, Open Quantum Systems: An Introduction, Springer, 2012.
• D. F. Walls and G. J. Milburn, Quantum Optics, 2nd ed., Springer, 2008.
• U. Weiss, Quantum Dissipative Systems, 4th ed., World Scientific, 2012.