Born Approximation for System–Bath Coupling

The Born approximation is the weak-coupling step in many open-system derivations. It says that, to the perturbative order being kept, the environment can be treated as remaining near a fixed reference state and system-environment correlations do not need to be promoted to independent dynamical variables.

In a common interaction-picture form,

$$\rho_{SE}^{I}(t) \approx \rho_S^{I}(t)\otimes\rho_E,$$

inside the second-order memory kernel. Here $S$ denotes the system and $E$ the environment. The approximation is not the Born rule for measurement probabilities. It is also not the claim that the exact state never becomes entangled with the environment.

The point is more delicate: weak coupling generates correlations, but their feedback on the reduced state is approximated by bath correlation functions computed in a stationary reference state.

Helpful background. System–bath Hamiltonians define the interaction picture; bath reference states define the correlations used below.

Weak coupling in the interaction picture

Start from a microscopic Hamiltonian, as in System–Bath Hamiltonians:

$$H = H_S+H_E+\lambda H_I, \qquad H_I = \sum_\alpha S_\alpha\otimes B_\alpha.$$

Take $\lambda$ dimensionless, $H_I$ with energy units, and a decomposition into Hermitian system and bath operators. A finite-dimensional or suitably regularized model is assumed so that the perturbation expansion and correlation integrals exist. The bath reference state $\rho_E$ is usually assumed stationary:

$$[H_E,\rho_E]=0.$$

It is also convenient to center the bath operators:

$$\operatorname{Tr}_E(B_\alpha\rho_E)=0.$$

If this condition does not hold, write

$$B_\alpha = \delta B_\alpha + \operatorname{Tr}_E(B_\alpha\rho_E)I_E$$

and absorb the mean term into the system Hamiltonian. Otherwise a first-order coherent shift can be mistaken for a dissipative effect.

In the interaction picture with respect to $H_S+H_E$,

$$H_I(t) = \sum_\alpha S_\alpha(t)\otimes B_\alpha(t),$$

where

$$S_\alpha(t) = e^{iH_St/\hbar} S_\alpha e^{-iH_St/\hbar}, \qquad B_\alpha(t) = e^{iH_Et/\hbar} B_\alpha e^{-iH_Et/\hbar}.$$

The exact total state obeys

$$\frac{d}{dt}\rho_{SE}^{I}(t) = -\frac{i\lambda}{\hbar} \left[ H_I(t), \rho_{SE}^{I}(t) \right].$$

The Born approximation enters only after this exact equation has been expanded perturbatively.

Exact integral identity before truncation

Integrating the interaction-picture equation once gives

$$\rho_{SE}^{I}(t) = \rho_{SE}^{I}(0) - \frac{i\lambda}{\hbar} \int_0^t ds\, \left[ H_I(s), \rho_{SE}^{I}(s) \right].$$

Substitute this expression back into the equation for $\dot\rho_S^{I}(t)$ and trace over the environment. The result is exact at this stage:

$$\frac{d}{dt}\rho_S^{I}(t) = -\frac{i\lambda}{\hbar} \operatorname{Tr}_E \left[ H_I(t), \rho_{SE}^{I}(0) \right] -\frac{\lambda^2}{\hbar^2} \int_0^t ds\, \operatorname{Tr}_E \left[ H_I(t), \left[ H_I(s), \rho_{SE}^{I}(s) \right] \right].$$

For an initially factorized state,

$$\rho_{SE}^{I}(0) = \rho_S(0)\otimes\rho_E,$$

with centered bath operators, the first-order term vanishes. Without centering, it produces a Hamiltonian correction.

The second-order term is still exact in form because it contains the true $\rho_{SE}^{I}(s)$. It is not closed on $\rho_S^{I}$.

Born Closure

The Born approximation replaces the total state inside the second-order term by

$$\rho_{SE}^{I}(s) \approx \rho_S^{I}(s)\otimes\rho_E.$$

This gives the second-order Born equation

$$\frac{d}{dt}\rho_S^{I}(t) = -\frac{\lambda^2}{\hbar^2} \int_0^t ds\, \operatorname{Tr}_E \left[ H_I(t), \left[ H_I(s), \rho_S^{I}(s)\otimes\rho_E \right] \right],$$

after the centered first-order term has been removed.

This equation is not yet Markovian. The earlier state $\rho_S^{I}(s)$ still appears under the integral. The Born approximation is the weak-coupling factorization step; the Markov Approximation is the later memory-shortening step.

Correlation-Function Form

For a stationary bath, define

$$C_{\alpha\beta}(t-s) = \operatorname{Tr}_E \left[ B_\alpha(t)B_\beta(s)\rho_E \right].$$

Then the Born equation can be written as

$$\frac{d}{dt}\rho_S^{I}(t) = -\frac{\lambda^2}{\hbar^2} \sum_{\alpha,\beta} \int_0^t ds\, \big( C_{\alpha\beta}(t-s) \left[ S_\alpha(t), S_\beta(s)\rho_S^{I}(s) \right] + C_{\beta\alpha}(s-t) \left[ \rho_S^{I}(s)S_\beta(s), S_\alpha(t) \right] \big).$$

This expression shows what the approximation keeps. The bath affects the system through two-point correlation functions computed in $\rho_E$. It does not require the bath to be featureless. Long-lived, oscillatory, or structured correlations can still appear here. Keeping that memory is more general than a Markov approximation, but does not make a second-order closure accurate at arbitrary coupling or time.

What Is Being Neglected

The Born approximation neglects dynamical variables for the correlated part of the total state. Write schematically

$$\rho_{SE}(t) = \rho_S(t)\otimes\rho_E + \chi_{SE}(t),$$

where

$$\operatorname{Tr}_E\chi_{SE}(t)=0.$$

Because $\rho_E$ is a fixed reference, $\chi_{SE}$ includes changes of the bath marginal as well as system–bath correlations. It is not necessarily the same as $\rho_{SE}-\rho_S\otimes\rho_E(t)$.

The approximation does not assert $\chi_{SE}(t)=0$ exactly. Rather, $\chi_{SE}$ is generated perturbatively and its leading influence is represented by bath correlation functions. Corrections involving the bath’s changed state, higher cumulants, repeated backaction, and nonperturbative dressing are neglected.

At a fixed time and for a regular perturbation expansion, the correction to the factorized state is typically $O(\lambda)$. Substitution into the already second-order kernel therefore changes the derivative first at $O(\lambda^3)$, or later when symmetry removes odd orders. This order count is not a uniform long-time error bound: terms can grow with time, and the relaxation times of interest often scale as $\lambda^{-2}$. A controlled weak-coupling limit requires additional assumptions on bath correlations and the time scaling (Davies, 1974).

Neither persistent entanglement by itself nor a small bare coupling settles validity. The question is whether neglected correlations and bath changes feed back at the accuracy and time scale being predicted. Initial correlations also require extra source terms unless the preparation removes them.

Validity Criteria

There is no universal single inequality called “the Born condition.” In practice, one checks a family of small parameters and time scales.

A useful schematic requirement is that the rate induced by the bath be slow compared with microscopic bath dynamics. If $\tau_E$ is a bath correlation time and $\Gamma$ is a relaxation or dephasing rate generated at order $\lambda^2$, one wants

$$\Gamma\tau_E\ll1.$$

This condition says that the system changes little over one bath memory time. It does not by itself justify every later step, but it is often part of the same hierarchy.

Other checks are equally important:

• the dimensionless coupling is small compared with the relevant system and bath scales;
• the bath is large or mixing enough that its reference state is not appreciably depleted;
• bath correlations decay or average in a way compatible with the prediction being made;
• the time interval of interest does not accumulate strong backaction on the environment;
• the preparation is compatible with a factorized or weakly correlated initial state.

The Approximation Checklist compares these assumptions with the distinct Markov and secular conditions.

Born, Markov, and Secular Are Distinct

Weak-coupling master-equation derivations often combine three steps:

Step	Main assumption	Typical mathematical move
Born	higher-order feedback is negligible	replace $\rho_{SE}^{I}(s)$ by $\rho_S^{I}(s)\otimes\rho_E$ in the second-order kernel
Markov	short bath memory relative to resolved system evolution	replace $\rho_S^{I}(s)$ by $\rho_S^{I}(t)$ and extend a memory integral
Secular	well-separated Bohr-frequency sectors	average rapidly rotating cross terms

These steps are logically separate. A Born equation may remain non-Markovian. A Markovian Redfield Equation may fail to be completely positive. A secular GKSL equation may be too coarse near nearly degenerate transitions.

The Lindblad–GKSL Equation describes a completely positive Markovian generator; obtaining it from this expansion requires additional justified steps. Complete positivity is discussed separately in Completely Positive Maps.

Worked example: weak coupling to one resonant mode

For a resonant qubit and one oscillator, take the interaction-picture Hamiltonian

$$H_I^I=\hbar g(\sigma_+b+\sigma_-b^\dagger), \qquad g>0,$$

and initial state $|e,0\rangle$. The two-dimensional sector $\{|e,0\rangle,|g,1\rangle\}$ is closed, with interaction matrix $\hbar gX$. Thus

$$|\Psi(t)\rangle=\cos(gt)|e,0\rangle-i\sin(gt)|g,1\rangle, \qquad p_e(t)=\cos^2(gt).$$

The bath’s photon population is $\sin^2(gt)$. Even if $g/\omega_0=10^{-3}$, at $t=\pi/(2g)$ the oscillator is certainly excited, so a fixed-vacuum bath approximation cannot describe its subsequent feedback reliably. The small ratio supports an energy-scale separation; it does not supply short bath memory.

At early times $p_e(t)=1-g^2t^2+O(g^4t^4)$. The order-$g^2$ expansion is accurate for $gt\ll1$ but becomes negative if extrapolated to $gt>1$. Solving a second-order master equation resums some terms, but its perturbative label alone does not guarantee positive or accurate results at later times.

A broad outgoing field can behave differently: emitted excitation disperses and the relevant incoming-field correlations remain close to their reference values. Whether that happens is a property of the spectrum and geometry, not merely of the word “bath.” Also, for $H_S\propto Z$, a purely transverse $X\otimes B$ coupling has no zero-frequency system component and gives no leading second-order pure-dephasing term; low-frequency dephasing requires a diagonal coupling component or a different mechanism.

Exercises

Centered bath operators

Let

$$H_I = S\otimes B$$

and let the bath reference state satisfy $\langle B\rangle_E=\operatorname{Tr}_E(B\rho_E)$. Show that the interaction can be rewritten as a centered interaction plus a system Hamiltonian correction.

Solution

Write

$$B = \delta B+\langle B\rangle_E I_E, \qquad \delta B = B-\langle B\rangle_E I_E.$$

Then

$$S\otimes B = S\otimes\delta B + \langle B\rangle_E S\otimes I_E.$$

The second term acts only on the system and can be absorbed into

$$H_S \mapsto H_S+\lambda\langle B\rangle_E S.$$

The remaining bath operator is centered because

$$\operatorname{Tr}_E(\delta B\rho_E)=0.$$

Where Born enters

Starting from

$$\frac{d}{dt}\rho_{SE}^{I}(t) = -\frac{i\lambda}{\hbar} \left[ H_I(t), \rho_{SE}^{I}(t) \right],$$

integrate once and identify the precise step at which the Born approximation is made.

Solution

Integration gives

$$\rho_{SE}^{I}(t) = \rho_{SE}^{I}(0) - \frac{i\lambda}{\hbar} \int_0^t ds\, \left[ H_I(s), \rho_{SE}^{I}(s) \right].$$

Substitution into the equation for $\dot\rho_S^{I}$ gives a second-order expression containing

$$\operatorname{Tr}_E \left[ H_I(t), \left[ H_I(s), \rho_{SE}^{I}(s) \right] \right].$$

The Born approximation is the replacement

$$\rho_{SE}^{I}(s) \approx \rho_S^{I}(s)\otimes\rho_E$$

inside this term. Keeping $\rho_S^{I}(s)$ rather than replacing it by $\rho_S^{I}(t)$ means the equation is not yet Markovian.

Born or Markov?

Classify each statement as a Born assumption, a Markov assumption, both, or neither.

1. The bath remains approximately in a stationary reference state.
2. The memory integral can be extended from $t$ to $\infty$.
3. The system and bath never become entangled in the exact dynamics.
4. The bath correlation time is much shorter than the relaxation time.

Solution

1. Born.
2. Markov, provided the bath correlations decay sufficiently fast.
3. Neither. It is stronger than the Born approximation and generally false.
4. It supports the Markov approximation and is also a useful consistency check for weak coupling, but by itself it is not the Born factorization step.

A questionable bath

A qubit is coupled resonantly to one high-quality oscillator mode. The oscillator starts in its ground state. Is the Born approximation reliable for the oscillator treated as the environment?

Solution

Not generically. A single high-quality mode can store excitation and return it to the qubit. Its state can change appreciably, and correlations with the qubit can persist. A better split may include the oscillator in the system and treat the oscillator’s damping reservoir as the bath.

References

• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002). Publisher.
• E. B. Davies, “Markovian master equations,” Communications in Mathematical Physics 39, 91–110 (1974). DOI.
• C. W. Gardiner and P. Zoller, Quantum Noise, 3rd ed., Springer (2004). Publisher.
• Á. Rivas and S. F. Huelga, Open Quantum Systems: An Introduction, Springer (2012). Author manuscript.