Pure Dephasing Model

A pure dephasing model is an open-system model in which the environment destroys relative phases between preferred alternatives without changing their populations. It is exactly solvable when the system operator coupled to the environment commutes with the system Hamiltonian.

The model is valuable because it separates three ideas that are often blended together:

• loss of coherence in a chosen basis;
• absence of energy relaxation in that basis;
• non-Markovian memory in the coherence factor.

The Markovian generator version is Pure Dephasing Master Equation. The finite-time channel version is Dephasing Channel. This page is the exactly solvable model behind both when the microscopic assumptions are simple enough.

Minimal Hamiltonian

Let the system Hamiltonian and coupling operator share an eigenbasis:

$$H_S\lvert m\rangle = E_m\lvert m\rangle, \qquad A\lvert m\rangle = a_m\lvert m\rangle.$$

The pure dephasing Hamiltonian is

$$H = H_S + H_B + A\otimes B,$$

with

$$[A,H_S]=0.$$

Because the interaction is diagonal in the energy basis, it cannot by itself move the system from $\lvert m\rangle$ to $\lvert n\rangle$. It can, however, make different basis states imprint different phases and different bath states.

Equivalently, the full Hamiltonian can be written as a conditional bath Hamiltonian:

$$H = \sum_m \lvert m\rangle\langle m\rvert \otimes \left( E_m I_B+H_B+a_m B \right).$$

Each system alternative makes the bath evolve under a different Hamiltonian.

Exact Reduced Dynamics

Assume an initially factorized state,

$$\rho(0) = \rho_S(0)\otimes\rho_B.$$

For a system basis state $\lvert m\rangle$, define the conditional bath unitary

$$U_m(t) = \exp \left[ -\frac{i}{\hbar} \left( H_B+a_m B \right)t \right].$$

Then the reduced density-matrix elements evolve as

$$\rho_{mn}(t) = e^{-i(E_m-E_n)t/\hbar} \chi_{mn}(t) \rho_{mn}(0),$$

where the coherence factor is

$$\chi_{mn}(t) = \operatorname{Tr}_B \left[ U_m(t)\rho_B U_n^\dagger(t) \right].$$

Diagonal elements have $m=n$, so $\chi_{mm}(t)=1$ and

$$\rho_{mm}(t)=\rho_{mm}(0).$$

Off-diagonal elements decay when the two conditional bath histories become distinguishable. The factor $\chi_{mn}(t)$ is a Loschmidt-echo-like overlap of bath evolutions. Its magnitude measures retained coherence; its phase contributes a bath-induced energy shift.

Qubit Form

For a qubit in the dephasing basis, a common Hamiltonian is

$$H = \frac{\epsilon}{2}\sigma_z + H_B + \frac{\sigma_z}{2}B.$$

The two conditional bath Hamiltonians are

$$H_\pm = H_B \pm \frac12 B.$$

The populations in the $\sigma_z$ basis are constant, while the coherence obeys

$$\rho_{+,-}(t) = e^{-i\epsilon t/\hbar} \chi(t)\rho_{+,-}(0),$$

with

$$\chi(t) = \operatorname{Tr}_B \left[ e^{-iH_+t/\hbar} \rho_B e^{iH_-t/\hbar} \right].$$

If $|\chi(t)|=e^{-\Gamma_\phi t}$ over the time window of interest, the model reduces to a Markovian exponential dephasing law. If $|\chi(t)|$ is Gaussian, algebraic, oscillatory, or partially reviving, a one-rate Markovian description is not faithful.

Classical Gaussian Phase Noise

A classical version replaces the bath operator by a stochastic frequency fluctuation:

$$H(t) = H_S + A\,\xi(t).$$

For one noise realization, the evolution is unitary. After averaging over realizations,

$$\rho_{mn}(t) = e^{-i(E_m-E_n)t/\hbar} \left\langle e^{-i\phi_{mn}(t)} \right\rangle \rho_{mn}(0),$$

where

$$\phi_{mn}(t) = \frac{a_m-a_n}{\hbar} \int_0^t ds\,\xi(s).$$

If $\xi(t)$ is zero-mean Gaussian noise with correlation

$$C(\tau) = \langle \xi(\tau)\xi(0)\rangle,$$

then

$$\left\langle e^{-i\phi_{mn}(t)} \right\rangle = \exp[-\Gamma_{mn}(t)],$$

with

$$\Gamma_{mn}(t) = \frac{(a_m-a_n)^2}{2\hbar^2} \int_0^t ds \int_0^t ds'\, C(s-s').$$

This formula is often the simplest bridge between noise measurements and dephasing envelopes.

Spectral Form

For stationary classical Gaussian noise, write

$$C(\tau) = \int_{-\infty}^{\infty} \frac{d\omega}{2\pi}\, e^{-i\omega\tau} S_{\xi\xi}(\omega).$$

Then

$$\Gamma_{mn}(t) = \frac{(a_m-a_n)^2}{2\hbar^2} \int_{-\infty}^{\infty} \frac{d\omega}{2\pi}\, S_{\xi\xi}(\omega) \left| \int_0^t ds\,e^{-i\omega s} \right|^2.$$

Since

$$\left| \int_0^t ds\,e^{-i\omega s} \right|^2 = \frac{4\sin^2(\omega t/2)}{\omega^2},$$

the coherence envelope samples noise through a filter centered near zero frequency. Control pulses replace the constant integrand by a sign-changing modulation function; that is the entry point to Dynamical Decoupling.

For white noise,

$$C(\tau)=D\,\delta(\tau),$$

one finds exponential decay:

$$\Gamma_{mn}(t) = \frac{(a_m-a_n)^2D}{2\hbar^2}t.$$

For quasi-static Gaussian noise with $\xi$ constant during each run but distributed with variance $\sigma^2$,

$$\Gamma_{mn}(t) = \frac{(a_m-a_n)^2\sigma^2}{2\hbar^2}t^2.$$

The first is a Markovian-looking irreversible envelope. The second is inhomogeneous broadening and can often be refocused by echo sequences.

Oscillator-Bath Pure Dephasing

The pure-dephasing limit of the spin-boson model sets the tunneling amplitude to zero. A common qubit-bath Hamiltonian is

$$H = \frac{\epsilon}{2}\sigma_z + \sum_k \hbar\omega_k b_k^\dagger b_k + \frac{\sigma_z}{2} \sum_k g_k \left( b_k+b_k^\dagger \right).$$

This is exactly solvable because $\sigma_z$ is conserved. In one common continuum convention, define

$$J(\omega) = \pi \sum_k g_k^2 \delta(\omega-\omega_k), \qquad \omega>0.$$

For a thermal Gaussian oscillator bath, the qubit coherence has the form

$$\rho_{+,-}(t) = \rho_{+,-}(0) e^{-i\epsilon t/\hbar} e^{-\Phi(t)} e^{i\Theta(t)}.$$

The dephasing exponent is

$$\Phi(t) = \frac{1}{\pi} \int_0^\infty d\omega\, \frac{J(\omega)}{\omega^2} \left[ 1-\cos(\omega t) \right] \coth \left( \frac{\beta\hbar\omega}{2} \right).$$

The phase $\Theta(t)$ depends on counterterm and bias-renormalization conventions, so the robust decoherence content is usually the magnitude $e^{-\Phi(t)}$. The same spectral density controls how low-frequency bath weight dephases the qubit.

Ohmic Example

Take an Ohmic zero-temperature spectrum in the convention

$$J(\omega) = 2\pi\alpha\,\omega e^{-\omega/\omega_c}.$$

Then

$$\Phi(t) = 2\alpha \int_0^\infty d\omega\, e^{-\omega/\omega_c} \frac{1-\cos(\omega t)}{\omega}.$$

Using

$$\int_0^\infty d\omega\, e^{-a\omega} \frac{1-\cos(b\omega)}{\omega} = \frac12 \ln \left( 1+\frac{b^2}{a^2} \right),$$

with $a=1/\omega_c$ and $b=t$, one obtains

$$\Phi(t) = \alpha \ln \left( 1+\omega_c^2t^2 \right).$$

Thus the zero-temperature Ohmic pure-dephasing envelope is algebraic:

$$e^{-\Phi(t)} = \left( 1+\omega_c^2t^2 \right)^{-\alpha}.$$

This is not a Markovian exponential. At finite temperature, the long-time envelope becomes exponential in many Ohmic models, with a rate set by temperature and coupling convention.

Markovian Limit

A pure-dephasing model reduces to a Lindblad pure-dephasing master equation when the coherence factor is approximately exponential over the times being modeled:

$$\chi_{mn}(t) \approx e^{-i\delta\omega_{mn}t} e^{-\gamma_{mn}t}.$$

Then

$$\rho_{mn}(t) \approx e^{-i(E_m-E_n)t/\hbar} e^{-i\delta\omega_{mn}t} e^{-\gamma_{mn}t} \rho_{mn}(0).$$

For a Hermitian Lindblad operator $L$ with eigenvalues $\ell_m$, the Markovian generator

$$\dot\rho = \gamma\mathcal D[L]\rho$$

gives

$$\gamma_{mn} = \frac{\gamma}{2} (\ell_m-\ell_n)^2.$$

This is a useful effective model, but it should not be assumed when the measured coherence has Gaussian decay, power-law decay, revivals, or strong dependence on pulse sequence.

Physical Interpretation

Pure dephasing is environmental distinguishability without population transfer. If two system alternatives make the environment evolve to nearly identical states, their coherence remains. If they make the environment evolve to nearly orthogonal states, their coherence is lost in the reduced state.

This interpretation does not require a human observer or a classical apparatus. It follows from tracing over degrees of freedom that carry which-alternative information.

The basis is part of the model. Pure dephasing in the $\sigma_z$ basis may be relaxation in the energy basis if the Hamiltonian contains a transverse term. This is why the full Spin-Boson Model distinguishes localized and energy bases.

Common Mistakes

• Calling every loss of Ramsey contrast irreversible dephasing.
• Forgetting that pure dephasing is basis-dependent.
• Replacing a nonexponential coherence envelope by a constant rate without checking the time window.
• Treating quasi-static disorder as a Markovian bath.
• Ignoring the imaginary phase of $\chi_{mn}(t)$ when it shifts transition frequencies.
• Applying the $\Delta=0$ spin-boson solution when the Hamiltonian has a transverse term that causes relaxation.
• Comparing formulas for $\Phi(t)$ without checking the convention for $J(\omega)$.
• Assuming classical noise and quantum bath models are physically identical just because they give the same coherence envelope.

Cross-Links

• Pure Dephasing Master Equation for the exponential Markovian generator.
• Dephasing Channel for the finite-time channel action.
• Spin-Boson Model for the full two-state bosonic-bath model.
• Noise Spectra for spectral conventions and low-frequency noise.
• One-Over-F Noise for cutoff-dependent low-frequency dephasing models.
• Correlation Functions for time-domain bath correlations.
• Fluctuation–Dissipation Relation for equilibrium constraints on thermal oscillator baths.
• Dephasing vs Dissipation for the physical distinction from relaxation.
• Dynamical Decoupling for echo and filter-function control of dephasing.
• Approximation Checklist for Markov and weak-coupling warnings.

References

• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002).
• U. Weiss, Quantum Dissipative Systems, 4th ed., World Scientific (2012).
• A. J. Leggett, S. Chakravarty, A. T. Dorsey, M. P. A. Fisher, A. Garg, and W. Zwerger, “Dynamics of the dissipative two-state system,” Reviews of Modern Physics 59, 1–85 (1987).
• M. Schlosshauer, Decoherence and the Quantum-to-Classical Transition, Springer (2007).
• G. Ithier et al., “Decoherence in a superconducting quantum bit circuit,” Physical Review B 72, 134519 (2005).
• E. Paladino, Y. M. Galperin, G. Falci, and B. L. Altshuler, “1/f noise: Implications for solid-state quantum information,” Reviews of Modern Physics 86, 361–418 (2014).
• L. Cywiński, R. M. Lutchyn, C. P. Nave, and S. Das Sarma, “How to enhance dephasing time in superconducting qubits,” Physical Review B 77, 174509 (2008).

Exercises

Conditional bath evolution

Starting from

$$H = \sum_m \lvert m\rangle\langle m\rvert \otimes \left( E_m I_B+H_B+a_m B \right),$$

derive the coherence factor $\chi_{mn}(t)$.

Solution

The full unitary is block diagonal:

$$U(t) = \sum_m e^{-iE_mt/\hbar} \lvert m\rangle\langle m\rvert \otimes U_m(t),$$

with

$$U_m(t) = \exp \left[ -\frac{i}{\hbar} \left( H_B+a_mB \right)t \right].$$

For an initially factorized state, the $mn$ reduced matrix element is

$$\rho_{mn}(t) = e^{-i(E_m-E_n)t/\hbar} \rho_{mn}(0) \operatorname{Tr}_B \left[ U_m(t)\rho_B U_n^\dagger(t) \right].$$

Thus

$$\chi_{mn}(t) = \operatorname{Tr}_B \left[ U_m(t)\rho_B U_n^\dagger(t) \right].$$

Populations are constant

Use the exact solution to show that $\rho_{mm}(t)=\rho_{mm}(0)$.

Solution

Set $m=n$ in the exact solution. The Hamiltonian phase is unity, and

$$\chi_{mm}(t) = \operatorname{Tr}_B \left[ U_m(t)\rho_B U_m^\dagger(t) \right].$$

By cyclicity of the trace and unitarity,

$$\chi_{mm}(t) = \operatorname{Tr}_B(\rho_B) = 1.$$

Therefore $\rho_{mm}(t)=\rho_{mm}(0)$.

White noise versus quasi-static noise

For classical Gaussian phase noise, compare $C(\tau)=D\delta(\tau)$ with a quasi-static random variable of variance $\sigma^2$.

Solution

The Gaussian dephasing exponent is

$$\Gamma_{mn}(t) = \frac{(a_m-a_n)^2}{2\hbar^2} \int_0^t ds \int_0^t ds'\, C(s-s').$$

For white noise,

$$\int_0^t ds \int_0^t ds'\, D\delta(s-s') = Dt,$$

so the envelope is exponential. For quasi-static noise, $C(s-s')=\sigma^2$ over one run, so

$$\int_0^t ds \int_0^t ds'\, \sigma^2 = \sigma^2t^2,$$

and the envelope is Gaussian in time. The quasi-static case is reversible in principle by echo techniques, while ideal white-noise decay is not.

Zero-temperature Ohmic integral

Show that for

$$J(\omega) = 2\pi\alpha\,\omega e^{-\omega/\omega_c},$$

the zero-temperature exponent is $\Phi(t)=\alpha\ln(1+\omega_c^2t^2)$ in the convention used on this page.

Solution

At zero temperature, $\coth(\beta\hbar\omega/2)\to1$. Substitute the Ohmic $J(\omega)$ into

$$\Phi(t) = \frac{1}{\pi} \int_0^\infty d\omega\, \frac{J(\omega)}{\omega^2} \left[ 1-\cos(\omega t) \right].$$

This gives

$$\Phi(t) = 2\alpha \int_0^\infty d\omega\, e^{-\omega/\omega_c} \frac{1-\cos(\omega t)}{\omega}.$$

Using

$$\int_0^\infty d\omega\, e^{-a\omega} \frac{1-\cos(b\omega)}{\omega} = \frac12 \ln \left( 1+\frac{b^2}{a^2} \right),$$

with $a=1/\omega_c$ and $b=t$ yields

$$\Phi(t) = \alpha \ln \left( 1+\omega_c^2t^2 \right).$$