Pure Dephasing Master Equation

The pure dephasing master equation describes loss of coherence in a preferred basis without population transfer in that basis. It is the continuous-time Markovian generator behind the phase-damping channel, unread monitoring of an observable, elastic environmental scattering, and many simple models of frequency noise. For the exactly solvable non-Markovian envelope behind this limit, see Pure Dephasing Model.

For a qubit with Hamiltonian

$$H = \frac{\hbar\omega_0}{2}\sigma_z,$$

one common convention is

$$\frac{d\rho}{dt} = - \frac{i}{\hbar}[H,\rho] + \frac{\Gamma_\phi}{2} \left( \sigma_z\rho\sigma_z-\rho \right).$$

Equivalently,

$$\frac{d\rho}{dt} = - \frac{i}{\hbar}[H,\rho] + 2\Gamma_\phi\mathcal D[P_e]\rho,$$

where $P_e=\lvert e\rangle\langle e\rvert$ is diagonal in the energy basis. The important invariant statement is physical: populations in the dephasing basis are unchanged, while coherences decay exponentially. Different authors move factors of $2$ between the rate and the Lindblad operator.

Helpful background. Lindblad operators define the dissipator, and dephasing versus dissipation separates coherence from energy changes.

Populations and coherence in the dephasing basis

Take $\Gamma_\phi\ge0$ and $Z=\operatorname{diag}(1,-1)$ in the displayed $0,1$ basis. With the chosen positive Hamiltonian sign, $|0\rangle$ is the higher-energy state for $\omega_0>0$; the labels $0,1$ here do not silently designate ground and excited states. Either rank-one Z projector gives the same projector dissipator.

A relaxation channel moves populations. Pure dephasing leaves populations in its specified basis fixed while damping their relative coherence. It need not imply a readable record in the actual environment; several different physical mechanisms have the same reduced generator.

For the qubit equation above,

$$\rho = \begin{pmatrix} \rho_{00} & \rho_{01}\\ \rho_{10} & \rho_{11} \end{pmatrix}$$

evolves as

$$\begin{aligned} \dot\rho_{00}&=0,\\ \dot\rho_{11}&=0,\\ \dot\rho_{01}&=-(i\omega_0+\Gamma_\phi)\rho_{01},\\ \dot\rho_{10}&=(i\omega_0-\Gamma_\phi)\rho_{10}. \end{aligned}$$

Thus

$$\rho_{01}(t) = e^{-i\omega_0 t} e^{-\Gamma_\phi t} \rho_{01}(0),$$

while $\rho_{00}$ and $\rho_{11}$ remain fixed. The corresponding finite-time channel is the Markovian case of the Dephasing Channel with coherence factor $\lambda(t)=e^{-\Gamma_\phi t}$ after separating the Hamiltonian phase.

For a numerical comparison with ODE integration and a vectorized Liouvillian, see Solving Lindblad Equations.

Bloch-Vector Form

Write

$$\rho = \frac12 \left( I+r_x\sigma_x+r_y\sigma_y+r_z\sigma_z \right).$$

The pure dephasing master equation gives

$$\begin{aligned} \dot r_x&=-\Gamma_\phi r_x-\omega_0 r_y,\\ \dot r_y&=\omega_0 r_x-\Gamma_\phi r_y,\\ \dot r_z&=0. \end{aligned}$$

In a frame rotating with the Hamiltonian, the transverse Bloch components decay as

$$r_x(t),r_y(t)\propto e^{-\Gamma_\phi t},$$

and the longitudinal component $r_z$ is constant. The Bloch ball contracts toward the $z$ axis rather than toward a thermal state.

Relation to T-Times

In spectroscopy and qubit experiments, $T_1$ is the population-relaxation time, $T_2$ is the transverse coherence-decay time, and $T_\phi$ is the pure-dephasing time. In the standard weak-coupling Markovian qubit model,

$$\frac{1}{T_2} = \frac{1}{2T_1} + \frac{1}{T_\phi}.$$

The factor $1/(2T_1)$ appears because population relaxation also reduces coherence. Pure dephasing is the additional contribution that remains when energy relaxation is absent or has been separated.

With the convention used in the opening equation,

$$T_\phi^{-1}=\Gamma_\phi.$$

Other communities may put factors of $2$ into the Lindblad operator or into the symbol called $\gamma_\phi$. Always translate the equation, not just the name of the rate.

For how this rate relation fits with Ramsey $T_2^*$, echo times, filter functions, and spatial decoherence estimates, see Decoherence Timescales.

In a driven two-level optical or microwave transition, this same pure-dephasing contribution enters the transverse rate of the Optical Bloch Equations.

General Diagonal Lindblad Operator

The qubit model is a special case of dephasing by a Hermitian Lindblad operator. Suppose

$$L=L^\dagger, \qquad L\lvert m\rangle = \ell_m\lvert m\rangle.$$

For

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

the matrix elements in the $L$ eigenbasis obey

$$\left(\dot\rho\right)_{mn} = - \frac{\gamma}{2} (\ell_m-\ell_n)^2 \rho_{mn}.$$

Diagonal entries have $m=n$ and are unchanged. Coherences between states with different $\ell_m$ decay. Coherences inside a degenerate eigenspace are protected from this dissipator.

For example, $L=\operatorname{diag}(0,1,2)$ gives decay rates $\gamma/2$ for $\rho_{01}$ and $\rho_{12}$, but $2\gamma$ for $\rho_{02}$. A separation twice as large in eigenvalue produces four times the rate. If instead $L=\operatorname{diag}(0,0,2)$, the entire $0,1$ block is unchanged by this dissipator. Hamiltonian coupling out of that block can still destroy protection.

Microscopic Origin

A simple microscopic source of pure dephasing is longitudinal system–bath coupling:

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

with

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

Assume Hermitian $A,B$ and a well-defined Hamiltonian. Then $[H,H_S]=0$, so bare system energy is exactly conserved, not merely to leading order. Choose a simultaneous eigenbasis of $A$ and $H_S$; its populations are fixed. Degenerate energy subspaces need not preserve populations in an arbitrary different basis. The resulting coherence envelope need not be exponential before a Markov approximation.

For a qubit, one often writes

$$H_I = \frac{\hbar}{2}\sigma_z\otimes B.$$

Here $B$ has angular-frequency units. Define the centered, ordered two-sided spectrum

$$S_{BB}(\omega)=\int_{-\infty}^{\infty}dt\,e^{i\omega t} \langle B(t)B(0)\rangle.$$

When the weak-coupling, short-memory, zero-frequency limit is valid, the dissipator coefficient is $S_{BB}(0)/4$, giving exactly $\Gamma_\phi=S_{BB}(0)/2$ in this coupling convention. Mean frequency shifts are included in $\omega_0$. Colored or quasistatic noise cannot generally be replaced by this constant rate. Sampling zero Bohr frequency means no energy transition; it does not require the physical noise to be static.

Pulse sequences can reshape this low-frequency sensitivity; see Dynamical Decoupling for spin echo, CPMG, and filter-function language.

Classical Phase Noise

Pure dephasing can also arise from a classical stochastic Hamiltonian,

$$H(t) = \frac{\hbar}{2} \left[ \omega_0+\xi(t) \right]\sigma_z.$$

For a single noise realization, the state evolves unitarily. After averaging over noise realizations, the off-diagonal element becomes

$$\rho_{01}(t) = e^{-i\omega_0 t} \left\langle e^{-i\int_0^t \xi(s)\,ds} \right\rangle \rho_{01}(0).$$

If the phase has zero mean and is Gaussian with variance $2\Gamma_\phi t$, the average produces exponential decay:

$$\left\langle e^{-i\phi(t)} \right\rangle = e^{-\Gamma_\phi t}.$$

Classical phase noise is not the only origin of dephasing. A quantum environment can produce the same reduced master equation while retaining quantum correlations with the system.

Measurement Interpretation

An unread measurement of a diagonal observable causes dephasing. In the Markovian continuous-measurement limit, the unconditional state often obeys a dephasing equation of the form

$$\dot\rho = \Gamma_m\mathcal D[A]\rho,$$

where $A$ is the monitored observable. The measurement record contains information about $A$; if the record is ignored, the ensemble state loses coherence between distinct $A$ eigenvalues.

This does not mean that every dephasing process is literally a performed measurement or that its actual environment contains a readable eigenvalue record. The master equation alone does not determine that microscopic interpretation.

For outcome-resolved measurements, see Quantum Instruments. For the unconditional channel, see Selective and Nonselective Measurements.

Oscillator Number Dephasing

For a harmonic oscillator, on the appropriate generator domain, phase diffusion in the energy basis is often modeled by

$$\dot\rho = \gamma_n\mathcal D[n]\rho, \qquad n=a^\dagger a.$$

In the number basis,

$$n\lvert m\rangle=m\lvert m\rangle,$$

so

$$\dot\rho_{mn} = - \frac{\gamma_n}{2} (m-n)^2\rho_{mn}.$$

Number states are unaffected by this dissipator, while superpositions of different number states lose relative phase. This model appears in cavity and circuit-QED contexts as number dephasing, dispersive measurement backaction, or phase diffusion, depending on the physical origin.

Do not confuse number dephasing with photon loss. Photon loss uses $L=a$ and changes number populations; number dephasing uses $L=n$ and leaves number populations fixed.

Exponential free decay does not determine an echo experiment

Static detuning illustrates why the noise model matters even when the free-decay envelope is known. Suppose each run has a constant random $\xi$ drawn from the Cauchy distribution

$$p(\xi)=\frac{\Gamma_\phi}{\pi(\xi^2+\Gamma_\phi^2)}.$$

Its Fourier transform is $\langle e^{-i\xi t}\rangle=e^{-\Gamma_\phi|t|}$. Thus its one-time free-evolution maps for $t\ge0$ form exactly the same dephasing semigroup as the constant-rate equation above. Nevertheless, an ideal midpoint echo cancels each run’s phase, $\xi T/2-\xi T/2=0$, restoring full contrast.

For Gaussian white noise with $\langle\xi(t)\xi(s)\rangle=2\Gamma_\phi\delta(t-s)$, changing the sign at the midpoint does not alter the phase variance, so the echo envelope remains $e^{-\Gamma_\phi T}$. The identical free-decay curves therefore do not establish identical multi-time physics. The Cauchy distribution has no finite variance; do not apply the Gaussian variance formula to it. See decoherence timescales for finite-variance Ramsey and echo examples.

Exercises

Off-diagonal decay

For

$$\dot\rho = \frac{\Gamma_\phi}{2} \left( \sigma_z\rho\sigma_z-\rho \right),$$

show that $\dot\rho_{01}=-\Gamma_\phi\rho_{01}$.

Solution

Since $\sigma_z=\operatorname{diag}(1,-1)$,

$$(\sigma_z\rho\sigma_z)_{01} = -\rho_{01}.$$

Therefore

$$\dot\rho_{01} = \frac{\Gamma_\phi}{2} \left( -\rho_{01}-\rho_{01} \right) = -\Gamma_\phi\rho_{01}.$$

Bloch-vector contraction

Use the same master equation to derive $\dot r_x=-\Gamma_\phi r_x$ and $\dot r_y=-\Gamma_\phi r_y$ in the interaction picture.

Solution

The density matrix is

$$\rho = \frac12 \begin{pmatrix} 1+r_z & r_x-ir_y\\ r_x+ir_y & 1-r_z \end{pmatrix}.$$

The off-diagonal relation gives

$$\frac{d}{dt}(r_x-ir_y) = -\Gamma_\phi(r_x-ir_y).$$

Equating real and imaginary parts gives

$$\dot r_x=-\Gamma_\phi r_x, \qquad \dot r_y=-\Gamma_\phi r_y.$$

General Hermitian dephasing

Let $L=L^\dagger$ and $L\lvert m\rangle=\ell_m\lvert m\rangle$. Derive the decay rate of $\rho_{mn}$ under $\dot\rho=\gamma\mathcal D[L]\rho$.

Solution

The positive term gives

$$(L\rho L)_{mn} = \ell_m\ell_n\rho_{mn}.$$

The anticommutator gives

$$\frac12(L^2\rho+\rho L^2)_{mn} = \frac12 (\ell_m^2+\ell_n^2) \rho_{mn}.$$

Thus

$$(\dot\rho)_{mn} = \gamma \left[ \ell_m\ell_n - \frac12(\ell_m^2+\ell_n^2) \right] \rho_{mn} = - \frac{\gamma}{2} (\ell_m-\ell_n)^2 \rho_{mn}.$$

T-time relation

An undriven qubit has population relaxation rate $T_1^{-1}$ and pure-dephasing rate $T_\phi^{-1}$. What is the standard weak-coupling expression for $T_2^{-1}$?

Solution

Population relaxation contributes half its rate to transverse coherence decay, and pure dephasing contributes directly:

$$\frac{1}{T_2} = \frac{1}{2T_1} + \frac{1}{T_\phi}.$$

This relation assumes the usual Markovian qubit model with relaxation and pure dephasing separated in the same basis.

References

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