Dephasing vs Dissipation

Dephasing and dissipation are both open-system effects, but they answer different physical questions.

Dephasing suppresses relative phase coherence in a chosen basis. Dissipation involves energy, excitation, or population flow between the system and its environment.

Here energy relaxation refers to irreversible population exchange in an energy basis. Population changes produced solely by a controlled unitary are not, by themselves, dissipation.

Helpful background. Coherence and Preferred Bases explains why phase coherence needs a specified basis. Density operators describe both energy populations and interference.

Effect	State change in the energy basis	Timescale in a simple qubit model
pure dephasing	suppresses off-diagonal terms, preserves populations	$T_\phi$
energy relaxation	drives populations toward equilibrium and usually changes coherence	$T_1$
transverse decay	combines relaxation and extra phase decay	$T_2$

Two-Level Reference Model

Consider a qubit with energy basis $\{|0\rangle,|1\rangle\}$ and density matrix

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

The diagonal entries are populations in this basis. The off-diagonal entries are coherences:

$$\rho_{01} = \langle0|\rho|1\rangle.$$

For a Hamiltonian

$$H = \hbar\omega_0|1\rangle\langle1| =\frac{\hbar\omega_0}{2}(I-Z),\qquad\omega_0>0,$$

closed evolution rotates the phase:

$$\rho_{01}(t) = e^{i\omega_0t}\rho_{01}(0)$$

with $Z=\operatorname{diag}(1,-1)$, so $|0\rangle$ is the ground state and $|1\rangle$ is excited. The conjugate element $\rho_{10}$ has the opposite phase. There is no decay in this closed two-level model. Dephasing or dissipation appears only when uncontrolled degrees of freedom are included or averaged over.

Pure Dephasing

Pure dephasing in the $Z$ basis leaves populations fixed while damping the off-diagonal terms:

$$\rho(t) = \begin{pmatrix} \rho_{00}(0)&e^{-\Gamma_\phi t}\rho_{01}(0)\\ e^{-\Gamma_\phi t}\rho_{10}(0)&\rho_{11}(0) \end{pmatrix}$$

in the rotating frame, with constant $\Gamma_\phi\ge0$.

A Lindblad form for this process is

$$\frac{d\rho}{dt} = \frac{\Gamma_\phi}{2} \left( Z\rho Z-\rho \right).$$

This gives

$$\frac{d\rho_{00}}{dt}=0, \qquad \frac{d\rho_{11}}{dt}=0,$$

and

$$\frac{d\rho_{01}}{dt} = -\Gamma_\phi\rho_{01}.$$

Thus pure dephasing destroys phase coherence without changing the energy populations.

Fluctuating energy splittings and unread energy-basis measurements can give this form. Quasi-static frequency noise instead generally gives a nonexponential ensemble envelope. Scattering can distinguish positions with little energy transfer to a heavy object, but finite recoil still produces momentum diffusion.

Dissipation and Energy Relaxation

Dissipation usually means irreversible energy exchange with uncontrolled degrees of freedom. For a zero-temperature qubit relaxation model, the excited state decays:

$$|1\rangle\longrightarrow |0\rangle.$$

The corresponding amplitude-damping master equation is

$$\frac{d\rho}{dt} = \Gamma_1\mathcal D[\sigma_-]\rho,$$

where

$$\mathcal D[L]\rho = L\rho L^\dagger - \frac12 \{L^\dagger L,\rho\}, \qquad \sigma_-=|0\rangle\langle1|.$$

It gives

$$\frac{d\rho_{11}}{dt} = -\Gamma_1\rho_{11}, \qquad \frac{d\rho_{00}}{dt} = \Gamma_1\rho_{11}.$$

For nonzero excited population, the average energy decreases. The ground state is unchanged. The coherence also decays:

$$\frac{d\rho_{01}}{dt} = -\frac{\Gamma_1}{2}\rho_{01}$$

apart from Hamiltonian phase rotation.

Thus relaxation is dissipative and decohering. It is not pure dephasing.

T1, T2, and Pure Dephasing Time

The relaxation time $T_1$ characterizes population relaxation. In the zero-temperature qubit model,

$$\rho_{11}(t) = e^{-t/T_1}\rho_{11}(0), \qquad T_1=\frac1{\Gamma_1}.$$

The transverse coherence time $T_2$ characterizes decay of off-diagonal coherence in the energy basis:

$$\rho_{01}(t) \sim e^{-t/T_2}e^{i\omega_0t}\rho_{01}(0).$$

When energy relaxation and independent pure dephasing both act, a common Markovian relation is

$$\frac1{T_2} = \frac1{2T_1} + \frac1{T_\phi}.$$

Here $T_\phi=1/\Gamma_\phi$ is the pure-dephasing time in the convention above.

This formula is not a definition of all possible open-system dynamics. It assumes a simple weak-coupling Markovian qubit model with a clear energy basis and separable relaxation and pure-dephasing contributions.

For how this relation fits into broader Ramsey, echo, noise-spectrum, and spatial-decoherence estimates, see Decoherence Timescales. Its operational use in CW linewidths, free-induction decays, and echo protocols is mapped in Magnetic Resonance Overview.

Finite Temperature

At finite temperature, upward and downward transitions can both occur:

$$|0\rangle\to |1\rangle, \qquad |1\rangle\to |0\rangle.$$

Let their rates be $\Gamma_\uparrow$ and $\Gamma_\downarrow$. Then

$$\frac1{T_1} = \Gamma_\uparrow+\Gamma_\downarrow.$$

The equilibrium population ratio satisfies detailed balance in a thermal bath:

$$\frac{\Gamma_\uparrow}{\Gamma_\downarrow} = e^{-\beta\hbar\omega_0}.$$

The coherence decay rate in the same simple Markovian setting is

$$\frac1{T_2} = \frac{\Gamma_\uparrow+\Gamma_\downarrow}{2} + \Gamma_\phi.$$

This assumes a fixed energy basis and nonnegative rates in a suitable weak-coupling thermal model. For the Hamiltonian above, the energy current is

$$\frac{d\langle H\rangle}{dt} =\hbar\omega_0(\Gamma_\uparrow\rho_{00}-\Gamma_\downarrow\rho_{11}).$$

It vanishes at the stationary population, although upward and downward transitions can both continue. Zero net mean-energy flow therefore does not prove that the system is uncoupled from a bath. Pure dephasing commuting with $H$ gives zero energy current for every input state, a stronger statement.

Bloch-Sphere Picture

For a qubit,

$$\rho = \frac12 \left( I+xX+yY+zZ \right).$$

Pure dephasing in the $Z$ basis acts as

$$(x,y,z) \longmapsto \left( e^{-t/T_\phi}x, e^{-t/T_\phi}y, z \right)$$

in the simplest convention. It shrinks transverse components and leaves the $Z$ population imbalance fixed.

Zero-temperature amplitude damping gives exactly, in the same rotating frame,

$$(x,y,z) \longmapsto \left( e^{-t/(2T_1)}x, e^{-t/(2T_1)}y, 1+e^{-t/T_1}(z-1) \right),$$

where $z=+1$ is the ground state. It both shrinks transverse coherence and pulls the state toward the ground-state pole.

For a concrete comparison, begin in $|+\rangle$ and evolve for a time equal to the stated decay constant. Pure dephasing at $t=T_\phi$ gives Bloch vector $(e^{-1},0,0)$ and leaves the excited population $1/2$. Amplitude damping at $t=T_1$ gives $(e^{-1/2},0,1-e^{-1})$ and excited population $e^{-1}/2$. The transverse components alone do not reveal the population drift; measuring $Z$ distinguishes the mechanisms.

Damped Oscillator

For a harmonic oscillator coupled to a thermal bath, dissipation is often modeled by

$$\frac{d\rho}{dt} = \kappa(\bar n+1)\mathcal D[a]\rho + \kappa\bar n\,\mathcal D[a^\dagger]\rho.$$

The $a$ term removes quanta, and the $a^\dagger$ term adds them. With $\kappa>0$, $\bar n\ge0$, and finite initial moments, they give $\langle n(t)\rangle=\bar n+[\langle n(0)\rangle-\bar n]e^{-\kappa t}$ and $\langle a(t)\rangle=e^{-\kappa t/2}\langle a(0)\rangle$ in the rotating frame.

The model is not merely random phase noise. At zero temperature, a coherent-state input remains a pure coherent state with attenuated amplitude, while a superposition of separated coherent states generally becomes mixed. Dissipation need not reduce the purity of every input or destroy every useful notion of coherence.

Measurement-Induced Dephasing

An unread ideal Lüders measurement in the $Z$ basis produces

$$\rho \longmapsto P_0\rho P_0+P_1\rho P_1.$$

For a qubit, this removes $\rho_{01}$ and $\rho_{10}$ while leaving $\rho_{00}$ and $\rho_{11}$ fixed. It is therefore dephasing in the measurement basis.

No energy need be dissipated if the measured observable commutes with the Hamiltonian and the apparatus interaction can be idealized as nondemolition. In a real detector, of course, amplification and readout may dissipate energy in the apparatus. The system-level channel and the apparatus thermodynamics are different descriptions.

For the measurement-theory version, see Projective Measurements.

The word dissipator does not imply energy loss

A Lindblad term $\mathcal D[L]$ is called a dissipator even when it describes energy-preserving dephasing. For $L=Z$ and the Hamiltonian used here, populations and mean energy are constant. For $L=|0\rangle\langle1|$, excited population relaxes. Evaluate the generator’s action on the relevant observable rather than inferring energy flow from its name.

The energy cost of amplifying or storing a measurement record is also distinct from the reduced system’s energy change. A measurement may preserve system energy while its detector consumes power.

Exercises

1. Pure dephasing from the Lindblad equation. For

$$\frac{d\rho}{dt} = \frac{\Gamma_\phi}{2} \left( Z\rho Z-\rho \right),$$

derive the equations of motion for $\rho_{00}$, $\rho_{11}$, and $\rho_{01}$.

Solution

In the $Z$ basis,

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

Therefore

$$Z\rho Z-\rho = \begin{pmatrix} 0&-2\rho_{01}\\ -2\rho_{10}&0 \end{pmatrix}.$$

Thus

$$\frac{d\rho_{00}}{dt}=0, \qquad \frac{d\rho_{11}}{dt}=0, \qquad \frac{d\rho_{01}}{dt}=-\Gamma_\phi\rho_{01}.$$

The process is pure dephasing in this basis.

2. Relaxation contribution to T2. For zero-temperature amplitude damping with rate $\Gamma_1$, show that $\rho_{01}$ decays at rate $\Gamma_1/2$.

Solution

Use

$$\frac{d\rho}{dt} = \Gamma_1 \left( \sigma_-\rho\sigma_+ - \frac12\{\sigma_+\sigma_-,\rho\} \right),$$

where $\sigma_+\sigma_-=|1\rangle\langle1|$. The jump term $\sigma_-\rho\sigma_+$ contributes only to the $|0\rangle\langle0|$ element. For the off-diagonal element,

$$\frac{d\rho_{01}}{dt} = - \frac{\Gamma_1}{2} \langle0| \left( |1\rangle\langle1|\rho + \rho|1\rangle\langle1| \right) |1\rangle.$$

The first term inside the parentheses gives zero, and the second gives $\rho_{01}$. Hence

$$\frac{d\rho_{01}}{dt} = -\frac{\Gamma_1}{2}\rho_{01}.$$

3. Classify the process. Classify each process as primarily dephasing, primarily dissipative, or both: random fluctuations of a qubit transition frequency; spontaneous emission into vacuum; an unread ideal measurement of $Z$; thermal relaxation of an oscillator.

Solution

Random transition-frequency fluctuations are primarily dephasing in the energy basis if they do not cause transitions.

Spontaneous emission into vacuum is dissipative because energy leaves the qubit, and it also decoheres superpositions involving the excited state.

An unread ideal measurement of $Z$ is dephasing in the $Z$ basis at the system-description level. The apparatus may dissipate energy during amplification, but the reduced system channel need not be energy relaxation.

Thermal relaxation of an oscillator is dissipative because quanta are exchanged with the bath. It also decoheres superpositions because environmental records and thermal fluctuations suppress phase coherence.

References

• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002). Publisher.
• C. W. Gardiner and P. Zoller, Quantum Noise, 3rd edition, Springer (2004). DOI.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary edition, Cambridge University Press (2010).
• C. P. Slichter, Principles of Magnetic Resonance, 3rd edition, Springer (1990). DOI.