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.

For why “coherence” always needs a specified basis or decomposition, see Coherence and Preferred Bases.

The practical distinction is:

Effect	Changes populations?	Changes coherences?	Typical timescale
pure dephasing	no, in the monitored basis	yes	$T_\phi$
energy relaxation	yes	yes, indirectly	$T_1$
total transverse decay	not a separate mechanism	yes	$T_2$

The slogan is useful:

$$\text{dissipation can cause decoherence,} \qquad \text{but decoherence need not be dissipation.}$$

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

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

closed evolution rotates the phase:

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

up to the sign convention for $Z$. There is no decay in an isolated system. 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 a simple Markovian convention.

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.

Common physical sources include fluctuating energy splittings, quasistatic frequency noise, which-path information, unread projective measurement in the energy basis, and elastic scattering that distinguishes alternatives without transferring energy.

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

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

The populations change, and the average energy decreases. 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

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

Finite-temperature damping is still dissipative because energy flows both ways until the system approaches a thermal state.

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 acts schematically as

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

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

This geometric picture is often the fastest diagnostic:

pure dephasing:      transverse shrinkage, no population drift
amplitude damping:   transverse shrinkage plus population drift
depolarizing noise:  isotropic shrinkage toward the center

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 oscillator quanta, and the $a^\dagger$ term adds thermal quanta. Together they relax the oscillator toward a thermal state.

This model also decoheres superpositions, but the mechanism is not merely random phase noise. Energy is exchanged with the bath. For oscillator coherent states, loss attenuates amplitude; for number-state superpositions, environmental records of emitted quanta suppress coherence between alternatives.

Measurement-Induced Dephasing

An unread projective 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.

Diagnostic Questions

To decide whether an effect is dephasing, dissipation, or both, ask:

• Are populations in the relevant basis changing?
• Is average system energy changing?
• Are off-diagonal terms decaying in a preferred basis?
• Does the environment learn which alternative occurred?
• Does the process drive the system toward a thermal or ground state?
• Is the observed decay of coherence explainable by $T_1$ alone, or is extra pure dephasing needed?
• Is the “dissipator” in a master equation describing energy flow, pure dephasing, or both?

The last question is important because the word “dissipator” in Lindblad notation is broader than thermodynamic dissipation. A Lindblad dissipator with $L=Z$ describes pure dephasing, even though no energy relaxation occurs for a Hamiltonian proportional to $Z$.

Common Mistakes

• Calling every decay of coherence dissipation.
• Calling every Lindblad dissipator energy loss.
• Assuming $T_2=T_1$. In simple qubit models, relaxation contributes only half its rate to transverse decay.
• Forgetting pure dephasing when $T_2$ is much shorter than $2T_1$.
• Using amplitude damping for a process that only randomizes phase.
• Using pure dephasing for a process that changes populations or heats the system.
• Ignoring the basis: dephasing in one basis can look like population mixing in another.
• Treating apparatus energy cost as identical to the reduced system’s dissipative dynamics.

References

• A. Abragam, The Principles of Nuclear Magnetism, Oxford University Press (1961).
• C. P. Slichter, Principles of Magnetic Resonance, Springer, 3rd ed. (1990).
• C. W. Gardiner and P. Zoller, Quantum Noise, Springer, 3rd ed. (2004).
• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press (2010).
• Y. Makhlin, G. Schön, and A. Shnirman, “Quantum-state engineering with Josephson-junction devices,” Reviews of Modern Physics 73, 357-400 (2001).

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.