advanced34 min studyLesson 416 in Physics

Pure dephasing and coherence times

Separate longitudinal relaxation, transverse decay, and reversible static phase spread.

On this page

What you’ll understand

  • Derive the coherence decay rate for a declared dephasing convention.
  • Extract a pure-dephasing time from measured T1 and T2.
  • Distinguish homogeneous Markov decay from inhomogeneous phase broadening.

A qubit can lose phase coherence without changing its energy populations. Conversely, energy relaxation already attenuates coherence even when no separate phase noise is present. The familiar times T1 and T2 measure different modes of a reduced equation, so their relation must be derived from a stated model. Static frequency spread, echo sequences, and memory can change which experimental decay deserves the same symbol.

A commuting noise direction leaves populations fixed

Take the energy basis as the eigenbasis of Z=diag⁡(1,−1)Z=\operatorname{diag}(1,-1). Conjugation by Z leaves diagonal density entries unchanged and reverses off-diagonal signs. A dissipator proportional to the difference between that conjugation and the original state therefore damps only coherence. The coefficient convention matters: the same symbol is sometimes used for the dissipator prefactor and sometimes for the actual off-diagonal decay rate, producing a factor-of-two ambiguity unless the equation is written explicitly.

Here define the pure-dephasing rate as the actual exponential coherence-decay rate. A Hamiltonian diagonal in the same basis contributes only a deterministic phase, removed in an interaction frame. Since the dissipator does not change populations, it exchanges no mean energy with the system Hamiltonian in this ideal model. This statement does not mean that all microscopic environment degrees of freedom are unchanged, nor that a more general noise coupling cannot cause both phase and energy effects.

ρ˙=γϕ2(ZρZ−ρ),ρ01(t)=e−γϕtρ01(0),ρ˙00=ρ˙11=0\dot\rho=\frac{\gamma_\phi}{2}(Z\rho Z-\rho),\qquad\rho_{01}(t)=e^{-\gamma_\phi t}\rho_{01}(0),\qquad\dot\rho_{00}=\dot\rho_{11}=0
γφ is the pure-dephasing rate in s⁻¹ and t is seconds. The displayed solution is in a frame removing a commuting energy splitting. This time-independent Markov model assumes γφ ≥ 0 and no population-changing jumps.↔ Scroll wide equations horizontally.

Population and transverse modes add differently

Add upward and downward amplitude damping with constant nonnegative rates. Their sum governs the return of excited population to its stationary value, and half their sum contributes to the coherence rate. Independent pure dephasing adds directly to that coherence rate. These statements follow from acting with each dissipator on a diagonal population mode and on an off-diagonal matrix unit. They do not require assigning the same decay constant to every matrix element.

Defining the longitudinal and transverse times by these exponentials yields the usual bound that T2 cannot exceed twice T1 within this model. A reported violation may signal inconsistent fitted definitions, nonexponential decay, correlated noise, experimental error, or a model outside these assumptions. It does not license a negative constant Lindblad dephasing rate. At infinite T1, pure dephasing still gives finite T2 while every initial energy population remains stationary.

Γ1=γ↓+γ↑=1T1,Γ2=Γ12+γϕ=1T2,1Tϕ=1T2−12T1≥0\Gamma_1=\gamma_\downarrow+\gamma_\uparrow=\frac1{T_1},\qquad \Gamma_2=\frac{\Gamma_1}{2}+\gamma_\phi=\frac1{T_2},\qquad \frac1{T_\phi}=\frac1{T_2}-\frac1{2T_1}\ge0
All Γ and γ quantities have units s⁻¹; all T quantities are seconds. Tφ is infinite when its inverse vanishes. The relation assumes additive, constant-rate Markov dissipators, no coherent drive, and T1 defined for deviations from stationary population.↔ Scroll wide equations horizontally.

An echo diagnoses a different kind of phase loss

An ensemble can show reduced transverse signal because its members precess at different but nearly static frequencies. Each member may remain pure, yet their averaged density matrix loses coherence. An ideal refocusing pulse can reverse this accumulated static phase spread, whereas a single deterministic phase correction cannot undo unknown fluctuating noise. The often-used free-induction time T2-star therefore need not equal the homogeneous transverse time extracted after appropriate refocusing.

Echo success is not a universal proof that the environment is classical or that all decoherence is reversible locally. Pulse errors, noise correlation times, and the measured ensemble determine the result. Pure-state Bloch geometry can display individual phases but cannot show an averaged vector shrinking without an explicit ensemble calculation. Diagnose a loss of coherence by measuring transverse observables or reconstructing the density matrix; computational-basis population measurements alone miss pure dephasing entirely.

Separate two mechanisms with equal measured times

A vacuum-relaxing qubit has T1=T2=10 μsT_1=T_2=10\,\mu\mathrm{s} and starts in ∣+⟩|+\rangle. Assume the additive constant-rate model. Find TϕT_\phi and the full interaction-frame state at t=10ln⁡2 μst=10\ln2\,\mu\mathrm{s}.

  1. The rates are Γ1=0.1 μs−1\Gamma_1=0.1\,\mu\mathrm{s}^{-1} and Γ2=0.1 μs−1\Gamma_2=0.1\,\mu\mathrm{s}^{-1}. Subtracting half the longitudinal rate gives γϕ=0.05 μs−1\gamma_\phi=0.05\,\mu\mathrm{s}^{-1}, hence Tϕ=20 μsT_\phi=20\,\mu\mathrm{s}.

  2. The vacuum stationary excited population is zero. Starting at one half, the population at this time is (1/2)e−t/T1=(1/2)(1/2)=1/4(1/2)e^{-t/T_1}=(1/2)(1/2)=1/4; ground population is therefore 3/43/4.

  3. The initial coherence is one half and decays with T2, so it also becomes 1/41/4. The determinant is 3/16−1/16=1/83/16-1/16=1/8, and purity is 1−2(1/8)=3/41-2(1/8)=3/4, confirming a positive mixed state.

The result

The pure-dephasing time is 20 μs. The state is [[3/4, 1/4], [1/4, 1/4]], with purity 3/4; equal T1 and T2 do not mean pure dephasing is absent.

Explore the model

Interactive lab

Interactive lab · quantum

Equal energy probabilities do not fix phase

Reset and set both controls to θ = 90° and φ = 60°, then record Bloch x = 0.5 and Bloch y ≈ 0.866. Reset and set both to θ = 90° and φ = −60°; predict the same x component and the opposite y component, with both basis probabilities 1/2. Keep playback paused. Their equal classical mixture would have Bloch length 0.5, but that mixture is a calculation outside this pure-state lab: neither dephasing nor ensemble averaging is rendered.

Equal energy probabilities do not fix phaseA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forIndividual phase rotations preserve purity; the average of different phases can reduce coherence.

Check your understanding

Within the constant-rate additive model, T2 equals 2T1. What does this imply?

Try it yourself

A pure-dephasing qubit has fixed populations 0.8 and 0.2, initial real coherence 0.4, and γϕ=2 s−1\gamma_\phi=2\,\mathrm{s}^{-1}. Find the matrix, purity, and plus-state probability at t=(ln⁡2)/2 st=(\ln2)/2\,\mathrm{s}.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Only the off-diagonal entries change; use the sum of squared diagonal entries plus twice the squared coherence.

Compare with the solution

The coherence factor is one half, so the output has diagonal entries 0.8 and 0.2 and off-diagonal entries 0.2. Its determinant is 0.16−0.04=0.120.16-0.04=0.12, and its purity is 0.64+0.04+2(0.04)=0.760.64+0.04+2(0.04)=0.76. The plus probability is (1+2(0.2))/2=0.7(1+2(0.2))/2=0.7. The energy populations and mean energy stay fixed; the initial matrix was pure because its determinant was zero.

Take it with you

  • A dephasing-rate convention must specify the dissipator coefficient.
  • Relaxation contributes half its population rate to transverse decay.
  • Static phase spread and homogeneous decoherence need different experimental diagnoses.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.