Relaxation and Dephasing: T1 and T2 Times

Hand any experimentalist a new qubit and the first two numbers they will quote back are $T_1$ and $T_2$ . They are the workhorse figures of merit for every physical qubit ever built, because they answer the only question that ultimately matters for computation: how long does the quantum information survive? A qubit is never truly isolated — it is coupled, however weakly, to a bath of electromagnetic modes, phonons, spins, and defects — and that coupling degrades its state in two physically distinct ways. Energy relaxation is the irreversible exchange of a quantum with the bath: the excited state $|1\rangle$ decays toward thermal equilibrium, emptying populations at rate $1/T_1$ . Dephasing is the loss of the definite phase relation between $|0\rangle$ and $|1\rangle$ with no energy transfer at all, erasing the off-diagonal coherences at rate $1/T_2$ . The two are not independent. Because a coherence is built from the amplitudes of two levels, energy relaxation automatically drags the coherence down at half the population rate, so even a perfectly phase-quiet qubit obeys $T_2 = 2T_1$ . Any genuine phase noise — the pure-dephasing channel with time $T_\phi$ — only adds to that rate, giving the master relation $1/T_2 = 1/(2T_1) + 1/T_\phi$ and the hard ceiling $T_2 \le 2T_1$ . This article takes those relations apart from three directions at once: the geometric Bloch-ball picture imported from nuclear magnetic resonance, the microscopic Lindblad master equation whose jump operators are $\sigma_-$ and $\sigma_z$ , and the laboratory pulse sequences — inversion recovery, Ramsey, and the Hahn echo — that actually measure the numbers and separate the intrinsic $T_2$ from the inhomogeneous $T_2^*$ .

Key takeaways

$T_1$ is the longitudinal (energy) relaxation time: how long the excited-state population survives before the qubit dumps a quantum into the bath and relaxes toward equilibrium — the amplitude-damping channel.
$T_2$ is the transverse (coherence) time: how long the off-diagonal density-matrix element, the phase memory of a superposition, survives before it is scrambled.
They are locked together by $1/T_2 = 1/(2T_1) + 1/T_\phi$ , where $T_\phi$ is the pure-dephasing time from energy-conserving phase noise, so complete positivity forces the ceiling $T_2 \le 2T_1$ — a reported $T_2 > 2T_1$ is always an error.
Both times fall straight out of the Lindblad master equation: the lowering operator $\sigma_-$ gives $T_1$ , the dephaser $\sigma_z$ gives $T_\phi$ , and reading off the matrix elements reproduces the master relation exactly.
Measurement separates the mechanisms: inversion recovery gives $T_1$ , a Ramsey fringe gives the inhomogeneous $T_2^*$ , and a Hahn echo refocuses static noise to recover the intrinsic $T_2$ , with $T_2^* \le T_2 \le 2T_1$ .
Key idea: energy relaxation ( $T_1$ ) damps populations and unavoidably halves coherence, while pure dephasing ( $T_\phi$ ) damps only coherence; the transverse rate is the sum $1/T_2 = 1/(2T_1) + 1/T_\phi$ , so $T_2 \le 2T_1$ always.

The core insight

Energy relaxation ( $T_1$ ) damps populations and unavoidably halves coherence; pure dephasing ( $T_\phi$ ) damps only coherence. The total transverse-decay rate is the sum, $1/T_2 = 1/(2T_1) + 1/T_\phi$ , so $T_2 \le 2T_1$ always.

Where this lives today

$T_1$ and $T_2$ are the currency of quantum-hardware progress, and the last two decades have been one long campaign to lengthen them. Superconducting-qubit coherence has climbed from the nanoseconds of the first Cooper-pair box to hundreds of microseconds in three-dimensional and tantalum-based transmons, and millisecond-scale $T_2$ in fluxonium — every gain won by attacking a specific term in the story above: better dielectrics to starve the two-level-system bath that limits $T_1$ , flux and charge sweet spots to suppress the $1/f$ noise that limits $T_\phi$ , and Purcell filters to close the readout-loss channel. The Hahn echo and its multi-pulse descendants, the CPMG and Carr–Purcell trains, now run in reverse as noise spectroscopy: sweeping the pulse spacing reshapes the filter that the sequence applies to the environment, reconstructing the noise spectral density $S(\omega)$ and diagnosing which bath dominates a device. Every cloud quantum processor publishes per-qubit $T_1$ and $T_2$ maps, and the ratio $T_2/t_{\mathrm{gate}}$ sets the circuit depth a device can run before decoherence swamps the signal — the number that must clear the surface-code threshold for fault tolerance. The open frontier is the noise itself: the microscopic origin of $1/f$ flux and charge noise, and the individual two-level defects that occasionally and unpredictably swallow a qubit's $T_1$ for minutes at a time, remain among the most stubborn unsolved problems in the field.

The physical picture

Picture the state of a single qubit as a point in the Bloch ball, with Bloch vector $\vec r = (r_x, r_y, r_z)$ . The north pole is the ground state $|0\rangle$ , the south pole the excited state $|1\rangle$ , and every superposition lives on the equatorial disc where the transverse components $(r_x, r_y)$ encode the phase. Decoherence acts on this vector along two orthogonal directions. Relaxation is vertical motion: the longitudinal component $r_z$ slides back toward its thermal-equilibrium value at rate $1/T_1$ , the qubit falling toward the ground state. Dephasing is radial collapse of the equatorial projection: the transverse components shrink at rate $1/T_2$ , an initially pure equatorial state spiralling inward until the vector reaches the centre and the state is maximally mixed. The first figure puts both motions, and both decay curves, side by side with real numbers.

Figure

A qubit in the Bloch ball loses energy vertically at rate 1/T₁ and phase radially at rate 1/T₂. With T₁ = 80 μs and T₂ = 100 μs the coherence outlives the population — legal because the only ceiling is T₂ ≤ 2T₁ = 160 μs — and the master relation 1/T₂ = 1/(2T₁) + 1/Tφ gives a pure-dephasing time Tφ ≈ 267 μs

Why the coherence can outlive the population — as it does in the figure, where $T_2 = 100\ \mu\mathrm{s}$ exceeds $T_1 = 80\ \mu\mathrm{s}$ — is worth pausing on, because it trips people up. A coherence $\rho_{01}$ is a product of the probability amplitude of being in $|0\rangle$ and the conjugate amplitude of being in $|1\rangle$ . When relaxation depletes the excited population at rate $1/T_1$ , the amplitude of $|1\rangle$ falls at half that rate, $1/(2T_1)$ , so the coherence inherits a decay of only $1/(2T_1)$ — half the population rate. That is why the true ceiling is not $T_2 \le T_1$ but the looser $T_2 \le 2T_1$ . Pure dephasing is a separate, energy-conserving channel: the environment effectively keeps reading out which level the qubit occupies, in the $\sigma_z$ basis, without ever absorbing the excitation. Each stolen glance randomizes the relative phase a little, and many small kicks add up to an exponential decay of coherence with no population change at all — its rate simply adds to $1/(2T_1)$ . Reading the numbers off a qubit is a matter of choosing the right pulse sequence, and the three canonical sequences map cleanly onto the three timescales. An inversion-recovery experiment flips the qubit and watches the population climb back, giving $T_1$ . A Ramsey sequence sandwiches a free-evolution interval between two $\pi/2$ pulses and watches the fringe decay, but that decay includes quasi-static, shot-to-shot frequency spread, so it returns the inhomogeneous time $T_2^*$ . A Hahn echo inserts a single $\pi$ pulse at the midpoint, which flips the accumulated phase and cancels any offset that was constant across the sequence, recovering the intrinsic $T_2$ . The ordering is universal: $T_2^* \le T_2 \le 2T_1$ . The second figure is the experimentalist's toolbox drawn as pulse timelines, with the resulting ladder of times on the right.

Figure

The measurement toolbox. Inversion recovery reads T₁ from population recovery; a Ramsey sequence between two π/2 pulses decays at the inhomogeneous rate 1/T₂; a Hahn echo inserts a midpoint π pulse to refocus static noise and recover the intrinsic T₂. The times obey T₂ ≤ T₂ ≤ 2T₁, shown here as 20 μs ≤ 100 μs ≤ 160 μs

The core equations

The single equation that organizes the whole subject is the master relation between the three times. It says the transverse-decay rate is a sum of two contributions — half the longitudinal rate, inherited from relaxation, plus the pure-dephasing rate — and it is the reason $T_2$ can never exceed twice $T_1$ .

\frac{1}{T_2} = \frac{1}{2T_1} + \frac{1}{T_\phi} \qquad (1)

where: $T_1$ — longitudinal (energy) relaxation time; $T_2$ — transverse (coherence) relaxation time; $T_\phi$ — pure-dephasing time, energy-conserving phase randomization. Reading: because $1/T_\phi \ge 0$ , the smallest the transverse rate can be is $1/(2T_1)$ , reached when there is no pure dephasing at all, giving $T_2 = 2T_1$ ; any real dephasing only shortens $T_2$ . Phenomenologically the dynamics are the decay part of the Bloch equations imported from magnetic resonance. In a frame rotating at the qubit frequency the coherent precession is removed, leaving pure exponential relaxation of the longitudinal and transverse components toward their equilibria.

\dot r_z = -\frac{1}{T_1}\left(r_z - r_z^{\mathrm{eq}}\right), \qquad \dot r_{x,y} = -\frac{1}{T_2}\, r_{x,y} \qquad (2)

where: $r_z$ — longitudinal Bloch component (population imbalance); $r_{x,y}$ — transverse components (coherence); $r_z^{\mathrm{eq}}$ — thermal-equilibrium value of $r_z$ . Reading: the longitudinal motion relaxes toward equilibrium, the transverse motion toward zero, at two independent rates whose only constraint is the master relation (1). Integrating equation (2) gives the two decay laws an experiment fits to directly: inversion recovery for the population, and a simple exponential for the coherence.

r_z(t) = r_z^{\mathrm{eq}} + \bigl(r_z(0) - r_z^{\mathrm{eq}}\bigr)e^{-t/T_1}, \qquad r_{x,y}(t) = r_{x,y}(0)\, e^{-t/T_2} \qquad (3)

where: $r_z(0)$ — initial longitudinal component, e.g. $-1$ just after an inversion pulse; $r_{x,y}(0)$ — initial transverse component, e.g. $1$ just after a $\pi/2$ pulse. Reading: $T_1$ is the $1/e$ time of the population's return to equilibrium, $T_2$ the $1/e$ time of the coherence's collapse. Where do the times come from microscopically? From the Lindblad (GKSL) master equation for the qubit, with two dissipators: the lowering operator $\sigma_-$ that carries energy to the bath, and the dephaser $\sigma_z$ that only scrambles phase.

\dot\rho = -\frac{i}{\hbar}[H,\rho] + \gamma\Bigl(\sigma_-\rho\,\sigma_+ - \tfrac12\{\sigma_+\sigma_-,\rho\}\Bigr) + \frac{\gamma_\phi}{2}\Bigl(\sigma_z\rho\,\sigma_z - \rho\Bigr) \qquad (4)

where: $\gamma = 1/T_1$ — energy-relaxation (amplitude-damping) rate; $\gamma_\phi = 1/T_\phi$ — pure-dephasing rate of the $\sigma_z$ channel; $\sigma_- = |0\rangle\langle 1|$ — lowering operator; $\sigma_z$ — Pauli- $z$ operator; $H = \tfrac{\hbar\omega_0}{2}\sigma_z$ . Reading: the first dissipator empties the excited state and half-damps the coherence; the second leaves populations alone and damps only the coherence — the two channels of the master relation, made concrete. The $T_1$ channel, expressed as a completely positive map, is amplitude damping: with probability $p$ the excited state emits and decays to the ground state, and the surviving no-jump branch scales the coherence by $\sqrt{1-p}$ .

K_0 = \begin{pmatrix}1 & 0 \\ 0 & \sqrt{1-p}\end{pmatrix}, \quad K_1 = \begin{pmatrix}0 & \sqrt{p} \\ 0 & 0\end{pmatrix}, \quad p = 1 - e^{-t/T_1} \qquad (5)

where: $p$ — probability of an emission event in time $t$ ; $K_0, K_1$ — Kraus operators for the no-decay and decay branches. Reading: the excited population is scaled by $1-p = e^{-t/T_1}$ , while the coherence is scaled by $\sqrt{1-p} = e^{-t/(2T_1)}$ — the half-rate, in one line. The $T_\phi$ channel is phase damping: it leaves the populations exactly invariant and multiplies the coherence by a decaying exponential. Together with the sequences that measure it, the Ramsey fringe and the Hahn echo carry the extra $T_2^*$ time set by static, refocusable spread.

\rho_{01} \mapsto e^{-t/T_\phi}\,\rho_{01}, \qquad S_{\mathrm{Ramsey}}(t) = e^{-t/T_2^*}\cos(\delta t), \qquad S_{\mathrm{Hahn}}(2\tau) = e^{-2\tau/T_2} \qquad (6)

where: $\rho_{01}$ — qubit coherence, the off-diagonal element; $\delta$ — detuning of the drive from the qubit frequency; $\tau$ — free-evolution time per arm of the echo; $T_2^*$ — Ramsey coherence time, including inhomogeneous broadening. Reading: pure dephasing touches only coherences; the Ramsey signal decays at the inhomogeneous rate $1/T_2^*$ , while the echo's midpoint $\pi$ pulse strips the static part and recovers the intrinsic $1/T_2$ .

Derivation AFrom the Lindblad equation to the T1 and T2 relation

The goal is to derive the master relation $1/T_2 = 1/(2T_1) + 1/T_\phi$ from first principles by solving the qubit Lindblad equation for its density-matrix elements, and to see exactly where the factor of two in the coherence rate comes from — namely, from the anticommutator of the amplitude-damping dissipator touching only one of the two indices of the coherence. We take the Hamiltonian $H = \tfrac{\hbar\omega_0}{2}\sigma_z$ and the dissipator of equation (4), write the density matrix in components, and evolve the diagonal and off-diagonal entries separately.

Step 1. Evolve the excited population. Acting on $\rho_{11} = \langle 1|\rho|1\rangle$ , the gain term $\sigma_-\rho\,\sigma_+$ deposits probability into $\rho_{00}$ , while the loss term $-\tfrac12\{\sigma_+\sigma_-,\rho\}$ removes it from $\rho_{11}$ , and the $\sigma_z$ dephaser commutes with diagonal projectors and contributes nothing, so $\dot\rho_{11} = -\gamma\,\rho_{11}$ and hence $\rho_{11}(t) = \rho_{11}(0)e^{-\gamma t}$ . Why: the excited-state decay is a pure exponential whose rate is $\gamma$ , which is the definition $T_1 = 1/\gamma$ .

Step 2. Separate the coherent part of the coherence. The Hamiltonian contributes $-\tfrac{i}{\hbar}[H,\rho]\big|_{01} = -i\omega_0\rho_{01}$ , a rotation at the qubit frequency. Why: this is free precession, not decay; moving to a frame rotating at $\omega_0$ removes it and leaves only the dissipative terms, which is exactly what the rotating-frame Bloch equation (2) does.

Step 3. Evolve the coherence under the dissipator. The amplitude-damping anticommutator gives $-\tfrac{\gamma}{2}\rho_{01}$ , because only one of the two indices of $\rho_{01}$ touches the excited state that $\sigma_+\sigma_- = |1\rangle\langle 1|$ projects onto; the dephaser gives $\tfrac{\gamma_\phi}{2}(\sigma_z\rho\,\sigma_z - \rho)\big|_{01} = -\gamma_\phi\rho_{01}$ , since $\sigma_z$ flips the sign of off-diagonals so $\sigma_z\rho\,\sigma_z\big|_{01} = -\rho_{01}$ . Adding them, $\dot\rho_{01} = -(\tfrac{\gamma}{2} + \gamma_\phi)\rho_{01}$ . Why: the two dissipators act additively on the coherence, one at half the population rate and one at the full dephasing rate.

Step 4. Read off the times. Integrating, $\rho_{01}(t) = \rho_{01}(0)\exp[-(\tfrac{\gamma}{2} + \gamma_\phi)t] \equiv \rho_{01}(0)e^{-t/T_2}$ , so $1/T_2 = \gamma/2 + \gamma_\phi = 1/(2T_1) + 1/T_\phi$ . Why: identifying the coherence-decay exponent with $1/T_2$ delivers the master relation, and because $\gamma_\phi \ge 0$ it immediately gives the ceiling $T_2 \le 2T_1$ . *Why this is consistent: with $\gamma_\phi = 0$ the map reproduces $\rho_{11} \to e^{-t/T_1}\rho_{11}$ and $\rho_{01} \to e^{-t/(2T_1)}\rho_{01}$ , i.e. $T_2 = 2T_1$ ; adding the dephaser multiplies the coherence by the extra factor $e^{-t/T_\phi}$ .*

Go deeper · Derivation B · 1 minCoherence decay of the amplitude-damping channel

Derivation BCoherence decay of the amplitude-damping channel

The goal is to confirm the factor of two directly from the completely positive Kraus form of the amplitude-damping channel, independently of the master equation, showing that energy relaxation alone forces the coherence to decay at exactly half the population rate and therefore pins the ceiling $T_2 = 2T_1$ whenever pure dephasing is absent. We apply the channel $\mathcal{E}(\rho) = K_0\rho K_0^\dagger + K_1\rho K_1^\dagger$ of equation (5), with jump probability $p = 1 - e^{-t/T_1}$ , to a general single-qubit density matrix and track the population and the coherence.

Step 1. Track the excited population. The no-jump operator $K_0$ scales $\rho_{11}$ by $(1-p)$ , while the jump operator $K_1$ moves probability $p\,\rho_{11}$ into the ground state, so the excited population becomes $\rho_{11} \mapsto (1-p)\rho_{11} = e^{-t/T_1}\rho_{11}$ . Why: this reproduces the $T_1$ exponential and confirms that $p$ is exactly the emission probability accumulated in time $t$ .

Step 2. Track the coherence. Only $K_0$ has a nonzero contribution to the $(0,1)$ entry; $K_1$ maps every input into the $|0\rangle\langle 0|$ corner and produces no off-diagonal term, so $\rho_{01} \mapsto (K_0\rho K_0^\dagger)_{01} = \sqrt{1-p}\,\rho_{01}$ . Why: the coherence picks up a single factor of $\sqrt{1-p}$ , the square root of the population factor, because the ground-state slot of $K_0$ is unity while the excited slot carries the $\sqrt{1-p}$ .

Step 3. Compare the rates. Since $\sqrt{1-p} = \sqrt{e^{-t/T_1}} = e^{-t/(2T_1)}$ , the coherence decays exactly twice as slowly as the population, so with no extra dephasing $1/T_2 = 1/(2T_1)$ , i.e. $T_2 = 2T_1$ . Why: the bilinear structure of a coherence — one amplitude from each level — is what turns the population factor into its square root, giving the universal half-rate. *Why the numbers check out: numerically $p(t=40\ \mu\mathrm{s},\, T_1 = 80\ \mu\mathrm{s}) = 0.3935$ gives $\sqrt{1-p} = 0.7788 = e^{-0.25}$ , confirming the half-rate coherence decay in the $\gamma_\phi = 0$ limit.*

Go deeper · Derivation C · 2 minWhy the Hahn echo recovers T2 from T2 star

Derivation CWhy the Hahn echo recovers T2 from T2 star

The goal is to show why a single $\pi$ pulse placed at the midpoint of a free-evolution interval cancels the dephasing from slow, quasi-static frequency spread — recovering the intrinsic $T_2$ from the shorter Ramsey time $T_2^*$ — while leaving genuinely fast noise and all of relaxation untouched. We model the qubit as precessing at a frequency that carries a random but slowly varying offset, compute the phase accumulated with and without the midpoint pulse, and average over the offset distribution.

Step 1. Accumulate phase in a Ramsey sequence. After the first $\pi/2$ pulse the qubit precesses for a time $t$ , accumulating a phase $\varphi = \int_0^t \delta\omega(t')\,dt' \approx \delta\omega\, t$ for an offset $\delta\omega$ that is essentially constant over one run but varies from run to run. Why: because $\delta\omega$ differs shot to shot, averaging $e^{i\varphi}$ over its distribution damps the fringe, and for a spread of width $\sigma_\omega$ this gives the inhomogeneous decay $e^{-t/T_2^*}$ with $1/T_2^* \sim \sigma_\omega$ .

Step 2. Insert the midpoint $\pi$ pulse. Split the evolution into two arms of length $\tau$ with a $\pi$ pulse between them; the pulse maps the accumulated phase $\varphi \to -\varphi$ , so the total phase becomes $-\delta\omega\,\tau + \delta\omega\,\tau = 0$ for any offset that is constant across the two arms. Why: the sign flip makes the second arm exactly undo the first whenever the frequency error did not change, so quasi-static spread is refocused into an echo regardless of its value.

Step 3. Keep only what survives. What remains uncancelled is noise that changes appreciably between the two arms — the genuinely fluctuating, higher-frequency part of the environment — so the echo decays at the intrinsic rate $1/T_2$ rather than the inhomogeneous $1/T_2^*$ , giving $T_2^* \le T_2 \le 2T_1$ . Why: the echo strips the refocusable static contribution and measures only irreversible decoherence, which is why $T_2$ from an echo is the meaningful intrinsic coherence time. *Why the echo works completely in the static limit: when offsets are purely static the echo signal is offset-independent, recovering full coherence at $2\tau$ ; the residual decay is set by the non-refocusable part of the spectrum, consistent with the filter-function treatment of dynamical decoupling.*

Limiting cases

The interplay of the two rates produces a handful of clean regimes that between them cover almost every real qubit.

Case	Result	Meaning
No pure dephasing	$T_\phi \to \infty \;\Rightarrow\; T_2 = 2T_1$	Coherence loss is entirely a byproduct of energy relaxation — the relaxation-limited ceiling; highly engineered superconducting and spin qubits approach this bound.
Dephasing-dominated	$T_\phi \ll T_1 \;\Rightarrow\; T_2 \approx T_\phi \ll 2T_1$	Random frequency fluctuations scramble phase long before any energy is exchanged; $T_2$ is set almost entirely by $T_\phi$ , typical of charge- or flux-noise-limited qubits.
Balanced	$\tfrac{1}{T_2} = \tfrac{1}{2T_1} + \tfrac{1}{T_\phi}$	The general Markovian case; both mechanisms contribute and $T_2$ alone cannot separate them — an independent $T_1$ measurement is needed to extract $T_\phi$ .
Inhomogeneous (Ramsey)	$\tfrac{1}{T_2^*} = \tfrac{1}{T_2} + \tfrac{1}{T_{\mathrm{inh}}}$	Quasi-static frequency spread across repetitions adds a further rate $1/T_{\mathrm{inh}}$ , so $T_2^* \le T_2$ ; a Hahn echo refocuses the static offset and recovers $T_2$ .

Common trapT₂ and T₂ star are not the same number

Do not confuse the intrinsic coherence time $T_2$ with the measured Ramsey time $T_2^*$ . The Ramsey decay includes inhomogeneous, quasi-static dephasing — slow frequency drifts averaged over many repetitions — so it always satisfies $T_2^* \le T_2$ . A Hahn echo applies a $\pi$ pulse at the midpoint to refocus that static spread, recovering the intrinsic $T_2$ . Two further traps: the ceiling is $T_2 \le 2T_1$ , not $T_2 \le T_1$ , so a qubit with $T_2 > T_1$ is perfectly legal (only $T_2 > 2T_1$ is impossible and signals a fit or measurement error); and $T_2 = 2T_1$ is reached only in the strict limit $T_\phi \to \infty$ , so any finite pure dephasing makes the inequality strict. Always quote the sequence — $T_2^{\mathrm{Ramsey}}$ versus $T_2^{\mathrm{echo}}$ — because echo can exceed Ramsey by one to two orders of magnitude when slow noise dominates.

Worked examples

1 · Extracting the pure-dephasing time (confidence builder)

A qubit is measured to have $T_1 = 80\ \mu\mathrm{s}$ and, in a Hahn-echo experiment, $T_2 = 100\ \mu\mathrm{s}$ . Find the pure-dephasing time $T_\phi$ and check the numbers are self-consistent.

Step 1. Rearrange the master relation (1) for the dephasing rate:

\frac{1}{T_\phi} = \frac{1}{T_2} - \frac{1}{2T_1} = \frac{1}{100} - \frac{1}{160} = 0.01000 - 0.00625 = 0.00375\ \mu\mathrm{s}^{-1}.

Step 2. Invert: $T_\phi = 1/0.00375 \approx \mathbf{267\ \mu s}$ . The pure-dephasing time is several times $T_1$ , so this qubit is close to — but not at — its relaxation ceiling.

Step 3. Consistency check: the ceiling is $2T_1 = 160\ \mu\mathrm{s}$ , and the measured $T_2 = 100\ \mu\mathrm{s}$ sits comfortably below it, so $1/T_\phi > 0$ and the result is physical. Note that $T_2 = 100 > T_1 = 80$ is allowed — the bound is $2T_1$ , not $T_1$ . $T_\phi \approx 267\ \mu s$ , and $T_2 = 100 < 2T_1 = 160\ \mu s$ , so the measurement is self-consistent.

2 · The amplitude-damping channel step by step (fully worked)

A qubit with $T_1 = 80\ \mu\mathrm{s}$ and no pure dephasing starts in the equatorial state $\tfrac{1}{\sqrt2}(|0\rangle + |1\rangle)$ , whose density matrix has $\rho_{11} = \tfrac12$ and $\rho_{01} = \tfrac12$ . After a free evolution of $t = 40\ \mu\mathrm{s}$ , find the emission probability, the excited population, and the coherence, and confirm $T_2 = 2T_1$ .

Step 1. Emission probability from equation (5): $p = 1 - e^{-t/T_1} = 1 - e^{-40/80} = 1 - e^{-0.5} = 1 - 0.6065 = \mathbf{0.3935}$ .

Step 2. Excited population is scaled by $1-p$ : $\rho_{11}(t) = (1-p)\rho_{11}(0) = 0.6065 \times 0.5 = \mathbf{0.3033}$ . Equivalently $\rho_{11}(0)e^{-t/T_1} = 0.5 \times 0.6065$ , the same number — the $T_1$ decay.

Step 3. Coherence is scaled by $\sqrt{1-p}$ : $\rho_{01}(t) = \sqrt{0.6065}\times 0.5 = 0.7788 \times 0.5 = \mathbf{0.3894}$ . Check the rate: $\sqrt{1-p} = 0.7788 = e^{-0.25} = e^{-t/(2T_1)}$ , so the coherence decays with time constant $2T_1 = 160\ \mu\mathrm{s}$ .

Step 4. Since there is no pure dephasing, the coherence time is exactly the relaxation ceiling. $p = 0.3935$ , population $0.303$ , coherence $0.389$ , and the coherence decays at $1/(2T_1)$ so $T_2 = 2T_1 = 160\ \mu s$ .

3 · Recovering the intrinsic T₂ from a Ramsey measurement

The same qubit ( $T_2 = 100\ \mu\mathrm{s}$ from echo) shows a much faster Ramsey decay, $T_2^* = 20\ \mu\mathrm{s}$ , because of shot-to-shot frequency drift. What inhomogeneous dephasing time $T_{\mathrm{inh}}$ does the drift contribute?

Step 1. Use the inhomogeneous relation $1/T_2^* = 1/T_2 + 1/T_{\mathrm{inh}}$ and solve for the added rate: $1/T_{\mathrm{inh}} = 1/T_2^* - 1/T_2 = 1/20 - 1/100 = 0.05 - 0.01 = 0.04\ \mu\mathrm{s}^{-1}$ .

Step 2. Invert: $T_{\mathrm{inh}} = 1/0.04 = \mathbf{25\ \mu s}$ . The quasi-static drift dominates the Ramsey signal, which is why $T_2^*$ is five times shorter than the intrinsic $T_2$ .

Step 3. The Hahn echo refocuses exactly this static contribution, removing the $1/T_{\mathrm{inh}}$ term and returning the intrinsic $T_2 = 100\ \mu\mathrm{s}$ . The drift contributes $T_{\mathrm{inh}} = 25\ \mu s$ , and the echo recovers $T_2 = 100\ \mu s$ from the Ramsey $T_2^* = 20\ \mu s$ .

Physics of the portrait

The very symbols $T_1$ and $T_2$ are borrowed. They did not originate in quantum computing at all but in nuclear magnetic resonance, where a physicist trying to describe an ensemble of precessing nuclear spins wrote down two phenomenological relaxation times and never imagined they would one day rank the world's qubits.

Origins · 1 minFelix Bloch and how this was discovered

Felix Bloch

1905–1983 · the relaxation times T₁ and T₂ from magnetic resonance

Felix Bloch, a Swiss-American theoretical physicist at Stanford, gave the field its language in a single 1946 paper. Seeking to describe how an ensemble of nuclear spins responds to a magnetic field, he wrote down a set of phenomenological equations of motion — now universally called the Bloch equations — in which the magnetization relaxes toward equilibrium along the field at a rate $1/T_1$ and its transverse components decay at a rate $1/T_2$ . He named $T_1$ the longitudinal or spin–lattice relaxation time, the timescale for the spins to shed energy into the surrounding lattice, and $T_2$ the transverse or spin–spin relaxation time, the timescale for the spins to lose their mutual phase coherence. For this work, and the experimental methods that accompanied it, he shared the 1952 Nobel Prize in Physics with Edward Purcell. Bloch's genius was to see that one need not know the microscopic bath in detail to make quantitative predictions: two numbers suffice to capture relaxation and dephasing, and everything harder can be folded into them. Three-quarters of a century later, when a quantum engineer reports a transmon's $T_1$ and $T_2$ , they are speaking Bloch's dialect exactly — the same two times, the same geometry of a vector relaxing longitudinally and shrinking transversely, now applied to a single artificial atom instead of an ensemble of nuclei.

History

1946

— Bloch introduces the longitudinal and transverse relaxation times T₁ and T₂ for nuclear magnetic resonance

1948

— Bloembergen, Purcell and Pound relate the relaxation rates to spectral densities of molecular motion

1950

— Hahn discovers the spin echo, refocusing inhomogeneous dephasing with a midpoint pulse

1957

— Redfield derives the master equation for a spin weakly coupled to a bath

1976

— Lindblad and, independently, Gorini, Kossakowski and Sudarshan give the general Markovian master equation

Going further

Texts. Nielsen and Chuang, Quantum Computation and Quantum Information (10th anniversary ed.), Sec. 8.3 (quantum noise, amplitude and phase damping) · Preskill, Lecture Notes on Quantum Computation, Ch. 3 (measurement and decoherence) · Watrous, The Theory of Quantum Information, Ch. 2 (channels and completely positive maps) · Wilde, Quantum Information Theory (2nd ed.), Ch. 4 (quantum channels) · Breuer and Petruccione, The Theory of Open Quantum Systems, Ch. 3 (Markovian master equations) · Slichter, Principles of Magnetic Resonance (the NMR origin of $T_1$ and $T_2$ ) · Krantz et al., "A Quantum Engineer's Guide to Superconducting Qubits", Appl. Phys. Rev. 6, 021318 (2019). The master relation $1/T_2 = 1/(2T_1) + 1/T_\phi$ and the ceiling $T_2 \le 2T_1$ follow from solving the Lindblad equation (equation 4): $\dot\rho_{11} = -\gamma\rho_{11}$ and $\dot\rho_{01} = -(\gamma/2 + \gamma_\phi)\rho_{01}$ . The amplitude-damping Kraus operators give the $\sqrt{1-p}$ coherence factor by evaluating $K_0\rho K_0^\dagger + K_1\rho K_1^\dagger$ . The Bloch-equation decay structure and echo refocusing are standard magnetic-resonance results. The worked-example numbers are arithmetic: Example 1 gives $T_\phi \approx 267\ \mu$ s from $T_1 = 80\ \mu$ s and $T_2 = 100\ \mu$ s; Example 2 gives $p = 0.3935$ and coherence factor $0.7788 = e^{-0.25}$ at $t = 40\ \mu$ s with $T_1 = 80\ \mu$ s; Example 3 gives $T_{\mathrm{inh}} = 25\ \mu$ s from $T_2^* = 20\ \mu$ s and $T_2 = 100\ \mu$ s. No platform-specific record coherence numbers are asserted, as state-of-the-art values evolve.

Prerequisites. Open Quantum Systems and Reduced Dynamics · Decoherence · Lindblad / GKSL Master Equation · Amplitude Damping Channel · Phase Damping Channel. Leads to. Process Tomography and the Choi–Jamiołkowski Isomorphism · Pauli and Bit/Phase-Flip Channels · Markovian versus Non-Markovian Dynamics · Master Equation Derivations: Born–Markov and Secular Approximations.

Ask the tutor: Give me your qubit's $T_1$ and $T_2$ and the sequence you used, and I will tell you its $T_\phi$ , whether it is relaxation- or dephasing-limited, and whether it is bumping the $2T_1$ ceiling.

What is the difference between T₁ and T₂, in one breath?

$T_1$ is about energy, $T_2$ is about phase. $T_1$ measures how long the qubit stays excited before dumping its energy quantum into the environment and relaxing toward the ground state — a change in the populations, the diagonal of the density matrix. $T_2$ measures how long a superposition keeps a definite phase before the relative phase between $|0\rangle$ and $|1\rangle$ is scrambled — a decay of the off-diagonal coherences. Energy loss inevitably costs phase too, which is why $T_1$ feeds into $T_2$ through $1/T_2 = 1/(2T_1) + 1/T_\phi$ , and the leftover term $T_\phi$ is the phase you lose without losing any energy at all.

Why is there a factor of 2 in the master relation?

Because a coherence is an amplitude product, and amplitudes decay at half the rate of the probabilities they build. The off-diagonal element $\rho_{01}$ is bilinear in the probability amplitude of being in $|0\rangle$ and the conjugate amplitude of being in $|1\rangle$ . When relaxation depletes the excited population at rate $1/T_1$ , the excited amplitude falls at half that rate, $1/(2T_1)$ , and the coherence inherits exactly that half-rate. Pure dephasing then adds its rate $1/T_\phi$ directly, because it acts on the phase and not the populations, giving $1/T_2 = 1/(2T_1) + 1/T_\phi$ . The amplitude-damping channel shows the same thing algebraically: the population scales by $1-p$ but the coherence only by $\sqrt{1-p}$ .

Can T₂ ever be larger than T₁?

Yes — this surprises people, but it is perfectly physical. The only hard ceiling is $T_2 \le 2T_1$ , so a qubit with, say, $T_1 = 80\ \mu$ s and $T_2 = 100\ \mu$ s is completely legal, as the worked example shows. What is forbidden is $T_2 > 2T_1$ , which would require a negative pure-dephasing rate and hence a non-physical, non-completely-positive channel. The confusion usually comes from misremembering the bound as $T_2 \le T_1$ . The correct statement is that coherence decays at at least half the relaxation rate, so $T_2$ tops out at twice $T_1$ , and it lands there exactly when there is no pure dephasing at all.

What exactly is the difference between T₂ and T₂*?

$T_2^*$ is what a Ramsey experiment measures, and it is generally shorter than the intrinsic $T_2$ because it includes inhomogeneous, quasi-static dephasing — slow drifts of the qubit frequency that are roughly constant within one run but vary from run to run, so averaging over many repetitions washes out the fringe faster. $T_2$ is the intrinsic coherence time you get once that refocusable spread is removed by a Hahn echo, whose midpoint $\pi$ pulse cancels any offset that stayed constant across the sequence. The universal ordering is $T_2^* \le T_2 \le 2T_1$ . Because $T_2^*$ depends on the averaging time and the drift, always state which sequence produced your number.

How do you actually measure T₁ and T₂ in the lab?

Three sequences do the work. For $T_1$ , an inversion-recovery experiment applies a $\pi$ pulse to put the qubit in $|1\rangle$ , waits a variable time, and reads the population; fitting the exponential recovery gives $T_1$ . For $T_2^*$ , a Ramsey sequence applies two $\pi/2$ pulses separated by a free-evolution time, producing a decaying oscillation whose envelope gives $T_2^*$ and whose frequency gives the detuning. For the intrinsic $T_2$ , a Hahn echo adds a $\pi$ pulse at the midpoint to refocus static noise, and its decay gives $T_2$ . Extending the echo to many pulses — CPMG — pushes $T_2$ higher still and, by varying the pulse spacing, doubles as a spectrometer for the environmental noise.

Why does relaxation care about the qubit frequency but dephasing does not?

Relaxation is an energy-exchanging transition, and energy conservation is a resonance condition: to fall from $|1\rangle$ to $|0\rangle$ the qubit must hand a quantum $\hbar\omega_0$ to the bath, which can only accept it if the bath has spectral weight at exactly that frequency, so $1/T_1$ is set by the noise power $S(\omega_0)$ . Pure dephasing exchanges no energy — it only jitters the qubit's transition frequency — so it is driven by the slow, near-DC part of the spectrum, $S(\omega \to 0)$ . This is why ubiquitous $1/f$ noise, which piles enormous weight at low frequency and almost none at gigahertz, dephases a qubit ferociously while barely relaxing it, and why the microscopic content of $T_\phi$ lives in the phase-damping rather than the amplitude-damping channel.