T1 Relaxation Time

Prepare a qubit in its excited state and walk away, and you will not find it there when you return. A real two-level system is never truly isolated: it is bolted to a vast, restless environment — the electromagnetic vacuum, thermal photons, lattice phonons, tunnelling defects in amorphous oxides, stray quasiparticles — and any of those channels can carry off the qubit's energy quantum $\hbar\omega_{01}$ . The longitudinal relaxation time $T_1$ is the single number that says how fast: the timescale on which the excited-state population drains away toward thermal equilibrium. The name is inherited from nuclear magnetic resonance, where $T_1$ relaxes the magnetization along the static field — the "longitudinal" direction — and the same symbol has followed the physics from spins in a magnet to superconducting circuits on a chip. What makes $T_1$ more than a decay constant is what sets it. Relaxation is a population-transfer process, and Fermi's golden rule ties its rate to one specific thing: how much environmental noise exists at exactly the qubit frequency. $1/T_1$ is proportional to the noise spectral density $S(\omega_{01})$ , so lengthening $T_1$ is the concrete task of quieting the bath at one frequency — by filtering, by detuning the qubit from a lossy resonator (the Purcell effect), or by removing the dielectric defects that host the noise. $T_1$ is also prior to the dephasing time $T_2$ : because losing energy also erases phase, relaxation puts a hard ceiling $T_2 \le 2T_1$ on any coherence. It is the clock every operation races against — the number of good gates scales as $T_1/t_{\mathrm{gate}}$ — which is why long $T_1$ sits at the heart of the DiVincenzo criteria and headlines the datasheet of every qubit ever built.

Key takeaways

$T_1$ is the timescale on which a qubit irreversibly loses its energy to the environment, decaying from $|1\rangle$ toward thermal equilibrium.
The excited-state population follows a single exponential, $P_1(t)=P_1(\infty)+[P_1(0)-P_1(\infty)]\,e^{-t/T_1}$ , relaxing toward a thermal floor $P_1(\infty)$ , not always to zero.
The rate is fixed by Fermi's golden rule, $1/T_1 \propto S(\omega_{01})$ — the noise spectral density at the qubit frequency, the one knob $T_1$ engineering turns.
Energy loss also destroys phase, so $T_1$ caps coherence through $1/T_2 = 1/(2T_1)+1/T_\phi$ , giving the hard bound $T_2 \le 2T_1$ .
The hardware-independent face of $T_1$ is the amplitude-damping channel, a single Kraus parameter $\gamma = 1-e^{-t/T_1}$ .
In today's superconducting qubits $T_1$ runs from tens of microseconds to a millisecond, limited by dielectric two-level-system defects, Purcell decay, and quasiparticles.
Key idea: $T_1$ is the inverse of the energy-relaxation rate set by the environmental noise spectral density at the qubit frequency, $1/T_1 = \Gamma_1 \propto S(\omega_{01})$ — it measures how fast a qubit forgets which state it is in, while $T_2$ measures how fast it forgets the phase between states.

Where this lives today

$T_1$ is the number every superconducting-qubit group lives and dies by, and its history is a story of chasing down loss channels one at a time. The first Cooper-pair-box qubits relaxed in nanoseconds; the transmon and 3D cavities pushed $T_1$ into the tens of microseconds by the mid-2010s; the current frontier reaches hundreds of microseconds to a millisecond. The gains came from identifying the dominant bath: amorphous oxide at interfaces hosts a dense forest of two-level-system (TLS) defects whose noise sits right at $\omega_{01}$ , so replacing niobium with tantalum and cleaning the substrate — Place and coworkers at Princeton, 2021 — roughly tripled transmon $T_1$ to the 0.3–0.5 ms range, while low-frequency fluxonium qubits have reached $T_1$ beyond a millisecond. Purcell filters are now standard, decoupling fast readout from qubit decay exactly as Worked Example 2 demands. And $T_1$ is not even constant in time: it fluctuates as TLS defects drift through resonance, and — startlingly — ionizing radiation matters. Cosmic rays and ambient radioactivity dump bursts of quasiparticles that cause correlated, chip-wide dips in $T_1$ (Vepsäläinen and coworkers, 2020; McEwen and coworkers at Google, 2022), a real obstacle for error correction that has sent groups to underground labs and shielded cryostats. The abstract $1/T_1\propto S(\omega_{01})$ has become, in the lab, a materials-science program.

The physical picture

Picture the qubit as a tiny pendulum tied to a vast bath of springs — every mode of the environment it can talk to. If any bath spring resonates near the pendulum's own frequency, energy drains into that mode and does not come back; if the bath is silent there, the pendulum keeps swinging. That is the whole content of $1/T_1 \propto S(\omega_{01})$ : the rate is how "crowded" the environment is with modes at the qubit frequency. Relaxation is one-directional bookkeeping on populations — probability moves from $|1\rangle$ to $|0\rangle$ by emission, and at finite temperature back by absorption — until the two rates reach detailed balance and leave a thermal mixture, not a pure ground state. The first figure pins the one relation the whole article turns on.

Figure

Energy relaxation of a qubit: the excited state loses one quantum ħω₀₁ to a continuum of bath modes at a rate fixed by the noise spectral density there, 1/T₁ = Γ₁ ∝ S(ω₀₁). For a 5 GHz transmon at 20 mK the bath is nearly empty, n_th ≈ 6×10⁻⁶, so emission dominates and the qubit decays toward the ground state

The second thing to hold onto is that $T_1$ is not just a rate but a ceiling. A relaxation event does not only move population — it also collapses any phase the qubit was carrying, because a decayed qubit and a surviving one are distinguishable to the environment. So relaxation alone decoheres a superposition at half the population rate, and no engineering can keep a coherent state alive longer than twice the time the energy takes to leak away: $T_2 \le 2T_1$ . Pure dephasing $T_\phi$ , which scrambles phase without moving energy, only pulls $T_2$ down further. The second figure is what an experimentalist actually measures — the excited-state decay curve — with the numbers of a good modern transmon and the amplitude-damping and $T_2$ -ceiling readings marked.

Figure

Measured excited-state population P₁(t) = e^(−t/T₁) for a superconducting transmon with T₁ = 100 μs and ω₀₁/2π = 5 GHz. After one T₁ the survival is 1/e = 0.368 while the amplitude-damping probability rises to γ = 1 − 1/e = 0.632; the two curves cross at P₁ = γ = 0.5 near t = 69 μs, and the relaxation ceiling on coherence is T₂ ≤ 2T₁ = 200 μs

The core equations

The starting point is kinematic: the excited-state population relaxes on the single timescale $T_1$ , and because populations obey a linear rate equation the decay is a pure exponential toward a thermal floor. Everything downstream is a computation of, or a consequence of, this one law.

P_1(t) = P_1(\infty) + \big[P_1(0) - P_1(\infty)\big]\,e^{-t/T_1} \qquad (1)

where: $P_1(t)$ — excited-state population at time $t$ ; $P_1(0)$ — initial population (typically $1$ after a $\pi$ pulse); $P_1(\infty)$ — thermal-equilibrium excited population; $T_1$ — longitudinal (energy) relaxation time. Reading it: the population decays not to zero but to $P_1(\infty)$ , on the one timescale $T_1$ ; at $T=0$ the floor vanishes and the curve is the familiar $e^{-t/T_1}$ . The rate itself is not free — it is fixed by the environment. Fermi's golden rule connects $1/T_1$ to a bath operator $\hat A$ that the qubit couples to and to the noise spectral density of that bath evaluated at the qubit frequency, summing the emission ( $+\omega_{01}$ ) and absorption ( $-\omega_{01}$ ) contributions.

\frac{1}{T_1} = \Gamma_1 = \frac{|\langle 0|\hat A|1\rangle|^2}{\hbar^2}\big[S(\omega_{01}) + S(-\omega_{01})\big] \;\propto\; S(\omega_{01}) \qquad (2)

where: $\Gamma_1$ — energy-relaxation rate; $\hat A$ — qubit operator coupling to the bath; $|\langle 0|\hat A|1\rangle|$ — transition matrix element; $S(\omega)$ — noise spectral density, the Fourier transform of the bath correlation function. Reading it: only noise at the qubit frequency relaxes the qubit, which is why $T_1$ engineering is spectral — you quiet one frequency, not the whole bath. Because energy loss also erases phase, $T_1$ enters the coherence budget directly. The transverse (total dephasing) rate is the sum of the relaxation contribution — exactly half the population rate — and the pure-dephasing rate.

\frac{1}{T_2} = \frac{1}{2T_1} + \frac{1}{T_\phi} \qquad (3)

where: $T_2$ — total dephasing time; $T_\phi$ — pure-dephasing time (frequency noise, no energy exchange); $T_1$ — relaxation time. Reading it: with no pure dephasing ( $T_\phi\to\infty$ ) one hits the relaxation-limited maximum $T_2 = 2T_1$ , so a measured $T_2 > 2T_1$ is a red flag for a calibration error. The thermal floor is set by how many quanta the bath already holds at the qubit frequency — the Bose occupation — which fixes both the steady-state population and the balance between absorption and emission.

n_{\mathrm{th}} = \frac{1}{e^{\hbar\omega_{01}/k_B T} - 1},\qquad P_1(\infty) = \frac{n_{\mathrm{th}}}{2n_{\mathrm{th}}+1} \qquad (4)

where: $n_{\mathrm{th}}$ — mean bath occupation at $\omega_{01}$ ; $T$ — effective bath temperature; $k_B$ — Boltzmann constant. Reading it: for a 5 GHz qubit in a 20 mK fridge $\hbar\omega_{01}/k_BT\approx 12$ and $n_{\mathrm{th}}\approx 6\times10^{-6}$ , so a measured $P_1(\infty)$ near a percent signals a hotter effective environment than the fridge. The hardware-independent face of all this is the amplitude-damping channel: the entire $T_1$ decay is packaged into two Kraus operators and a single damping parameter $\gamma$ .

\mathcal{E}(\rho) = E_0\rho E_0^\dagger + E_1\rho E_1^\dagger,\quad E_0=\begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix},\ E_1=\begin{pmatrix}0&\sqrt{\gamma}\\0&0\end{pmatrix},\quad \gamma = 1-e^{-t/T_1} \qquad (5)

where: $\gamma$ — damping (decay) probability; $E_0$ — no-jump Kraus operator; $E_1$ — quantum-jump (decay) operator; $\rho$ — qubit density matrix. Reading it: $E_1$ is the emission event $|1\rangle\to|0\rangle$ , while $E_0$ is the "no-click" evolution that shrinks the surviving excited amplitude — together they map laboratory relaxation onto the theory of quantum channels. Finally, one dominant and controllable relaxation channel in circuit QED has a closed form: a qubit dispersively coupled to a lossy readout resonator inherits the resonator's loss, scaled by the coupling-to-detuning ratio.

\Gamma_1^{\mathrm{Purcell}} = \left(\frac{g}{\Delta}\right)^2 \kappa \qquad (6)

where: $g$ — qubit–resonator coupling; $\Delta=\omega_q-\omega_r$ — qubit–resonator detuning; $\kappa$ — resonator energy decay rate. Reading it: the detuned qubit hybridizes with the lossy mode by the small amount $(g/\Delta)^2$ and leaks at that fraction of $\kappa$ — large detuning, or a filter at $\omega_{01}$ , protects $T_1$ .

Derivation AFrom the population master equation to the exponential decay law

The goal is to show that the single-exponential relaxation law of equation (1) is not an assumption but the forced solution of a linear rate equation for the excited-state population, and that the same equation delivers the thermal floor $P_1(\infty)$ and identifies $1/T_1$ as the sum of the emission and absorption rates. We balance the upward and downward transitions, read off the steady state, and solve the resulting first-order ODE.

Step 1. Balance the population flow. In time $\mathrm{d}t$ the excited state loses probability at the emission rate $\Gamma_\downarrow$ and gains it from the ground state at the absorption rate $\Gamma_\uparrow$ , giving $\dot P_1 = -\Gamma_\downarrow P_1 + \Gamma_\uparrow(1-P_1)$ . Why: relaxation is bookkeeping on diagonal density-matrix elements — nothing but two competing one-way transition rates, with the ground-state population written as $1-P_1$ .

Step 2. Read the total rate off the ODE. Rewriting, $\dot P_1 = -(\Gamma_\downarrow+\Gamma_\uparrow)P_1 + \Gamma_\uparrow$ , so the coefficient of $P_1$ is the full relaxation rate: $1/T_1 = \Gamma_\downarrow+\Gamma_\uparrow$ . Why: it is the sum, not the difference — both emission and absorption push the population toward its steady value, so both add to $1/T_1$ .

Step 3. Set $\dot P_1=0$ for the steady state, $P_1(\infty)=\Gamma_\uparrow/(\Gamma_\downarrow+\Gamma_\uparrow)$ . Detailed balance fixes the ratio through the Bose factor, $\Gamma_\uparrow/\Gamma_\downarrow = n_{\mathrm{th}}/(n_{\mathrm{th}}+1)$ , giving $P_1(\infty)=n_{\mathrm{th}}/(2n_{\mathrm{th}}+1)$ . Why: the ratio of up- to down-rates is pinned by temperature, which is what makes $T_1$ measurements double as thermometry.

Step 4. Solve the linear ODE. With constant coefficients, $P_1(t)=P_1(\infty)+[P_1(0)-P_1(\infty)]e^{-t/T_1}$ ; at $T=0$ , $P_1(\infty)=0$ , recovering pure spontaneous decay $e^{-t/T_1}$ . Why: a first-order linear equation can only give exponential approach to its fixed point, so the single-exponential shape is a theorem — a non-exponential decay is evidence of physics beyond this simple bath. *The solution recovers $P_1(\infty)=n_{\mathrm{th}}/(2n_{\mathrm{th}}+1)$ and, at $T=0$ , the pure amplitude-damping law $e^{-t/T_1}$ , consistent with the Lindblad master equation for a two-level system coupled to a thermal bath.*

Go deeper · Derivation B · 1 minThe relaxation rate from Fermi's golden rule

Derivation BThe relaxation rate from Fermi's golden rule

The goal is to derive $1/T_1 \propto S(\omega_{01})$ from first principles — to show that the qubit's decay rate is literally the environmental noise spectral density at the transition frequency, times a matrix element — by treating the environment as a continuum of weakly coupled modes. We apply the golden rule, package the bath's mode sum as a spectral density, and use detailed balance to fix emission against absorption.

Step 1. Write the interaction as $\hat H_{\mathrm{int}}=\hat A\otimes\hat B$ , with $\hat A$ a qubit operator and $\hat B$ a bath operator. Fermi's golden rule gives the decay rate into all bath final states, $\Gamma_{1\to0}=\frac{2\pi}{\hbar^2}|\langle 0|\hat A|1\rangle|^2\sum_f|\langle f|\hat B|i\rangle|^2\,\delta(\omega_{01}-\omega_{fi})$ . Why: weak coupling to a dense continuum is the golden rule's home turf, and the delta function enforces energy conservation — the qubit decays only by handing its quantum to a bath mode of matching frequency.

Step 2. Recognize the bath sum as a spectral density. The weighted sum over final states is the Fourier transform of the bath correlation function, $S(\omega)=\int_{-\infty}^{\infty}\langle\hat B(t)\hat B(0)\rangle\,e^{i\omega t}\,\mathrm{d}t$ , evaluated at $\omega=\omega_{01}$ . Why: the microscopic clutter of bath modes collapses into one measurable function of frequency, and the qubit only ever "sees" its value at $\omega_{01}$ .

Step 3. Separate emission and absorption. Positive frequency releases energy, $\Gamma_\downarrow\propto S(+\omega_{01})$ ; negative absorbs it, $\Gamma_\uparrow\propto S(-\omega_{01})$ . Summing, $1/T_1=\frac{|\langle 0|\hat A|1\rangle|^2}{\hbar^2}[S(\omega_{01})+S(-\omega_{01})]$ . Why: a quantum bath has an asymmetric spectrum — it gives and takes energy at different rates — and the two signed frequencies bookkeep the directions cleanly.

Step 4. Impose detailed balance. The quantum fluctuation–dissipation theorem fixes $S(-\omega_{01})/S(+\omega_{01})=e^{-\hbar\omega_{01}/k_BT}$ , reproducing the thermal steady state and, at $T=0$ , killing absorption so $1/T_1=|\langle 0|\hat A|1\rangle|^2 S(\omega_{01})/\hbar^2$ . Why: dissipation and fluctuation are two faces of one spectral density — the same asymmetry that lets the bath absorb is what forces a thermal, not pure, ground state. *The derivation reproduces $1/T_1\propto S(\omega_{01})$ and the detailed-balance ratio, in agreement with Bloch–Redfield theory and the standard circuit-QED noise-spectroscopy relation $\Gamma_1=|\langle 0|\hat A|1\rangle|^2 S(\omega_{01})/\hbar^2$ .*

Go deeper · Derivation C · 1 minAmplitude damping and the T₂ ≤ 2T₁ bound

Derivation CAmplitude damping and the T₂ ≤ 2T₁ bound

The goal is to prove that energy relaxation alone decoheres a superposition at exactly half the population-decay rate, which fixes the hard ceiling $T_2\le 2T_1$ , and to show where the pure-dephasing term of equation (3) adds on top. We push a coherence through the amplitude-damping channel, see how the off-diagonal element shrinks, translate that to a rate, and append independent phase noise.

Step 1. Track the coherence under damping. Applying the channel of equation (5) to a general $\rho$ , the off-diagonal element transforms as $\rho_{01}\to\sqrt{1-\gamma}\,\rho_{01}$ , while the excited population transforms as $\rho_{11}\to(1-\gamma)\rho_{11}$ . Why: $E_0$ carries $\sqrt{1-\gamma}$ on the excited amplitude, so a coherence (linear in that amplitude) decays with the square root, whereas a population (quadratic) decays with the full factor.

Step 2. Convert to rates. Since $1-\gamma=e^{-t/T_1}$ , the coherence factor is $\sqrt{1-\gamma}=e^{-t/2T_1}$ , so relaxation decays $|\rho_{01}|$ at rate $1/(2T_1)$ — exactly half the population rate. Why: this factor of two is the whole origin of the $2T_1$ bound — a kinematic consequence of amplitude being the square root of probability, not an extra assumption.

Step 3. Add pure dephasing. Independent, energy-conserving frequency noise multiplies the coherence by $e^{-t/T_\phi}$ , so $|\rho_{01}(t)|=|\rho_{01}(0)|\,e^{-t/2T_1}\,e^{-t/T_\phi}$ . Why: the two mechanisms are statistically independent, so their envelopes multiply and their rates add.

Step 4. Read off $T_2$ . The combined exponent defines $1/T_2=1/(2T_1)+1/T_\phi$ ; with $T_\phi\to\infty$ this saturates at $T_2=2T_1$ , and since $1/T_\phi\ge 0$ one always has $T_2\le 2T_1$ . Why: the pure-dephasing rate is non-negative, so it can only shorten $T_2$ — there is no beating the relaxation ceiling, which is why a reported $T_2>2T_1$ must be an artifact. *The result gives $T_2=2T_1$ at $T_\phi\to\infty$ and $T_2\le 2T_1$ in general; the factor-of-two coherence-versus-population relation matches the standard Bloch–Redfield and NMR definitions of $T_1$ and $T_2$ .*

Limiting cases

How $T_1$ and its consequences behave in the regimes that matter for real devices.

Case	Result	Meaning
Zero temperature ( $n_{\mathrm{th}}\to 0$ )	$P_1(t)=P_1(0)\,e^{-t/T_1}$	Pure spontaneous emission; the steady state is the ground state $\vert 0\rangle$ and the channel is pure amplitude damping with rate $\Gamma_1=1/T_1$ .
Finite temperature	$P_1(\infty)=\dfrac{n_{\mathrm{th}}}{2n_{\mathrm{th}}+1}$ , $\dfrac{1}{T_1}\propto(2n_{\mathrm{th}}+1)$	Decay relaxes toward a nonzero thermal floor and speeds up by $(2n_{\mathrm{th}}+1)$ ; a resolved floor measures the effective bath temperature.
No pure dephasing ( $T_\phi\to\infty$ )	$T_2 = 2T_1$	The relaxation-limited maximum coherence time; any pure dephasing only lowers $T_2$ , so $T_2\le 2T_1$ always.
Purcell-dominated	$\Gamma_1=\left(\dfrac{g}{\Delta}\right)^2\kappa$	A qubit dispersively coupled to a resonator of linewidth $\kappa$ inherits decay set by coupling and detuning; large $\Delta$ protects $T_1$ .
Short time ( $t\ll T_1$ )	$P_1(t)\approx P_1(0)\,(1-t/T_1)$ , $\gamma\approx t/T_1$	Linear decay; the per-shot error of an idle qubit grows as $t/T_1$ , the form used in error budgets.

Common trapThe excited state always decays cleanly to zero, and a long T₂ just means a long T₁

Two half-truths, both wrong. First, the excited population does not generally decay to zero: at finite temperature it relaxes toward $P_1(\infty)=n_{\mathrm{th}}/(2n_{\mathrm{th}}+1)$ , and in real superconducting qubits that floor is often around a percent — far above the fridge-equilibrium value — because non-equilibrium quasiparticles and stray infrared photons make the qubit's effective temperature hotter than the mixing chamber. Fitting to a bare $e^{-t/T_1}$ with a forced zero offset then biases $T_1$ . Second, $T_1$ and $T_2$ are not the same clock: $T_1$ measures energy loss, $T_2$ measures phase loss, and they demand different fixes. A $T_1$ -limited qubit needs a quieter bath at $\omega_{01}$ (better dielectrics, Purcell filtering); a $T_\phi$ -limited qubit needs frequency-noise mitigation (flux sweet spots, charge-noise-insensitive design). And $T_2$ can never exceed $2T_1$ — if your fit says otherwise, suspect the measurement, not the physics.

Worked examples

1 · A 5 GHz transmon with T₁ = 100 μs (fully worked)

A superconducting transmon has transition frequency $\omega_{01}/2\pi = 5$ GHz and a measured $T_1 = 100\ \mu$ s. The dilution fridge sits at 20 mK. Find the equilibrium excited population, the amplitude-damping probability after one $T_1$ , the ceiling on $T_2$ , and roughly how many 20 ns gates fit inside the coherence window. Then explain why the measured residual population is nearer 1%.

Step 1. Thermal occupation at the fridge temperature. Form $\hbar\omega_{01}/k_BT = hf/k_BT = (6.626\times10^{-34})(5\times10^{9})/[(1.381\times10^{-23})(0.020)] = \mathbf{12.0}$ . Then $n_{\mathrm{th}} = 1/(e^{12}-1) = \mathbf{6.2\times10^{-6}}$ , so the ideal equilibrium population is $P_1(\infty)\approx n_{\mathrm{th}} = \mathbf{6\times10^{-6}}$ — utterly negligible.

Step 2. Amplitude damping after one $T_1$ . At $t=T_1$ , $\gamma = 1-e^{-1} = \mathbf{0.632}$ , and the survival is $P_1(T_1)=e^{-1}=\mathbf{0.368}$ . So after $100\ \mu$ s an initially excited qubit has a 63% chance of having emitted its quantum — this is the $1/e$ point marked in the second figure.

Step 3. Coherence ceiling and gate budget. The relaxation-limited maximum is $T_2 \le 2T_1 = \mathbf{200\ \mu s}$ ; reaching it requires pure dephasing $T_\phi \gg T_1$ . With a gate time $t_{\mathrm{gate}}=20$ ns, the number of gates before relaxation matters is $T_1/t_{\mathrm{gate}} = 100\ \mu\mathrm{s}/20\ \mathrm{ns} = \mathbf{5000}$ — the rough depth ceiling this $T_1$ imposes on a circuit.

Step 4. Why the measured floor is nearer a percent. The 6×10⁻⁶ prediction assumes the qubit sees 20 mK, but non-equilibrium quasiparticles and stray thermal photons give an effective temperature closer to 50 mK. Repeating Step 1 at 50 mK: $\hbar\omega_{01}/k_BT = 4.80$ , $n_{\mathrm{th}} = 1/(e^{4.80}-1) = 8.3\times10^{-3}$ , so $P_1(\infty) = n_{\mathrm{th}}/(2n_{\mathrm{th}}+1) = \mathbf{0.82\%}$ . The floor is not the fridge temperature — it is a thermometer for the qubit's true environment, here about 50 mK.

2 · Purcell-limited T₁ and how far to detune

A transmon couples to its readout resonator with $g/2\pi = 100$ MHz at a detuning $\Delta/2\pi = 1$ GHz; the resonator loses photons at $\kappa/2\pi = 1$ MHz. Find the Purcell-limited $T_1$ . Then find the detuning needed to push the Purcell ceiling above 1 ms.

Step 1. Purcell rate at the given detuning. The hybridization fraction is $(g/\Delta)^2 = (100/1000)^2 = 0.01$ , so $\Gamma_1^{\mathrm{Purcell}} = 0.01\times(2\pi\times10^6\ \mathrm{s}^{-1}) = 6.3\times10^{4}\ \mathrm{s}^{-1}$ .

Step 2. Convert to a time. $T_1^{\mathrm{Purcell}} = 1/\Gamma_1^{\mathrm{Purcell}} = \mathbf{16\ \mu s}$ . If nothing else limited the qubit, this single channel would cap $T_1$ near 16 μs — modest, and a real problem for readout design.

Step 3. Detune to reach 1 ms. Writing $T_1^{\mathrm{Purcell}} = (\Delta/g)^2/\kappa$ , solve at $T_1=1$ ms: $(\Delta/g)^2 = \kappa T_1 = (2\pi\times10^6)(10^{-3}) = 6.3\times10^{3}$ , so $\Delta/g = 79$ and $\Delta/2\pi \approx \mathbf{7.9\ GHz}$ . A nearly eightfold-larger detuning buys a Purcell ceiling above 1 ms — but it weakens dispersive readout, which is why real devices add a Purcell filter to block $\kappa$ at $\omega_{01}$ instead of detuning alone.

Physics of the portrait

The symbols $T_1$ and $T_2$ did not begin in a quantum computer. They were written down in 1946 to describe nuclear spins relaxing in a magnet, by the physicist who turned magnetic resonance into a quantitative science — and have survived intact all the way to the superconducting chip.

Origins · 1 minFelix Bloch and how this was discovered

Felix Bloch

1905–1983 · nuclear magnetic resonance and the theory of relaxation

Felix Bloch was a Swiss-born physicist who, after foundational work on electrons in crystals — Bloch waves and Bloch's theorem are his — turned to the magnetism of atomic nuclei. In 1946 at Stanford he introduced the phenomenological equations that still bear his name, describing how a nuclear magnetization, tipped away from equilibrium, returns to it. He named two distinct timescales. The longitudinal relaxation time $T_1$ ("spin-lattice") governs recovery of the magnetization along the static field — energy exchange with the lattice — while the transverse time $T_2$ ("spin-spin") governs decay of the magnetization across it — loss of phase coherence among precessing spins. That very distinction, longitudinal energy relaxation versus transverse dephasing, is the one this article rests on, transplanted from spins to qubits with the physics unchanged. Bloch shared the 1952 Nobel Prize with Edward Purcell — whose group detected nuclear resonance in bulk matter the same year, and whose name attaches to the cavity-enhanced emission that limits $T_1$ in circuit QED. The microscopic theory of what sets these times followed from Bloembergen, Purcell and Pound and then Wangsness, Bloch and Redfield, whose master equation is the direct ancestor of the golden-rule rate $1/T_1\propto S(\omega_{01})$ .

History

1946

— Bloch introduces the phenomenological Bloch equations, defining the longitudinal T₁ and transverse T₂ relaxation times

1946

— Purcell, Torrey and Pound detect nuclear magnetic resonance in bulk matter, and Purcell predicts cavity-enhanced spontaneous emission

1948

— Bloembergen, Purcell and Pound relate nuclear relaxation rates to the spectral density of fluctuating fields

1957

— Wangsness, Bloch and Redfield cast relaxation as a microscopic master equation, the Bloch–Redfield theory

2005

— Ithier and coworkers apply the noise-spectral-density picture to decoherence in a superconducting qubit

Going further

Texts. Nielsen and Chuang, Quantum Computation and Quantum Information, §8.3 (quantum operations and the amplitude-damping channel) · Preskill, Lecture Notes on Quantum Computation, Ch. 3 (decoherence, master equations, and golden-rule relaxation) · Wilde, Quantum Information Theory, Ch. on quantum channels (amplitude damping) · Watrous, The Theory of Quantum Information (channels and CPTP maps) · Kitaev, Shen and Vyalyi, Classical and Quantum Computation (noise models) · Krantz et al., A Quantum Engineer's Guide to Superconducting Qubits, Appl. Phys. Rev. 6, 021318 (2019) (T₁, T₂, and noise spectroscopy in practice). Key results to keep in mind: the thermal steady state of the population rate equation is $P_1(\infty)=n_{\mathrm{th}}/(2n_{\mathrm{th}}+1)$ ; the detailed-balance ratio is $S(-\omega)/S(+\omega)=e^{-\hbar\omega/k_BT}$ ; the coherence relation $1/T_2=1/(2T_1)+1/T_\phi$ follows from $\sqrt{1-\gamma}=e^{-t/2T_1}$ ; the Worked Example 1 numbers are $n_{\mathrm{th}}\approx 6\times10^{-6}$ at 20 mK, $P_1(\infty)\approx 0.82\%$ at 50 mK, and $\gamma(T_1)=0.632$ ; the Worked Example 2 Purcell values are $T_1^{\mathrm{Purcell}}\approx 16\ \mu$ s and $\Delta/2\pi\approx 7.9$ GHz for a 1 ms ceiling. Device $T_1$ values are representative modern figures, not universal constants, since they are strongly material- and design-dependent.

Prerequisites. DiVincenzo Criteria · Density Matrix and Open Systems · Fermi's Golden Rule. Leads to. The Transmon Qubit · Circuit Quantum Electrodynamics · Dispersive Readout · Sideband Cooling.

Ask the tutor: Ask me to walk the golden-rule derivation of $1/T_1\propto S(\omega_{01})$ line by line, or to work out the effective temperature behind a measured 1% excited-state floor, and I will take it step by step.

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

$T_1$ is about energy; $T_2$ is about phase. $T_1$ is how long an excited qubit takes to lose its energy quantum and fall back toward the ground state — it acts on populations, the diagonal of the density matrix. $T_2$ is how long a superposition keeps a well-defined phase between $|0\rangle$ and $|1\rangle$ — it acts on the off-diagonal coherences. They are linked but not identical: because any energy-loss event also destroys phase, relaxation contributes to $T_2$ through $1/T_2=1/(2T_1)+1/T_\phi$ , giving the ceiling $T_2\le 2T_1$ . Pure dephasing $T_\phi$ — random modulation of the qubit frequency with no energy exchange — is the extra ingredient that pulls $T_2$ below that ceiling.

Why does only noise at the qubit frequency matter for T₁?

Because relaxation is a resonant energy exchange. To decay, the qubit must hand its quantum $\hbar\omega_{01}$ to some environmental mode, and energy conservation — the delta function in Fermi's golden rule — means the receiving mode must sit at the qubit frequency. Noise at other frequencies cannot accept the quantum, so it does not contribute to $T_1$ . That is why the rate is $1/T_1\propto S(\omega_{01})$ evaluated at that one frequency, and why $T_1$ engineering is spectral surgery: you filter or detune to open a quiet spot in the bath exactly at $\omega_{01}$ , leaving every other frequency untouched. Low-frequency noise, by contrast, mostly causes pure dephasing.

Does the excited state really not decay all the way to zero?

Only at absolute zero does it. At finite temperature the environment holds thermal quanta at the qubit frequency, so absorption competes with emission and the population settles at a nonzero floor $P_1(\infty)=n_{\mathrm{th}}/(2n_{\mathrm{th}}+1)$ set by detailed balance. For a 5 GHz qubit in a 20 mK fridge this floor is theoretically about $6\times10^{-6}$ — essentially zero. But measured floors in real qubits are often near a percent, because stray infrared photons and non-equilibrium quasiparticles push the effective temperature closer to 50 mK. Resolving that floor is a genuine thermometer for the qubit's environment, which is why careful experiments fit the offset rather than forcing it to zero.

What is the amplitude-damping channel, and why do information theorists like it?

It is the hardware-independent description of $T_1$ decay as a quantum channel. Rather than tracking a specific bath, it summarizes the entire effect of relaxation with two Kraus operators and a single number $\gamma=1-e^{-t/T_1}$ , the probability the qubit has emitted its quantum. One operator, $E_1$ , is the actual quantum jump $|1\rangle\to|0\rangle$ ; the other, $E_0$ , is the "no-jump" evolution that quietly shrinks the surviving excited amplitude by $\sqrt{1-\gamma}$ . Information theorists like it because it is exactly solvable, it is the canonical lossy channel, and its parameters feed directly into channel capacities and error-correction thresholds — you reason about relaxation abstractly, then plug in $\gamma(t)$ from a measured $T_1$ .

What is the Purcell effect and why does it hurt T₁?

The Purcell effect is the modification of a system's spontaneous-emission rate by its electromagnetic environment. In circuit QED a qubit is deliberately coupled to a readout resonator so you can measure it, but that resonator is lossy — it leaks photons at rate $\kappa$ — and so opens a decay path. Even detuned by $\Delta$ , the qubit hybridizes with the resonator by the small fraction $(g/\Delta)^2$ and inherits that fraction of the loss, giving a rate $(g/\Delta)^2\kappa$ . Large detuning suppresses it but weakens readout, so the standard fix is a Purcell filter: a frequency-selective element that blocks the resonator's loss at the qubit frequency while passing it at the readout frequency.

How is T₁ actually measured in the lab?

By inversion recovery, the direct implementation of equation (1). You prepare $|1\rangle$ with a calibrated $\pi$ pulse, wait a variable delay $\tau$ , then read out the excited-state population, repeating over many shots at each delay. Sweeping $\tau$ traces the decay curve, and fitting $P_1(\tau)=A\,e^{-\tau/T_1}+P_1(\infty)$ returns $T_1$ as the time constant, with amplitude $A$ absorbing state-preparation error and offset $P_1(\infty)$ capturing the thermal floor. Because individual defects drift, $T_1$ is not perfectly stable, so careful characterization repeats the measurement over hours to report a distribution rather than a single number.