Relaxation and Dephasing (T₁, T₂)

Two clocks for one qubit

A qubit coupled to a bath loses information along two distinct channels, each with its own characteristic time:

$T_1$ — the relaxation (or longitudinal) time. How fast the populations equilibrate, i.e. how fast an excited qubit decays toward its steady state along the $z$ -axis of the Bloch sphere.
$T_2$ — the coherence (or transverse) time. How fast the off-diagonal coherences decay, i.e. how fast a superposition loses its phase, measured in the $x$ – $y$ plane of the Bloch sphere.

In NMR and superconducting-qubit language these are the longitudinal and transverse relaxation times of the Bloch equations.

Where they come from in the Lindblad equation

Combine the amplitude-damping and pure-dephasing channels of the previous lessons. Amplitude damping with rate $\gamma_1 = 1/T_1$ uses $L_1=\sqrt{\gamma_1}\,\sigma_-$ ; pure dephasing with rate $\gamma_\phi$ uses $L_\phi=\sqrt{\gamma_\phi/2}\,\sigma_z$ . Solving the resulting equations of motion for the Bloch components gives

\langle\sigma_z\rangle(t) - \langle\sigma_z\rangle_{\infty} \propto e^{-t/T_1},\qquad \langle\sigma_{x,y}\rangle(t) \propto e^{-t/T_2}.

The fundamental relation

Reading off the decay of the coherence $\rho_{01}$ under both channels: amplitude damping contributes $\gamma_1/2$ to its decay (the half from the anticommutator), and pure dephasing contributes its full rate. Defining $1/T_\phi = \gamma_\phi$ ,

\boxed{\;\frac{1}{T_2} = \frac{1}{2T_1} + \frac{1}{T_\phi}\;}

The relaxation channel feeds the transverse decay at half its rate, because killing an excited population necessarily scrambles half of the phase; pure dephasing adds on top of that without touching the populations.

The inequality $T_2 \le 2T_1$

Since $1/T_\phi \ge 0$ , the relation immediately implies

T_2 \le 2 T_1,

with equality only in the absence of pure dephasing ( $T_\phi \to \infty$ ). A measured $T_2$ longer than $2T_1$ would signal an error or a non-Markovian effect; a $T_2$ much shorter than $2T_1$ signals that pure dephasing dominates. This single inequality is one of the most-used diagnostics in experimental quantum computing.

Try it

This is a numerical exercise — your code should return a number. Given $T_1 = 80\,\mu s$ and a pure-dephasing time $T_\phi = 120\,\mu s$ , use $1/T_2 = 1/(2T_1) + 1/T_\phi$ to compute and return $T_2$ in microseconds.

Run your code to see the quantum state.