I learned that T1 is relaxation time (time from $|1\rangle$ to $|0\rangle$) and T2 is coherence time. The relaxation is a specific case of decoherence. What's the difference between them and what's the exact meaning of coherence time T2?


2 Answers 2


T2 is so-called dephasing time.

It describes how long the phase of a qubit stays intact. In your words, it is time from $|+\rangle= \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ to $|-\rangle= \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$, or conversely.

Just note that both T1 and T2 are not actually "time from state x to state y" but rather decay constants. Probability that a qubit will stay in state $|1\rangle$ after time $t$ is given by formula

$$ P(|1\rangle) = \mathrm{e}^{-\frac{t}{T1}}. $$

Similarly for T2.

Both times T1 and T2 are together called decoherence times.


Slight correction to Martin Vesely's answer: $T_2$ is not the (decay constant) time after which an initial state $|+\rangle$ will necessarily switch to the state $|-\rangle$. If it were, then error correction would be easy. Instead, it's the (decay constant) time after which an initial state $|+\rangle$ will evolve into an equal classical probabilistic mixture of the $|+\rangle$ and $|-\rangle$ states, so that you can no longer confidently predict the state. That is, it's the autocorrelation time after which the initial and final states become uncorrelated, not negatively correlated.

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    $\begingroup$ Thanks for clarification. $\endgroup$ Commented Feb 6, 2020 at 15:17
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    $\begingroup$ +1 to this answer - a good explanation can be found here as well, I think it is helpful to see the "how do you measure it" and "how the curves typically look like": ocw.mit.edu/courses/mathematics/… $\endgroup$ Commented Oct 22, 2020 at 18:41

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