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?
$\begingroup$ Old question, but I'm adding this as a comment. This online doc very nicely describes T1 and T2 times: mri-q.com/bloch-equations.html $\endgroup$
Commented Apr 9 at 8:13T2 is so-called dephasing time.
It describes how long the phase of a qubit stays intact. In your words, it is time from $|+\rangle= \frac>(|0\rangle + |1\rangle)$ to $|-\rangle= \frac>(|0\rangle - |1\rangle)$ , 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
Similarly for T2.
Both times T1 and T2 are together called decoherence times.
24.1k 3 3 gold badges 37 37 silver badges 90 90 bronze badges answered Feb 6, 2020 at 9:47 Martin Vesely Martin Vesely 14.8k 4 4 gold badges 29 29 silver badges 69 69 bronze badges $\begingroup$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.
answered Feb 6, 2020 at 14:16 2,861 12 12 silver badges 26 26 bronze badges $\begingroup$ Thanks for clarification. $\endgroup$ Commented Feb 6, 2020 at 15:17$\begingroup$ +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/… $\endgroup$