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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?

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T2 is so-called dephasing time.

It describes how long a phase of qubit stay 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 togetger called decoherence times.

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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. $\endgroup$ – Martin Vesely Feb 6 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/… $\endgroup$ – Balint Pato 2 days ago
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Just some more information on $T_1$ and $T_2$ times:

$T_1$ and $T_2$ are known as two measurements of decoherence. In general, $T_1 > T_2$.

$T_1$ = Relaxation Time = Average Transition Time between two states. For example, the time it takes for a $|1\rangle$ state to become a $|0\rangle$ state. Imagine the Bloch sphere, the arrow pointing at the north pole will move along the longitudinal axis to reach the ground state at the south pole. This is why it is also known as depolarisation.

$T_2$ = Dephasing Time = The time for the superposition state to lose its coherence (as stated in the first answer). Again in the Bloch sphere example, imagine the arrow pointing at some point along the equator, i.e. the latitudinal axis. Over time the arrow rotates along the equator and its 'phase' information becomes lost i.e. becomes undetermined or uncorrelated (second answer).

I think this paper gives a good example using spin: https://arxiv.org/pdf/cond-mat/0509395.pdf

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  • $\begingroup$ Thanks for your explanation! $\endgroup$ – peachnuts Jun 28 at 18:04
  • $\begingroup$ I don't think I agree with explanation. As I explain in my answer, $T_1$ is not the time that it takes for the $|0\rangle$ state to become a $|1\rangle$ state. It's the time that it takes the $|0\rangle$ state to become an (equal) incoherent mixture of both the $|0\rangle$ and the $|1\rangle$ state. I think a better analogy is that the state moves from the north pole of the Bloch sphere to the center, not to the south pole. As for dephasing, I would picture that as a state moving toward the $z$-axis while maintaining a constant $z$-coordinate. $\endgroup$ – tparker Jul 20 at 20:18

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