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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- 2$\begingroup$ 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 $\endgroup$user3237992– user32379922024-04-09 08:13:55 +00:00Commented Apr 9, 2024 at 8:13
- $\begingroup$ See quantumcomputing.stackexchange.com/questions/2432/… for a detailed discussion on how to define T2 times (in different ways, called $T_2$ and $T_2^\ast$). $\endgroup$Norbert Schuch– Norbert Schuch2025-08-22 18:16:39 +00:00Commented Aug 22 at 18:16
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2 Answers
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4 $T_2$ is the so-called dephasing time.
It describes how long it takes for the phase information of a qubit to get lost, in the "preferred basis" of the system (which is typically taken to be the computational basis.)
More formally, if your system is initially in the state $$\lvert\psi(t=0)\rangle = \sqrt{p_0}\lvert0\rangle + e^{i\phi} \sqrt{p_1} \lvert1\rangle \ , $$ this is the time after which the phase information $e^{i\phi}$ is lost, i.e., the state is (approximately) independent of $\phi$. If the time $T_1$ is much larger than $T_2$, this means that $p_0$ is approximately unaffected, i.e., the system is converging to the mixed state $$\rho=p_0\lvert 0\rangle\langle 0\rvert + p_1 \lvert 1 \rangle\langle 1\rvert\ ,$$ with a time scale given by $T_1$.
As always, this convergence is exponential, so for times $t$ with $T_1\gg t$, we have that $$ \|\lvert\psi(0)\rangle\langle\psi(0) - \rho\rvert\| \propto e^{-t/T_1}\ . $$
- 1$\begingroup$ Is it accurate to say dephasing is the qubit state transforming from $|+\rangle$ to $|-\rangle$? The qubit gradually transitions from $|+\rangle$ to the maximally mixed state $\rho =\dfrac{1}{2} |0\rangle \langle 0| + \dfrac{1}{2} |1\rangle \langle 1|$. $\endgroup$Soroush khoubyarian– Soroush khoubyarian2025-01-25 21:11:58 +00:00Commented Jan 25 at 21:11
- 2$\begingroup$ @Soroushkhoubyarian: Yes, you are right that my answer is not fully accurate. Please see below an answer by user tparker which clarifies further. $\endgroup$Martin Vesely– Martin Vesely2025-01-25 21:54:26 +00:00Commented Jan 25 at 21:54
- 2$\begingroup$ Given that this is the accepted answer and highly voted (why?), I thought I'd better edit it to be correct. E.g. on the phone, many people might not even notice there is a second - differing! - correct answer. $\endgroup$Norbert Schuch– Norbert Schuch2025-08-22 13:09:26 +00:00Commented Aug 22 at 13:09
- 1$\begingroup$ As to "why": Ah, it was HNQ. $\endgroup$Norbert Schuch– Norbert Schuch2025-08-22 13:32:16 +00:00Commented Aug 22 at 13:32
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2 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.
- 2$\begingroup$ Thanks for clarification. $\endgroup$Martin Vesely– Martin Vesely2020-02-06 15:17:03 +00:00Commented Feb 6, 2020 at 15:17
- 9$\begingroup$ +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– Balint Pato2020-10-22 18:41:49 +00:00Commented Oct 22, 2020 at 18:41