4
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

| improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for clarification. $\endgroup$ – Martin Vesely Feb 6 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.