Decoherence graph of $T_1$, $T_2$, $T_2^*$ in IBMQ

Question

I would like to ask you a question because I cannot clearly understand the difference between $T_2$ and $T_2^*$ in the decoherence graph provided by IBM. Let me summarize the questions I have as follows.

[Circuit] The gate-based circuit for measuring the Decoherence characteristics of the device is as follows. It's a common method, so it doesn't seem necessary to explain it.

[Decoherence Graph] Below is a graph of the measurements released by IBM. Of course, there are slight differences between devices, but I think the patterns will be similar.

The results of $T_1$ are very intuitive. As the amplitude decreases gradually, it changes from $|1\rangle$ to $|0\rangle$, and this pattern seems to be fitable in the form of a function.

The results of $T_2^*$ are understandable to some extent. It seems that the states of $|1\rangle$ and $|0\rangle$ can be measured in the form of sine curves as shown in the graph when measuring after the second H according to the phase change by frequency fluctuation in the state of $|+\rangle$. In other words, $T_2^*$ can be said to detect a phase change in the qubit.

• Q1) In the case of $T_2$, I'm not sure what kind of change we're observing. Unlike $T_1$ and $T_2^*$, which are interpreted intuitively, I would like to know how to interpret the graph of $T_2$. In other words, what causes affect $T_2$, and how does it change the state?

• Q2) In addition, if you look at the Legend of each graph, the numerical values are specified, and I am curious about the criteria. For example, $T_1$, $T_2$, and $T_2^*$ for qubit0 are 24.1, 21.7, and 22.8, respectively, and I wonder how the constants of $T_1$, $T_2$, and $T_2^*$ are determined from the measurement results of the graph.

Answer 1

$T_1$ is know as the transverse relaxation time and it corresponds to a characteristic time in which an excitation to $|1\rangle$ is "lost", causing the qubit to fall back down to the ground state $|0\rangle$ (really the thermal state, but the importance of that difference depends on your qubit). Specifically, this process typically follows a decaying exponential, so fitting $e^{-\Delta t/T_1}$ to your data vs $\Delta t$ plot lets you extract $T_1$.

The $T_2$'s are known as longitudinal relaxation times, because they measure how long it takes for a superposition state on the equator of the Bloch sphere to decay (vs an excited state decaying for $T_1$). $T_2^*$ refers to the time constant of such a decay process when we don't do anything special to try to keep our coherence, and we get it by fitting our data vs $\Delta t$ to a function of $e^{-\Delta t/T_2^*} \cos(\omega \Delta t)$ since there's some natural sinusoidal behavior in the presence of no decay. Now $T_2$ corresponds to a similar decay time of a superposition state on the surface of the Bloch sphere. The one you mentioned is know as a Hahn Echo sequence with the $\pi$ pulse in the middle. That pulse helps deal with constant and low frequency noise/decoherence effects and this almost always gives a better (longer) time constant for this kind of relaxation. The Wikipedia page for spin echo has some great visualizations for what's actually happening there.

Answer 2

I would like to suggest you my point of view regarding your second question:

Q2) In addition, if you look at the Legend of each graph, the numerical values are specified, and I am curious about the criteria. For example, T1, T2, and T2* for qubit0 are 24.1, 21.7, and 22.8, respectively, and I wonder how the constants of T1, T2, and T2* are determined from the measurement results of the graph.

$T_1$ is the relaxtion time, i.e the amount of time it takes for the qubit to go back from the excited state ($|1\rangle$) to the ground state ($|0\rangle$). In This graph:

The blue dots are the actual results of running the circuit with a delay of $x\ \mu s$. The orange curve is a model that best fits the results, which can mathematically be described as: $P = e^{-\frac{t}{T}}$. Not every value of $T$ gives a good approximation, as you can see:

Indeed, $T = 24.1 \ \mu s$ gives much better curve than $T = 2 \ \mu s$, for example. Of couse the approximation is being computed by mathematical methods and not by trial and error (curve fitting). OK, we have found a $T$ that gives a good fit, but why $T = T_1$? You probably ask yourself.

I like to think about the area bounded under the graph of the curve $P$. Again, we are looking for the amount of time it takes for the qubit to decay from the excited $|1\rangle$ state to the ground $|0\rangle$ state - So for $t = 0$ we are completely at $|1\rangle$, and for large value of $t$ we are completely at $|0\rangle$. So if we take an integral of $P(t)$ over the interval $[0, \infty)$ we get the relaxation time $T_1$. If we compute the integral we see that $T = T_1$:

$$T_1 = \int_{0}^{\infty} e^{-\frac{t}{T}}\,dt = \lim_{c \to \infty} \Big[\int_{0}^{c} e^{-\frac{t}{T}}\,dt \Big] = \lim_{c \to \infty} \Big[ -Te^{-\frac{t}{T}} \Big|_0^{c} \Big] = \lim_{c \to \infty} \Big[ T - Te^{-\frac{c}{T}} \Big] = T$$