# Decoherence graph of T1, T2, T2* in IBMQ

I would like to ask you a question because I cannot clearly understand the difference between T2 and T2* 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 T1 are very intuitive. As the amplitude decreases gradually, it changes from |1> to |0> and this pattern seems to be fitable in the form of a function.

The results of T2* are understandable to some extent. It seems that the states of |1> and |0> 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 |+>. In other words, T2* can be said to detect a phase change in qubit.

• Q1) In the case of T2, I'm not sure what kind of change we're observing. Unlike T1 and T2*, which are interpreted intuitively, I would like to know how to interpret the graph of T2. In other words, what causes affect T2 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, 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.

• While I think the this Stack Exchange really likes separate questions to be separate posts, these are closely related enough for me to take a stab at both Aug 31, 2022 at 3:20

$$T_1$$ 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.

• I just realized I wrote a good bit about T2 vs T2* for a previous question here: quantumcomputing.stackexchange.com/questions/26325/…, but the question about T1 vs T2* vs T2 and how to extract them from the plots makes this unique enough to stand on its own (by my thinking). Aug 31, 2022 at 3:41

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 $$T = T_1$$:

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