# Need mathematical Calculation of input(T1,T2,f) and output(Readout Error)

I would like to know about the mathematical relations among T1, T2, frequency, readout error, and single-qubit error? This screenshot is from the ibmq-16 Melbourne excel data file.

$$T_1$$ and $$T_2$$ are two measurement of decoherence on a qubit.
$$T_1$$ is known as the "relaxation time" or "longitudinal coherence time" or "amplitude damping".... It measures the loss of energy from the system. You can calibrate/measure/determine the $$T_1$$ time by first initialize the qubit in the $$|0\rangle$$ then apply the $$X$$ gate, where $$X = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} = |0\rangle\langle 1| + |1\rangle \langle 0|$$ and wait for certain amount of time and measure the probability of the state being in the $$|1\rangle$$ eigenstate.
$$T_2$$ is known as the "dephasing time" or "transverse coherence time" or "phase coherence time" or "phase damping" ... and $$T_2$$ can be determined by again initialize the qubit in the state $$|0\rangle$$ then apply the Hadamard gate $$H = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}$$ to the inital qubit state $$|0\rangle$$. We will also wait for some time, $$t$$, and then apply another Hadamard gate, then measure the probability of the qubit being in the state $$|0\rangle$$. Here, as you can see, if we have no decoherence then the qubit will ended up back to the state $$|0\rangle$$ with 100% probability, as $$HH|0\rangle = |0\rangle$$. But of course this is not the case with qubit, the longer the wait time, the closer this probability will get to $$1/2$$ as the qubit will go/dephase from the state $$\dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$$ to $$|0\rangle$$ or $$|1\rangle$$ before the second Hadamard gate. Which will then put the qubit back in the superposition state.