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.
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$\begingroup$ This can be helpful: quantumcomputing.stackexchange.com/questions/10082/… $\endgroup$ – Martin Vesely Sep 4 '20 at 7:06
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$T_1$ 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.
