Every qubit is a realization of some two-level quantum system with $\left| 0 \right\rangle$ and $\left| 1 \right\rangle$ states. These states have their energies $E_0$ and $E_1$ respectively. By solving the Schrödinger equation for the two-level quantum system one can show time evaluation of the $\left| 0 \right\rangle$, $\left| 1 \right\rangle$ and $\left| + \right\rangle$ states:

$$ \left| \psi_0(t) \right\rangle = e^{-i \frac{E_0}{\hbar} t} \left| 0 \right\rangle \\ \left| \psi_1(t) \right\rangle = e^{-i \frac{E_1}{\hbar} t} \left| 1 \right\rangle \\ \left| \psi_+(t) \right\rangle = \frac{1}{\sqrt{2}}\left( e^{-i \frac{E_0}{\hbar} t} \left| 0 \right\rangle + e^{-i \frac{E_1}{\hbar} t} \left| 1 \right\rangle \right)= \frac{e^{-i \frac{E_0}{\hbar} t}}{\sqrt{2}} \left(\left| 0 \right\rangle + e^{-i \frac{E_1 - E_0}{\hbar} t} \left| 1 \right\rangle \right) $$

My question is about the relative phase $\varphi = e^{-i \frac{E_1 - E_0}{\hbar} t}$. This $\phi$ acts as a phase gate and changes the quantum state. Is this relative phase important and how in real quantum computers we should deal with it? One problem that I see is the following circuit:

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After the measurement, the state should be $\left| 0 \right\rangle$, because $HH = I$. But If $t$ is such that the $\varphi = \pi$ then we will measure $\left| 1 \right\rangle$. We can forget about it if $\varphi(t) << \pi$ in the whole computational time. Otherwise, we should make sure that the time between gates should be $2\pi$.

Is this a problem and if yes how IBM, Google or etc. deal with this problem when constructing the circuits? How I understand the frequency $(E_1 - E_0)/\hbar$ value is important and for example, the ibmq_armonk qubit's frequency is equal to ~4.97428 GHz (qiskit_textbook), so should we take into account this value when we play with ibmq_armonk by using OpenPulse?


1 Answer 1


The could be a problem, but it depends on how you're realising your qubits. Some realisations are configured so that $E_0=E_1$, and then there's no problem.

There is (at least from the theoretical perspective) a simple fix: if you're supposed to be waiting a time $t$, then, instead:

  • wait time $t/2$
  • apply bit flip
  • wait time $t/2$
  • apply bit flip.

This exchanges the roles of 0 and 1 for half the time so that they end up with equal phases.

Alternatively, you just take this into account and apply compensating phase gates inside your circuit.

  • $\begingroup$ Thanks for the answer. Can you mention an example of a qubit realization that has $E_0 = E_1$? $\endgroup$ Feb 28, 2020 at 11:38
  • 1
    $\begingroup$ IIRC, something like the hyperfine ground states of a Rubidium atom (e.g. trapped in an optical lattice) are a very good approximation to that because the timescales of the gates are much shorter than the timescale due to $1/(E_0-E_1)$. $\endgroup$
    – DaftWullie
    Feb 28, 2020 at 11:58
  • $\begingroup$ You may wish to save your answer tick in case someone can answer your specific question about the ibm hardware, which is not something I kno anything about. $\endgroup$
    – DaftWullie
    Feb 28, 2020 at 11:59
  • $\begingroup$ Thanks a lot. I will keep the answer tick for a while, but still, your answer was very helpful. $\endgroup$ Feb 28, 2020 at 12:11

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