How to deal with the relative phase from the evaluation of the not interacting qubit?

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:

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?

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.

• Thanks for the answer. Can you mention an example of a qubit realization that has $E_0 = E_1$? – Davit Khachatryan Feb 28 '20 at 11:38
• 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 $1/(E_0-E_1)$. – DaftWullie Feb 28 '20 at 11:58
• 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. – DaftWullie Feb 28 '20 at 11:59
• Thanks a lot. I will keep the answer tick for a while, but still, your answer was very helpful. – Davit Khachatryan Feb 28 '20 at 12:11