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A qubit in the state $|1\rangle + |0\rangle$, normalized, is time-dependent, right? So it sweeps around the equator of the Bloch sphere at a frequency proportional to the energy gap according to Schrodinger's equation?

With no applied field and energy levels $E_1$ and $E_0$, with $\psi=\begin{bmatrix}a\\b\end{bmatrix}$, you have Schrodinger's equation

$\frac{d}{dt}a=-iE_0/\hbar a$ and $\frac{d}{dt}b=-iE_1/\hbar$ so $\phi=arctan2(b,a)$ is time dependent with frequency $(E_1-E_0)/\hbar$, and that is the angle of rotation about the z axis in the Bloch sphere.

Am I missing something?

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  • $\begingroup$ If these are your qubits, you have to measure in a rotating X axis that rotates at frequency $\Delta E$ $\endgroup$ Feb 5, 2022 at 2:56

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You are mixing up the physical realization of a qubit and the logical abstraction of a qubit.

In the logical abstraction, the $|1\rangle$ state doesn't have dynamics or a higher energy or anything like that. It's just $|1\rangle$ and it stays $|1\rangle$.

In the physical realization the state being used to represent $|1\rangle$ may or may not be constantly changing relative to the state being used to represent $|0\rangle$. For example, in superconducting qubits, we have a concept of a rotating frame that we use to cancel out the dynamics. But if the $|1\rangle$ state was "photon is vertically polarized instead of horizontally polarized" or it was "photon is in the left lane instead of the right lane" then no such correction would be needed.

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  • $\begingroup$ Thanks, @CraigGidney. So if I have a solution with components $c_0$ and $C_1$ to Schrodinger's equation and I want to extract a Bloch vector from it, I multiply $c_i$ by $e^{i E_i/\hbar t}$? $\endgroup$
    – Anna Naden
    Feb 4, 2022 at 21:15
  • $\begingroup$ @AnnaNaden Yes, I think so. There's details in chapter 3 of web.physics.ucsb.edu/~martinisgroup/theses/Neeley2010b.pdf $\endgroup$ Feb 4, 2022 at 21:33
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The state $|0\rangle + |1\rangle$ normalized $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ is an x-basis vector, which is given by $$ \left\{|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle),\ |-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)\right\}. $$ I wouldn't say that these vectors, in a closed quantum system, are time-dependent. Absent some driving force, or interactions with the environment, these states should remain fixed. Any precession around the Bloch sphere equator (or more generally - any time dependence) would be determined by the specific Hamiltonian you are considering.

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  • $\begingroup$ Thanks for your feedback, @Bebotron. See the edited question where I explain why I concluded that the state was time-dependent $\endgroup$
    – Anna Naden
    Feb 4, 2022 at 20:33

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