= Qubit precession under a constant Hamiltonian
{title2=$H=\hbar\omega\sigma_z/2$}
Choose the <Pauli matrices> with $\sigma_z=\operatorname{diag}(1,-1)$, and let $\omega>0$ be an <angular frequency>. The <computational basis> is an <orthonormal basis> of <energy eigenstates> for the <qubit> <Hamiltonian operator>
$$
H=\frac{\hbar\omega}{2}\sigma_z,\qquad
E_0=\frac{\hbar\omega}{2},\quad E_1=-\frac{\hbar\omega}{2},\qquad
U(t)=\begin{pmatrix}e^{-i\omega t/2}&0\\0&e^{i\omega t/2}\end{pmatrix}.
$$
In this convention $|1\rangle$ is the lower <energy eigenstate>; the labels describe a <basis> choice, not a universal <ground state>/<excited state> ordering. Start in the <Hadamard basis> <pure state> $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$. Then
$$
|\psi(t)\rangle=\frac{e^{-i\omega t/2}|0\rangle+e^{i\omega t/2}|1\rangle}{\sqrt2}
=e^{-i\omega t/2}\frac{|0\rangle+e^{i\omega t}|1\rangle}{\sqrt2},\qquad
\rho(t)=\frac12\begin{pmatrix}1&e^{-i\omega t}\\e^{i\omega t}&1\end{pmatrix}.
$$
The first factor in the second expression is a <global phase>. The ratio of the two <probability amplitudes> has relative <quantum phase> $\omega t$, so its sign follows directly from $E_1-E_0=-\hbar\omega$.
Multiplying this <density matrix> by each <Pauli matrix> and taking its <trace> gives the <Bloch vector>:
$$
\langle\sigma_x\rangle_t=\cos(\omega t),\qquad
\langle\sigma_y\rangle_t=\sin(\omega t),\qquad
\langle\sigma_z\rangle_t=0.
$$
The <Bloch vector> travels counterclockwise around the equator of the <Bloch sphere>, from $+x$ toward $+y$ when viewed from $+z$. A <computational-basis measurement> always has $P_Z(0)=P_Z(1)=1/2$, while the <Hadamard basis> and $\sigma_y$ <eigenbasis> measurements give
$$
P_X(+)=\frac{1+\cos(\omega t)}2,\qquad
P_Y(+)=\frac{1+\sin(\omega t)}2.
$$
At $\omega t=\pi/2$, the <quantum state> is $|+i\rangle=(|0\rangle+i|1\rangle)/\sqrt2$ up to a <global phase>, giving $P_Y(+)=1$. At $\omega t=\pi$, it is $|-\rangle$ up to a <global phase>, giving $P_X(+)=0$. Both retain equal <energy measurement> <probabilities>.
\Image[/unitary-qubit-precession.png]
{title=Relative phase rotates a qubit while energy populations stay fixed}
{description=Original ideal-model equatorial Bloch-vector diagram and exact Born-rule probabilities for $H=\hbar\omega\sigma_z/2$ and initial $|+\rangle$. The horizontal variable is the dimensionless phase $\omega t$; no curve is measured data.}
At $t=2\pi/\omega$, the <time-evolution operator> is $U(t)=-I$, so $|\psi(t)\rangle=-|+\rangle$. The physical <quantum state ray>, its <density operator>, and every measurement's <probabilities> have returned. The representative <vector> returns exactly only at $t=4\pi/\omega$, when $U(t)=I$. The minus sign is a <global phase> for this isolated evolution; comparing an evolved branch with an unevolved branch through a <controlled unitary gate> can make it a relative <quantum phase>, as explained in <global phase of an uncontrolled quantum gate is unobservable>.
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