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For a nucleus of spin $I>0$ and electronic $J=1/2$, the magnetic-dipole hyperfine interaction is $A\mathbf{I}\cdot \mathbf{J}$, using dimensionless angular-momentum operators. The electric-quadrupole hyperfine contribution vanishes for $J=1/2$. In zero field, $F=I\pm 1/2$ labels the levels:Their signed separation is $\Delta =A(I+1/2)$. For $A>0$, the larger-$F$ multiplet lies higher; negative $A$ reverses this order. If $I=0$ there is no such hyperfine splitting.

Let the field define $z$ and writeHere $g_I$ is the signed nuclear Zeeman coefficient expressed in Bohr-magneton units, including its nuclear-to-Bohr-magneton factor; it is not an unconverted nuclear $g$ factor. Total projection $m=m_I+m_J$ is conserved, while $F$ need not be. In the basis $|m-1/2,+1/2⟩$, $|m+1/2,-1/2⟩$, the two-state block isFor $A>0$, define $x=(g_J-g_I){\mu }_{B}B/\Delta$. Diagonalization gives the Breit-Rabi formula,For either sign of $A$, the block itself remains valid; its ordered eigenvalues are $-A/4+g_I{\mu }_{B}Bm\pm \frac{1}{2}\sqrt{{\Delta }^2+2Am(g_J-g_I){\mu }_{B}B+(g_J-g_I{)}^2{\mu }_B^2{B}^2}$.

For a stretched hyperfine Zeeman state at projection $m=\pm (I+1/2)$ there is only one state each. Their energies are exactlyAssigning two branches to each stretched projection would double-count states.

In weak fields the shift is $g_F{\mu }_{B}mB$, with $g_F=g_I\pm (g_J-g_I)/(2I+1)$ for $F=I\pm 1/2>0$; an $F=0$ state has no first-order Zeeman shift. At strong hyperfine-decoupling fields the eigenstates approach product states withThis is the hyperfine regime of the Paschen-Back effect, still assuming the field is weak enough not to invalidate the electronic $J=1/2$ manifold. States of the same $m$ mix and avoid crossing when their off-diagonal coupling is nonzero; different-$m$ levels can cross.

Use the ordered basis $(|1⟩,|2⟩)$ and ${\sigma }_z=\mathrm{diag}(1,-1)$, withA pure quantum superposition $|+x⟩=(|1⟩+|2⟩)/\sqrt{2}$ has $\mathbf{R}=(1,0,0)$ on the Bloch sphere. In contrast, a mixed state $\rho =p|1⟩⟨1|+(1-p)|2⟩⟨2|$ has $\mathbf{R}=(0,0,2p-1)$, inside the sphere for $0<p<1$. At equal populations its vector is zero. Equal populations alone therefore do not distinguish a coherent superposition from a mixture; their off-diagonal density matrix elements differ.

Choose the drive phase so the rotating-frame Hamiltonian is $H_{\mathrm{rot}}=\hslash (\Omega {\sigma }_x+\delta {\sigma }_z)/2$, where $\delta =\omega -{\omega }_0$ as specified. The Rabi vector is the angular-velocity vectorHere $\Omega$ is the on-resonance population oscillation angular frequency, not the unhalved coefficient of a cosine coupling. From $\mathbf{R}(0)=\hat{z}$, rotation about $\mathbf{W}$ at $W=\sqrt{{\Omega }^2+{\delta }^2}$ givesFor $\Omega \gg |\delta |$, the rotation axis is nearly transverse, the trajectory nearly a great circle, and transfer is almost complete. For $\Omega =\delta >0$, the axis bisects $+x$ and $+z$, the orbit lies in $R_x+R_z=1$, and $p_2=\frac{1}{2}{\sin }^2(\sqrt{2}\Omega t/2)$ never exceeds $1/2$. Each member remains a pure state in this coherent evolution.

For the Ramsey free interval, let atom $j$ have detuning ${\delta }_j=\delta +{\eta }_j$. Instantaneous ideal pulses give the Ramsey ensemble coherence factor $\chi (T)=⟨e^{i{\eta }_{j}T}⟩$. Averaging the free-evolution vector obtained in step (ii) givesDifferent atoms fan out around the equator, shortening the ensemble Bloch vector even though every individual's vector remains unit length. A broad smooth detuning distribution usually makes the average spiral towards the centre and damps the Ramsey fringes towards $1/2$. The dephasing time is of order $1/\Delta {\omega }_0\sim 5/\delta$. inhomogeneous ensemble dephasing is not necessarily irreversible decoherence of individual atoms; an appropriate spin echo can rephase static frequency offsets.

The given spread does not specify a probability distribution or whether it denotes an RMS width. An exact envelope is therefore not determined. The figure illustrates, rather than infers, a Gaussian distribution with RMS ${\sigma }_{\delta }=0.2\delta$, for which $\chi =e^{-{\sigma }_{\delta }^2{T}^2/2}$. A uniform distribution would instead give a sinc envelope and side lobes. Both have the same qualitative initial loss of contrast on the quoted inverse-spread scale.

In the rubidium-87 ground-state hyperfine structure, $J=1/2$ and $I=3/2$, so $F=1,2$. Taking $H_{\mathrm{hf}}=A\mathbf{I}\cdot \mathbf{J}/{\hslash }^2$ givesThe zero-field splitting is ${\Delta }_{\mathrm{hf}}=2A\simeq h\times 6.8\mathrm{GHz}$. Ignoring the small nuclear magnetic term, the weak-field Landé g-factors are $g_{F=1}\simeq -1/2$ and $g_{F=2}\simeq +1/2$. The diagram labels all three $F=1$ and five $F=2$ substates, with linear shifts $g_F{\mu }_B{m}_{F}B$.

The pair $|1⟩=|F=1,m_F=0⟩$ and $|2⟩=|F=2,m_F=0⟩$ is connected by a magnetic dipole transition with $\Delta F=1$, $\Delta m_F=0$, driven with its oscillating magnetic component along the quantization axis. Both first-order shifts vanish, giving the rubidium-87 clock transition first-order insensitivity near zero field. Its insensitivity does not imply absence of quadratic shifts.

For these two $m_F=0$ levels the Breit-Rabi formula gives the quadratic Zeeman shift of a clock transition:where the last expression neglects the nuclear term and sets $g_J\simeq 2$. HenceThis uses ${\mu }_B/h\simeq 1.40MHz/G$ and the supplied $A/h\simeq 3.4\mathrm{GHz}$. With physical moments defined by ${\mu }_i=-\partial E_i/\partial B$, the differential magnetic moment is ${\mu }_2-{\mu }_1\simeq -8.2\times 10^{-3}{\mu }_B$. If effective magnetic moment denotes the positive energy slope instead, that slope difference is positive. The magnitude and physical frequency sensitivity are the same. Retaining the full atomic factors gives the familiar quadratic clock-frequency coefficient about $575Hz/{\mathrm{G}}^2$; the simplified supplied parameters give about $576Hz/{\mathrm{G}}^2$.^[ref]

Use a far-detuned optical dipole trap, such as a red-detuned focused-beam trap, to confine both clock states. Their first-order magnetic moments vanish; their residual quadratic magnetic slopes are opposite, so a simple static magnetic-field minimum does not confine both in the same way. An optical potential acts through the electronic polarizability and can trap both ground hyperfine states. It is not perfectly state independent: differential AC Stark shifts, intensity noise and residual magnetic gradients still need control during Ramsey evolution.

For the two $m=0$ branches of the Breit-Rabi formula, their splitting is $\sqrt{{\Delta }_{\mathrm{hf}}^2+[(g_J-g_I){\mu }_{B}B{]}^2}$. Expanding at small field gives the quadratic frequency shift displayed above. Vanishing linear frequency sensitivity at zero field therefore does not eliminate field-gradient dephasing at a finite bias field.

The ground-hyperfine magnetic-dipole transition between $F=1,m_F=0$ and $F=2,m_F=0$ has vanishing first-order Zeeman shift near zero field. It obeys $\Delta F=1,\Delta m_F=0$. Its residual quadratic shift follows from the Breit-Rabi formula, rather than from assigning both states permanently zero magnetic dipole moment.

For electronic $J=1/2$, an extremal total projection has only one product state rather than a two-dimensional spin block. With dimensionless angular-momentum operators and nuclear Zeeman coefficient $g_I$ in Bohr-magneton units, its energy is $AI/2\pm {\mu }_{B}B(g_J/2+g_{I}I)$. Extending both signs of a two-state Breit-Rabi formula square root to each stretched projection would invent an extra state.