Insert the interaction-picture expansion into the Schrödinger equation. The phases multiplying the basis states already solve the unperturbed evolution, leavingHermiticity requires $H_{21}^{\prime }=\hslash ({\omega }^{\prime }{)}^{\ast }$, not $\hslash {\omega }^{\prime }$ unless the coupling is real. Since the diagonal matrix elements vanish and ${\omega }_0={\omega }_2-{\omega }_1$, expanding the cosine givesThe conjugate pair of couplings conserves $|c_1{|}^2+|c_2{|}^2$. Near resonance, the difference frequency is slow while the sum frequency is approximately $2{\omega }_0$. Rapid terms alternate sign before an appreciable slow transfer accumulates. The rotating-wave approximation retains the slow terms when $|{\omega }^{\prime }|,|\omega -{\omega }_0|\ll \omega +{\omega }_0$; resonance alone is insufficient for a strong drive. Counter-rotating corrections include small shifts of the resonance and need not remain negligible at arbitrary strength or arbitrarily long times.

For the spin preparation, use $H=-\mathbf{\mu }\cdot \mathbf{B}=-\gamma \mathbf{s}\cdot \mathbf{B}$ and $\mathbf{s}=\hslash \mathbf{\sigma }/2$. Choose $B>0$ and $\gamma >0$ to fix helicity, with $|\uparrow ⟩=(1,0{)}^T$, $|\downarrow ⟩=(0,1{)}^T$ and ${\omega }_0=\gamma B$. The initial $|\uparrow ⟩$ is the lower state. For negative $\gamma$, reverse the RF rotation sense; pulse areas use $|\gamma |$, and signed Larmor precession determines the corresponding phase adjustment.

A circularly rotating magnetic drive uses two transverse fields in quadrature during $0<t<\tau$:With $U_0(t)=e^{+i{\omega }_{0}t{\sigma }_z/2}$, their interaction Hamiltonian is exactly constant,The minus sign in the second RF component selects co-rotation for the specified positive gyromagnetic ratio. Exponentiating this Pauli matrix givesThe laboratory phase of a resonant spin pulse includes free precession. Returning to the laboratory frame and removing a common phase,Unlike a linearly polarized RF field, this exactly co-rotating field has no counter-rotating component. A single linearly polarized field $B_{x,\mathrm{peak}}\cos ({\omega }_{0}t+\phi )$ gives the same slow rotation in the weak-drive rotating-wave approximation with $\Omega =|\gamma |B_{x,\mathrm{peak}}/2$. For the circular field above, $\Omega =|\gamma |B_1$. Keeping these amplitude conventions distinct avoids a factor-of-two pulse-duration error.

For a single harmonic pulse detuned by $\Delta ={\omega }_0-\omega$, the effective rotating-frame field tilts out of the transverse plane. Its rotation rate becomes $\sqrt{{\Omega }^2+{\Delta }^2}$, and its maximum population transfer is ${\Omega }^2/({\Omega }^2+{\Delta }^2)$. Exact inversion to $-z$ is impossible with one fixed nonzero-detuning pulse. Half transfer, needed for a transverse eigenstate, gives the equatorial reachability of a detuned spin pulse criterion $|\Delta |\le \Omega$. With that condition, changing the duration and RF phase can still prepare $+x$ or $+y$ at the chosen laboratory stopping time. The mere statement $\omega \approx {\omega }_0$ does not ensure this condition, because the drive can be much weaker than either frequency. If $|\Delta |>\Omega$, neither transverse target is reached by that single pulse. These restrictions concern fixed-frequency, fixed-amplitude single-pulse control; retuning, chirping or phase-controlled composite pulses change the available controls and can remove the restriction. No dissipative spin relaxation is included in this coherent two-state model.

Introduce a bookkeeping parameter $\lambda$ in $H=H_0+\lambda H_1$ and expand $|{\varphi }_n⟩=|{\psi }_n⟩+\lambda |{\eta }_n⟩+O({\lambda }^2)$ and ${\mathcal{E}}_n=E_n+\lambda \Delta E_n+O({\lambda }^2)$. Fix phase and normalization at this order by $⟨{\psi }_n|{\eta }_n⟩=0$. The first-order Schrödinger equation isangular momentum projection onto $⟨{\psi }_n|$ gives $\Delta E_n=⟨{\psi }_n|H_1|{\psi }_n⟩$. angular momentum projection onto a different eigenstate givesCompleteness therefore yields the first-order nondegenerate perturbation theory resultNonzero energy denominators are essential. The displayed state has unit norm through first order; exact normalization would change terms of second order. A continuum, if present, contributes its corresponding integral as well as any discrete sum.

For the oscillator in a static electric field, electrostatic potential energy is $H_1=-qϵx$ and $\omega =\sqrt{k/m}$. parity gives $⟨0|x|0⟩=0$, and the supplied Gaussian wavefunctions giveSince $E_0-E_1=-\hslash \omega$, the two-state first-order ground state isThus $⟨x⟩=2\eta x_{\mathrm{zp}}+O({ϵ}^2)$, and the induced electric dipole isIt is aligned with the field for either sign of $q$. The harmonic-oscillator electric polarizability is $q^2/k$. The position operator connects $|0⟩$ only to $|1⟩$, so the requested two-state calculation contains the complete first-order dipole response, rather than dropping other first-order terms.

Complete the square to find the exact electric-field displacement of a quantum harmonic oscillator:The exact normalized ground wavefunction is ${\psi }_0(x-x_0)$, with energy $\hslash \omega /2-q^2{ϵ}^2/(2k)$. Its mean displacement is $x_0$, soThe first-order dipole equals the exact dipole, because the restoring force is exactly linear and translating a harmonic well does not change its width or frequency. This agreement of one expectation value does not make the truncated wavefunction exact. The Hellmann–Feynman theorem provides an independent check: $d=-\partial E_0/\partial ϵ$.

For perturbation theory to approximate the state, $|\eta |\ll 1$, or $|x_0|\ll 2x_{\mathrm{zp}}$. The exact translated ground state has the original-oscillator expansionIt is a coherent state with mean occupation ${\eta }^2$. Large fields require many oscillator levels, whereas the first-order state retains only two. The normalized two-state truncation of a displaced oscillator instead giveswhich incorrectly tends towards zero for large $|\eta |$ instead of following the displaced well. In position space the truncated polynomial times Gaussian also develops a node absent from the exact ground state. Taking a linearized formula outside its small-state-correction regime is therefore invalid even though its dipole coefficient happens to be exact.

The quoted expansion is nondegenerate perturbation theory for a known isolated eigenstate of a time-independent Hamiltonian. A useful small-mixing condition is $|⟨k|H_1|i⟩|\ll |E_i-E_k|$ for coupled states. First order is a diagonal expectation value; second order describes virtual mixing. A degenerate or nearly degenerate coupled subspace must first be treated by diagonalization there. Symmetry can eliminate terms, and convergence still needs checking. Ground-state second-order shifts are nonpositive because all denominators are negative.^[ref]

For the rotational Stark effect, $H_1=-pE\cos \theta$, where $\theta$ is the polar angle to the field. The source calls it azimuthal, which is inconsistent with this coupling. Rotor states have parity $(-1{)}^J$, while $\cos \theta$ is odd, so every first-order matrix element within one fixed-$J$ manifold vanishes. The $m$ degeneracy therefore causes no first-order splitting.

Only $J\pm 1$ with the same $m$ contribute in second order. Using their energy differences $2BJ$ and $-2B(J+1)$,For $J\ge 1$ this simplifies to the quadratic Stark shift of a polar linear rotor,For $J=0$, only $J^{\prime }=1,m=0$ contributes, with squared angular matrix element $1/3$ and denominator $-2B$:Here $B$ is the rotational energy constant, unlike the magnetic field in Question3.

The unperturbed line gives $B/h=6.08\mathrm{GHz}$. The $J=1$ shifts are $(pE{)}^2/(10B)$ for $m=0$ and $-(pE{)}^2/(20B)$ for $m=\pm 1$. Subtracting the common lower-state shift gives transition shifts $4(pE{)}^2/(15Bh)$ and $7(pE{)}^2/(60Bh)$. Their separation determines the dipole moment from a rotational Stark doublet:The $\mathrm{D}$ unit is the debye. Using external $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{D}=3.33564\times 10^{-30}\mathrm{C}\mathrm{m}$ gives $pE/B\simeq 0.0593$, consistent with weak-field perturbation theory. The two shifted frequencies are about $12.165707$ and $12.162497\mathrm{GHz}$; polarization selects $\Delta m=0$ or $\pm 1$. The inferred value is a magnitude; these shifts do not determine the dipole's sign.^[ref][ref]

The square-root branches avoid one another when $\Delta \ne 0$; crossings with a parallel-spin branch are allowed because total spin projections differ. At zero field the positive-$\Delta$ sketch has one lower singlet and a triplet above it. The source supplies no numerical moment ratio, so the two panels show explicit illustrative signed choices, including the exactly compensated case.

For spin one-half, $s_{\pm }=s_x\pm i{s}_y=(\hslash /2)({\sigma }_x\pm i{\sigma }_y)$, henceThey raise or lower the spin projection by one unit of $\hslash$ and annihilate the corresponding endpoint. The exchange identity ${\mathbf{\sigma }}_1\cdot {\mathbf{\sigma }}_2={\sigma }_{1z}{\sigma }_{2z}+2({\sigma }_{1+}{\sigma }_{2-}+{\sigma }_{1-}{\sigma }_{2+})$ preserves total spin projection.

Write $\Delta =\Delta E$, $u={\mu }_{0}B(g_1+g_2)$ and $w={\mu }_{0}B(g_1-g_2)$. In the listed order $(\uparrow \uparrow ,\downarrow \downarrow ,\uparrow \downarrow ,\downarrow \uparrow )$, direct action givesThe printed matrix instead uses $(\uparrow \uparrow ,\uparrow \downarrow ,\downarrow \uparrow ,\downarrow \downarrow )$. Permuting the listed indices to $(1,3,4,2)$ recovers its central block and diagonal entries. This is a basis-order mismatch in the source, not a different Hamiltonian.

The two-spin exchange spectrum with unequal signed magnetic moments isUse the products $g_j{\mu }_0$ as the signed magnetic coefficients in the stated Hamiltonian. Opposite physical charges must be represented in those signs; they do not by themselves imply equal moment magnitudes. Here ${\mu }_0$ denotes the source's magneton, not vacuum permeability.