--- title: The Paschen-Back and Intermediate-Field Regimes module: Atoms in External Fields moduleNumber: 6 lessonNumber: 2 order: 602 summary: > When the magnetic interaction grows past the fine-structure coupling, spin and orbital angular momentum decouple and precess independently about the field. The anomalous Zeeman pattern reverts to a simple triplet, the Paschen-Back effect. Between the two limits neither coupling dominates and the level positions follow from diagonalizing the combined spin-orbit and Zeeman Hamiltonian. We build the two-by-two problem for a single valence electron, solve it in closed form, and show both limits emerge from one expression. topics: [Atoms in External Fields] sources: - book: Bransden & Joachain ref: "Ch. 9 — Atoms in External Fields; §9.2 The Paschen-Back effect and the intermediate field" - book: Foot ref: "Ch. 5 — The LS-coupling scheme; §5.5 The Zeeman effect and strong fields" - book: Griffiths & Schroeter ref: "Ch. 7 — Time-Independent Perturbation Theory; §7.4.3 Weak and strong-field Zeeman" - book: Demtröder ref: "Ch. 5 — The Hydrogen Atom; §5.2 Atoms in magnetic fields" draft: false --- The [weak-field Zeeman effect](/atomic-physics/atoms-in-external-fields/zeeman-effect) treats the magnetic interaction as a small perturbation on the fine-structure eigenstates. Nothing forces the field to stay small. As $B$ grows, the Zeeman energy $\mu_B B$ eventually overtakes the spin-orbit splitting, and the hierarchy of couplings inverts: the field now sets the fast precession and the spin-orbit interaction becomes the perturbation. In this **strong-field** or **Paschen-Back** regime the good quantum numbers change from $(j, m_j)$ to $(m_\ell, m_s)$, and the anomalous multi-line pattern collapses back to a normal-looking triplet. The genuinely hard case is the **intermediate field**, where neither coupling dominates and the level positions require an exact diagonalization. For a single valence electron that diagonalization is a two-by-two problem with a closed-form solution. ## The competing energy scales Two interactions compete for control of the spin. The spin-orbit coupling, $$ H_{\text{so}} = A\,\vec L\cdot\vec S, \qquad A = \frac{\zeta_{n\ell}}{\hbar^2}, $$ with $\zeta_{n\ell} = \dfrac{\mu_0}{4\pi}\,g_s\,\dfrac{e^2\hbar^2}{2m_e^2}\big\langle r^{-3}\big\rangle$ the radial spin-orbit constant, locks $\vec L$ and $\vec S$ into a fixed $\vec J$. The Zeeman coupling, $$ H_Z = \frac{\mu_B B}{\hbar}\big(L_z + 2S_z\big), $$ torques $\vec L$ and $\vec S$ separately about $\hat z$. Which one wins is set by the dimensionless ratio $$ x \equiv \frac{\mu_B B}{\Delta_{\text{fs}}}, \qquad \Delta_{\text{fs}} = A\hbar^2\big(\ell + \tfrac12\big), $$ where $\Delta_{\text{fs}}$ is the zero-field splitting between the $j = \ell+\tfrac12$ and $j = \ell-\tfrac12$ levels. The weak-field theory is the expansion in $x \ll 1$; the Paschen-Back theory is the expansion in $x \gg 1$. $$ % caption: Weak field: $\vec L$ and $\vec S$ lock into $\vec J$, which precesses % about B. Strong field: $\vec L$ and $\vec S$ decouple and precess separately. \begin{tikzpicture}[scale=1.0, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} % --- weak field (left) --- \draw[->, black!70, very thick] (-3.7,-1.6) -- (-3.7,1.8) node[above] {B}; \draw[->, black, very thick] (-2.4,-1.4) -- (-2.4,1.2) node[above] {J}; \draw[->, acc, thick] (-2.4,-1.4) -- (-3.3,0.35); \node[acc, anchor=east] at (-3.35,0.3) {L}; \draw[->, black, thick] (-3.3,0.35) -- (-2.4,1.2); \node[black, anchor=east] at (-3.0,0.95) {S}; \node[black, anchor=north] at (-2.9,-1.8) {weak: L,S lock to J}; % --- strong field (right) --- \draw[->, black!70, very thick] (1.2,-1.6) -- (1.2,1.8) node[above] {B}; \draw[->, acc, very thick] (2.2,-1.4) -- (2.6,0.9) node[right] {L}; \draw[->, black, very thick] (2.2,-1.4) -- (1.5,0.5) node[left] {S}; \draw[dashed, black] (2.2,0.2) ellipse (0.9 and 0.25); \node[black, anchor=north] at (2.1,-1.7) {strong: L,S separate}; \end{tikzpicture} $$ ## Strong field: the Paschen-Back limit When $x \gg 1$, drop $H_{\text{so}}$ at zero order. The eigenstates are the uncoupled product states $\lvert \ell\,s\,m_\ell\,m_s\rangle$, and the Zeeman energy is exact and additive, $$ E_Z = \mu_B B\,\big(m_\ell + 2m_s\big). $$ The spin-orbit term is then a first-order correction. In an $(m_\ell, m_s)$ eigenstate the transverse pieces average to zero, $\langle L_x S_x + L_y S_y\rangle = 0$, because $L_\pm$ and $S_\pm$ change $m_\ell$ or $m_s$ and connect to orthogonal states, leaving only $$ \langle H_{\text{so}}\rangle = A\,\langle L_z S_z\rangle = A\hbar^2\,m_\ell m_s = \zeta_{n\ell}\,m_\ell m_s . $$ The complete strong-field energy is therefore $$ E = E_{n\ell} + \mu_B B\,(m_\ell + 2m_s) + \zeta_{n\ell}\,m_\ell m_s . $$ The dipole selection rules in this basis are $\Delta m_\ell = 0, \pm1$ (the photon acts on the orbital motion) and $\Delta m_s = 0$ (radiation does not flip the spin). A transition frequency shifts by $$ h\nu = h\nu_0 + \mu_B B\,\Delta m_\ell + \big(\text{small spin-orbit term}\big), $$ so to leading order only three groups survive at $\nu_0$ and $\nu_0 \pm \mu_B B/h$: the anomalous pattern has reverted to the **normal triplet**. This reversion is the defining experimental signature of the Paschen-Back effect, and its onset marks the field at which $\mu_B B$ passes $\Delta_{\text{fs}}$. $$ % caption: The uncoupled strong-field states arranged by $m_\ell + 2m_s$, with % group spacing $\mu_B B$ and a small spin-orbit split $\zeta\,m_\ell m_s$ % within each group; the pattern collapses to a normal triplet. \begin{tikzpicture}[scale=1.0, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} % energy axis \draw[->, black] (-3.6,-2.0) -- (-3.6,2.4) node[above] {energy}; % strong-field levels of a p electron by ml + 2ms % ml in {+1,0} with ms=+1/2, and {0} etc -- draw six representative levels \foreach \y/\lab in {2.0/{top group}, 0.0/{middle group}, -1.8/{bottom group}}{ \draw[acc, very thick] (-1.2,\y) -- (1.2,\y); \node[acc, anchor=west] at (1.35,\y) {\lab}; } % split within groups by spin-orbit (small) \foreach \y in {2.0,0.0,-1.8}{ \draw[black] (-1.2,\y+0.18) -- (1.2,\y+0.18); \draw[black] (-1.2,\y-0.18) -- (1.2,\y-0.18); } \node[black, anchor=north] at (0,-2.3) {groups set by field; small split from spin-orbit}; \end{tikzpicture} $$ ## The intermediate field: an exact two-by-two Between the limits, neither basis diagonalizes the Hamiltonian $H = H_{\text{so}} + H_Z$. What survives is the projection quantum number $$ m_j = m_\ell + m_s, $$ conserved because $[J_z, H_{\text{so}}] = 0$ and $[J_z, H_Z] = 0$ (both interactions are rotationally invariant about $\hat z$). For a single electron $s = \tfrac12$, at most two uncoupled states share a given $m_j$: $$ \lvert 1\rangle = \big\lvert m_\ell = m_j - \tfrac12,\ m_s = +\tfrac12\big\rangle, \qquad \lvert 2\rangle = \big\lvert m_\ell = m_j + \tfrac12,\ m_s = -\tfrac12\big\rangle. $$ The Hamiltonian is block-diagonal in $m_j$, with each block a $2\times2$ matrix (a $1\times1$ block for the two stretched states $m_j = \pm(\ell + \tfrac12)$, which have no partner). The Zeeman part is diagonal: $$ \langle 1\lvert H_Z\rvert 1\rangle = \mu_B B\,\big(m_j + \tfrac12\big), \qquad \langle 2\lvert H_Z\rvert 2\rangle = \mu_B B\,\big(m_j - \tfrac12\big), $$ reading $m_\ell + 2m_s$ in each state. The spin-orbit part has both diagonal and off-diagonal pieces. Writing $\vec L\cdot\vec S = L_z S_z + \tfrac12(L_+ S_- + L_- S_+)$, the diagonal entries come from $L_z S_z$, $$ \langle 1\lvert H_{\text{so}}\rvert 1\rangle = A\hbar^2\,\tfrac12\big(m_j - \tfrac12\big), \qquad \langle 2\lvert H_{\text{so}}\rvert 2\rangle = -A\hbar^2\,\tfrac12\big(m_j + \tfrac12\big), $$ and the off-diagonal entry from $\tfrac12 L_- S_+$ connecting $\lvert 2\rangle$ to $\lvert 1\rangle$, $$ V \equiv \langle 1\lvert H_{\text{so}}\rvert 2\rangle = \frac{A\hbar^2}{2}\sqrt{\big(\ell + \tfrac12\big)^2 - m_j^2}, $$ using the ladder normalizations $L_-\lvert m_\ell\rangle = \hbar\sqrt{\ell(\ell+1) - m_\ell(m_\ell-1)}\,\lvert m_\ell-1\rangle$ and the analogous $S_+$. $$ % caption: Within a fixed $m_j$ only two uncoupled states mix; spin-orbit is % diagonal in the coupled basis, Zeeman in the uncoupled one, and they compete. \begin{tikzpicture}[scale=1.0, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} \node[draw, acc, minimum width=30mm, minimum height=10mm] (s1) at (-2.4,1.2) {state 1}; \node[draw, acc, minimum width=30mm, minimum height=10mm] (s2) at (-2.4,-1.2) {state 2}; \draw[<->, black!70, very thick] (s1) -- (s2) node[midway, right] {coupled by V}; \node[black, anchor=west] at (-4.0,1.2) {}; \node[anchor=south, black] at (-2.4,2.0) {same $m_j$}; % right: the 2x2 block, hand-drawn brackets (pmatrix delimiters garble in nodes) \draw[black!70, thick] (1.1,0.9) -- (0.9,0.9) -- (0.9,-0.9) -- (1.1,-0.9); \draw[black!70, thick] (3.1,0.9) -- (3.3,0.9) -- (3.3,-0.9) -- (3.1,-0.9); \node at (0.5,0) {$H =$}; \node at (1.55,0.5) {$d_1$}; \node at (2.75,0.5) {$V$}; \node at (1.55,-0.5) {$V$}; \node at (2.75,-0.5) {$d_2$}; \node[black, anchor=west] at (0.6,-1.5) {diagonal from Zeeman and $L_zS_z$}; \end{tikzpicture} $$ ### Diagonalizing the block The eigenvalues of a symmetric $2\times2$ matrix are the mean of the diagonal plus or minus the radius, $E_\pm = \bar d \pm \sqrt{\big(\tfrac{d_1 - d_2}{2}\big)^2 + V^2}$. The mean diagonal is $$ \bar d = -\frac{A\hbar^2}{4} + \mu_B B\,m_j, $$ and the diagonal difference is $d_1 - d_2 = A\hbar^2 m_j + \mu_B B$. Substituting $V$ and simplifying the discriminant (the $m_j^2$ terms cancel) gives the closed-form **Breit-Rabi expression** for fine structure, $$ \; E_\pm = E_{n\ell} - \frac{A\hbar^2}{4} + \mu_B B\,m_j \;\pm\; \frac{\Delta_{\text{fs}}}{2} \sqrt{\,1 + \frac{2m_j}{\ell + \tfrac12}\,x + x^2\,}\; \qquad x = \frac{\mu_B B}{\Delta_{\text{fs}}}. $$ The single square root interpolates between the two limits. > **Worked example.** Take the two limits of the boxed formula. For > $x \to 0$, expand the root as $1 + \dfrac{m_j\,x}{\ell + \frac12}$: > $$ > E_\pm \approx E_{n\ell} - \frac{A\hbar^2}{4} \pm \frac{\Delta_{\text{fs}}}{2} > + \mu_B B\,m_j\Big(1 \pm \frac{1}{2\ell+1}\Big). > $$ > The constant $\pm\Delta_{\text{fs}}/2$ reproduces the two fine-structure > levels $j = \ell \pm \tfrac12$, and the coefficient of $\mu_B B\,m_j$ is > exactly the Landé factor $g_J$ for those levels. For $x \to \infty$, expand as > $x\big(1 + \dfrac{m_j}{(\ell + \frac12)x}\big)$: > $$ > E_\pm \approx E_{n\ell} - \frac{A\hbar^2}{4} > + \mu_B B\Big(m_j \pm \tfrac12\Big) > \pm \frac{A\hbar^2}{2}\,\frac{m_j}{\ell + \frac12}. > $$ > The term $\mu_B B(m_j \pm \tfrac12) = \mu_B B(m_\ell + 2m_s)$ is the > Paschen-Back energy, and the residual reproduces $\zeta\,m_\ell m_s$. One > formula, both regimes. The two stretched states carry no square-root structure: with $m_j = \pm(\ell + \tfrac12)$ there is a single basis state, and its energy is linear in $B$ at all field strengths, $$ E = E_{n\ell} + \frac{A\hbar^2\ell}{2} + \mu_B B\,(m_\ell + 2m_s), \qquad m_\ell = \pm\ell,\ m_s = \pm\tfrac12 . $$ ## The correlation diagram Plotting the eigenvalues against $x$ threads every $j$-multiplet sublevel of the weak-field limit onto an $(m_\ell, m_s)$ level of the strong-field limit. States of the same $m_j$ repel (the off-diagonal $V$ produces an **avoided crossing**), while states of different $m_j$ cross freely, since no matrix element connects them. $$ % caption: Level energies versus field. Weak-field $j$ sublevels on the left % connect to strong-field $(m_\ell,m_s)$ levels on the right; same-$m_j$ curves avoid. \begin{tikzpicture}[scale=1.0, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} % axes \draw[->, black] (0,-2.4) -- (0,2.6) node[above] {energy}; \draw[->, black] (0,0) -- (6.0,0) node[right] {x}; % upper j = l+1/2 fans up, lower j = l-1/2 fans down at left, cross to lines at right % four curves \draw[acc, very thick] (0.2,1.4) .. controls (2.5,2.0) and (4.0,2.1) .. (5.6,2.2); \draw[acc, very thick] (0.2,0.7) .. controls (2.5,1.0) and (4.0,0.7) .. (5.6,0.5); \draw[black, very thick, dashed] (0.2,-0.7) .. controls (2.5,-0.4) and (4.0,0.2) .. (5.6,0.9); \draw[black, very thick, dashed] (0.2,-1.4) .. controls (2.5,-1.6) and (4.0,-1.9) .. (5.6,-2.0); % avoided crossing marker \draw[dashed, black] (3.6,0.75) circle (0.45); \node[black, anchor=south] at (3.6,1.25) {avoided crossing}; \node[acc, anchor=west] at (0.05,1.9) {weak: j, $m_j$}; \node[black, anchor=north] at (3.2,-2.3) {strong: $m_l$, $m_s$}; \end{tikzpicture} $$ > **Worked example.** Take a single $p$ electron, $\ell = 1$, so > $\Delta_{\text{fs}} = \tfrac32 A\hbar^2$ separates $^2P_{3/2}$ from $^2P_{1/2}$. > The projection $m_j$ ranges over $\pm\tfrac32, \pm\tfrac12$. The stretched > states $m_j = \pm\tfrac32$ are single and shift linearly, > $E = \tfrac12 A\hbar^2 + \mu_B B(m_\ell + 2m_s)$ with $m_\ell = \pm1$, > $m_s = \pm\tfrac12$. Each $m_j = \pm\tfrac12$ block is a genuine $2\times2$ with > $$ > V = \frac{A\hbar^2}{2}\sqrt{\big(\tfrac32\big)^2 - \big(\tfrac12\big)^2} > = \frac{A\hbar^2}{2}\sqrt{2}, > $$ > and the boxed formula gives, for $m_j = +\tfrac12$, > $$ > E_\pm = E_{n p} - \frac{A\hbar^2}{4} + \frac{\mu_B B}{2} > \;\pm\; \frac{\Delta_{\text{fs}}}{2}\sqrt{\,1 + \tfrac23 x + x^2\,}, > \qquad x = \frac{\mu_B B}{\Delta_{\text{fs}}} . > $$ > At $x = 0$ the two roots are the $^2P_{3/2}$ and $^2P_{1/2}$ sublevels; as > $x \to \infty$ they become the uncoupled $(m_\ell, m_s)$ levels $(0, +\tfrac12)$ > and $(1, -\tfrac12)$. The six sublevels of the $^2P$ term (two stretched, four > from the two $2\times2$ blocks) thread continuously from six weak-field > $m_j$ states to six strong-field $(m_\ell, m_s)$ states. The avoided crossing is the direct fingerprint of the off-diagonal spin-orbit coupling: its minimum gap equals $2V = A\hbar^2\sqrt{(\ell + \tfrac12)^2 - m_j^2}$ at the field where the two diagonal entries coincide, $A\hbar^2 m_j + \mu_B B = 0$. That the gap is nonzero for every non-stretched $m_j$ is why the transition from $LS$ to $jj$-like labelling is smooth rather than a set of hard level crossings. ## Regime map | Regime | Condition | Good numbers | Energy | | --- | --- | --- | --- | | Weak (anomalous Zeeman) | $\mu_B B \ll \Delta_{\text{fs}}$ | $j, m_j$ | $g_J\,\mu_B B\,m_j$ | | Intermediate | $\mu_B B \sim \Delta_{\text{fs}}$ | $m_j$ only | Breit-Rabi root | | Strong (Paschen-Back) | $\mu_B B \gg \Delta_{\text{fs}}$ | $m_\ell, m_s$ | $\mu_B B(m_\ell + 2m_s) + \zeta\,m_\ell m_s$ | Because $\Delta_{\text{fs}}$ scales as $Z^4/n^3$, the field needed to reach the Paschen-Back regime rises steeply with nuclear charge and falls with excitation: hydrogen $n = 3$ reaches it in a fraction of a tesla, while the sodium D doublet demands tens of tesla. The same competition of a small internal coupling against an external field reappears for the hyperfine interaction, where the analogous crossover is the original Breit-Rabi problem, and for the electric case treated in the [next lesson](/atomic-physics/atoms-in-external-fields/stark-effect-polarizability).[^bj-pb] [^bj-pb]: **Bransden & Joachain**, _Physics of Atoms and Molecules_, §9.2 — the Paschen-Back effect and the exact intermediate-field diagonalization for a single valence electron; the fine-structure Breit-Rabi formula and its two limiting expansions. See also **Foot**, _Atomic Physics_, §5.5.