The Zeeman eﬀect: hydrogen in an external magnetic ﬁeld

3.4 The Zeeman eﬀect: hydrogen in an external magnetic ﬁeld

(Since we will not ignore spin, this whole section is about the so-called anomalous Zeeman eﬀect. The so-called normal Zeeman eﬀect cannot occur for hydrogen, but is the special case for certain multi-electron atoms for which the total spin is zero.)

With an external magnetic ﬁeld along the $z$ -axis, the perturbing Hamiltonian is ${\hat{H}}^{(1)} = - μ \cdot B = (e B ∕ 2 m) ({\hat{L}}_{Z} + 2 {\hat{S}}_{z})$ . The factor of 2 multiplying the spin is of course the famous $g$ -factor for spin, as predicted by the Dirac equation. Clearly this is diagonalised in the ${| n l m_{l} m_{s} ⟩}$ basis ( $s = \frac{1}{2}$ suppressed in the labelling as usual.) Then $E_{n l m_{l} m_{s}}^{(1)} = - (e B ℏ ∕ 2 m) (m_{l} + 2 m_{s})$ . If, for example, $l = 2$ there are 7 possible values of $m_{l} + 2 m_{s}$ between $- 3$ and $3$ , with $- 1, 0$ and $1$ being degenerate ( $5 \times 2 = 10$ states in all).

This is ﬁne if the magnetic ﬁeld is strong enough that we can ignore the ﬁne structure discussed in the last section. But typically it is not. For a weak ﬁeld the ﬁne structure eﬀects will be stronger, so we will consider them part of ${\hat{H}}^{(0)}$ for the Zeeman problem; our basis is then ${| n l j m_{j} ⟩}$ and states of the same $j$ but diﬀerent $l$ are degenerate. This degeneracy however is not a problem, because the operator $({\hat{L}}_{z} + 2 Ŝ_{z})$ does not connect states of diﬀerent $l$ . So we can use non-degenerate perturbation theory, with

E_{n l j m j}^{(1)} = \frac{e B}{2 m} ⟨ n l j m_{j} | {\hat{L}}_{z} + 2 {\hat{S}}_{z} | n l j m_{j} ⟩ .

If ${\hat{J}}_{z}$ is conserved but ${\hat{L}}_{z}$ and ${\hat{S}}_{z}$ are not, the expectation values of the latter two might be expected to be proportional to the ﬁrst, modiﬁed by the average degree of alignment: $⟨ {\hat{S}}_{z} ⟩ = ℏ m_{j} ⟨ \hat{S} \cdot \hat{J} ⟩ ∕ {⟨ \hat{J} ⟩}^{2}$ and similarly for $L_{z}$ . (This falls short of a proof but is in fact correct; see Mandl 7.5.3 for details.) Using $2 \hat{S} \cdot \hat{J} = {\hat{S}}^{2} + {\hat{J}}^{2} - {\hat{L}}^{2}$ and the equivalent with $\hat{S} \leftrightarrow \hat{L}$ gives

E_{n l j m j}^{(1)} = \frac{e B ℏ m_{j}}{2 m} (1 + \frac{j (j + 1) - l (l + 1) + s (s + 1)}{2 j (j + 1)}) = \frac{e B ℏ m_{j}}{2 m} g_{j l s} .

Of course for hydrogen $s (s + 1) = \frac{3}{4}$ , but the expression above, which deﬁnes the Landé $g$ factor, is actually more general and hence I’ve left it with an explicit $s$ . For hydrogen, $j = l \pm \frac{1}{2}$ and so $g = (1 \pm \frac{1}{2 l + 1})$ .

Thus states of a given $j$ (already no longer degenerate due to ﬁne-structure eﬀects) are further split into $(2 j + 1)$ equally-spaced levels. Since spectroscopy involves observing transitions between two states, both split but by diﬀerent amounts, the number of spectral lines can be quite large.

Gasiorowicz ch 12.3
Mandl ch 7.5
(Shankar ch 14.5)
Townsend ch 11.8

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