Nuclear Moments and Hyperfine Interactions

Abstract

Until now the nucleus has been often considered as a point charge with infinite mass, when compared to the electron mass.

Appendix 5.1 Fine and Hyperfine Structure in Hydrogen

Having introduced the various interaction terms (spin-orbit, relativistic corrections and hyperfine interaction) to be taken into account for one-electron states in atoms, it is instructive to reconsider the Hydrogen atom and to look at the detailed energy diagram (Fig. 5.6).

Fig. 5.6

Energy levels in Hydrogen including the effects contributing to its detailed structure. The scale is increased from left to right and some energy splittings are numerically reported to give an idea of the energy separations. The fine structure of the n = 2 level is detailed in Fig. 5.7

The solution of the non-relativistic Schrodinger equation (Sect. 1.4) provided the eigenvalues $E_{n,l} \,=\, - R_Hhc / n^2$. Then the spin-orbit Hamiltonian $\mathcal {H}_{so} \,=\, (e^2/2m^2c^2r^3)\mathbf {l} \cdot \mathbf {s}$ was introduced (Sect. 1.6). However we did not really discuss at that point the case of Hydrogen (where other relativistic effects are of comparable strength) dealing instead with heavier atoms (Sect. 2.2 and Chap. 3) where the most relevant contribution to the fine structure arises from $\mathcal {H}_{so}$. At Sect. 2.2 and Problem 1.38 it was pointed out that a more refined relativistic description would imply a shift of the s-states (where $l\,=\,0$ while at the same time a divergent behaviour for $r\rightarrow 0$ is related to the positional part of $\mathcal {H}_{so}$). Finally the hyperfine magnetic interaction was introduced (Sect. 5.2) where $\mathcal {H}_{hyp.}\,=\, a_{j}\mathbf {I} \cdot \mathbf {j}$, with $I \,=\,1/2$, $j \,=\, l \pm 1/2$ and $a_j$ given by Eqs. (5.12) and (5.13).

A simple relativistic correction which could remove the accidental degeneracy in l was already deduced in the old quantum theory. As a consequence of the relativistic mass $m\,=\, m(v)$, for elliptical orbits in the Bohr model, Sommerfeld derived for the energy levels

$$\begin{aligned} E_{n,k}\,=\,-\frac{R_Hhc}{n^2}\left[ 1+\frac{\alpha ^2}{n^2}\left( \frac{n}{k}-\frac{3}{4}\right) +...\right] \end{aligned}$$

where k is a second quantum number related to the quantization of the angular momentum $\int p_{\theta }d\theta \,=\, k h$ ($\theta$ polar angle) (see Problem 1.4), while $\alpha \,=\, (e^2/\hbar c ) \simeq 1 / 137$ is the fine-structure constant .

The Dirac electrodynamical theory , which includes spin-orbit interaction and classical relativistic effects (the relativistic kinetic energy being $c(p^2 + m^2c^2)^{1/2}- mc^2\simeq (p^2/2m)-(p^4/8m^3c^2)$ (see Problem 1.38), provided the fine-structure eigenvalues

$$\begin{aligned} E^{fs}_{n,j}\,=\,-\frac{R_Hhc\alpha ^2}{n^3}\left[ \frac{1}{j+1/2}-\frac{3}{4n}\right] \,=\, E_n^0\frac{\alpha ^2}{n^2}\left[ \frac{n}{j+1/2}-\frac{3}{4}\right] , \end{aligned}$$

with the relevant findings that the quantum number j and not l is involved and the shift for the s-states is explicit. Accordingly, the ground state of Hydrogen atom is shifted by $-1.8 \times 10^{-4}$ eV and the $n\,=\,2$ energy level is splitted in a doublet, the $p_{3/2}$ and $p_{1/2}$ states (this latter degenerate with $s_{1/2}$) being separated by an amount of 0.3652 cm$^{-1}$. The H$_{\alpha }$ line of the Balmer series (at 6562.8 Å) was then detected in the form of a doublet of two lines, since the Doppler broadening in optical spectroscopy prevented the observation of the detailed structure.

Giulotto and other spectroscopists, through painstaking measurements, noticed that the relativistic Dirac theory had to be modified and that a more refined description was required in order to account for the detailed structure of the H$_{\alpha }$ line. A few years later (1947) Lamb , by means of microwave spectroscopy (thus inducing magnetic dipole transitions between the levels) could directly observe the energy separation between terms at the same quantum number j. The energy difference between 2S and 2P states turned out 0.03528 cm$^{-1}$ and the line had a fine structure of five lines, some of them broadened. Later on, by Doppler-free spectroscopy using dye lasers (Hansch et al., see Problem 5.20 for an example) the seven components of the H$_{\alpha }$ line consistent with the Lamb theory could be inferred. It was also realized that this result had to be generalized and the states with the same n and j quantum numbers, but different l, have different energy.

The Lamb shift (reported in detail in Fig. 5.7 for the $n\,=\,2$ states) triggered the development of the quantum electrodynamical theory, which fully account for the fine structure of the levels on the basis of physical grounds that electrons are continuously emitting and adsorbing photons by transitions to virtual states. These states are poorly defined in energy due to their very short lifetimes. Qualitatively the Lamb shift can be considered the result of zero-point fluctuations of the set of harmonic oscillators describing the electromagnetic radiation field. These fluctuations induce analogous effects on the motion of the electron. Since the electric field in the atom is not uniform, the effective potential becomes different from the one probed by the electron in the average position.

The shift of the ground state due to the Lamb correction is about six times larger than the magnetic hyperfine splitting.

As regards the hyperfine splitting in the Hydrogen atom, at Sect. 5.2 it has been shown how the structure depicted in Fig. 5.6 is originated.

Fig. 5.7

Lamb shift for the $n\,=\,2$ levels in Hydrogen

Problems

Problem 5.8

The electric quadrupole moment of the deuteron is $Q\,=\, 2.8 \cdot 10^{-3}$ barn. By referring to an ellipsoid of uniform charge, evaluate the extent of departure of the nuclear charge distribution from the sphere. Assume for average nuclear radius $R_n\simeq 1.89 \cdot 10^{-13}$cm.

Solution: From

$$\begin{aligned} Q\,=\,\frac{1}{Ze}\rho \int _V(3z^2-r^2) d\tau , \end{aligned}$$

for the ellipsoid, defined by the equation $(x^2+y^2)/a^2+ (z^2/b^2)\,=\,1$, one obtains $Q\,=\,(2/5)(b^2-a^2)$. If the average nuclear radius is taken to be $R^3_n\,=\,a^2b$ (the volume of the ellipsoid is $\frac{4}{3}\pi a^2b)$, with $R_n+\delta R_n\,=\,b$, then, for $\delta R_n\ll R_n$

$$ a^2\,=\,\frac{R^3_n}{R_n+\delta R_n}\,=\,\frac{R^2_n}{1+\frac{\delta R_n}{R_n}}\approx R^2_n\left( 1-\frac{\delta R_n}{R_n}\right) $$

and

$$\begin{aligned} b^2-a^2&\approx R^2_n\left[ 1+2\left( \frac{\delta R_n}{R_n}\right) +\left( \frac{\delta R_n}{R_n}\right) ^2\right] -R^2_n\left( 1-\frac{\delta R_n}{R_n}\right) \\&= R^2_n\left[ 3\left( \frac{\delta R_n}{R_n}\right) +\left( \frac{\delta R_n}{R_n}\right) ^2\right] \approx 3R^2_n\left( \frac{\delta R_n}{R_n}\right) . \end{aligned}$$

Hence

$$ Q\,=\,\frac{6}{5}R^2_n\left( \frac{\delta R_n}{R_n}\right) $$

corresponding to $\left( \frac{\delta R_n}{R_n}\right) \approx 6.5 \cdot 10^{-2}.$

Problem 5.9

The $D_2$ line of the Na doublet (see Fig. 5.3) displays an hyperfine structure in form of triplet, with separation between pairs of adjacent lines in the ratio not far from 1.5. Justify this experimental finding from the hyperfine structure of the energy levels and the selection rules (see Problem 5.20 for some detail on the experimental method).

Solution: From the splittings in Fig. 5.3 and the selection rule $\varDelta F \,=\, 0, \pm 1$ one can deduce that the hyperfine spectrum consists of three lines $\nu _{\{3,2,1\}\leftrightarrow 2}$ corresponding to the transitions $^2P_{\frac{3}{2}}(F\,=\,3,2,1)\leftrightarrow {} ^2S_{\frac{1}{2}}(F\,=\,2)$ and of three lines $\nu _{\{2,1,0\}\leftrightarrow 1}$ corresponding to the transitions $^2 P_{\frac{3}{2}}(F\,=\,2,1,0)\leftrightarrow {} ^2S_{\frac{1}{2}} (F\,=\,1)$. From the interval rule

$$\begin{aligned} \frac{\nu _{3,2}-\nu _{2,2}}{\nu _{2,2}-\nu _{1,2}}\,=\,\frac{3}{2} \,\,\,\, \mathrm {and} \,\,\,\, \frac{\nu _{2,1}-\nu _{1,1}}{\nu _{1,1}-\nu _{0,1}}\,=\,2. \end{aligned}$$

The lines in Fig. 5.3 correspond to the transitions $^2P_{\frac{3}{2}}(F\,=\,3,2,1)\leftrightarrow {} ^2S_{\frac{1}{2}}(F\,=\,2)$ (see Problem 5.20).

Problem 5.10

Plot the magnetic hyperfine levels for an atom in the electronic state $^2S_{1/2}$ and nuclear spin $I\,=\,1$. Then derive the corrections due to a magnetic field, in the weak and strong field regimes (with respect to the hyperfine energy). Classify the states in the two cases and draw a qualitative correlation between them.

Solution: In the weak-field regime the effective magnetic moment is along $\mathbf {F}$. By neglecting the contribution from the nuclear magnetic moment one writes

and the hyperfine correction is

$$\begin{aligned} \varDelta E\,=\,g_{F} \, \mu _B \, H_0 \, m_F. \end{aligned}$$

$g_F$ is calculated by projecting along $\mathbf {F}$: , with $cos \widehat{FJ}\,=\, [J(J+1)+F(F+1)-I(I+1)]/2\sqrt{J(J+1)}\sqrt{F(F+1)}$.

Fig. 5.8

Hyperfine structure of the $S_{1/2}$ state with $I\,=\,1$: a in zero field; b in weak field, Zeeman regime; c in strong field, Paschen-Back regime

Thus

$$ g_F\,=\,g_J\frac{F(F+1)+J(J+1)-I(I+1)}{2F(F+1)}. $$

A relatively small field breaks up the $\mathbf {I}\cdot \mathbf {J}$ coupling and the hyperfine Zeeman effect is replaced by the hyperfine Paschen-Back effect. The oscillating components in the x and y directions average to zero and the final result is that the nuclear angular momentum vector $\mathbf {I}$ is oriented along $\mathbf {H}_{0}$. The quantum number F is no longer defined while the quantum numbers $m_I$ and $m_J$ describe $\mathbf {I}$ and $\mathbf {J}$. The splitting involves three terms, one being $g_J \, \mu _B \, H_0 \,m_J$, already considered in the Zeeman effect (Sect. 4.3.2), the other is $a m_I m_J$ and the third one, $-\mu _N g_I m_I H_0$, is negligible. See Fig. 5.8 for a pictorial view of the angular momenta and of the correspondent magnetic moments.

Problem 5.11

  In the Na atom the hyperfine interaction for the P state is much smaller than the one in the S ground state. In poor resolution the hyperfine structure is observed in the form of a doublet, with relative intensities 5 and 3. From this observation derive the nuclear spin (see also Problem 5.20).

Solution: From

The intensity being $\propto (2F+1)$ and the ratio $(I+1)/I\,=\, 5/3$, then $I\,=\, 3/2$.

Problem 5.12

For a solid ideally formed by a mole of non-interacting deuterons in an electric field gradient, derive the contributions to the entropy and to the specific heat, in the high temperature limit (see Problem 5.6) .

Solution: The quadrupolar interaction $e^2qQ[3M^2-I(I+1)]/4$ yields two energy levels, one doubly degenerate ($M\,=\,0,\pm 1$).

By indicating with $\epsilon$ the separation between the levels, the partition function is

$$ Z(\beta )\,=\,\left( 1+2e^{-\beta \epsilon }\right) ^{N_A}, \mathrm {\,\,with\,\,} \beta \,=\,1/k_BT . $$

From the free energy

$$ F(T)\,=\,-k_BT \ln Z\,=\, -RT\ln \left( 1+2e^{-\epsilon /k_BT}\right) , $$

the entropy turns out

$$ S\,=\,-\frac{\partial F}{\partial T}\,=\,R\ln \left( 1+2e^{-\epsilon /k_BT}\right) +\frac{2N_A\epsilon }{T}\frac{e^{-\epsilon /k_BT}}{1+2e^{-\epsilon /k_BT}}. $$

The internal energy is

$$ U\,=\,2N_A\epsilon \frac{1}{e^{\epsilon /k_BT}+2}. $$

In the high temperature limit

$$ U\simeq \frac{2}{3}N_A\epsilon \left( 1-\frac{\epsilon }{3k_BT}\right) , $$

so that

$$ C\,=\,\frac{\partial U}{\partial T}\,=\,\frac{2}{9}R\left( \frac{\epsilon }{k_BT}\right) ^2\propto T^{-2}, $$

namely the high-temperature tail of a Schottky anomaly (a “bump” in the specific heat versus temperature), typical of two-levels systems.

Problem 5.13

Consider the Hydrogen atom, in the ground state, in a magnetic field H${}_0$ and write the Hamiltonian including the hyperfine interaction. First derive the eigenvalues and the spin eigenvectors in the limit H${}_0 \rightarrow$ 0 and estimate the frequencies of the transitions induced by an oscillating magnetic field (perpendicular to the quantization axis).

Then derive the correction to the eigenvalues due to a weak magnetic field.

Finally consider the opposite limit of strong magnetic field. Draw the energy levels with the appropriate quantum numbers, again indicating the possibility of inducing magnetic dipole transitions between the hyperfine levels (this is essentially the EPR experiment, see for details Chap. 6) and from the resulting lines show how the hyperfine constant can be extracted.

Figure out a schematic correlation diagram connecting the eigenvalues for variable external field.

Solution: From the Hamiltonian

$$ \mathcal {H}_s\,=\,2\mu _B\mathbf {S}\cdot \mathbf {H}_0-\gamma \hbar \mathbf {I}\cdot \mathbf {H}_0+a \mathbf {I}\cdot \mathbf {S} $$

($\gamma$ nuclear gyromagnetic ratio, a hyperfine interaction constant, with $a\,=\,hc/\lambda$ and $\lambda \,=\,21$ cm).

For $\mathrm {H_0} \rightarrow 0$ the eigenstates are classified by S, I, F and $M_F$ and for $I \,=\, S \,=\,1/2$ two magnetic hyperfine states, $F \,=\, 0$ and $F \,=\, 1$, occur. Then $E_{\frac{1}{2}, \frac{1}{2}, 1} \,=\, {a}/{4}$ and $E_{\frac{1}{2}, \frac{1}{2}, 0} \,=\, -{3}/{4} a$.

The spin eigenvectors are the same of any two spins system, i.e.

$$ \left. \begin{array}{c} \vert \alpha _e\alpha _p>\\ \vert \beta _e\ \beta _p>\\ \frac{1}{\sqrt{2}}\vert \alpha _e\beta _p+\alpha _p\beta _e>\end{array}\right\} \text{ defining } \text{ the } \text{ triplet } T_{1,1}\ T_{1,0}\ T_{1,-1} $$

and

$$ \frac{1}{\sqrt{2}}\vert \alpha _e\beta _p-\alpha _p\beta _e> \text{ defining } \text{ the } \text{ singlet } S_{0,0}. $$

The oscillating magnetic field acts as a perturbation involving the operator $\mu _x\,=\,2\mu _BS_x-\gamma \hbar I_x$ (for details see Sect. 6.2). The matrix elements for the triplet and singlet states turn out

$$\begin{aligned}&<1,1\vert \mu _x\vert 1,0>\,=\,\frac{1}{2}\frac{1}{\sqrt{2}}(g\mu _B-\gamma \hbar )\\&<1,1\vert \mu _x\vert 0,0>\,=\,\frac{1}{2}\frac{1}{\sqrt{2}}(-g\mu _B-\gamma \hbar )\\&<1,0\vert \mu _x\vert 1,-1>\,=\,\frac{1}{2}\frac{1}{\sqrt{2}}(g\mu _B-\gamma \hbar )\\&<0,0\vert \mu _x\vert 1,-1>\,=\,\frac{1}{2}\frac{1}{\sqrt{2}}(g\mu _B+\gamma \hbar )\\&<1,0\vert \mu _x\vert 0,0>\,=\, <1,1\vert \mu _x\vert 1, -1>\,=\,0. \end{aligned}$$

Therefore the allowed transitions are $S\rightarrow T_1$ and $S\rightarrow T_{-1}$ corresponding to the transition frequency $\nu \,=\,\frac{a}{h}\,=\,1420~\text{ MHz }$ (and formally $T_{-1}\rightarrow T_{0},\, \, T_0\rightarrow T_{+1}$ at $\nu \,=\,0$).

For weak field H${}_0$, neglecting the interaction with the proton magnetic moment and considering that the perturbation acts on the basis where $F^2$, $F_z$, $I^2$ and $S^2$ are diagonal, the matrix for $\mathcal {H}\,=\, a\mathbf {I} \cdot \mathbf {S} + 2\mu _B \mathbf {S} \cdot \mathbf {H}_0$ is

$$\begin{aligned} \left( \begin{array}{cccc} {a}/{4}+ \mu _B\mathrm {H_0} &{}0 &{}0 &{}0\\ 0 &{}{a}/{4} &{}0 &{}-\mu _B\mathrm {H_0}\\ 0 &{}0 &{}{a}/{4}- \mu _B\mathrm {H_0} &{}0\\ 0 &{}-\mu _B\mathrm {H_0}&{}0 &{} -{3a}/{4} \end{array}\right) \\ \end{aligned}$$

From the secular equation the eigenvalues are found by solving

$$ \frac{a}{4}+ \mu _B\mathrm {H_0}- E\,=\,0 \,\,\,\,\,\,\,\,\,\,\, \frac{a}{4}- \mu _B\mathrm {H_0}- E\,=\,0 $$

$$ \left( \frac{a}{4} - E\right) \left( -\frac{3a}{4} -E\right) - \mu _B^2 \mathrm {H_0}^2\,=\,0 $$

yielding $E_{1,2}\,=\,a/4 \pm \mu _B\mathrm {H_0}$ and $E_{3,4}\,=\,-a/4 \pm (a/2)[1 + 4 \mu _B^2\mathrm {H_0}^2/a^2]^{1/2}$ (the states $F\,=\,1, M_F\,=\,0$ and $F\,=\,0, M_F\,=\,0$ being little affected by a weak magnetic field).

The Breit-Rabi diagram , as reported below, holds

In the strong field regime the eigenvalues are the ones for $S_z$, $I_z \, S_z$ and $I_z$:

$$ E\,=\,2\mu _B\mathrm {H_0} m_S+a m_S M_I-\gamma \hbar M_I \mathrm {H_0} $$

The first term is dominant and the diagram is

with the electronic transitions $\varDelta m_S\,=\,\pm 1$ (and $\varDelta M_I \,=\, 0)$ at the frequencies

$$ \nu _{1,2}\,=\,\frac{2\mu _B\mathrm {H_0}\pm a/2}{h}. $$

The nuclear transitions $\varDelta M_I\,=\,\pm 1$ (and $\varDelta m_S \,=\, 0)$ occur at a / 2h.

Since the internal field due to the electron is usually much larger than $\mathrm {H_0}$ the third term can be neglected (see the figure below).

Problem 5.14

The $^{209}Bi$ atom has an excited $^2 D_{5/2}$ state, with 6 sublevels due to hyperfine interaction. The separations between the hyperfine levels are $0.23,\, 0.31, \, 0.39, \, 0.47$ and 0.55 cm$^{-1}$. Evaluate the nuclear spin and the hyperfine constant.

Solution: From $E(F,I,J)\,=\,(a/2)\left[ F(F+1)-I(I+1)-J(J+1)\right]$ and

$E_{F+1}-E_F\,=\,a(F+1)$, one finds $a \,=\, 0.08~\text{ cm }^{-1}\,\, \text {and} \,\, F_{\max } \,=\, 7$.

Therefore $F \,=\, 2,3,4,5,6,7$ and since $J \,=\, 5/2$ the nuclear spin must be $I \,=\, \frac{9}{2}.$

Problem 5.15

A proton and an anti-proton, at a given distance d, interact through the magnetic dipole-dipole interaction. Derive the total spin eigenstates and eigenvalues in term of the proton magnetic moment(it is reminded that the magnetic moment of the antiproton is the same of the proton, with negative gyromagnetic ratio) .

Solution: From

with and , by choosing the z axis along $\mathbf {r}$

$$ \mathcal {H}\,=\,-4 \, \frac{\mu _p^2}{d^3}\, \mathbf {s}_1\cdot \mathbf {s}_2+12\, \frac{\mu _p^2}{d^3}\, s^z_1s^z_2. $$

Since $\mathbf {s}_1\cdot \mathbf {s}_2\,=\, {S(S+1)}/{2}- {3}/{4}$ and $s^z_1s^z_2 \,=\,(1/2)M^2_S - ({1}/{2}) \cdot ({1}/{2})$, one finds

$$ \begin{array}{cccc} \text{ Eigenstates } &{}S &{}M_S &{}\text{ Energies }\\ \text {singlet} &{}0 &{}0 &{}0\\ &{} &{}1 &{}\,\,\,\,\, 2\mu _p^2/d^3\\ \text {triplet} &{}1 &{}0 &{}\,\,\,\,\, -4\mu _p^2/d^3\\ &{} &{}-1 &{}\,\,\,\,\, 2\mu _p^2/d^3 \end{array} $$

i.e.

Problem 5.16

Two electrons interact through the dipolar Hamiltonian. A strong magnetic field is applied along the z-direction, at an angle $\theta$ with the line connecting the two electrons. Find the eigenvalues and the corresponding eigenfunctions for the two spins system, in terms of the basis functions $\alpha _{1,2}$ and $\beta _{1,2}$.

Solution: In the light of Eq. (5.8) for the dipolar field (see also Problem 5.2) the total Hamiltonian is

$$\begin{aligned} \mathcal {H}&= \mathcal {H}_0+ \mathcal {H}_d\\&=2\mu _B H_0(s_z^{(1)}+s_z^{(2)})+\frac{4\mu _B^2}{r^3}\left\{ \mathbf {s}_1\cdot \mathbf {s}_2-\frac{3}{r^2}[(\mathbf {s}_1\cdot \mathbf {r})(\mathbf {s}_2\cdot \mathbf {r})]\right\} . \end{aligned}$$

In order to evaluate the matrix elements it is convenient to write the perturbation Hamiltonian in the form (called dipolar alphabet )

$$ \mathcal {H}_d\,=\,\frac{4 \mu _B^2}{r^3}[A+B+C+D+E+F] $$

where

$$\begin{aligned} A\,=\, s_z^{(1)}s_z^{(2)}\left[ 1-3\cos ^2\theta \right] ,\,\,\, B\,=\, -\frac{1}{4}\left[ s^{(1)}_+s^{(2)}_-+s^{(1)}_-s^{(2)}_+\right] \left( 1-3\cos ^2\theta \right) , \end{aligned}$$

$\theta$ angle between $\mathbf {H}_0$ and r. The terms C, D, E and F involve operators of the form $s^{(1)}_+s^{(2)}_z$, $s^{(1)}_-s^{(2)}_z$, $s^{(1)}_+ s^{(2)}_+$, $s_-^{(1)}s_-^{(2)}$ and can be neglected. In fact these terms are off-diagonal and produce admixtures of the zero-order states to an amount of the order of $\left( {\mu _B}/{r^3}\right) /H_0$ (i.e. ${\sim }10^{-4}$ for $H_0\,=\, 1$ T).

Thus the dipolar Hamiltonian is written in the form

$$ \mathcal {H}_d\,=\,\underbrace{\frac{4\mu _B^2}{r^3}\left( 1-3\cos ^2\theta \right) }_{\mathcal A}\left[ s_z^{(1)}s_{z}^{(2)}-\frac{1}{4}\left( s_+^{(1)}s^{(2)}_- + s^{(1)}_-s^{(2)}_+\right) \right] , $$

most commonly used.

The complete set of the basis functions is $\alpha _1 \alpha _2 ,\, \, \alpha _1 \beta _2 , \, \, \alpha _2 \beta _1 \, \, \text{ and } \, \, \beta _1 \beta _2$ and the matrix elements are

$$\begin{aligned}<\alpha \alpha {\vert } {\mathcal {H}_T}\vert \alpha \alpha>&= 2~\mu _B\,H_0 + \frac{1}{4}{\mathcal A}\\<\alpha \beta \vert \mathcal {H} _T\vert \alpha \beta>&=\,<\beta \alpha \vert \mathcal { H}_T\vert \beta \alpha>\,=\,-\frac{{\mathcal A}}{4}\\<\alpha \beta \vert \mathcal {H}_T\vert \beta \alpha>&=\,<\beta \alpha \vert \mathcal { H}_T\vert \alpha \beta>\,=\,-\frac{\mathcal A}{4}\\ <\beta \beta \vert \mathcal {H} _T\vert \beta \beta >&=-2 \mu _BH_0+{\mathcal A}/4 \end{aligned}$$

It is noted that while the term A is completely diagonal, the term B only connects $\vert m^{(1)}_s m^{(2)}_s>$ to states $<m^{(1)}_s+1, m^{(2)}_s-1\vert$ or $<m^{(1)}_s-1, m^{(2)}_s+1\vert$. B simultaneously flips one spin up and the other down.

The secular equation is

$$ \left| \begin{array}{cccc} (+2\mu _BH_0+\frac{\mathcal A}{4})-E &{}0 &{}0 &{}0\\ 0 &{}-\frac{\mathcal A}{4}-E &{}-\frac{\mathcal A}{4} &{}0\\ 0 &{}-\frac{\mathcal A}{4} &{}-\frac{\mathcal A}{4}-E &{}0\\ 0 &{}0 &{}0 &{}\left( -2\mu _BH_0+\frac{\mathcal A}{4}\right) -E\end{array}\right| \,=\,0. $$

and the eigenvalues turn out

$$\begin{aligned} E_1&=-2\mu _B\left( H_0-\frac{\mu _B}{2 r^3}(1-3\cos ^2\theta )\right) \\ E_2&= 0\\ E_3&= -\frac{2\mu _B^2}{r^3}(1-3\cos ^2\theta )\\ E_4&= +2\mu _B\left( H_0-\frac{\mu _B}{2r^3}(1-3\cos ^2\theta )\right) . \end{aligned}$$

The correspondent eigenfunctions being $\alpha _1 \alpha _2 , \beta _1 \beta _2 \,\,\, \text {and} \,\,\, \frac{1}{\sqrt{2}}\left[ \alpha _1\beta _2 \pm \alpha _2\beta _1\right]$, as expected.

Problem 5.17

In the Ba atom the line due to the transition from the $6s\,6p$ $J\,=\,1$ to the $(6s)^2$ ground state in high resolution is evidenced as a triplet, with line intensities in the ratio 1, 2 and 3. Evaluate the nuclear spin.

Solution: Since $\mathbf {F}\,=\,\mathbf {I}+\mathbf {J}$

$$ \text {for} \,\,\, J\,=\,0 \,\,\, \text {one has} \,\,\, I \,=\, F \,\,\, \Longrightarrow \,\,\, \text {no splitting} $$

$$ \text {for} \,\,\, J \,=\,1 \,\,\, \Longrightarrow \,\,\, \text {splitting in} \,\,\, (2I+1) \,\,\, \text {or in} \,\,\, (2J+1) \mathrm {\,\,terms.} $$

$$ I\,=\,0 \,\,\, \Longrightarrow \,\,\, \text {no splitting,} $$

$$ \text {for} \,\,\, I \,=\, \frac{1}{2} \,\,\, \mathrm {and} \,\,\, J\,=\,1 \,\,\, \varDelta F\,=\,0,\pm 1 \,\,\, \Longrightarrow \,\,\, \text {two lines} $$

$$ I\,=\,1 \,\,\, \mathrm {or} \,\,\, I>1 \,\,\, \Longrightarrow \,\,\, \text {three lines}. $$

Looking at the intensities, proportional to $e^{-E/k_B T} (2F+1)$, where the energy E is about the same

$$ \text {for} \,\,\,\, I\,=\,1 \,\,\, F\,=\, 0, 1, 2 \,\,\, \Longrightarrow \,\,\, \text {intensities:}\,\, 1, 3, 5 $$

$$ \text {for} \,\,\, I\,=\,\frac{3}{2} \, \,\,\, F\,=\, \frac{1}{2}, \frac{3}{2}, \frac{5}{2} \,\,\, \Longrightarrow \,\,\, \text {intensities:}\,\, 2, 4, 6 . $$

Therefore $I\,=\,\frac{3}{2}.$

Problem 5.18

In the assumption that in a metal the magnetic field on the electron due to the hyperfine interaction with $I\,=\,1/2$ nuclei is $H_z\,=\,(a/N)\varSigma _n I_n ^z$ (a constant and same population on the two states) prove that the odd moments of the distribution are zero and evaluate $<H^2_z>$. Then evaluate $<H^4_z>$ and show that for large N the distribution tends to be Gaussian, the width going to zero for $N\rightarrow \infty$.

Solution:

$\langle \hat{H}_z ^{2n+1}\rangle \,=\,0$ for symmetry. Since $(I^z_n)^2\,=\,\frac{1}{4}$

$$ \left\langle \left( \sum _n I^z_n\right) ^2\right\rangle \,=\,\sum _n\left\langle (I^z_n)^2\right\rangle +\sum _{n\ne m}\left\langle I^z_n I^z_m\right\rangle \,=\,\frac{1}{4}N, $$

and then $\langle H_z ^2\rangle \,=\,(\frac{a}{2N})^2 N$.

$$\begin{aligned} \langle H_z ^4\rangle&=\left( \frac{a}{N}\right) ^4\sum _{i,j,k,l}\langle I^z_iI^z_jI^z_kI^z_l\rangle \,=\, \left( \frac{a}{N}\right) ^43\sum _{i,j}\left\langle (I^z_i)^2(I^z_j)^2\right\rangle -\left( \frac{a}{N}\right) ^4\sum _i\left\langle (I^z_i)^4\right\rangle \\&= \left( \frac{a}{2N}\right) ^4(3N^2-N). \end{aligned}$$

In the thermodynamical limit one has $\langle H_z ^4\rangle \simeq 3\left( \frac{a}{2N}\right) ^4 N^2$: the first two even moments correspond to the Gaussian moments.

Problem 5.19

Evaluate the transition probability from the state $M\,=\, -1/2$ to $M \,=\, +1/2$ by spontaneous emission, for a proton in a magnetic field of 7500 Oe.

Solution: From the expression for $A_{21}$ derived in Appendix 1.3 and extending it to magnetic dipole transitions, one can write

with . From $I_{\pm }\,=\, I_x \pm iI_y$ one derives $A_{21}\,=\, (2/3)({\gamma ^2\hbar }/{c^3})\omega ^3 _L$ and for $\gamma \,=\, 42.576 \cdot 2\pi \cdot 10^2$ Hz/G, $\omega _L \,=\,\gamma H_0 \,=\, 2\pi \cdot 31.9 \, \text {MHz}$, yielding

$$A_{21}\simeq 1.5\times 10^{-25}\,\mathrm{s}^{-1}.$$

Problem 5.20

High-resolution laser spectroscopy allows one to evidence the hyperfine structure in the optical lines with almost total elimination of the Doppler broadening.

The figure below

shows the hyperfine structure of the $^2S_{1/2}- ^2P_{3/2}$ $D_2$ line of Na at 5890 Å (transitions $\varDelta F\,=\, 0, \pm 1$ from the $F\,=\,2$ level of the electronic ground state). (This spectrum is obtained by irradiating a collimated beam of sodium atoms at right angles by means of a narrow-band single-mode laser and detecting the fluorescent light after the excitation. This and other high resolution spectroscopic techniques are described in the book by Svanberg) .

From the figure, discuss how the magnetic and electric hyperfine constants could be derived and estimate the life-time of the $^2P_{3/2}$ state (in the assumption that is the only source of broadening).

Then compare the estimated value of the life time with the one known (from other experiments), $\tau \,=\,1.6$ ns. In the assumption that the extra-broadening is due to Doppler second-order relativistic shift , quadratic in (v / c), estimate the temperature of the oven from which the thermal atomic beam is emerging, discussing the expected order of magnitude of the broadening (see Problem 1.30).

Solution: For the ground-state $^2S_{1/2}$, $I\,=\,3/2$ and $J\,=\, S\,=\, 1/2$, the quadrupole interaction being zero, from Eq. (5.15) the separation between the $F\,=\,2$ and $F\,=\,1$ states yields the magnetic hyperfine constant $a\,=\, \varDelta _{1,2}/(F+1)\,=\, 886$ MHz, corresponding to an effective magnetic field of about 45 T.

The sequence of the hyperfine levels for the $^2P_{3/2}$ state does not follow exactly the interval rule. In the light of Eq. (5.20) an estimate of the quadrupole coupling constant b can be derived (approximate, the correction being of the order of the intrinsic line-widths).

In the assumption that the broadening (12 MHz) is due only to the life-time one would have $\tau \,=\,1/2\pi \varDelta \nu \simeq 13.3 \times 10^{-9}$ s, a value close to the one ($\tau \simeq 16\times 10^{-9}$s) pertaining to the $3^2P_{3/2}$ state ($\varDelta \nu \simeq 10$ MHz).

The most probable velocity of the beam emerging from the oven is $v\,=\, \sqrt{ 3k_BT/M_{Na}}$, that for $T\simeq 500$ K corresponds to about $7\times 10^4$ cm/s.

While the first-order Doppler broadening is in the range of a few GHz, scaling by a term of the order of v / c leads to an estimate of the second-order Doppler broadening in the kHz range. Thus the extra-broadening of a few MHz is likely to be due to the residual first-order broadening (for a collimator ratio of the beam around 100 being typically around some MHz) .

Problem 5.21

From the perturbation generated by nuclear magnetic moment on the electron, derive the effective magnetic field in the hyperfine Hamiltonian (Eq. (5.6)).

Solution: From the vector potential (see Fig. 5.1 and Eq. (5.4)) the magnetic Hamiltonian for the electron is

Since

while $div(\mathbf {r}/ r^3)\,=\, 4\pi \delta (\mathbf {r})$, one writes

$$ \mathcal {H}_{hyp}\,=\, 2\mu _B g_nM_n \frac{\mathbf {I} \cdot \mathbf {l}}{r^3} - 2\mu _B g_nM_n \left\{ \frac{\mathbf {s} \cdot \mathbf {I}}{r^3} - \frac{3(\mathbf {I} \cdot \mathbf {r})(\mathbf {s} \cdot \mathbf {r})}{r^5}\right\} + 2\mu _B g_nM_n(\mathbf {s} \cdot \mathbf {I}) 4\pi \delta (\mathbf {r})\equiv $$

$$ \equiv A+ B+ C , $$

To deal with the singularities at the origin involved in B and C, let us define with $V_{\varepsilon }$ a little sphere of radius $\varepsilon$ centered at $r\,=\,0$. Then in the integral for the expectation values

$$ I\,=\, \int _{V_{\varepsilon }} B\phi ^*(\mathbf {r})\phi (\mathbf {r}) d\tau \equiv \int _{V_{\varepsilon }} B f(\mathbf {r}) d\tau $$

one can expand $f(\mathbf {r})$ in Taylor series, within the volume $V_{\varepsilon }$

In I there are two types of terms, one of the form

$$ s_xI_x\frac{\partial ^2}{\partial x^2}\left( \frac{1}{r}\right) \,\,\,\,\,(a) $$

the other of the form

$$ \left( s_xI_y+s_yI_x\right) \frac{\partial ^2}{\partial x\partial y}\left( \frac{1}{r}\right) \,\,\,\,\,(b) $$

In the expansion is odd while (a) terms are even, thus yielding zero in I. The product of (a) terms with the third term in $f(\mathbf {r})$ when even, contributes with a term quadratic in $\varepsilon$.

The terms of type (b) are odd in the two variables, while includes odd terms in a single variable. In the same way are odd (and do not give contribution) the terms (b)f(0). Finally the terms (b) times the third term in the expression again contribute to I only to the second order in $\varepsilon$. Therefore, one can limit I to

$$ I\,=\, 2g_nM_n\mu _B f(0) \frac{1}{3} \int _{V_{\varepsilon }} (\mathbf {s} \cdot \mathbf {I})\nabla ^2\left( \frac{1}{r}\right) d\tau . $$

Since $\nabla ^2(1/r)\,=\, -4\pi \delta (\mathbf {r})$ the magnetic hyperfine hamiltonian can be rewritten²

$$ \mathcal {H}_{hyp}\,=\, 2\mu _Bg_nM_n \frac{\mathbf {I} \cdot \mathbf {l}}{r^3} - 2\mu _Bg_nM_n \biggl [ \frac{\mathbf {s} \cdot \mathbf {I}}{r^3} - \frac{3(\mathbf {s} \cdot \mathbf {r})(\mathbf {I} \cdot \mathbf {r}) }{r^5}\biggr ]^* + \frac{16\pi }{3} \mu _B g_nM_n \mathbf {s}\,\cdot \,\mathbf {I} \delta (\mathbf {r}) $$

Thus the effective field $\mathbf {h}_{eff}$ in the form given in Eq. (5.6) is justified.

A model which allows one to derive similar results for the dipolar and the contact terms is to consider the nucleus as a small sphere with a uniform magnetization $\mathbf {M}$, namely a magnetic moment . For $r> R$ the magnetic field is the one of a point magnetic dipole. Inside the sphere $\mathbf {H}_{int}\,=\, (8\pi /3)\mathbf {M}$. By taking the limit $R\rightarrow 0$, keeping $\mu _n$ constant and then assuming that $\mathrm {M}\rightarrow \infty$, so that $\int _{r<R} H_{int} d\mathbf {r}_n \,=\, {8\pi \mu _n}/{3}$, the complete expression of the field turns out

Problem 5.22

From the energy of the nuclear charge distribution in the electric potential due to the electron (Eq. (5.16)) derive the hyperfine quadrupole Hamiltonian (Eq. (5.20)).

Solution: By starting from Eq. (5.18) a new tensor $Q_{ij}$ so that $\sum _l Q_{ll}\,=\,0$ is defined

$$ Q_{ij}\,=\, 3 Q'_{ij} -\delta _{ij}\sum _l Q'_{ll} $$

and in terms of $Q'_{ij}$ one has

$$ E_Q\,=\, \frac{1}{6}\sum _{ij} Q_{ij}V_{ij} + \frac{1}{6}\sum _{l} Q'_{ll}\sum _jV_{jj} $$

The second term can be neglected since $\sum _jV_{jj}\simeq 0$. Thus

$$ \mathcal {H}_{hyp}^Q\,=\, \sum _{ij} \hat{Q_{ij}} \frac{{\hat{V}}_{ij}}{6}\,\,\ $$

where the operators are

$$ \hat{Q_{ij}}\,=\, e \sum _n \left( 3x_i^nx_j^n - \delta _{ij} r_n^2\right) \,\,\ $$

$$ {\hat{V}}_{ij}\,=\, -e \sum _e \frac{\left( 3x_ix_j - \delta _{ij} r_e^2\right) }{r_e^5}\,\,\ $$

This Hamiltonian can be simplified by expressing the five independent components of $Q_{ij}$ in terms of one. Semiclassically this simplification originates from the precession of the nuclear charges around $\mathbf {I}$, yielding a charge distribution with cylindrical symmetry around the z direction of the nuclear spin.

Then $Q_{ij}\,=\,0$ for $i\ne j$ and being $\sum _lQ_{ll}\,=\,0$, one has $Q_{11}\,=\, Q_{22}\,=\, - Q_{33}/2$ with $Q_{33}\,=\,\int \rho _n(\mathbf {r})(3z^2 -r^2) d\tau _n$.

In the quantum description the reduction of $\mathcal {H}_{hyp}^Q$ is obtained by considering that only the dependence from the orientation is relevant. Thus, for the matrix elements $<I, M'_I|\hat{Q_{ij}} | I,M_I>$ (other quantum numbers for the nuclear state being irrelevant), by using Wigner-Eckart theorem one writes

$$<I, M'_I|{\hat{Q}}_{ij} | I,M_I>\,=\, C <I, M'_I\left| \frac{3}{2}(I_iI_j + I_jI_i)- \delta _{ij}I^2 \right| I,M_I>. $$

By defining, in analogy to the classical description, the quadrupole moment Q in proton charge units as

$$ Q\,=\,<II\left| \frac{{\hat{Q}}_{zz}}{e}\right| II>\equiv <II\left| \sum _n \left( 3z_n^2 -r_n^2\right) \right| II> $$

the constant C is obtained:

$$ C<II\left| 3I_z^2- I^2\right| II>\,=\, C\left[ 3I^2 -I(I+1)\right] \,=\, eQ $$

Therefore all the components $Q_{ij}$ are expressed in terms of Q, which has the classical physical meaning (see Eqs. (5.2) and (5.22)). Then the quadrupole operator is

$$ {\hat{Q}}_{ij}\,=\, \frac{eQ}{I(2I-1)}\left\{ \frac{3}{2}(I_iI_j + I_jI_i) -\delta _{ij}I^2\right\} $$

Analogous procedure can be carried out for the electric field gradient operator:

$$ {\hat{V}}_{ij}\,=\, \frac{eq_J}{J(2J-1)}\left\{ \frac{3}{2}(J_iJ_j + J_jJ_i) -\delta _{ij}J^2\right\} $$

where

$$ q_J\,=\,<JJ\left| \frac{{\hat{V}}_{zz}}{e}\right| JJ>\,=\, <JJ\left| -\frac{\sum _e(3z_e^2 -r_e^2}{r_e^5}\right| JJ> $$

Finally, since

$$ \sum _{ij}I_iI_jJ_iJ_j\,=\, \left( \sum _i I_iJ_i \right) ^2\,=\, (\mathbf {I} \cdot \mathbf {J})^2 $$

$$ \sum _{ij}I_iI_j\delta _{ij}J^2\,=\, \left( \sum _i I_i\right) ^2J^2\,=\, I^2J^2 $$

$$ \sum _{ij}I_iI_jJ_jJ_i\,=\, \left( \mathbf {I} \cdot \mathbf {J}\right) ^2 + (\mathbf {I} \cdot \mathbf {J}) $$

the quadrupole hyperfine Hamiltonian is written

$$ \mathcal {H}_{hyp}^Q\,=\, \frac{eq_JQ}{2I(2I-1)J(2J-1)}\left\{ 3(\mathbf {I} \cdot \mathbf {J})^2 + \frac{3}{2}(\mathbf {I} \cdot \mathbf {J}) - I^2J^2\right\} , $$

as in Eq. (5.20) (see also Eq. (5.24)).

Problem 5.23

At Sect. 1.5 the isotope effect due to the reduced mass correction has been mentioned. Since two isotopes may differ in the nuclear radius R by an amount $\delta R$, once that a finite nuclear volume is taken into account a further shift of the atomic energy levels has to be expected. In the assumption of nuclear charge Ze uniformly distributed in a sphere of radius $R\,=\, r_FA^{1/3}$ (with Fermi radius $r_F\,=\,1.2\times 10^{-13}$ cm) estimate the volume shift in an hydrogenic atom and in a muonic atom. Finally discuss the effect that can be expected in muonic atoms with respect to ordinary atoms in regards of the hyperfine terms.

Solution: The potential energy of the electron is $V(r)\,=\,-Ze^2/r$ for $r\ge R$, while (see Problem 1.6)

$$ V(r)\,=\, -3 \frac{Ze^2}{2R}\left( 1- \frac{r^2}{3R^2}\right) \,\,\,\mathrm {for\,\,} r\le R $$

The first-order correction, with respect to the nuclear point charge hydrogenic Hamiltonian, turns out

$$ \varDelta E\,=\, \frac{Ze^2}{2R}\int _0^R |{R}_{nl}(r)|^2\left( -3 + \frac{r^2}{R^2} + \frac{2R}{r}\right) r^2 dr \simeq \frac{Ze^2}{10}R^2|{R}_{nl}(0)|^2 $$

The correction is negligible for non-s states, where ${R}_{nl}(0)\simeq 0$, while for s states one has

$$ \varDelta E\,=\, \frac{2}{5} e^2R^2 \frac{Z^4}{a_0^3n^3} $$

In terms of the difference $\delta R$ in the radii (to the first order) the shift turns out

$$ \delta E\simeq \frac{4}{5} e^2R^2 \frac{Z^4}{a_0^3n^3} \frac{\delta R}{R} $$

In muonic atoms (see Sect. 1.5) because of the change in the reduced mass and in the Bohr radius $a_0$, the volume isotope effect is dramatically increased with respect to ordinary hydrogenic atoms.

As regards the hyperfine terms one has to consider the decrease in the Bohr radius and in the Bohr magneton ($\mu _B\propto 1/m$) (see Problem 5.4). For the hyperfine quadrupole correction small effects have to be expected, since only states with $l\ne 0$ are involved.

Finally it is mentioned that an isomeric shift , analogous to the volume isotope shift, occurs when a radiative decay (e.g. from $^{57}$Co to $^{57}$Fe) changes the radius of the nucleus. The isomeric shift is experimentally detected in the Mössbauer resonant absorption spectrum (see Sect. 14.6).