Hyperfine Structure

The proton in a hydrogen atom is a spin one-half charged particle, and therefore possesses a spin magnetic moment. By analogy with Equation (5.44), we can write

$$\mbox{\boldmath$\mu$} _p = \frac{g_p\,e}{2\,m_p}\,{\bf S}_p,$$ (7.141)

where $\mu$ $_p$ is the proton magnetic moment, ${\bf S}_p$ the proton spin, $m_p$ the proton mass, and $g_p$ the proton $g$ -factor. The proton $g$ -factor is found experimentally to take that value $5.59$ [117]. [In writing the previous equation, we have made use of the fact that the proton is essentially stationary (in the center of mass frame), and, therefore, possesses zero orbital angular momentum.] Note that the spin magnetic moment of a proton is much smaller (by a factor of order $m_e/m_p$ ) than that of an electron.

According to classical electromagnetism, the vector potential due to a point magnetic dipole ${\bf M}$ located at the origin is [49]

$${\bf A}({\bf r}) = \frac{\mu_0}{4\pi}\,\frac{{\bf M}\times {\bf x}}{r^{\,3}},$$ (7.142)

where $r=\vert{\bf x}\vert$ . The associated magnetic field takes the form [49]

$${\bf B}=\nabla\times {\bf A} =\frac{\mu_0}{4\pi}\left[ \frac{3\,({\bf M}\cdot{\bf e}_r)\,{\bf e}_r-{\bf M}}{r^{\,3}}\right],$$ (7.143)

where ${\bf e}_r={\bf x}/r$ . Suppose that ${\bf M}=M\,{\bf e}_z$ . The Cartesian components of ${\bf B}$ are thus

$$B_x(r,\theta,\varphi) = \frac{\mu_0\,M}{4\pi\,r^{\,3}}\,3\,\cos\theta\,\sin\theta\,\cos\varphi,$$ (7.144)

$$B_y(r,\theta,\varphi) = \frac{\mu_0\,M}{4\pi\,r^{\,3}}\,3\,\cos\theta\,\sin\theta\,\sin\varphi,$$ (7.145)

$$B_z(r,\theta,\varphi) =\frac{\mu_0\,M}{4\pi\,r^{\,3}}\,(3\,\cos^2\theta - 1),$$ (7.146)

where ($r$ , $\theta$ , $\varphi$ ) are conventional spherical coordinates. It is easily demonstrated that

$$\int_V d^{\,3}{\bf x} \,B_x = \int_V d^{\,3}{\bf x}\, B_y=\int_V d^{\,3}{\bf x} \,B_z=0,$$ (7.147)

where $V$ is a spherical volume of radius $R$ , centered on the origin. However, we can also write [67]

$$\int_V d^{\,3}{\bf x} \,{\bf B} = \int_V d^{\,3}{\bf x} \,\nabla\... ...bf S}\times {\bf A}= R^{\,2}\oint_{r=R} d{\mit\Omega}\,{\bf e}_r\times {\bf A},$$ (7.148)

where $S$ is the bounding surface of volume $V$ , and $d{\mit\Omega}$ an element of solid angle. According to Equation (7.145),

$$\int_V d^{\,3}{\bf x} \,{\bf B} =\frac{\mu_0}{4\pi} \oint d{\mit\Omega}\,{\bf e}_r\times ({\bf M}... ...u_0}{4\pi} \oint d{\mit\Omega} \,[{\bf M} - ({\bf M}\cdot{\bf e}_r)\,{\bf e}_r]$$

$$= \mu_0\,{\bf M} -\frac{\mu_0}{4\pi}\oint d{\mit\Omega} \,({\bf M}\cdot{\bf e}_r)\,{\bf e}_r.$$ (7.149)

Let ${\bf M}=M\,{\bf e}_z$ , and ${\bf F} = \int d{\mit\Omega} \,({\bf M}\cdot{\bf e}_r)\,{\bf e}_r$ . It follows that

$$F_x = M\oint d{\mit\Omega}\,\cos\theta\,\sin\theta\,\cos\varphi =0,$$ (7.150)

$$F_y = M\oint d{\mit\Omega}\,\cos\theta\,\sin\theta\,\sin\varphi =0,$$ (7.151)

$$F_z = M\oint d{\mit\Omega} \cos^2\theta = \frac{4\pi}{3}\,M,$$ (7.152)

which implies that ${\bf F} = (4\pi/3)\,{\bf M}$ . Hence, we obtain

$$\int_V d^{\,3}{\bf x} \,{\bf B}= \frac{2}{3}\,\mu_0\,{\bf M}.$$ (7.153)

However, the previous expression is inconsistent with Equations (7.147)-(7.149). Note that the right-hand side of Equation (7.156) is independent of the radius, $R$ , of the integration volume $V$ . Consequently, we can take the limit $R\rightarrow 0$ without changing the value of $\int_V d^{\,3}{\bf x} \,{\bf B}$ . We deduce that the non-zero contribution to this integral originates from the origin. Hence, we can reconcile the previously mentioned inconsistency by modifying Equation (7.146) to read

$${\bf B} =\frac{\mu_0}{4\pi}\left[ \frac{3\,({\bf M}\cdot{\bf e}_r... ...-{\bf M}}{r^{\,3}}\right] + \frac{2\,\mu_0}{3}\,\delta^{\,3}({\bf x})\,{\bf M}.$$ (7.154)

Here, $\delta^{\,3}({\bf x})\equiv \delta(x)\,\delta(y)\,\delta(z)$ is a three-dimensional Dirac delta function. This function has the property that

$$\int_V d^{\,3}{\bf x} \,F({\bf x})\,\delta^{\,3}({\bf x}-{\bf x}_0) = F({\bf x}_0),$$ (7.155)

where $F({\bf x})$ is a general function that is well-behaved in the vicinity of ${\bf x}={\bf x}_0$ (which is assumed to lie in the volume $V$ ) [92].

According to the previous formula, the proton's magnetic moment, $\mu$ $_p$ , generates a magnetic field of the form

$${\bf B} = \frac{\mu_0}{4\pi\,r^{\,3}}\,\left[3\,(\mbox{\boldmath$... ...p\right] + \frac{2\,\mu_0}{3}\,\delta^{\,3}({\bf x})\,\mbox{\boldmath$\mu$}_p\,$$ (7.156)

where ${\bf x}$ measures position relative to the proton. Now, the Hamiltonian of the electron in the magnetic field generated by the proton is simply [49]

$$H_1 = - \mbox{\boldmath$\mu$} _e\cdot {\bf B},$$ (7.157)

where

$$\mbox{\boldmath$\mu$} _e = - \frac{e}{m_e}\,{\bf S}_e.$$ (7.158)

Here, $\mu$ $_e$ is the electron magnetic moment [see Equation (7.98)], and ${\bf S}_e$ the electron spin. Thus, the perturbing Hamiltonian is written

$$H_1=\frac{\mu_0\,g_p\,e^{\,2}}{8\pi\,m_p\,m_e}\left[\frac{3\,({\b... ..._0\,g_p\,e^{\,2}}{3\,m_p\,m_e}\,{\bf S}_p\cdot{\bf S}_e\,\delta^{\,3}({\bf x}).$$ (7.159)

Note that, because we have neglected coupling between the proton spin and the magnetic field generated by the electron's orbital motion, the previous expression is only valid for $l=0$ states.

According to standard first-order perturbation theory, the energy-shift induced by spin-spin coupling between the proton and the electron is the expectation value of the perturbing Hamiltonian. Hence,

$${\mit\Delta} E = \frac{\mu_0\,g_p\,e^{\,2}}{8\pi\,m_p\,m_e}\left\langle\frac{3\,... ..._r)\,({\bf S}_e\cdot{\bf e}_r) - {\bf S}_p\cdot{\bf S}_e}{r^{\,3}}\right\rangle$$

$$\phantom{=}+ \frac{\mu_0\,g_p\,e^{\,2}}{3\,m_p\,m_e}\,\langle{\bf S}_p\cdot{\bf S}_e\rangle\,\vert\psi({\bf0})\vert^{\,2}.$$ (7.160)

In the final term on the right-hand side, the expectation value is taken over the overall spin state. For the ground state of hydrogen, which is spherically symmetric, the first term in the previous expression vanishes by symmetry. Moreover, it is easily demonstrated that $\vert\psi_{000}({\bf0})\vert^{\,2}= 1/(\pi\,a_0^{\,3})$ . Thus, we obtain

$${\mit\Delta} E_{000} = \frac{\mu_0\,g_p\,e^{\,2}}{3\pi\,m_p\,m_e\,a_0^{\,3}}\,\langle{\bf S}_p\cdot{\bf S}_e\rangle.$$ (7.161)

Let

$${\bf S} = {\bf S}_e + {\bf S}_p$$ (7.162)

be the total spin. We can show that

$${\bf S}_p\cdot{\bf S}_e = \frac{1}{2}\,(S^2-S_e^{\,2}-S_p^{\,2}).$$ (7.163)

Thus, the simultaneous eigenstates of the perturbing Hamiltonian and the main Hamiltonian are the simultaneous eigenstates of $S_e^{\,2}$ , $S_p^{\,2}$ , and $S^2$ . (The use of simultaneous eigenstates of the perturbing and main Hamiltonian avoids the possibility of singular terms arising in the perturbation expansion to second order--see Section 7.5.) However, both the proton and the electron are spin one-half particles. According to Section 6.4, when two spin one-half particles are combined (in the absence of orbital angular momentum) the resulting state has either spin 1 or spin 0. In fact, there are three spin 1 states, known as triplet states, and a single spin 0 state, known as the singlet state. For all states, the eigenvalues of $S_e^{\,2}$ and $S_p^{\,2}$ are $(3/4)\,\hbar^{\,2}$ . The eigenvalue of $S^2$ is 0 for the singlet state, and $2\,\hbar^{\,2}$ for the triplet states. Hence,

$$\langle {\bf S}_p\cdot{\bf S}_e\rangle = - \frac{3}{4}\,\hbar^{\,2}$$ (7.164)

for the singlet state, and

$$\langle {\bf S}_p\cdot{\bf S}_e\rangle = \frac{1}{4}\,\hbar^{\,2}$$ (7.165)

for the triplet states.

It follows, from the previous analysis, that proton-electron spin-spin coupling breaks the degeneracy of the two $(1s)_{1/2}$ states of the hydrogen atom, lifting the energy of the triplet configuration, and lowering that of the singlet. This splitting is known as hyperfine structure. The net energy difference between the singlet and the triplet states is

$${\mit\Delta} E_{000} = \frac{8}{3}\,g_p\,\frac{m_e}{m_p}\,\alpha^{\,2}\,\vert E_0\vert = 5.88\times 10^{-6}\,{\rm eV},$$ (7.166)

where $\vert E_0\vert=13.6\,{\rm eV}$ is the (magnitude of the) ground-state energy, and $\alpha=1/137$ the fine structure constant. Note that the hyperfine energy-shift is much smaller, by a factor $m_e/m_p$ , than a typical fine structure energy-shift. (See Exercise 14.) If we convert the previous energy into a wavelength (using $\lambda = c/\nu= c\,h/{\mit\Delta}E_{000}$ ) then we obtain

$$\lambda = 21.1\,{\rm cm}.$$ (7.167)

This is the wavelength of the radiation emitted by a hydrogen atom that is collisionally excited from the singlet to the triplet state, and then decays back to the lower energy singlet state. The 21cm line is famous in radio astronomy because it was used to map out the spiral structure of our galaxy in the 1950's [114].