Zeeman Effect

Consider a hydrogen atom placed in a uniform $z$-directed external magnetic field of strength $B$. The modification to the Hamiltonian of the system is

$$H_1 = -\mbox{\boldmath$\mu$}\cdot{\bf B},$$ (992)

where

$$\mbox{\boldmath$\mu$}= - \frac{e}{2 m_e} ({\bf L} + 2 {\bf S})$$ (993)

is the total electron magnetic moment, including both orbital and spin contributions [see Eqs. (758)-(760)]. Thus,

$$H_1 = \frac{e B}{2 m_e} (L_z+ 2 S_z).$$ (994)

Suppose that the applied magnetic field is much weaker than the atom's internal magnetic field (977). Since the magnitude of the internal field is about 25 tesla, this is a fairly reasonable assumption. In this situation, we can treat $H_1$ as a small perturbation acting on the simultaneous eigenstates of the unperturbed Hamiltonian and the fine structure Hamiltonian. Of course, these states are the simultaneous eigenstates of $L^2$, $S^2$, $J^2$, and $J_z$ (see previous section). Hence, from standard perturbation theory, the first-order energy-shift induced by a weak external magnetic field is

$$\Delta E_{l,1/2;j,m_j} = \langle l,1/2;j,m_j\vert H_1\vert l,1/2;j,m_j\rangle$$

$$= \frac{e B}{2 m_e} \left(m_j \hbar + \langle l,1/2;j,m_j\vert S_z\vert l,1/2;j,m_j\rangle\right),$$ (995)

since $J_z=L_z+S_z$. Now, according to Eqs. (825) and (826),

$$\psi^{(2)}_{j,m_j} = \left(\frac{j+m_j}{2 l+1}\right)^{1/2}... ...t(\frac{j-m_j}{2 l+1}\right)^{1/2} \psi^{(1)}_{m_j+1/2,-1/2}$$ (996)

when $j=l+1/2$, and

$$\psi^{(2)}_{j,m_j} = \left(\frac{j+1-m_j}{2 l+1}\right)^{1/... ...\frac{j+1+m_j}{2 l+1}\right)^{1/2} \psi^{(1)}_{m_j+1/2,-1/2}$$ (997)

when $j=l-1/2$. Here, the $\psi^{(1)}_{m,m_s}$ are the simultaneous eigenstates of $L^2$, $S^2$, $L_z$, and $S_z$, whereas the $\psi^{(2)}_{j,m_j}$ are the simultaneous eigenstates of $L^2$, $S^2$, $J^2$, and $J_z$. In particular,

$$S_z \psi^{(1)}_{m,\pm 1/2} = \pm \frac{\hbar}{2} \psi^{(1)}_{m,\pm 1/2}.$$ (998)

It follows from Eqs. (996)-(998), and the orthormality of the $\psi^{(1)}$, that

$$\langle l,1/2;j,m_j\vert S_z\vert l,1/2;j,m_j\rangle = \pm \frac{m_j \hbar}{2 l+1}$$ (999)

when $j=l\pm 1/2$. Thus, the induced energy-shift when a hydrogen atom is placed in an external magnetic field--which is known as the Zeeman effect--becomes

$$\Delta E_{l,1/2;j,m_j} = \mu_B B m_j\left[1\pm \frac{1}{2 l+1}\right]$$ (1000)

where the $\pm$ signs correspond to $j=l\pm 1/2$. Here,

$$\mu_B = \frac{e \hbar}{2 m_e} = 5.788\times 10^{-5} {\rm eV/T}$$ (1001)

is known as the Bohr magnetron. Of course, the quantum number $m_j$ takes values differing by unity in the range $-j$ to $j$. It, thus, follows from Eq. (1000) that the Zeeman effect splits degenerate states characterized by $j=l+1/2$ into $2 j+1$ equally spaced states of interstate spacing

$$\Delta E_{j=l+1/2} = \mu_B B \frac{2 l+2}{2 l+1}.$$ (1002)

Likewise, the Zeeman effect splits degenerate states characterized by $j=l-1/2$ into $2 j+1$ equally spaced states of interstate spacing

$$\Delta E_{j=l-1/2} = \mu_B B \frac{2 l}{2 l+1}.$$ (1003)

In conclusion, in the presence of a weak external magnetic field, the two degenerate $1S_{1/2}$ states of the hydrogen atom are split by $2 \mu_B B$. Likewise, the four degenerate $2S_{1/2}$ and $2P_{1/2}$ states are split by $(2/3) \mu_B B$, whereas the four degenerate $2P_{3/2}$ states are split by $(4/3) \mu_B B$. This is illustrated in Fig. 24. Note, finally, that since the $\psi^{(2)}_{l,m_j}$ are not simultaneous eigenstates of the unperturbed and perturbing Hamiltonians, Eqs. (1002) and (1003) can only be regarded as the expectation values of the magnetic-field induced energy-shifts. However, as long as the external magnetic field is much weaker than the internal magnetic field, these expectation values are almost identical to the actual measured values of the energy-shifts.

Figure 24: The Zeeman effect for the $n=1$ and $2$ states of a hydrogen atom. Here, $\epsilon = \mu _B B$. Not to scale.