The 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

$$H_1 = -\mbox{\boldmath$\mu$}\cdot{\bf B},$$ (971)

where

$$\mbox{\boldmath$\mu$}= - \frac{e}{2\,m_e}\,({\bf L} + 2\,{\bf S})$$ (972)

is the total electron magnetic moment, including both orbital and spin contributions [see Eqs. (737)-(739)]. Thus,

$$H_1 = \frac{e\,B}{2\,m_e}\,(L_z+ 2\,S_z).$$ (973)

Suppose that the applied magnetic field is much weaker than the atom's internal magnetic field (956). 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 subsection). Hence, from standard perturbation theory, the first-order energy-shift induced by a weak external magnetic field is

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

$$= \frac{e\,B}{2\,m_e}\,\left(m_j\,\hbar + \langle l,1/2;j,m_j\vert S_z\vert l,1/2;j,m_j\rangle\right),$$ (974)

since $J_z=L_z+S_z$. Now, according to Eqs. (804) and (805),

$$\psi^{(2)}_{j,m_j} = \left(\frac{j+m_j}{2\,l+1}\right)^{1/2}... ...t(\frac{j-m_j}{2\,l+1}\right)^{1/2}\,\psi^{(1)}_{m_j+1/2,-1/2}$$ (975)

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

$$\psi^{(2)}_{j,m_j} = \left(\frac{j+1-m_j}{2\,l+1}\right)^{1/... ...\frac{j+1+m_j}{2\,l+1}\right)^{1/2}\,\psi^{(1)}_{m_j+1/2,-1/2}$$ (976)

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,

$$S_z\,\psi^{(1)}_{m,\pm 1/2} = \pm \frac{\hbar}{2}\,\psi^{(1)}_{m,\pm 1/2}.$$ (977)

It follows from Eqs. (975)-(977), and the orthormality of the $\psi^{(1)}$, that

$$\langle l,1/2;j,m_j\vert S_z\vert l,1/2;j,m_j\rangle = \pm \frac{m_j\,\hbar}{2\,l+1}$$ (978)

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

$$\Delta E_{l,1/2;j,m_j} = \mu_B\,B\,m_j\left[1\pm \frac{1}{2\,l+1}\right]$$ (979)

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

$$\mu_B = \frac{e\,\hbar}{2\,m_e} = 5.788\times 10^{-5}\,{\rm eV/T}$$ (980)

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. (979) that the Zeeman effect splits degenerate states characterized by $j=l+1/2$ into $2\,j+1$ equally spaced states of interstate spacing

$$\Delta E_{j=l+1/2} = \mu_B\,B\,\frac{2\,l+2}{2\,l+1}.$$ (981)

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

$$\Delta E_{j=l-1/2} = \mu_B\,B\,\frac{2\,l}{2\,l+1}.$$ (982)

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. 21. Note, finally, that since the $\psi^{(2)}_{l,m_j}$ are not simultaneous eigenstates of the unperturbed and perturbing Hamiltonians, Eqs. (981) and (982) 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 21: The Zeeman effect for the $n=1$ and $2$ states of a hydrogen atom. Here, $\epsilon = \mu _B\,B$. Not to scale.