Principles of Quantum Mechanics · Problem 15.1.2

Addition of Angular Momenta — R. Shankar

Physics — Quantum Mechanics · hyperfine-interaction-hydrogen

Problem

In addition to the Coulomb interaction, there exists another, called the hyperfine interaction, between the electron and proton in the hydrogen atom. The Hamiltonian describing this interaction, which is due to the magnetic moments of the two particles, is,

H_{hf} = A\mathbf{S}_1\cdot\mathbf{S}_2 \qquad (A > 0) \tag{15.1.22}

(This formula assumes the orbital state of the electron is

|1, 0, 0\rangle

.) The total Hamiltonian is thus the Coulomb Hamiltonian plus

H_{hf}

. (1) Show that

H_{hf}

splits the ground state into two levels:

E_+ = -\text{Ry} + \frac{\hbar^2 A}{4}, \qquad E_- = -\text{Ry} - \frac{3\hbar^2 A}{4} \tag{15.1.23}

and that corresponding states are triplets and singlet, respectively. (2) Try to estimate the frequency of the emitted radiation as the atom jumps from the triplet to the singlet. To do so, you may assume that the electron and proton are two dipoles

\mu_e

and

\mu_p

separated by a distance

a_0

, with an interaction energy of the order

\mathscr{H}_{hf} \cong \frac{\mu_e \cdot \mu_p}{a_0^3}

Show that this implies that the constant in Eq. (15.1.22) is

A \sim \frac{2e}{2mc}\frac{(5.6)e}{2Mc}\frac{1}{a_0^3}

(where 5.6 is the

g

factor for the proton), and that

\Delta E = E_+ - E_- = A\hbar^2

is a correction of order

(m/M)a^2

relative to the ground-state energy. Estimate that the frequency of emitted radiation is a few tens of centimeters, using the mnemonics from Chapter 13. The measured value is 21.4 cm. This radiation, called the 21-cm line, is a way to detect hydrogen in other parts of the universe. (3) Estimate the probability ratio

P(\text{triplet})/P(\text{singlet})

of hydrogen atoms in thermal equilibrium at room temperature.

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