Introduction to Quantum Mechanics · Problem 11.33

Quantum Dynamics — David J. Griffiths

Physics — Quantum Mechanics · 21cm-hyperfine-spontaneous-emission

Problem

The spontaneous emission rate for the 21-cm hyperfine line in hydrogen (Section 7.5) can be obtained from Equation 11.63, except that this is a magnetic dipole transition, not an electric one: ℘→1cM=1c⟨1 ∣ (μe+μp) ∣ 0⟩,\wp \to \frac{1}{c}\mathbf{M} = \frac{1}{c}\langle 1\,|\,(\boldsymbol{\mu}_e + \boldsymbol{\mu}_p)\,|\,0\rangle, where μe=−emeSe,μp=5.59 e2mpSp\boldsymbol{\mu}_e = -\frac{e}{m_e}\mathbf{S}_e, \quad \boldsymbol{\mu}_p = \frac{5.59\,e}{2m_p}\mathbf{S}_p are the magnetic moments of the electron and proton (Equation 7.89), and ∣0⟩|0\rangle, ∣1⟩|1\rangle are the singlet and triplet configurations (Equations 4.175 and 4.176). Because mp≫mem_p \gg m_e, the proton contribution is negligible, so A=ω03e23πϵ0ℏc5me2∣⟨1∣ Se ∣0⟩∣2.A = \frac{\omega_0^3 e^2}{3\pi\epsilon_0 \hbar c^5 m_e^2}|\langle 1|\,\mathbf{S}_e\,|0\rangle|^2. Work out ∣⟨1∣ Se ∣0⟩∣2|\langle 1|\,\mathbf{S}_e\,|0\rangle|^2 (use whichever triplet state you like). Put in the actual numbers, to determine the transition rate and the lifetime of the triplet state. *Answer:* 1.1×1071.1 \times 10^7 years.

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