QB_PH_SEC6_030: Lasers and Quantum Optics

A two-level atomic system has degenerate energy levels with statistical weights $g_1 = 1$ (lower level) and $g_2 = 3$ (upper level). What is the relation between the Einstein absorption coefficient $B_{12}$ and the stimulated emission coefficient $B_{21}$?

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Detailed Balance in Radiation Field**: $N_1 B_{12} \rho(\nu) = N_2 [A_{21} + B_{21} \rho(\nu)]$ 2. **Boltzmann Distribution with Degeneracy**: $\frac{N_2}{N_1} = \frac{g_2}{g_1} e^{-h\nu / k_B T}$ 3. **Matching Planck's Law**: $\rho(\nu) = \frac{A_{21}/B_{21}}{\frac{g_1 B_{12}}{g_2 B_{21}} e^{h\nu / k_B T} - 1}$ Comparing with Planck's formula $\rho(\nu) = \frac{8\pi h \nu^3}{c^3} \frac{1}{e^{h\nu / k_B T} - 1}$ forces the coefficient of the exponential to equal unity: $\frac{g_1 B_{12}}{g_2 B_{21}} = 1 \implies g_1 B_{12} = g_2 B_{21}$ 4. **Substitution**: $B_{12} = \frac{g_2}{g_1} B_{21} = \frac{3}{1} B_{21} = 3 B_{21}$ 5. **Conclusion**: $B_{12} = 3 B_{21}$. Option (A) is correct.