By appropriate modification of the hydrogen formula, determine the hyperfine splitting in the ground state of 

(a) Muonic hydrogen (in which a muon—same charge and g-factor as the electron, but 207 times the mass— substitutes for the electron), 

(b) Positronium (in which a positron—same mass and g-factor as the electron, but opposite charge—substitutes for the proton), and 

(c) Muonium (in which an anti-muon—same mass and g-factor as a muon, but opposite charge—substitutes for the proton). Don’t forget to use the reduced mass (Problem 5.1) in calculating the “Bohr radius” of these exotic “atoms,” but use the actual masses in the gyromagnetic ratios. Incidentally, the answer you get for positronium (4.82 x 10^-4 eV) is quite far from the experimental value (8.41 x 10^-4 eV); the large discrepancy is due to pair annihilation (e⁺ + e^- → ϒ+ϒ), which contributes an extra (3/4) ΔE, and does not occur (of course) in ordinary hydrogen, muonic hydrogen, or muonium.

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and R = (miri + m2r2)/(m1+m2) (the center of mass). (a) Show that r = R + (p/m)r, r2 = R-(/m2)r, and V₁ = (u/m₂) VR + V₁ V₂ = (u/m₁) VR - V₁, where mm 2 mi + m₂ is the reduced mass of the system. (b) Show that the (time-independent) Schrödinger equation (5.7) becomes ħ² 2 (m1 + m₂) -V₂/V ħ² 2μ (5.15) V²V + V(r) = Ev. (c) Separate the variables, letting (R, r) = R(R), (r). Note that R satisfies the one-particle Schrödinger equation, with the total mass (m1 + m2) in place of m, potential zero, and energy ER, while , satisfies the one-particle Schrödinger equation with the reduced mass in place of m, potential V (r), and energy Er. The total energy is the sum: E = ER + Er. What this tells us is that the center of mass moves like a free particle, and the relative motion (that is, the motion of particle 2 with respect to particle 1) is the same as if we had a single particle with the reduced mass, subject to the potential V. Exactly the same decomposition occurs in classical mechanics; it reduces the two-body problem to an equivalent one-body problem.