Muonic hydrogen is a negative muon (rest energy ) bound to a fixed (infinitely heavy)…

PHYSICS · Subjective · Fermi University Physics · Mathematical Methods for Physics · Quantum Mechanics · Central potentials and canonical three-dimensional systems

The Structure of Quantum Mechanics, Ch Hydrogen, and the symmetry that explains it

Problem

Muonic hydrogen is a negative muon (rest energy mμc2=105.7 MeVm_\mu c^2 = 105.7\,\text{MeV}) bound to a fixed (infinitely heavy) proton by the Coulomb potential V(r)=−k/rV(r)=-k/r, k=e2/4πϵ0k=e^2/4\pi\epsilon_0, exactly as for the electron (rest energy mec2=0.511 MeVm_ec^2=0.511\,\text{MeV}) but with the muon's mass mμm_\mu in place of mem_e throughout. Define M⃗=1mμp⃗×L⃗−kr^\vec M=\frac{1}{m_\mu}\vec p\times\vec L - k\hat r classically and, on the bound subspace, K⃗^=−mμ/2H^ M⃗^\hat{\vec K}=\sqrt{-m_\mu/2\hat H}\,\hat{\vec M} built from the Hermitian M⃗^=12mμ(p⃗^×L⃗^−L⃗^×p⃗^)−kr⃗^r.\hat{\vec M}=\frac{1}{2m_\mu}(\hat{\vec p}\times\hat{\vec L}-\hat{\vec L}\times\hat{\vec p}) - k\frac{\hat{\vec r}}{r}. You may use L⃗^⋅K⃗^=0\hat{\vec L}\cdot\hat{\vec K}=0 and K^2+L^2+ℏ2=−mμ2H^k2.\hat K^2+\hat L^2+\hbar^2=-\frac{m_\mu}{2\hat H}k^2. (a) Show, from p⃗˙=−kr^/r2\dot{\vec p}=-k\hat r/r^2 and zero torque, that dM⃗/dt=0d\vec M/dt=0 and state the consequence for the muon's orbit. (b) Show the symmetrised M⃗^\hat{\vec M} above is Hermitian while the unsymmetrised p⃗^×L⃗^\hat{\vec p}\times\hat{\vec L} is not, and state on which states K⃗^\hat{\vec K} is defined. (c) Given [K^i,K^j]=iℏϵijkL^k[\hat K_i,\hat K_j]=i\hbar\epsilon_{ijk}\hat L_k and [L^i,K^j]=iℏϵijkK^k[\hat L_i,\hat K_j]=i\hbar\epsilon_{ijk}\hat K_k, form A⃗^±=12(L⃗^±K⃗^)\hat{\vec A}_\pm=\tfrac12(\hat{\vec L}\pm\hat{\vec K}) and show each satisfies an angular-momentum algebra with [A^+i,A^−j]=0[\hat A_{+i},\hat A_{-j}]=0. (d) Derive EnE_n from the Casimir relation, noting explicitly how the mass mμm_\mu enters, and evaluate E2E_2 numerically (in eV). State the degeneracy of that level.

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