Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 14A (Quantum Mechanics)

The anti-muon has the same charge as a proton, so the Coulomb potential seen by the electron is the same as in the hydrogen atom. The main difference is the source mass. Separating the centre-of-mass motion replaces $m_e$ by the Coulomb two-body reduced massSince $m_{\mu }\gg m_e$, this is close to $m_e$, though less close than the electron-proton reduced mass. The smaller reduced mass gives a slightly larger Bohr radius and energies of slightly smaller magnitude. Under the assumptions in the question, the Coulomb form, angular eigenfunctions and quantum numbers are otherwise unchanged; anti-muon decay and spin effects are being neglected.

For fixed $l$, the radial equation isSubstituteCancellation of the coefficient of $1/r$ requiresthe reduced-mass Bohr radius. The remaining constant term givesEach such no-radial-node state is a circular Coulomb bound state with principal quantum number $n=l+1$.

Comparing the prepared energy with the formula in part (ii),so $l+1=3$ and $l=2$. The radial probability density, apart from normalization, isUsing the given factorial integral, its mean is the ratioFor $l=2$ this gives the mean radius of a circular Coulomb bound state

The energy fixes $l=2$, irrespective of the magnetic quantum number $m$. Therefore every prepared state is an eigenstate of ${\hat{L}}^2$ withAn immediate measurement of the total orbital-angular-momentum magnitude cannot yield $\hslash$, so the requested probability is