Measurement of the magnetic moment of the positive muon by a stroboscopic muon-spin-rotation technique

Abstract

A new determination of the magnetic moment of the positive muon in units of the magnetic moment of the proton is presented. The Larmor precession of positive muons in liquid bromine was observed by a stroboscopic technique in a field of 0.75 T and combined with concomitant proton NMR measurements in the same chemical environment. The stroboscopic method allows use of the full muon stopping rate available at the Schweizerisches Institut f\"ur Nuklearforschung (SIN) muon channel. Moreover, it permits an intrinsically precise determination of muon Larmor frequency and proton NMR frequency measuring the magnetic field by comparison with the stable reference frequency of the SIN accelerator ($\frac{\ensuremath{\Delta}\ensuremath{\Omega}}{\ensuremath{\Omega}}\ensuremath{\approx}{10}^{\ensuremath{-}8}$). Two different bromine targets were used which allowed an unambiguous determination of the chemical field shift experienced by the muons. One target contained pure and water-free liquid bromine (${\mathrm{Br}}_{2}$), where stopped muons form (${\ensuremath{\mu}}^{+}{e}^{\ensuremath{-}}$)Br molecules. The other target was slightly contaminated with water; there a chemical reaction chain places the muons into (${\ensuremath{\mu}}^{+}{e}^{\ensuremath{-}}$)HO molecules. The diamagnetic shielding of protons in the analogous molecules HBr and ${\mathrm{H}}_{2}$O in liquid bromine was measured by high-resolution NMR. Values for the isotopic shift of the diamagnetic shielding, when protons are replaced by muons, are available from quantum chemical calculations. After application of the chemical-shift corrections, the results from the two different bromine targets are consistent. The final result is $\frac{{\ensuremath{\mu}}_{\ensuremath{\mu}}}{{\ensuremath{\mu}}_{p}}=3.1833441(17)(or \ifmmode\pm\else\textpm\fi{}0.53 \mathrm{ppm})$. This value agrees with other recent precision determinations of $\frac{{\ensuremath{\mu}}_{\ensuremath{\mu}}}{{\ensuremath{\mu}}_{p}}$. For the muon mass the present result implies $\frac{{m}_{\ensuremath{\mu}}}{{m}_{e}}=206.76835(11)(\ifmmode\pm\else\textpm\fi{}0.53 \mathrm{ppm})$.