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One-Particle Measurement of the Antiproton Magnetic Moment

The antiproton $(\bar{p})$ magnetic moment $\mu \bar{p} = \mu_{\bar{p}} S/(\bar{h}/2)$ is proportional to its spin $S$. A single trapped $\bar{p}$ is used for the first time to measure the $\bar{p}$ magnetic moment in nuclear magnetons, giving $\mu_\bar{p}/\mu N = -2.792 845 \pm 0.000 012$. The 4.4...

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Detalles Bibliográficos
Autores principales: DiSciacca, J., Marshall, M., Marable, K., Gabrielse, G., Ettenauer, S., Tardiff, E., Kalra, R., Fitzakerley, D.W., George, M.C., Hessels, E.A., Storry, C.H., Weel, M., Grzonka, D., Oelert, W., Sefzick, T.
Lenguaje:eng
Publicado: 2013
Materias:
Acceso en línea:https://dx.doi.org/10.1103/PhysRevLett.110.130801
http://cds.cern.ch/record/1510452
Descripción
Sumario:The antiproton $(\bar{p})$ magnetic moment $\mu \bar{p} = \mu_{\bar{p}} S/(\bar{h}/2)$ is proportional to its spin $S$. A single trapped $\bar{p}$ is used for the first time to measure the $\bar{p}$ magnetic moment in nuclear magnetons, giving $\mu_\bar{p}/\mu N = -2.792 845 \pm 0.000 012$. The 4.4 parts per million (ppm) uncertainty is 680 times smaller than previously realized. Comparing to the proton moment $\mu_{p} = \mu_{p} S/\bar{h}/2)$ measured using the same method and trap electrodes gives $\mu_{\bar{p}}/\mu_{p} = -1.000 000 \pm 0.000 005$ to 5 ppm, consistent with the prediction of the CPT theorem.