I can't find (Google, PDG etc.) an experimental value for the magnetic moment of the Kaons $K^{⁺/-}$. They are u/s mesons so shouldn't there be a non-zero value even if total spin is zero? Lifetime is ~ $10^{-8}$s, good enough to measure magnetic moment in case of the baryons. Is there a particular problem measuring this?
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$\begingroup$ As you state, no values are recorded in the PDG , which tries to give with errors all measurements in particle physics. IMO it is not only the lifetime that allows or not measurement, but the decay products too. $\endgroup$– anna vCommented Nov 17, 2022 at 5:00
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2 Answers
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Yes, it appears that no one one has made an explicit measurement of the charged kaon magnetic moment as they have for various hyperons. A limit on any $K^-$ magnetic moment could be set using existing kaon exotic atom spectroscopy measurements, but again it seems that no one has bothered to do so. This lack of interest is presumably because kaons are spin-0 pseudoscalar mesons, so no magnetic moment is expected.
If the $K^-$ did have a magnetic moment, then the spectral lines of kaonic atoms (where a $K^-$ has replaced an electron) would have fine structure, just as regular atoms have fine structure lines corresponding to whether the electron and proton magnetic moments are parallel or antiparallel. No one has reported any such fine structure, despite there being many kaonic atom spectroscopy experiments, so there is an implicit limit on any magnetic moment that could be made explicit if somebody wanted to analyze the many published spectra. There is, however, one old limit on the $\pi^-$ magnetic moment from pionic atomic spectroscopy.
answered Nov 17, 2022 at 19:31
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They are up/strange mesons, so shouldn't there be a non-zero value even if total spin is zero?
No. A spin-zero quantum system is not orientable: it remains unchanged under rotations. A nonzero magnetic dipole moment, on the other hand, cannot remain unchanged under rotations. The kaon can’t have a magnetic moment because the kaon field and the magnetic field have different symmetries.
This rule is a consequence of the Wigner-Eckhart theorem. A two-state spinor field, like an electron, has enough complexity to carry a dipole moment, but not a quadrupole moment. A vector field has enough complexity for a quadrupole moment, but not an octupole moment. In another answer, I suggest some ways to think of these restrictions.
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1$\begingroup$ Rob, I realize I have never had the Wigner-Eckhart theorem's implications for composite systems clear in my head, so I just posted a question asking When can spinless "particles" have magnetic moments?. $\endgroup$ Commented Nov 17, 2022 at 23:08
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$\begingroup$ @David Bailey „Composite systems“ is a good point. When I referred to „up/strange“ I had constituent quarks on my mind which seem still the best way to describe magnetic moments of particles. So I don’t know if the given arguments such as the Wigner–Eckart theorem are strictly applicable. Maybe it is just that conventional magnetic fields can not break the symmetry of $K^+=-\frac{1}{\sqrt{2}}\left(u\uparrow\bar{s}\downarrow-u\downarrow\bar{s}\uparrow\right)$? $\endgroup$– qatchCommented Nov 19, 2022 at 13:46
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$\begingroup$ @qatch Yes, that is what I am wondering. $\endgroup$ Commented Nov 19, 2022 at 14:53