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For Charmonium, why does the spin-spin interaction mostly affect the $L = 0$ states?

My textbook states that this is because "only then is the wave function at the origin non-vanishing". Could anyone expand on this and explain why this should be the case?

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  • $\begingroup$ The wavefunction in a central potential goes like $r^L$ near $r=0$. Charmonium is a two-body system, but you transform into an equivalent one-body system. $\endgroup$ – jwimberley Mar 9 '15 at 18:02
  • $\begingroup$ @Jwimberley That is an answer, not a comment. $\endgroup$ – rob May 10 '15 at 3:42
  • $\begingroup$ @rob I thought it was a comment, since it didn't explain why this prevents spin-spin interaction. $\endgroup$ – jwimberley May 11 '15 at 12:13
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A spin-spin interaction is really a magnetic moment - magnetic moment interaction, where the magnetic moment of each particle is proportional to spin. [Of course, it might be a chromomagnetic moment - chromomagnetic moment interaction if two quarks are interacting, as they are here.] In any case, the interaction term goes like $\vec S_1 \cdot \vec S_2$. Since magnetic moments generate short-range fields ($\sim 1/r^3$) we want the two particles to be awfully close to each other to interact. In a typical non-relativistic situation (as in charmonium) this will happen if the overlap of the wavefuntions is significant. In the hydrogen atom for instance a quick glance at the orbitals (look up "atomic orbital" in Wikipedia and see some shapes) will convince you that orbitals with $L\ne 0$ give virtually zero overlap.

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