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?

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.