One simple way of detailing the very basic structure of the nuclear shell model involves placing the nucleons in a 3D isotropic oscillator. It's easy to show that the energy eigenvalues are $E = \hbar\omega\left(n+\frac{3}{2}\right)$ with a degeneracy of $D(n) = \frac{1}{2}(n+1)(n+2)$, where $n = n_1 + n_2 + n_3 = a^\dagger_1 a_1 + a^\dagger_2 a_2 + a^\dagger_3 a_3$.

However, in the book I am using presently (Ring, The Nuclear Many-Body Problem) and in several other texts, the authors write the quantum numbers as $n=2(n_r-1)+\ell$ (or, some places, $n=2n_r+\ell$). This seems odd, since we're trading 3 quantum numbers ($n_1, n_2, n_3$) for 2 ($n_r, \ell$). I suppose this is fine if we're only using $n_r$ and $\ell$ to enumerate our hypothesized nuclear energy levels, but in any case it's not clear to me why you would make the choice $n = 2n_r + \ell$ to do this (besides, obviously, that it does an okay job of reproducing the lower magic numbers). These authors trot the expression out without any explanation, and I'd like to know why this expression is used. I'm still not clear on the significance of $n_r$ and $\ell$ in nuclei, so it's puzzling.

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    $\begingroup$ Just concerning the "loss" of a quantum number: the author is assuming that you know that $l_z \in \{-l, -l+1, \dots l-1, l\}$ is also a good quantum number of the system, but that states that differ only in $l_z$ are degenerate in the absence of a magnetic field. SO you are trading three for three, but one of the later set has been assumed because it is not important yet. $\endgroup$ – dmckee May 31 '11 at 14:33
  • $\begingroup$ See e.g., wikipedia page en.wikipedia.org/wiki/… or on-line notes quantummechanics.ucsd.edu/ph130a/130_notes/node244.html $\endgroup$ – Qmechanic May 31 '11 at 14:56
  • $\begingroup$ You should answer your own question. I think that we the links provided you have all the elements to elaborate it. $\endgroup$ – rmhleo Aug 18 '15 at 14:09

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