Physics
Spin orbit interaction in nucleus
Step-by-step physics solution: Spin orbit interaction in nucleus
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1. What the student is asking (in plain language)
The student wants a clear, worked‑out description of the nuclear shell model including the spin‑orbit interaction:
- How the single‑particle Hamiltonian is built (central potential + spin‑orbit term).
- How the energy levels are obtained analytically (or at least semi‑analytically) for a simple potential (harmonic‑oscillator or Woods‑Saxon).
- How the extra spin‑orbit term changes the ordering of the levels and therefore produces the observed magic numbers (2, 8, 20, 28, 50, 82, 126).
In addition the student asks for references (books or papers) that actually carry out the calculation rather than just showing a schematic diagram.
Below is a self‑contained derivation that follows the classic textbook treatment, together with a short bibliography of the most useful sources.
2. Step‑by‑step derivation of the shell‑model spectrum with spin‑orbit coupling
2.1 The single‑particle Hamiltonian
In the independent‑particle picture each nucleon moves in an average (mean‑field) potential generated by all the others.
A very common choice (simple enough for analytic work) is the three‑dimensional isotropic harmonic‑oscillator potential
[ V_{\rm HO}(r)=\frac12 m\omega^{2}r^{2}, ]
supplemented by a spin‑orbit term that mimics the strong, short‑range interaction discovered by Mayer and Jensen:
[ V_{\rm so}(r)= - \lambda\; \frac{1}{r}\,\frac{dV_{\rm HO}}{dr}\,\mathbf{l}!\cdot!\mathbf{s} = - \lambda \, \hbar\omega \; \mathbf{l}!\cdot!\mathbf{s}. ]
The minus sign gives the *inverted (i.e. opposite to the atomic case) spin‑orbit coupling that is required to reproduce the nuclear magic numbers.*
(\lambda) is a dimensionless strength parameter (empirically (\lambda\sim 0.1!-!0.3) for nuclei).
Thus the full Hamiltonian for a single nucleon reads
[ \boxed{H = \frac{p^{2}}{2m}+V_{\rm HO}(r) - \lambda \hbar\omega\;\mathbf{l}!\cdot!\mathbf{s}}. ]
Because the potential is central, the good quantum numbers are
- principal oscillator quantum number (N = 2n_{r}+l) (with (n_{r}=0,1,2,\dots)),
- orbital angular momentum (l),
- total angular momentum (j = l\pm \tfrac12) (arising from (\mathbf{j}=\mathbf{l}+\mathbf{s})),
- its projection (m_{j}) (degenerate for a given (j)).
| The eigenfunctions of the first two terms are the familiar harmonic‑oscillator states ( | N,l,m_{l}\rangle). |
2.2 Expectation value of (\mathbf{l}!\cdot!\mathbf{s})
For a given (l) and (j) the operator (\mathbf{l}!\cdot!\mathbf{s}) has a simple eigenvalue:
[ \boxed{\langle \mathbf{l}!\cdot!\mathbf{s}\rangle_{l,j} =\frac12\Bigl[j(j+1)-l(l+1)-\tfrac34\Bigr]}. ]
Thus
| (l) | (j=l+\frac12) (j‑up) | (\langle\mathbf{l}!\cdot!\mathbf{s}\rangle) |
|---|---|---|
| 0 | (1/2) | 0 |
| 1 | (3/2) | (+1/2) |
| 1 | (1/2) | (-1) |
| 2 | (5/2) | (+1) |
| 2 | (3/2) | (-3/2) |
| … | … | … |
The splitting between the two members of a given ((l)) doublet is therefore
[ \Delta E_{\rm so}(l)=\lambda \hbar\omega\, \Bigl[\,\langle\mathbf{l}!\cdot!\mathbf{s}\rangle_{j=l+1/2} -\langle\mathbf{l}!\cdot!\mathbf{s}\rangle_{j=l-1/2}\Bigr] = \lambda \hbar\omega\;(l+1). ]
Because the sign in front of (V_{\rm so}) is negative, the higher‑j member ( (j=l+1/2) ) is lower in energy.
2.3 Energy of a level ((N,l,j))
The harmonic‑oscillator part gives
[ E_{\rm HO}(N)=\hbar\omega\Bigl(N+\frac32\Bigr) =\hbar\omega\Bigl(2n_{r}+l+\frac32\Bigr). ]
Adding the spin‑orbit contribution:
[ \boxed{E(N,l,j)=\hbar\omega\Bigl(N+\frac32\Bigr) -\lambda\hbar\omega\;\langle\mathbf{l}!\cdot!\mathbf{s}\rangle_{l,j}}. ]
All states with the same ((N,l,j)) are ((2j+1))-fold degenerate (the magnetic quantum number (m_{j})).
The total degeneracy of a shell is the sum over all ((l,j)) belonging to that shell.
2.4 Building the spectrum shell by shell
We now list the first few oscillator shells, compute the spin‑orbit shifts, and order the resulting levels by increasing energy.
We take a typical value (\lambda = 0.2) (the exact number only scales the splittings; the ordering is robust for (0<\lambda\lesssim0.5)).
| Oscillator shell (N) | ((l,j)) label | (E_{\rm HO}) (in (\hbar\omega)) | (\langle\mathbf{l}!\cdot!\mathbf{s}\rangle) | (E) (in (\hbar\omega)) |
|---|---|---|---|---|
| 0 | (0s_{1/2}) | ( \frac32) | 0 | (1.5) |
| 1 | (1p_{3/2}) | ( \frac52) | (+1/2) | (2.5-0.2\times0.5=2.4) |
| (1p_{1/2}) | ( \frac52) | (-1) | (2.5+0.2\times1=2.7) | |
| 2 | (2s_{1/2}) | ( \frac72) | 0 | (3.5) |
| (1d_{5/2}) | ( \frac72) | (+1) | (3.5-0.2\times1=3.3) | |
| (1d_{3/2}) | ( \frac72) | (-3/2) | (3.5+0.2\times1.5=3.8) | |
| 3 | (2p_{3/2}) | ( \frac92) | (+1/2) | (4.5-0.2\times0.5=4.4) |
| (2p_{1/2}) | ( \frac92) | (-1) | (4.5+0.2\times1=4.7) | |
| (1f_{7/2}) | ( \frac92) | (+3) | (4.5-0.2\times3=3.9) | |
| (1f_{5/2}) | ( \frac92) | (-2) | (4.5+0.2\times2=4.9) | |
| 4 | (1g_{9/2}) | ( \frac{11}{2}) | (+4) | (5.5-0.2\times4=4.7) |
| (2d_{5/2}) | ( \frac{11}{2}) | (+1) | (5.5-0.2\times1=5.3) | |
| (2d_{3/2}) | ( \frac{11}{2}) | (-3/2) | (5.5+0.2\times1.5=5.8) | |
| (3s_{1/2}) | ( \frac{11}{2}) | 0 | (5.5) | |
| (1h_{11/2}) | ( \frac{11}{2}) | (+5) | (5.5-0.2\times5=4.5) | |
| … | … | … | … | … |
Now order the levels by their actual energies (lowest → highest):
- (0s_{1/2}) (2 nucleons)
- (1p_{3/2}) (4)
- (1p_{1/2}) (2) → first magic number 2+4+2 = 8
- (1d_{5/2}) (6)
- (2s_{1/2}) (2) → second magic number 20 (2+6+6+2 = 16? Actually we need to sum up to the closure of the (N=2) shell; the ordering gives 2+6+2+6 = 16? Let’s be systematic.)
Let’s compute cumulative occupancies using the ordered list:
| Level (ordered) | Degeneracy (2j+1) | Cumulative nucleons |
|---|---|---|
| (0s_{1/2}) | 2 | 2 |
| (1p_{3/2}) | 4 | 6 |
| (1p_{1/2}) | 2 | 8 ← magic 8 |
| (1d_{5/2}) | 6 | 14 |
| (2s_{1/2}) | 2 | 16 |
| (1d_{3/2}) | 4 | 20 ← magic 20 |
| (1f_{7/2}) | 8 | 28 ← magic 28 |
| (2p_{3/2}) | 4 | 32 |
| (1f_{5/2}) | 6 | 38 |
| (2p_{1/2}) | 2 | 40 |
| (1g_{9/2}) | 10 | 50 ← magic 50 |
| (1h_{11/2}) | 12 | 62 |
| (2d_{5/2}) | 6 | 68 |
| (3s_{1/2}) | 2 | 70 |
| (2d_{3/2}) | 4 | 74 |
| … | … |
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