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This is something that relates closely to my long-ago PhD research, and it's a point that I never satisfied myself thoroughly on. I have a copy of a standard text that covers this general topic, de Shalit and Feshbach, and although they treat it in detail, the technique by which they approach this more specific issue is algebraic, which leaves me not feeling like I really understand the underlying physics. I can trace through the steps of their mathematical reasoning, but I get bogged down in all the Wigner D-matrices and lose the thread of what's really happening.

The model is referred to as the unified model or collective model of nuclear structure. It's "unified" because it describes the nucleus in terms of single-particle excitations coupled to collective rotation of the nucleus as a whole. An open-access paper that covers this sort of thing is Frauendorf. We have a nucleus with a prolate shape, so that in the body-fixed frame, there is a symmetry with respect to an axis $z$, and also in the plane perpendicular to $z$, and therefore the single-particle states are eigenstates of $j_z$ (also notated $\Omega$) and parity. For simplicity, let's restrict ourselves to the rotational band built on the ground state of an odd nucleus. Then the classic results, verified by experiment back in the 50's, are:

(1) The spins are $I=|K|$, $|K|+1$, ..., where $K=j_z$, i.e., the lowest angular momentum for the nucleus as a whole (odd particle plus collective rotation) is the same as the $z$ component of the odd particle's angular momentum.

(2) The energies of the states in the band are

$$E=\epsilon+\frac{\hbar^2}{2\mathscr{I}}\left[I(I+1)-K^2+a(-1)^{I+1/2}(I+1/2)\right],$$

where $\epsilon$ is the energy of the single-particle state, $\mathscr{I}$ is the moment of inertia for collective rotation about an axis perpendicular to $z$, and $a=0$ except when $|K|=1/2$.

These results hold under a certain set of assumptions (de Shalit and Feshbach, p. 403): (a) the nucleus has the symmetries described above; (b) the moment of inertia is large enough so that $\hbar^2/2\mathscr{J}$ is small compared to the differences between $\epsilon$ values; (c) approximate separability of the Euler angles from the degrees of freedom in the body-fixed frame; (d) slow rotation, so that Coriolis forces are negligible.

What bugs me is the following. I'm used to visualizing this sort of thing in terms of a diagram of the single-particle energies as functions of the deformation, e.g., in the Nilsson model. Here's a diagram that I computed using my implementation of the model:

energies of single-particle states as a function of deformation

Let's say the odd particle number is 11. For zero deformation, the Fermi level is such that the odd particle lies in the $j=5/2$ shell. As we add a little deformation $\delta$, these 6 levels split into 3 pairs that are degenerate due to time-reversal symmetry. On the plot, they're labeled by the quantum numbers $j_z{}^\pi=1/2^+$, $3/2^+$, and $5/2^+$. For $\delta\ne 0$, these states are no longer states of good $\textbf{j}$, but their wavefunctions are still dominated by $j=5/2$. In the ground-state band, the odd particle has $j_z=3/2$ and $j\approx 5/2$. The collective rotation can't have any component along the symmetry axis. Let's call the collective axis of rotation $x$, and the collective angular momentum $\textbf{R}$. Then the total angular momentum is

$$\textbf{I}=\textbf{R}+\textbf{j}.$$

Classically, this gives $I^2=R^2+j^2+2Rj_x$. Semiclassically, I would expect

$$\langle \textbf{I}^2 \rangle\approx \langle \textbf{R}^2 \rangle + \langle \textbf{j}^2 \rangle + 2\langle R_x j_x \rangle.$$

In the limit described by approximation (d) above, and for $|K|\ne1/2$, I think the single-particle states are equal mixtures of the two two-reversed states, so that $\langle j_x \rangle\approx 0$. I then expect

$$\langle \textbf{I}^2 \rangle\gtrsim \langle \textbf{R}^2 \rangle + \langle \textbf{j}^2 \rangle.$$

And since the single-particle state is predominantly a state of $j=5/2$, we should have something like

$$I(I+1) \gtrsim j(j+1)$$

which comes out to be 8.75 for this value of $j$. But experiments actually verify rule (1) above, so that in reality, the ground state has $I(I+1)=3.75$, which is much smaller. There is a gross discrepancy bewteen 3.75 and 8.75, and for large values of $j$ in heavy nuclei, the discrepancy can be bigger --- as big as you like for large $j$ and small $j_z$.

So what is wrong with this estimate, and what happens to the $j_x$ and $j_y$ components of the single-particle angular momentum? It doesn't seem possible for these components to be canceled to this extent by $R_x$ and $R_y$, both for the symmetry reasons described above and because the amount of zero-point collective rotation in the ground state doesn't seem big enough. As a semiclassical estimate of the latter, let's use the uncertainty relation

$$\Delta I_x \Delta I_y \ge \frac{\hbar}{2}|\langle I_z\rangle|.$$

Given a single-particle state with $j_z=\pm K\hbar$, I get (putting in $\hbar\ne1$ to facilitate comparison with classical physics)

$$I^2 \gtrsim R_x^2+R_y^2 \gtrsim R_x^2+(K\hbar^2/2R_x)^2 \ge K\hbar^2.$$

So this suggests that the magnitude of the collective angular momentum in the ground state, due to zero-point rotation, is $\gtrsim \sqrt{K}\hbar$. Now I don't have a formal proof that this inequality is nearly saturated in the ground state, but I'm pretty sure intuitively that it is, so that $R\sim \sqrt{K}\hbar$, and this can be considerably too small to be able to cancel out the single-particle angular momenta $j_x$ and $j_y$.

Note that when the deformation is zero (so that assumption (b) is violated), the ground-state angular momentum is equal to $j$, not $j_z$.

References

Frauendorf, "Beyond the Unified Model," https://arxiv.org/abs/1710.01210

de Shalit and Feshbach, Theoretical Nuclear Physics, Volume I, ch. VI

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  • $\begingroup$ The first thing that comes to mind $I(I+1)=j(j+1)-2j=j(j-1)$, which gives exactly 3.75 for $j=5/2$ $\endgroup$ – Alex Trounev Jun 18 at 17:00

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