Past exam of the advanced quantum physics course of the University of Cambridge / 2015 / Paper 1 / 1

The Pauli exclusion principle allows each one-particle magnetic substate only once, since the fermions share the other quantum numbers. To maximize the total angular momentum projection, occupy $m_j=5/2,3/2,1/2,-1/2$, givingThe corresponding Slater determinant is a highest-weight vector: every attempted raising operation either tries to exceed $j$ or duplicates an occupied substate and vanishes. Therefore it belongs to a multiplet with $\boxed{J_{\max }=4}$, not the distinguishable-particle value $4j=10$. Equivalently, four particles in a six-state shell have two holes whose maximum angular momentum projection is $5/2+3/2=4$. The general maximum total angular momentum of equivalent fermions is $J_{\max }=Nj-N(N-1)/2$ for $N\le 2j+1$.

At incident wave number $k$, low kinetic energy means $ka\ll 1$: the de Broglie wavelength is much larger than the scattering region. A generic finite-range scatterer can then be described to leading order without resolving its angular structure. In partial-wave scattering, this is leading S-wave scattering. The $\ell =0$ spherical harmonic is constant, whereas the centrifugal term ${\hslash }^2\ell (\ell +1)/(2m{r}^2)$ impedes penetration to the potential for $\ell >0$.

For a central potential with regular threshold behaviour, the finite-range partial-wave threshold law gives ${\delta }_{\ell }=O(k^{2\ell +1})$. Thuswhich is independent of scattering angle. For finite nonzero scattering length $a_s$, $f\to -a_s$ and $d\sigma /d\Omega \to a_s^2$ is isotropic. The potential range $a$ and scattering length $a_s$ are different quantities. The claim describes the generic leading S-wave scattering limit; tuned vanishing S-wave scattering amplitude or exceptional threshold resonances require separate analysis rather than a universal angular conclusion.

Closed $1s^2$ and $2s^2$ subshells have $L=S=0$. In the open $2p^4$ subshell, three electrons can have parallel spins in $m_{\ell }=-1,0,1$, and the fourth must have the opposite spin, giving maximum $S=1$. Treating the shell as two $p$ holes gives the same allowed terms as $p^2$; at this spin, antisymmetry permits $L=1$, rather than $L=2$. One can also see $M_L=1$ by placing the fourth electron in $m_{\ell }=1$.

Hund's rules therefore select the ${}^{3}P$ term. Since four electrons exceed half filling of the six-state $p$ subshell, its lowest atomic fine-structure level has the largest $J=L+S=2$:Its zero-field magnetic degeneracy is $2J+1=5$. The experimentally tabulated neutral-oxygen ground level has precisely this term and $J$.^[ref]