12.8 Some important results

This section gives some results that are used frequently in quantum analysis, but usually not explicitly stated.

First a note on notations. It is fairly common to use the letter $j$ for orbital angular momentum, $s$ for spin, and $j$ for combinations of orbital and angular momentum. This subsection will follow these conventions were appropriate.

If all possible angular momentum states are filled with a fermion, the resulting angular momentum is zero and the wave function is spherically symmetric. For example, consider the simplified case that there is one spinless fermion in each spherical harmonic at a given azimuthal quantum number $l$. Then it is easy to see from the form of the spherical harmonics that the combined wave function is independent of the angular position around the $z$-​axis. And all spherical harmonics at that $l$ are filled whatever you take to be the $z$-​axis. This makes noble gasses into the equivalent of billiard balls. More generally, if there is one fermion for every possible direction of the angular momentum, by symmetry the net angular momentum can only be zero.

If a spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ fermion has orbital angular momentum quantum number $l$, net (orbital plus spin) angular momentum quantum number $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l+{\textstyle\frac{1}{2}}$, and net momentum in the $z$-​direction quantum number $m_j$, its net state is given in terms of the individual orbital and spin states as:

\big\vert j\:m_j\big\r...
+ \sqrt{\frac{j-m_j}{2j}}Y_l^{m_j+\frac12}{\downarrow}

If the net spin is $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l-{\textstyle\frac{1}{2}}$, assuming that $l$ $\raisebox{.3pt}{$>$}$ 0, that becomes

j=l-{\textstyle\frac{1}{2}}: \quad
\big\vert j\:m_j\big\...
+ \sqrt{\frac{j+1+m_j}{2j+2}}Y_l^{m_j+\frac12}{\downarrow}

Note that if the net angular momentum is unambiguous, the orbital and spin magnetic quantum numbers $m$ and $m_s$ are in general uncertain.

For identical particles, an important question is how the Clebsch-Gordan coefficients change under particle exchange:

\langle j_{ab} m_{ab} \vert\big\vert j_a\:m_a\big\rangle \...
...\big\vert j_b\:m_b\big\rangle \big\vert j_a\:m_a\big\rangle

For $j_a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j_b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, this verifies that the triplet states $j_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 are symmetric, and the singlet state $j_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 is antisymmetric. More generally, states with the maximum net angular momentum $j_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j_a+j_b$ and whole multiples of 2 less are symmetric under particle exchange. States that are odd amounts less than the maximum are antisymmetric under particle exchange.

When the net angular momentum state is swapped with one of the component states, the relation is

\lefteqn{\langle j_{ab} m_{ab} \vert\big\vert j_a\:m_a\big\r...
...g\rangle \big\vert j_b\:\rule[2.5pt]{5pt}{.5pt}m_b\big\rangle

This is of interest in figuring out what states produce zero net angular momentum, $j_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. In that case, the right hand side is zero unless $j_b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j_a$ and $m_b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-m_a$; and then $\langle{j}_am_a\vert\big\vert\:0\big\rangle \big\vert j_a\:m_a\big\rangle $ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. You can only create zero angular momentum from a pair of particles that have the same square angular momentum; also, only product states with zero net angular momentum in the $z$-​direction are involved.