12.9 Momentum of partially filled shells

One very important case of combining angular momenta occurs for both electrons in atoms and nucleons in nuclei. In these problems there are a number of identical fermions in single-particle states that differ only in the net (orbital plus spin) momentum in the chosen $z$-​direction. Loosely speaking, the single-particle states are the same, just at different angular orientations. Such a set of states is often called a shell. The question is then: what combinations of the states are antisymmetric with respect to exchange of the fermions, and therefore allowed? More specifically, what is their combined net angular momentum?


Table 12.1: Possible combined angular momentum of identical fermions in shells of single-particle states that differ in magnetic quantum number. The top shows odd numbers of particles, the bottom even numbers.
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...&2&1&1&&1\\
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The answer is given in table 12.1, {D.68}. In it, $I$ is the number of fermions in the shell. Further $j^{\rm p}$ is the net angular momentum of the single-particle states that make up the shell. (Or the azimuthal quantum number of that angular momentum really.) Similarly the values of $j$ indicate the possible net angular momentum quantum numbers of all $i$ fermions combined. The main body of the table lists the multiplicity of sets with the given angular momentum. Note that the table is split into odd and even numbers of particles. That simplifies the presentation, because odd numbers of particles produce only half-integer net angular momentum, and even numbers only integer net angular momentum.

For example, consider a single particle, $I$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, in a set of single-particle states with angular momentum $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. For a single particle, the combined momentum $j$ is simply the single particle momentum $j^{\rm p}$, explaining the single 1 in the $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ column. But note that the 1 stands for a set of states; the magnetic net quantum number $m^{\rm p}$ of the single particle could still be any one of $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, ..., $-\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. All the ten states in this set have net angular momentum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$.

Next assume that there are two particles in the same $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ single-particle states. Then if both particles would be in the $m^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ single-particle state, their combined angular momentum in the $z$-​direction $m$ would be 2 $\times$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ $\vphantom0\raisebox{1.5pt}{$=$}$ 9. Following the Clebsch-Gordan derivation shows that this state would have combined angular momentum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 9. But the two identical fermions cannot be both in the $m^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ state; that violates the Pauli exclusion principle. That is why there is no entry in the $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 9 column. If the first particle is in the $m^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ state, the second one can at most be in the $m^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ state, for a total of $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8. More precisely, the particles would have to be in the antisymmetric combination, or Slater determinant, of these two states. That antisymmetric combination can be seen to have combined angular momentum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8. There are other combinations of states that also have $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8, but values of $m$ equal to 7, 6, ..., $\vphantom0\raisebox{1.5pt}{$-$}$8, for a total of 17 states. That set of 17 states is indicated by the 1 in the $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8 column.

It is also possible for the two $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ particles to combine their angular momentum into smaller even values of the total angular momentum $j$. In fact, it is possible for the particles to combine their angular momenta so that they exactly cancel one another; then the net angular momentum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. That is indicated by the 1 in the $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 column. Classically you would say that the momentum vectors of the two particles are exactly opposite, producing a zero resultant. In quantum mechanics true angular momentum vectors do not exist due to uncertainty of the components, but complete cancelation is still possible.

The $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 set consists of just one state, because $m$ can only be zero for a state with zero angular momentum. The entire table row for two $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ particles could in principle be derived by writing out the appropriate Clebsch-Gordan coefficients. But that would be one very big table.

If there are five $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ particles. they can combine their angular momenta into quite a wide variety of net angular momentum values. For example, the 2 in the $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ column indicates that there are two sets of states with combined angular momentum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. Each set has 6 members, because for each set $m$ can be any one of $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, ..., $-\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. So there are a total of 12 independent combination states that have net angular momentum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$.

Note that a shell has $2j^{\rm p}+1$ different single-particle states, because the magnetic quantum number $m^{\rm p}$ can have the values $j^{\rm p}$, $j^{\rm p}{-}1$, ..., $\vphantom0\raisebox{1.5pt}{$-$}$$j^{\rm p}$. Therefore a shell can accommodate up to $2j^{\rm p}+1$ fermions according to the exclusion principle. However, the table only lists combined angular momentum values for up to $j^{\rm p}+\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ particles. The reason is that any more is unnecessary. A given number of holes in an otherwise filled shell produces the same combined angular momentum values as the same number of particles in an otherwise empty shell. For example, two fermions in a $j^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ shell, (zero holes), have the same combined angular momentum as zero particles: zero. Indeed, those two fermions must be in the antisymmetric singlet state with spin zero. In general, a completely filled shell has zero angular momentum and is spherically symmetric.


Table 12.2: Possible combined angular momentum of identical bosons.
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The same situation for identical bosons is shown in table 12.2. For identical bosons there is no limit to the number of particles that can go into a shell. The table was arbitrarily cut off at 9 particles and a maximum spin of 18.