D.69 Momentum of shells

Table 12.1 was originally taken from [34], who in turn took it from the book of Mayer and Jensen. However, the final table contains three typos, as can be seen from the fact that in three cases the numbers of states do not add up to the correct total. (The errors are: for 3 particles with spin 9/2, the 13/2 combined state is omitted, for 4 particles with spin 9/2, the spin 8 state should be double, and for 4 particles with spin 11/2, a spin 7 (double) state is missing. Similarly, [5, p. 140] has the same missing 13/2 combined state, and in addition for 3 particles with spin 7/2, there is a 1/2 state that should not be there.)

So table 12.1 was instead computer-generated, and should therefore be free of typos. Since the program had to be written anyway, some more values were generated and are in table D.1.

Deducing the table using Clebsch-Gordan coefficients would be a messy exercise indeed. A simpler procedure, [29], will here be illustrated for the example that the number of fermions is $I=3$ and the angular momentum of the single-particle states is $j^{\rm p}=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. Then the possibilities for the single-particle angular momentum in the $z$-direction are $m^{\rm p}=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $-\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $-\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, and $-\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. So there are 6 different one particle states, and these will give rise to $6!/3!(6-3)!=20$ different antisymmetric states for 3 particles, chapter 5.7.

Table D.1: Additional combined angular momentum values.

The combination states can be chosen to have definite values of the combined angular momentum $j$ and momentum in the $z$-direction $m$. In the absence of any antisymmetrization requirements, that can be seen from the way that states combine using Clebsch-Gordan coefficients. And these states of definite combined angular momentum must either be antisymmetric and allowable, or symmetric and not allowed. The reason is that exchanging fermions does not do anything physically, since the fermions are identical. So the angular momentum and particle exchange operators commute. Therefore, the eigenstates of the angular momentum operators can also be taken to be eigenstates of the particle exchange operators, which means either symmetric (eigenvalue 1) or antisymmetric (eigenvalue $-$1).

Let $m$ be the total magnetic quantum number of the 3 fermions in any combination of $j^{\rm p}=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ single-particle states. First note that $m$ is the sum of the three $m^{\rm p}$ values of the individual particles. Next, the highest that $m^{\rm p}$ can be is $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, but the fermions cannot all three be in the same $m^{\rm p}=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ state, only one can. Three fermions need three different states, so the highest the combined $m$ can be is $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em... ...riptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. This triplet of values of $m^{\rm p}$ gives exactly one antisymmetric combination of states with $m=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. (There is only one Slater determinant for three different given states, chapter 5.7). Since the combined angular momentum of this state in any arbitrary direction can never be observed to be more than $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, because that would violate the above argument in a rotated coordinate system, it must be a $j=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ state. The first conclusion is therefore that the angular momenta cannot combine into a total greater than $j=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. And since $j$ cannot be less than $m$, there must be states with $j=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$.

But note that if $j=m=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ is a valid combination of single-particle states, then so should be the states with $j=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ for the other values of $m$; these can be thought of as fully equivalent states simply oriented under a different angle. That means that there are a total of 10 combination states with $j=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, in which $m$ is any one of $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, ..., $-\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$.

Next consider what combinations have $m=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. The only combination of three different $m^{\rm p}$ values that adds up to $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ is $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em... ...riptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. So there is only one combined state with $m=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. Since it was already inferred above that there must be one such state with $j=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, that must be the only one. So apparently there is no state with $j=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$: such a state would show up as a second $m=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ state under the right orientation.

There are two independent possibilities to create a triplet of different states with $m=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$: $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em... ...riptfont0 3}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ or $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em... ...riptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. One combination of such a type is already identified as being a $j=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ state, so the second must correspond to a $j=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ state. Since the orientation should again not make a difference, there must be a total of 6 such states, one for each of the different values of $m$ in the range from $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ to $-\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$.

There are three ways to create a triplet of states with $m=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$: $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em... ...riptfont0 5}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em... ...riptfont0 3}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, and $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em /\kern-.2em... ...riptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. Two of these are already identified as being $j=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ and $j=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, so there must be one set of 4 states with $j=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$.

That makes a total of 20 states, so there must not be any states with $j=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. Indeed, there are only three ways to produce $m=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$: $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em... ...riptfont0 5}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 5}\kern-.2em /\kern-.2em... ...riptfont0 3}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, and $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em /\kern-.2em... ...riptfont0 3}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, and each of these three states is already assigned to a value of $j$.

It is tricky, but it works. And it is easily put on a computer.

For bosons, the idea is the same, except that states with equal values of $m^{\rm p}$ can no longer be excluded.