Number of Angular Momentum States (3 particles)

[Vanadium 50]

Thanks. I was thinking about when one has only one singlet.

 

[PeroK]

You're right, on that.

The fully antisymmetric state is unique, by inspection. Every other bit of my argument is invalid. (But I still suspect a theorem)

Here's a symmetry argument for the three spin-1 case.

First, we analyse by $j$ values (as before):

$j = 3$ has $7$ levels
$j = 2$ has $5$ levels
$j = 1$ has $3$ levels
$j = 0$ has $1$ level

There are numerous ways to meet the condition of degeneracies, assuming a common degeneracy across each level.

Then, we analyse by $m$ value:

$m = 3$ is $1$ dimensional
$m = 2$ is $3$ dimensional (cycles of $110$)
$m =1$ is $6$ dimensional (cycles of $100$ and $11-1$
$m = 0$ is $7$ dimensional (cycles of $10-1$ and $000$)

Now, $J^2$ is symmetric and preserves the $m$-value, hence preserves the above dimensionality, in terms of $|j m \rangle$ eigenstates. That might be your theorem:

There is only one $m = 3$ state, hence $j = 3$ has degeneracy $1$ - i.e. no degeneracy (as previously established).

$m = 2$ is $3$ dimensional. One dimension is for $j = 3$, leaving two for $j = 2$. Hence $j = 2$ has degeneracy $2$.

Looking at $m = 1$ implies $j = 1$ has degeneracy $3$.

And $m = 0$ implies $j =0$ has degeneracy $1$.

Looking at the negative values gives the same answer. So, we have, as expected:

$j = 3$ has $7$ levels with degeneracy $1$
$j = 2$ has $5$ levels with degeneracy $2$
$j = 1$ has $3$ levels with degeneracy $3$
$j = 0$ has $1$ level degeneracy $1$

The same argument also works for the established result in the four-electron case.

 

[Vanadium 50]

So the theorem is that for addition of 2 or 3 integral angular momenta j, there is exactly one singlet. For more angular momenta the number of singlets is larger: ten j = 1 have 603 singlets.

For four j = 1, the number of singlets is 2j + 1. For five, it's (5j² = 5j + 2)/2.