Say we have two identical spin 3/2 particles in the same orbital state. What are the possible total spin? I know that there is a simple formula for adding angular momenta, but this breaks down when the particles occupy the same orbital state. Is there a formula to solve this problem for fermions of any spin?


For fermions, the total state has to be antisymmetric:

$$\Psi(x_1, x_2) = -\Psi(x_2, x_1) $$

where $\Psi$ can be factored into a spatial part and a spin part:

$$\Psi(x_1, x_2) = \psi(x_1, x_2)\chi_{1,2} $$

You've specified that the spatial part have the same orbital function, so:

$$ \psi_{\pm}(x_1, x_2) \equiv \frac 1 {\sqrt 2}[f(x_1)f(x_2) \pm f(x_2)f(x_1)]$$

where I've explicitly shown the (anti)symmetric form. Note that the antisymmetric part vanishes for all $x_i$ so the spatial part must be symmetric:

$$ \psi_+ = \frac 1 {\sqrt 2}[f(x_1)f(x_2) + f(x_2)f(x_1)]$$

That mean $\chi_{1, 2}$ must be an antisymmetric.

Since the particles are spin 3/2, we need to look at antisymmetric combinations$^1$ of $|m_1\rangle$ and $|m_2\rangle$ in terms of $|J, M\rangle$, the total spin state(s):

$$ |\frac 3 2\rangle|\frac 1 2\rangle - |\frac 1 2\rangle|\frac 3 2\rangle = \sqrt 2 |2, 2\rangle $$

$$ |\frac 3 2\rangle|-\frac 1 2\rangle - |-\frac 1 2\rangle|\frac 3 2\rangle = \sqrt 2 |2, 1\rangle $$

$$ |\frac 3 2\rangle|-\frac 3 2\rangle - |-\frac 3 2\rangle|\frac 3 2\rangle = |2,0\rangle + |0, 0\rangle $$

$$ |\frac 1 2\rangle|-\frac 1 2\rangle - |-\frac 1 2\rangle|\frac 1 2\rangle = |2,0\rangle - |0, 0\rangle $$

The remaining 2 nontrivial combinations can be found by subbing $m_1 \rightarrow -m1$, $m_2 \rightarrow -m_2$, and $M \rightarrow -M$.

From here you invert the equations to find the pure $|J, M\rangle$ states, and those are the eigenstates of total spin.

[1] To find the antisymmetric combinations of $m_1$ and $m_2$, you start with Clebsch-Gordan coefficients, $C_{\frac 3 2 m_1 \frac 3 2 m_2 J M}$. For example, the various:

$$c_{\frac 3 2 +\frac 3 2 \frac 3 2 -\frac 3 2 J 0} $$


$$ |\frac 3 2\rangle|-\frac 3 2\rangle= \frac 1 2 |0, 0\rangle + \sqrt{\frac 9{20}}|1, 0\rangle + \frac 1 2 |2, 0\rangle + \sqrt{\frac 1 {20}}|3, 0\rangle $$

Note that if you interchange $m_1$ and $m_2$:

$$ |-\frac 3 2\rangle|+\frac 3 2\rangle= -\frac 1 2 |0, 0\rangle + \sqrt{\frac 9{20}}|1, 0\rangle - \frac 1 2 |2, 0\rangle + \sqrt{\frac 1 {20}}|3, 0\rangle $$

So it is clear that the (anti)symmetric combinations have J odd (even).

The general principle is as follows: when taking the tensor product of 2 identical representations, you use Young diagrams. This gives you a totally symmetric combination, and a totally antisymmetric combination. The Hook length formula can then be used to compute the dimensions of these two irreducible representations. For spinors, vectors, and spin 3/2 (or 4-vectors), and spin-2, you get the following (respectively):

$$ {\bf 2} \otimes {\bf 2} = {\bf 3}_S + {\bf 1}_A $$ $$ {\bf 3} \otimes {\bf 3} = {\bf 5}_S + {\bf 3}_A $$ $$ {\bf 4} \otimes {\bf 4} = {\bf 10}_S + {\bf 6}_A $$ $$ {\bf 5} \otimes {\bf 5} = {\bf 15}_S + {\bf 20}_A $$

So in our spin 3/2 case where we are looking only at antisymmetric combinations, we have six dimensions corresponding to:

$$ {\bf 6} = {\bf 5} \oplus {\bf 1} $$

corresponding to spin-2 and spin-0, aka tensor and scalar (even $J$).

The symmetric combinations are:

$$ {\bf 10} = {\bf 7} \oplus {\bf 3} $$

corresponding to the spin-3 and spin-1 (vector) (odd $J$).

The thing about the Hook length formula is that it is all-powerful, and will give the dimension of any irreducible subspace over any field in any number of dimensions.

Perhaps it is too much? By inspection, if you combine 2 identical $n$ dimensional representations, the symmetric combination has $T_n$ dimensions, and the antisymmetric combination has $T_{n-1}$ dimensions, where $T_n$ are the Triangular Numbers:

$$ T_n \equiv \frac 1 2 n(n+1) $$

Moreover, for an odd (even) dimensional spin, the (anti)symmetric combinations are all even $J$, and for an even (odd) dimensional spin, the (anti)symmetric combinations are all odd $J$.

  • $\begingroup$ I have two questions. First, how did you write the total spin vectors? Was that with the tensor product between the |m1> |m2> vectors? $\endgroup$ – Michael O'Brien Dec 7 '18 at 17:18
  • $\begingroup$ Second, if we aren’t interested in the eigenstates in the |J,M> basis, is there a quick and dirty way to deduce the possibilities for total spin? $\endgroup$ – Michael O'Brien Dec 7 '18 at 17:20

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