# Identical Spin Fermions in the same orbital state: Finding total spin

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:

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

give:

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

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 $$J$$.

• I have two questions. First, how did you write the total spin vectors? Was that with the tensor product between the |m1> |m2> vectors? – Michael O'Brien Dec 7 '18 at 17:18
• 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? – Michael O'Brien Dec 7 '18 at 17:20