A follow-up question of $\frac{1}{2}\otimes\frac{1}{2}=0⊕1 $ : If I have 4 spin-1/2 particles in my system, how can I use a series of direct sums to represent $\frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2}$ ? How to determine how many singlets, total spin 1 (I think that's 3), and total spin 2s are there? Thanks!
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There is a general formula for the Kronecker multiplication of an arbitrary number of doublets. Applied to your case, it yields $$ \frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2}= 2\oplus 1\oplus 1\oplus 1\oplus 0\oplus 0, $$ a total of 16 states. These multiplicities are specified by the Catalan triangle.
The general formula for n doublets is more transparent in a different notation to label your representations by their dimension 2s+1, instead of the spin s that you use. That is, instead of your s =1/2, use 2, instead of s=0, use 1, etc. $$ \mathbf{2}^{\otimes n} = \bigoplus_{k=0}^{\lfloor n/2 \rfloor}~ \left(\frac{n + 1 - 2k}{n + 1}{n + 1 \choose k}\right)~(\mathbf{n} + \mathbf{1} - \mathbf{2}\mathbf{k})~, $$ where $\lfloor n/2 \rfloor$ is the integer floor function, and the number preceding the boldface irreducible representation dimensionality label indicates multiplicity of that representation in the representation reduction (the Clebsch series).
So, in your case, the formula reads $$ \mathbf{2}^{\otimes 4} = \mathbf{5}\oplus (3) \mathbf{3} +(2)\mathbf{1}, $$ displayed above. The advantage of this notation is the arithmetical multiplicity of states check, automatically provided by taking away the boldface and circles, and providing a grade school arithmetic accounting of states!
Similar general formulas are available for arbitrary representation (spin) additions.
Cosmas Zachos
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