0
$\begingroup$

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!

$\endgroup$
2

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ @ Cosmas Zachos Thank you so much for the explanation! I'm wondering how many subspaces are there in this case? Are there 3 orthogonal subspaces (correspond to spin 0,1 and 2) of the total Hilbert space? Or are there 5 subspaces, each of which corresponds to a possible number of 'spin-up' particles (0,1,2,3,4)? Thanks!! $\endgroup$
    – ZR-
    Sep 11, 2020 at 16:06
  • $\begingroup$ As indicated, there are 6 subspaces in the Clebsch reduction:, each corresponding to each representation. So, one for the spin 2 quintet, three for the spin 1 triplets, and two for the spin singlets. Members for each representation transform among themselves completely ignorant of what happens to other subspaces. $\endgroup$ Sep 11, 2020 at 16:44

Not the answer you're looking for? Browse other questions tagged or ask your own question.