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Question

Consider a system of four non-identical spin 1/2 particles. Find the possible values for the total spin and state the number of eigenstates for each of these.

Attempt

So I coupled S1 and S2 to get S12 and I also coupled S3 and S4 to get S34. I will then couple S12 and S34 to get S1234: (states are in the form (S1, S2, S12, m))

Eigenstates for S12: {(1/2, 1/2, 1, 1),(1/2, 1/2, 1, 0),(1/2, 1/2, 1, -1),(1/2, 1/2, 0, 0)}

Eigenstates for S34: {(1/2, 1/2, 1, 1),(1/2, 1/2, 1, 0),(1/2, 1/2, 1, -1),(1/2, 1/2, 0, 0)}

Eigenstates for S1234: {(1,1,2,2),(1,1,2,1),(1,1,2,0),(1,1,2,-1),(1,1,2,-2),(1,1,1,1),(1,1,1,0),(1,1,1,-1),(1,1,0,0),(1,0,1,1),(1,0,1,0),(1,0,1,-1),(0,1,1,1),(0,1,1,0),(0,1,1,-1),(0,0,0,0)}

That would make 16 different states but i'm not sure about the last 7 states (it disagrees with the answers my friends have). Cheers!

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    $\begingroup$ Your answer is correct: $\frac{1}{2} \otimes \frac{1}{2} \otimes \frac{1}{2} \otimes \frac{1}{2} = \left( 0 \oplus 1 \right) \otimes \left( 0 \oplus 1 \right) = 0^2 \oplus 1^3 \oplus 2$. Two singlets, three triplets, and one quintet giving 16 states in total. Note that these are only eigenstates of non-identical particles because some of these states belong to higher-dimensional representations of the symmetric group; they transform to one another after exchange. This is called parastatistics. $\endgroup$
    – Praan
    Commented Dec 7, 2015 at 18:29
  • $\begingroup$ Possible duplicate of Problem counting spin states $\endgroup$
    – Cosmas Zachos
    Commented Apr 14, 2016 at 20:48

1 Answer 1

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As Praan confirmed, you are fine. Your result is the n=4 case of the general expression for composing spin 1/2 doublets, (so here denoted by their dimensionality, 2), $$ {\mathbf 2}^{\otimes n} = \bigoplus_{k=0}^{\lfloor n/2 \rfloor}~ \Bigl( {n+1-2k \over n+1} {n+1 \choose k}\Bigr)~~({\mathbf n}+{\mathbf 1}-{\mathbf 2}{\mathbf k})~, $$ where $\lfloor n/2 \rfloor$ is the integer floor function--the largest integer smaller than the argument, and the number preceding the boldface irreducible representation dimensionality (2 j +1) label (here just 5, 3 and 1 ) indicates multiplicity of that representation in the C-G reduction, here 1, 3, and 2, so that 16=5+9+2 states.

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