Suppose there are two electrons $e_1$ and $e_2$ with spin $s_1$ and $s_2$ respectively. The electron has spin $\frac{1}{2}$ thus the spin projection $m_s \in \{-\frac{1}{2},\frac{1}{2}\}$. I know that they can form a triplet and a singlet because $2 \otimes 2 = 3 \oplus 1$. But I'd like to see a derivation of this. I know one way to do this is to start with the up-up-state and use the ladder operator on it to get the other two triplet states. However, this seems a bit pedestrian and I think I would understand group theory much better if I was to see an actual group theory approach. I'm hoping something along the way of calculating a matrix that is the representation $D$ which can be reduced to $D=D^{(1)}\oplus D^{(2)}$. And from there it is hopefully possible to derive the triplet and singlet with the coresponding sign in the superposition state.

As I'm having difficulties understand this concept I think I also have problems expressing what I really want - so please, if I made myself unclear at any point, let me know.

In the link that was posted in the first comment, I found my basic understanding problem. We actually have 5 ways to represent the tensorial product:
$2\otimes 2 = 1 \oplus 1 \oplus 1 \oplus 1 = 2 \oplus 2 = 2 \oplus 1 \oplus 1 = 3 \oplus 1 = 4 $ Now one way to find out, that it is a triplet, is that, as mentioned before, we find that we can get spins with $m_s = 1,0,-1$. Therefore only the triplet makes sense. We can verify this by using $S_-$ twice on $|1 1>$.

I struggle with the following: Out of physical context, I think we should still be able to come to the same result without making use of the ladder operator trick or other clever suggestions. If we chose a representation for the spin-$\frac{1}{2}$ particles, the tensor product should yield the triplet and singulet structure. I just have no idea, what that representation looks like. Or does such thing not exist?

I tried looking at the $S_z$ projection operator. The compound oeprator is just the sum of the operators operating on each state. $S_z = S^1_z + S^2_z = S^1_z \otimes 1 + 1 \otimes S^2_z$
This gives me $\begin{pmatrix}2 & 0 & 0 &0 \\ 0 & 0 & 0&0\\0&0&0&0\\0&0&0&-2\end{pmatrix}$
From the looks of it, the first row is the up-up-state and the last row corresponds to the down-down state. The $m_s=0$ projections are in the second and third row. Now, how can I derive the triplet and singlet state from here?

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    $\begingroup$ I think you need to read an accessible text on group theory for physicists (Sattinger&Weaver, Fulton&Harris). There, you need to reach topics related to the tensor products of $su(2)$ representations. This is what the addition of angular momentum in QM is. However, for a warm up you can begin with something like this. $\endgroup$ – mavzolej Apr 5 '17 at 23:29
  • $\begingroup$ Thanks, that was an interesting read. The fact that he mentioned there are actually 5 possibilities for direct sums that the tensor product could be represented with, dropped the penny. I update my question above, can you or someone else verfiy this? $\endgroup$ – infinitezero Apr 6 '17 at 8:48

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