quantum mechanics - Possible spin states?

Given a system of two particles with spin up and down, I have troubles to understand the possible states of this system.

I would have normally thought, that the possible states are the tensor products of the states of each particle with one of the other. This would have given us the states that are listed in Wikipedia under: substituting in the four basis states. So why do we define these new states(three with total spin $1$ and one with total spin $0$) that are given by triplet and singlet representations? So maybe you don't know what I am asking so I repeat my question: What is the meaning of these triplet and singlet states, because I would have thought that the four states listed above are the possible states of a system? The reference to what I am talking about: wikipedia reference

Answer

If we consider two spin $1/2$ particles with spin up and spin down states, then there are four possibilities in total: \begin{equation} \begin{array}{cccc} |++\rangle \; ,& |+-\rangle \; ,& |-+\rangle \; ,& |--\rangle \end{array} \end{equation} where this notation means for instance: \begin{equation} |+-\rangle = |s_1=1/2,m_1=1/2;s_2=1/2,m_2=-1/2\rangle \end{equation} We will suppose that the system consisting of both particle has zero orbital angular momentum and let $S_z$ denote that operator acting on the system. Then it is very easy to evaluate the following eigenvalue equations: \begin{align} S_z|++\rangle &=\hbar|++\rangle \\ S_z|+-\rangle &=0|+-\rangle\\ S_z|-+\rangle &=0|-+\rangle \\ S_z|--\rangle &=-\hbar|--\rangle \\ \end{align} and so $S_z$ can be written as: \begin{equation} S_z = \hbar \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \end{equation} Now, if $s=1$ (remember $s$ denotes the quantum number for the total system), then the three states of the system are: \begin{align} |s=1,m=1\rangle & = |++\rangle \\ |s=1,m=0\rangle & = \sqrt{\frac{1}{2}}(|+-\rangle+|-+\rangle) \\ |s=1,m=-1\rangle & = |--\rangle \end{align} with eigenvalues: \begin{align} S_z |1,1\rangle & = \hbar |1,1\rangle \\ S_z |1,0\rangle & = 0|1,0\rangle \\ S_z |1,-1\rangle & = -\hbar |1,-1\rangle \end{align} where I have switched to a more compact notation: \begin{equation} |s=1,m=1\rangle \equiv |1,1\rangle \end{equation} However, we can also get an eigenstate with $s=0$: \begin{equation} |0,0\rangle = \sqrt{\frac{1}{2}}(|+-\rangle-|-+\rangle) \end{equation} with eigenvalue: \begin{equation} S_z |0,0\rangle = 0|0,0\rangle \end{equation} Now, we can also verify that: \begin{equation} S^2 \equiv S_x^2 + S_y^2 + S_z^2 \end{equation} satisfies: \begin{align} S^2 |1,1\rangle & = 2 \hbar^2 |1,1\rangle \\ S^2 |1,0\rangle & = 2 \hbar^2 |1,0\rangle \\ S^2 |1,-1\rangle & = 2 \hbar^2 |1,-1\rangle \\ S^2 |0,0\rangle & =0|0,0\rangle \end{align} Therefore, we see that the following two important equations are satisfied: \begin{equation} \begin{array}{cc} S^2 |s , m \rangle = \hbar^2 s (s+1) |s , m \rangle \; ,& S_z |s , m \rangle = \hbar m |s , m \rangle \end{array} \end{equation} To sum up, we have found the possible eigenvalues for the magnitude and $z$-component of the system and the eigenstates corresponding to these values: the allowed values for total spin are $s=1$ and $s=0$, while the allowed value of $s_z$ are $\hbar$, $0$, and $-\hbar$ and the corresponding eigenstates in the product basis $|1,1\rangle$, $|1,0\rangle$, $|1,-1\rangle$ and $|0,0\rangle$. Thus, the meaning of these triplet and singlet states is that they are the possible states of the system consisting of the two aforementioned particles. This is often written as: \begin{equation} \frac{1}{2} \otimes \frac{1}{2} = 1 \oplus 0 \end{equation} which means that the tensor product of two spin-$1/2$ Hilbert spaces is a direct sum of a spin-$1$ space and a spin-$0$ space.