Notation - angular momentum for composite systems In the discussion for the addition of angular momentum for composite systems, my lecturer uses the following notation in his notes when referring to a composite system of two spin-half particles:

$ \frac{1}{2} \otimes \frac{1}{2} = 1 \oplus 0$ 

There is no further explanations in the notes but I am just curious what this means in the context of composite systems. 
 A: This answer is less mathematical and more physical. If you want to understand more about this decomposition you should consult texts on representation theory. Nevertheless, you can understand this without the full mathematical knowledge.
You started with two spins, or two Hilbert spaces $\mathcal{H}_{1,2}={\rm span}\left\{\left|\uparrow\right>,\left|\downarrow\right>\right\}$ each describing one of your particles. In these bases the operators $\hat{S}^{2}_{1,2}$ and $\hat{S}^{z}_{1,2}$ were diagonal. Your next step was to build a new Hilbert space that describes the two-particle system. This is known as the tensor product space
$$\begin{matrix}\mathcal{H}=\mathcal{H}_{1}\otimes\mathcal{H}_{2}&=&{\rm span}\left\{\left|\uparrow\right>\otimes\left|\uparrow\right>,\left|\downarrow\right>\otimes\left|\uparrow\right>,\left|\uparrow\right>\otimes\left|\downarrow\right>,\left|\downarrow\right>\otimes\left|\downarrow\right>\right\}=\\\\&=&{\rm span}\left\{\left|\uparrow\uparrow\right>,\left|\downarrow\uparrow\right>,\left|\uparrow\downarrow\right>,\left|\downarrow\downarrow\right>\right\}\end{matrix}$$
This is what you write as $\dfrac{1}{2}\otimes\dfrac{1}{2}$. You know from addition of angular momentum that it is possible to define a new basis
$$\left\{\begin{matrix}\left|0,0\right>&=&\dfrac{\left|\uparrow\downarrow\right>-\left|\downarrow\uparrow\right>}{\sqrt{2}}\\\left|1,-1\right>&=&\left|\downarrow\downarrow\right>\\\left|1,0\right>&=&\dfrac{\left|\uparrow\downarrow\right>+\left|\downarrow\uparrow\right>}{\sqrt{2}}\\\left|1,1\right>&=&\left|\uparrow\uparrow\right>\end{matrix}\right.$$
This basis diagonalizes $\hat{S}^{2}=\hat{S}_{1}^{2}+\hat{S}_{2}^{2}$ and $\hat{S}^{z}=\hat{S}^{z}_{1}+\hat{S}^{z}_{2}$, with $\hat{S}^{2}$ taking the form
$$\hat{S}^{2}=\hbar^{2}\cdot 1 \cdot\left(1+1\right)\left(\begin{matrix}0&&&\\&1&&\\&&1&\\&&&1\end{matrix}\right)$$
This is interesting. For a single particle of a specific spin $S$ the operator $\hat{S}^{2}$ is just $\hbar^{2}S\left(S+1\right)\hat{I}$ where $\hat{I}$ is the identity matrix. It means that you can interpret the above matrix as a system of spin $S=0$ plus a system of spin $S=1$, which mathematically can be written as $0\oplus1$.
