Why do we have $V^{1/2} \otimes V^{1/2} = V^0 \oplus V^1$ (spin). I would like to understand why we have :
$V^{1/2} \otimes V^{1/2} = V^0 \oplus V^1$
I need to find a proof of this, and in fact I am looking for the proof of the general case where we have :
$$ V^{j} \otimes V^{k}= \sum_{i=|j-k|}^{j+k} \oplus V^i$$
If possible a proof with the less mathematical tools needed (I am doing physics, I know linear algebra, some basics of group & representation theory but this is not my domain of expertise).
In my notations : $V^{j}$ is a complex vector space of dimension $2*j+1$ (it is the space where the spin $j$ lives)
 A: Without loss of generality, assume that $j\geq k$.  The dimension of the weight space of the $\mathfrak{su}(2)$-module $M:=V^j\otimes V^k$ with weight $j+k-t$ is precisely $t+1$, where $t\leq 2k$ is a nonnegative integer.  Since each weight space of the simple $\mathfrak{su}(2)$-module $V^r$ is $1$ for each half integer $r\geq 0$, we conclude that the multiplicity of $V^{j+k}$ in $M$ is $1$, and so the weight space of $M/V^{j+k}$ with weight $j+k-1$ is of dimension $1$ (provided that $k>0$).  Thus, the multiplicity of $V^{j+k-1}$ in $M$ is $1$ in case $k>0$.  Then, you proceed by induction and show that the multiplicity of $V^r$ in $M$ is $1$ for each $r=j+k,j+k-1,j+k-2,\ldots,j-k$.  That is,
$$M=V^j\otimes V^k=\bigoplus_{r=j-k}^{j+k}\,V^r\,.$$
For this result, you need to know Weyl's Complete Reducibility Theorem.  In particular,
$$V^{1/2}\otimes V^{1/2}=V^0\oplus V^1\,.$$
However, to get a generator of the submodule isomorphic to $V^r$ in $M$, it is not very easy (well, not too difficult, but cumbersome in my opinion).
Let $\mathfrak{su}(2)_\mathbb{C}=\mathfrak{sl}(2,\mathbb{C})=\mathbb{C}x\oplus\mathbb{C}h\oplus\mathbb{C}y$, where $x$ is the creation operator, $y$ the annihilation operator, and $h=[x,y]$.  In the special case $j=k=\frac{1}{2}$, we observe that $$V^{1/2}=\mathbb{C}u^+\oplus \mathbb{C}u^-\,$$ where $u^+$ is a highest-weight vector of $V^{1/2}$ and $u^-=y\cdot u^+$ a lowest-weight vector.  Clearly, $V^1$ is generated by $u^+\otimes u^+$.  That is,
$$V^1=\mathbb{C}\left(u^+\otimes u^+\right) \oplus \mathbb{C}\left(u^+\otimes u^-+u^-\otimes u^+\right) \oplus \mathbb{C}\left(u^-\otimes u^-\right)\,.$$On the other hand, $V^0$ is $1$-dimensional, so you need a linear combination
$$a\left(u^+\otimes u^-\right)+ b\left(u^-\otimes u^+\right)$$ 
that vanishes when $x$ and $y$ are applied.  It can be easily seen that $a+b=0$ is required, so
$$V^0=\mathbb{C}\left(u^+\otimes u^--u^-\otimes u^+\right)\,.$$
