Tensor operator on nucleon nucleon interaction 
i'm trying to prove that from
$$
S_{12}=\left[3 \frac{\left(\vec{\sigma_{1}} \cdot \vec{r}\right)\left(\vec{\sigma_{2}} \cdot \vec{r}\right)}{r^{2}}-\vec{\sigma_{1}} \cdot \vec{\sigma_{2}}\right]
$$
then
$$
S_{12}=2\left[3 \frac{(\widehat{\vec{s}} \cdot \vec{r})^{2}}{r^{2}}-\widehat{\vec{s}}^{2}\right].
$$
Is true, given that $$\widehat{\vec{s}}=\frac{1}{2}\left(\sigma_{1}+\sigma_{2}\right)~?$$

Now i've found that from evaluating $(\vec{S} \cdot \vec{r})^2$ that
$$(\vec{S} \cdot \vec{r})^2= \frac{1}{2} \left( \vec{r}^2 +  (\vec{\sigma_1}\cdot \vec{r}) (\vec{\sigma_2}\cdot \vec{r})\right) \\
2(\vec{S} \cdot \vec{r})^2  - \vec{r^2} = (\vec{\sigma_1}\cdot \vec{r}) (\vec{\sigma_2}\cdot \vec{r})$$
However when i tried to evaluate    $\vec{S}^2$ i've found that
$$
 \vec{S} = \frac{1}{2} (\vec{\sigma_1} + \vec{\sigma_2}) \\
 \vec{S}^2  = \frac{1}{4} (\vec{\sigma_1}+\vec{\sigma_2})^2 \\
 =\frac{1}{4} (\vec{\sigma_1}^2 + \vec{\sigma_2}^2 + \underbrace{(\vec{\sigma_1} \vec{\sigma_2}) +( \vec{\sigma_2}  \vec{\sigma_1})}_{\{\sigma_1,\sigma_2\}= 0}) \\
 = \frac{1}{4} (2 I) \\
 = \frac{1}{2} I
$$
which i don't think is true? But i'm having a hard time figuring out what i did wrong
 A: The problem is that the spin operators don't anti-commute.  The spin operators, being operators corresponding to different particles, commute. You're getting this mixed up with the fact that the Pauli spin matrices of a single particle anti-commute (provided they are different spin matrices). But that's not what's going on here.  Instead, this the dot product between the full spin vectors of two different particles.  (Really, under the hood, we are taking the tensor product of the two operators, one for particle 1 and one for particle 2.)
Thus, when you compute $\hat{S}^2$, you should actually get
$$
\hat{S}^2
=
\frac{1}{4}
\left(
\hat{\boldsymbol{\sigma}}_1\cdot\hat{\boldsymbol{\sigma}}_1
+
\hat{\boldsymbol{\sigma}}_2\cdot\hat{\boldsymbol{\sigma}}_2
+2\hat{\boldsymbol{\sigma}}_1\cdot\hat{\boldsymbol{\sigma}}_2
\right)
=
\frac{1}{4}
\left(
3\hat{I}+3\hat{I}
+2\hat{\boldsymbol{\sigma}}_1\cdot\hat{\boldsymbol{\sigma}}_2
\right)
=\frac{3}{2}\hat{I}+\frac{1}{2}\hat{\boldsymbol{\sigma}}_1\cdot\hat{\boldsymbol{\sigma}}_2\,.
$$
Note that $\hat{\boldsymbol{\sigma}}_i\cdot\hat{\boldsymbol{\sigma}}_i = 3\hat{I}$ because the dot product actually yields the sum of the three squares of the components of the spin vector, each of which is a Pauli spin matrix whose square is the identity.
Combined with the first term, which you've computed correctly as
$$
\frac{1}{r^2}(\vec{\textbf{S}}\cdot\textbf{r})^2
=
\frac{1}{2}\frac{1}{r^2}
\left(
r^2\hat{I} - (\hat{\boldsymbol{\sigma}}_1\cdot\textbf{r})
(\hat{\boldsymbol{\sigma}}_2\cdot\textbf{r})
\right)\,,
$$
we get
\begin{align}
2\left(
3\frac{1}{r^2}(\hat{\textbf{S}}\cdot\textbf{r})^2
-\hat{{S}}^2
\right)
&=2
\left(
3
\frac{1}{2}\frac{1}{r^2}
\left(
r^2\hat{I} - (\hat{\boldsymbol{\sigma}}_1\cdot\textbf{r})
(\hat{\boldsymbol{\sigma}}_2\cdot\textbf{r})
\right)
-\left(
\frac{3}{2}\hat{I}+\frac{1}{2}\hat{\boldsymbol{\sigma}}_1\cdot\hat{\boldsymbol{\sigma}}_2
\right)
\right)
\\
&=3
\frac{(\hat{\boldsymbol{\sigma}}_1\cdot\textbf{r})
(\hat{\boldsymbol{\sigma}}_2\cdot\textbf{r})
}
{r^2}
-\frac{1}{2}\hat{\boldsymbol{\sigma}}_1\cdot\hat{\boldsymbol{\sigma}}_2
\end{align}
