Tensorpart of NN potential The potential of the nucleon nucleon interaction includes a tensor part which is given by:
    $S_{12}(\hat{\pmb{r}}) = \hat{\pmb{r}} \cdot \pmb{\sigma_1} \hat{\pmb{r}} \cdot \pmb{\sigma_2} - \frac{1}{3} \pmb{\sigma_1}\cdot\pmb{\sigma_2}$
Where $\hat{\pmb{r}}$ ist a unit coordinate space operator and $\pmb{\sigma_k}$ are vectors of the Pauli matrices acting on particle 1 and 2.
Why is the second term substracted?
 A: In view of your comments I have explained my answer. It is usual to define the non-central potential in such a way that its average over all directions is zero. However integrating over all angles only the first term would give, 
$ \frac{1}{4\pi}\int({\pmb{r}} \cdot \pmb{\sigma_1} )({\pmb{r}} \cdot \pmb{\sigma_2})d\omega = \frac{1}{3} r^2\pmb{\sigma_1}\cdot\pmb{\sigma_2}$
A: Consider the general symmetric rank 2 tensor:
$S_{ij} = \frac{1}{2}(T_{ij}+T_{ji})$
It has 6 independent terms. However, one of them is a scalar times the unit tensor:
$\frac{1}{3}{\rm Tr}({T})\delta_{ij}$.
so it doesn't really count as a rank 2 tensor part. The so-called natural form for a rank 2 tensor is:
$ T^{(2)}_{ij} = S_{ij} - \frac{1}{3}{\rm Tr}({T})\delta_{ij}$
which has 5 terms.
The antisymmetric (vector) part has 3 terms:
$ T^{(1)} = A_{ij} = \frac{1}{2}(T_{ij}-T_{ji}) $,
and the scalar part--the trace:
$ T^{(0)} = \frac{1}{3}T_{kk}\delta_{ij} $
has 1, for a total of 9, as required. Note how they match-up with the spherical harmonics: $Y^{m}_{l=2}$, $Y^{m}_{l=1}$, and $Y^0_0$.
@SAKhan is saying that the 1st term in your equation counts the scalar part, so that it must be subtracted off to make pure rank-2 tensor interaction.
