# Tensor force operator on singlet state

I am reading chapter 3 of Bertulani's Nuclear Physics in a Nutshell, and during his discussion of the tensor force operator,

$$S_{12} = 3(\vec{\sigma}_1 \cdot \hat{r})(\vec{\sigma}_2 \cdot \hat{r}) - \vec{\sigma}_1 \cdot \vec{\sigma}_2,$$

he shows that for the singlet state: $$S_{12} = 0$$. I understand the derivation of this result, but not the comment he makes regarding the intuition behind it:

This is an expected result since there is no preferential direction for the singlet state.

What does this mean? And why would the triplet have a preferred direction? (Note: this whole chapter is related to the construction of the phenomenological interaction potential for two spin-1/2 nucleons)

The singlet is completely invariant under rotations: if $$\vert 00\rangle$$ is the singlet, then $$R(\Omega)\vert 00\rangle=\vert 00\rangle$$ for any rotation $$R(\Omega)$$. Thus you can take your singlet and an orient it (with rotations) in any direction you want without changing the state.
The $$S=1$$ states, on the other hand, transform into each others under rotation and in particular it is possible to start - say - from the $$S=1,M=1$$ state and rotate it into a linear combination of the other $$\vert S=1,M\rangle$$ states and “orient” it by rotation so as to produce a state with non-zero average values of $$\langle S_k\rangle$$ in a specific direction.