Pauli matrix for triplet state? Question is, what would be the result of applying the operator $\hat A = [3I + \vec\sigma_1 . \vec\sigma_2]$ on the |singlet$\rangle$ and |triplet$\rangle$ states ($\vec\sigma_1$ acts on the 1st particle and $\vec\sigma_2$ acts on the second particle ONLY), ie, $$\hat A|singlet\rangle=?|singlet\rangle$$
and $$\hat A|triplet\rangle=?|triplet\rangle$$
I am stuck at the triplet part of the question.
For a system of 2 spin half particles, where $\vec\sigma_1$ acts on the 1st particle and $\vec\sigma_2$ acts on the second particle ONLY, (like adding angular momentum of two electrons) $$\vec\sigma=\vec\sigma_1+\vec\sigma_2$$
squaring both sides, $$\vec\sigma^2=(\vec\sigma_1+\vec\sigma_2)^2$$ 
from which we have$$\vec\sigma_1 . \vec\sigma_2 = (\sigma^2 - \sigma_1^{2} - \sigma_2^{2})/2$$
Now, $\sigma_1^{2}=\sigma_{1x}^{2}+\sigma_{1y}^{2}+\sigma_{1z}^{2}=3I$ and similarly, $\sigma_2^{2}=3I$.
and that for the singlet state, the value of $\sigma^2=0$, (which i gathered from the total spin being $0$ for the singlet state) which gives $$\vec\sigma_1 . \vec\sigma_2 = (0 - 3I - 3I)/2=-3I$$
I dont know what the value of $\sigma^2$ is for the triplet state (i do know that the total spin $S$ is $\sqrt2\hbar$)?
I am not able to relate the total spin with the $\vec\sigma$ properly
 A: As @rob asked you to, you are meant to simply write down 
$$
\hat{B}\equiv\vec{\sigma}_1\cdot\vec{\sigma}_2 =  {\sigma}_1^x  {\sigma}_2^x +{\sigma}_1^y  {\sigma}_2^y+{\sigma}_1^z  {\sigma}_2^z \\=  ({\sigma}_1^x+i {\sigma}_1^y)({\sigma}_2^x -i{\sigma}_2^y )/2 +({\sigma}_1^x-i{\sigma}_1^y ) ({\sigma}_2^y +i{\sigma}_2^y)/2+{\sigma}_1^z  {\sigma}_2^z\\  
\equiv  {\sigma}_1^+  {\sigma}_2^- +{\sigma}_1^-  {\sigma}_2^+ +{\sigma}_1^z  {\sigma}_2^z ~,
$$
where $\sigma^+ \uparrow=0$, and $\sigma^+ \downarrow=\uparrow \sqrt{2}$, etc... for both 1 and 2. Recall
$$
\sigma^+ =  \sqrt{2}
    \begin{pmatrix}
      0&1\\
      0&0
    \end{pmatrix} .
$$
Acting on the singlet, $\uparrow \downarrow- \downarrow \uparrow$ , 
this $\hat B$  has the obvious eigenvalue -3.
The triplet is $\uparrow \uparrow$;  $(\uparrow\downarrow+\downarrow\uparrow)/\sqrt{2}$; $\downarrow \downarrow$, and so it obviously has eigenvalue 1 under the action of $\hat{B}$. 
Your $\hat A= 3 1\!\!1 +\hat{B}$ has eigenvalues 0 and 4 respectively, given my normalizations. This is to say, of course, that, for the triplet, 
$\sigma^2/4=2=(1+1)1$, as expected.
A: Setting $\hbar=1$ for simplicity, the matrices you need are the $S=1$ matrices.  One easily obtains
$$
S_z=\left(\begin{array}{ccc}
1&0&0\\
0&0&0\\
0&0&-1\end{array}\right)\, ,\quad
S_+=\sqrt{2}\left(\begin{array}{ccc}
0&1&0\\
0&0&1\\
0&0&0\end{array}\right)\, ,\quad S_+=S_-^\dagger,
$$
from which one recovers $S_x$ and $S_y$ by inverting $S_\pm=S_x\pm i S_y$.
The matrix for $S^2$ will be $2\times I$ where $I$ is the $3\times 3$ unit matrix.
