Pauli matrix for triplet state?

Question

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

Answer 1

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.

Answer 2

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.