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The singlet state of two spin$\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}$particles is defined as follows:

$\mathbf{|}\mathbf{00}\mathbf{>}\mathbf{鈮�}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left(鈬�-鈬�\right)$

The triplet state of two spin role="math" localid="1658204414547" $\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}$ particles is defined as follows:

$$\begin{gathered} \mathbf{|}\mathbf{11}\mathbf{鉄�}\mathbf{鈮�}\mathbf{|}\mathbf{鈫�}\mathbf{鈫�}\mathbf{鉄�} \\ \mathbf{}\mathbf{|}\mathbf{10}\mathbf{鉄�}\mathbf{鈮�}\mathbf{1}\mathbf{/}\mathbf{鈭�}\mathbf{2}\mathbf{(}\mathbf{|}\mathbf{鈫�}\mathbf{鈫�}\mathbf{鉄�}\mathbf{+}\mathbf{|}\mathbf{鈫�}\mathbf{鈫�}\mathbf{鉄�}\mathbf{)}\mathbf{} \\ \mathbf{|}\mathbf{1}\mathbf{-}\mathbf{1}\mathbf{鉄�}\mathbf{鈮�}\mathbf{|}\mathbf{鈫�}\mathbf{鈫�}\mathbf{鉄�} \end{gathered}$$

The action of Pauli spin operators on the quantum states is defined as follows:

$$\begin{gathered} {\mathrm{S}}_{\mathrm{z}}|鈫�鉄�=\frac{\mathrm{魔}}{2}|鈫�鉄� \\ {\mathrm{S}}_{\mathrm{z}}|鈫�鉄�=鈹�-\mathrm{魔}/2|鈫�鉄� \\ {\mathrm{S}}_{\mathrm{x}}|鈫�鉄�=鈹�\mathrm{魔}/2|鈫�鉄� \\ {\mathrm{S}}_{\mathrm{x}}|鈫�鉄�=鈹�\mathrm{魔}/2|鈫�鉄� \end{gathered}$$

The previous spin states are orthonrmalized.

That is :

And

Assume, without loss of generality, that the component of the angular momentum vector of particle number$l,S_a^{\left(1\right)}$, is directed in the z direction while the component of the angular momentum vector of particle number$2,S_b^{\left(2\right)}$, is located in the zx -plane with an angle$胃$between the two normalized operators' vectors,$\hat{a}$and$\hat{b}$, respectively.

${\mathrm{S}}_{\mathrm{a}}^{\left(1\right)}={\mathrm{S}}_{\mathrm{z}}^{\left(1\right)}{\mathrm{andS}}_{\mathrm{b}}^{\left(2\right)}=\cos {\mathrm{胃厂}}_{\mathrm{z}}^{\left(2\right)}+\sin {\mathrm{胃厂}}_{\mathrm{x}}^{\left(2\right)}$

Calculate the expectation value of${\mathrm{S}}_{\mathrm{a}}^{\left(1\right)}{S}_{\mathrm{b}}^{\left(2\right)}$in the singlet state of the two spin-$-\frac{1}{2}$particles,$\overline{)00>}$.

role="math" localid="1658206640658"

Simplify the above expression.

Further evaluate the above expression.

The expectation value of the product of the operators ${\mathrm{S}}_{\mathrm{a}}^{\left(1\right)}$ and ${\mathrm{S}}_{\mathrm{b}}^{\left(2\right)}鉄�{\mathrm{S}}_{\mathrm{a}}^{\left(1\right)}{\mathrm{S}}_{\mathrm{b}}^{\left(2\right)}鉄�$ in the singlet state, $|0鈥�鈥�鈥�0鉄�$, of the two spin- $-\frac{1}{2}$particles is: $-\frac{{鈩�}^2}{4}\mathrm{肠辞蝉胃}$.

Thus,role="math" localid="1658204824299" $鉄�{\mathrm{S}}_{\mathrm{a}}^{\left(1\right)}{\mathrm{S}}_{\mathrm{b}}^{\left(2\right)}鉄�=-\frac{{鈩�}^2}{4}\mathrm{肠辞蝉胃}$where$胃$is the angle between$\hat{a}$and$\hat{b}$is 0_.