Question

Suppose two spin -1/2particles are known to be in the singlet configuration (Equation Let ${\mathbf{S}}_{\mathbf{a}}^{\mathbf{\left(}\mathbf{1}\mathbf{\right)}}$be the component of the spin angular momentum of particle number 1 in the direction defined by the unit vector$\hat{\mathbf{a}}$ Similarly, let${\mathbf{S}}_{\mathbf{b}}^{\mathbf{\left(}\mathbf{2}\mathbf{\right)}}$ be the component of 2’s angular momentum in the direction$\hat{\mathbf{b}}$ Show that

$\mathbf{⟨}{\mathbf{S}}_{\mathbf{a}}^{\mathbf{\left(}\mathbf{1}\mathbf{\right)}}{\mathbf{S}}_{\mathbf{b}}^{\mathbf{\left(}\mathbf{2}\mathbf{\right)}}\mathbf{⟩}\mathbf{=}\mathbf{-}\frac{{\mathbf{\hslash }}^{\mathbf{2}}}{\mathbf{4}}\mathbf{cos\theta }$

where $\mathit{\theta }$ is the angle between $\hat{\mathbf{a}}$ and$\hat{\mathbf{b}}$

Short answer

It is proved that $⟨{\mathrm{S}}_{\mathrm{a}}^{\left(1\right)}{\mathrm{S}}_{\mathrm{b}}^{\left(2\right)}⟩=-\frac{{\hslash }^2}{4}\mathrm{cos\theta }$.

Step-by-step solution

Expression for the Singlet, Triplet and Pauli spin operators

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

$\mathbf{|}\mathbf{00}\mathbf{>}\mathbf{\equiv }\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left(⇅-⇵\right)$

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

$$\begin{gathered} \mathbf{|}\mathbf{11}\mathbf{⟩}\mathbf{\equiv }\mathbf{|}\mathbf{\uparrow }\mathbf{\uparrow }\mathbf{⟩} \\ \mathbf{}\mathbf{|}\mathbf{10}\mathbf{⟩}\mathbf{\equiv }\mathbf{1}\mathbf{/}\mathbf{\surd }\mathbf{2}\mathbf{(}\mathbf{|}\mathbf{\uparrow }\mathbf{\downarrow }\mathbf{⟩}\mathbf{+}\mathbf{|}\mathbf{\downarrow }\mathbf{\uparrow }\mathbf{⟩}\mathbf{)}\mathbf{} \\ \mathbf{|}\mathbf{1}\mathbf{-}\mathbf{1}\mathbf{⟩}\mathbf{\equiv }\mathbf{|}\mathbf{\downarrow }\mathbf{\downarrow }\mathbf{⟩} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{S}}_{\mathrm{z}}|\uparrow ⟩=\frac{\mathrm{ħ}}{2}|\uparrow ⟩ \\ {\mathrm{S}}_{\mathrm{z}}|\downarrow ⟩=├-\mathrm{ħ}/2|\downarrow ⟩ \\ {\mathrm{S}}_{\mathrm{x}}|\uparrow ⟩=├\mathrm{ħ}/2|\downarrow ⟩ \\ {\mathrm{S}}_{\mathrm{x}}|\downarrow ⟩=├\mathrm{ħ}/2|\uparrow ⟩ \end{gathered}$$

The previous spin states are orthonrmalized.

That is :

And

Determination of the angle between  a^ and  b^

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$\theta$between the two normalized operators' vectors,$\hat{a}$and$\hat{b}$, respectively.

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

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 ${\mathrm{S}}_{\mathrm{b}}^{\left(2\right)}⟨{\mathrm{S}}_{\mathrm{a}}^{\left(1\right)}{\mathrm{S}}_{\mathrm{b}}^{\left(2\right)}⟩$ in the singlet state, $|0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}0⟩$, of the two spin- $-\frac{1}{2}$particles is: $-\frac{{\hslash }^2}{4}\mathrm{cos\theta }$.

Thus,role="math" localid="1658204824299" $⟨{\mathrm{S}}_{\mathrm{a}}^{\left(1\right)}{\mathrm{S}}_{\mathrm{b}}^{\left(2\right)}⟩=-\frac{{\hslash }^2}{4}\mathrm{cos\theta }$where$\theta$is the angle between$\hat{a}$and$\hat{b}$is 0_.