91影视

$\mathrm{s}=\frac{3}{2}$

Following formula is used to calculate projection of spin angular moment with respect to z-axis,$\mathbf{肠辞蝉胃}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{z}}}{\mathbf{\left|}\mathbf{S}\mathbf{\right|}}$ Angular momentum along z-axis is,${\mathbf{S}}_{\mathbf{z}}\mathbf{=}{\mathbf{m}}_{\mathbf{s}}\mathbf{h}$ spin angular momentum vector's magnitude is$\mathbf{\left|}\mathbf{S}\mathbf{\right|}\mathbf{=}\sqrt{\mathbf{S}\mathbf{(}\mathbf{S}\mathbf{+}\mathbf{1}\mathbf{)}{\mathbf{h}}^{\mathbf{2}}}$.

The omega particle has spin $\frac{3}{2}$. Its components of angular momentum along the $\mathrm{z}$-axis are therefore ${\mathrm{S}}_{\mathrm{z}}={\mathrm{m}}_{\mathrm{s}}\mathrm{h}$.

For ${\mathrm{m}}_{\mathrm{z}}=鈭�\frac{3}{2},鈭�\frac{1}{2},+\frac{1}{2},+\frac{3}{2}$

The magnitude of its spin angular momentum vector is calculated by substituting $\frac{3}{2}$for $\mathrm{S}$ in the equation $\left|\mathrm{S}\right|=\sqrt{\mathrm{S}(\mathrm{S}+1){\mathrm{h}}^2}$as follows.

$\left|\mathrm{S}\right|=\sqrt{\mathrm{S}(\mathrm{S}+1){\mathrm{鈩�}}^2}=\sqrt{\left(\frac{3}{2}\right)\left(\frac{3}{2}+1\right){\mathrm{鈩�}}^2}=\sqrt{\frac{15}{4}}\mathrm{h}$

The possible angles with respect to the z-axis are shown in the diagram.

For${\mathrm{m}}_{\mathrm{z}}=\frac{3}{2}$the angle is equal to

$\begin{array}{c}\mathrm{肠辞蝉胃}=\frac{{\mathrm{S}}_{\mathrm{z}}}{\left|\mathrm{S}\right|}=\frac{\frac{3}{2}\mathrm{h}}{\sqrt{\frac{15}{4}{\mathrm{h}}^2}}=\frac{3}{\sqrt{15}}=0.77460\\ \mathrm{胃}={39.2}^{掳}\end{array}$

For ${\mathrm{m}}_{\mathrm{z}}=\frac{1}{2}$the angle is equal to

$\begin{array}{c}\mathrm{肠辞蝉胃}=\frac{{\mathrm{S}}_{\mathrm{z}}}{\left|\mathrm{S}\right|}=\frac{\frac{1}{2}\mathrm{h}}{\sqrt{\frac{15}{4}{\mathrm{h}}^2}}=\frac{1}{\sqrt{15}}=0.25820\\ \mathrm{胃}={75.0}^{掳}\end{array}$

For ${\mathrm{m}}_{\mathrm{z}}=鈭�\frac{1}{2}$the angle is equal to

$\begin{array}{c}\mathrm{肠辞蝉胃}=\frac{{\mathrm{S}}_2}{\left|\mathrm{S}\right|}=\frac{鈭�\frac{1}{2}\mathrm{h}}{\sqrt{\frac{15}{4}{\mathrm{h}}^2}}=鈭�0.25820\\ \mathrm{胃}={105}^{掳}\end{array}$

For ${\mathrm{m}}_{\mathrm{z}}=鈭�\frac{3}{2}$ the angle is equal to

$\begin{array}{c}\mathrm{肠辞蝉胃}=\frac{{\mathrm{S}}_{\mathrm{z}}}{\left|\mathrm{S}\right|}=\frac{鈭�\frac{3}{2}\mathrm{h}}{\sqrt{\frac{15}{4}{\mathrm{h}}^2}}=鈭�\frac{3}{\sqrt{15}}=鈭�0.77460\\ \mathrm{胃}={140.8}^{掳}\end{array}$

Therefore, Angles that intrinsic angular momentum vector ${39.2}^{{}^{掳}},{75}^{{}^{掳}},{105}^{{}^{掳}}\text{and}{140.8}^{{}^{掳}}$.