91Ó°ÊÓ

Spin is a body's entire angular momentum, also known as intrinsic angular momentum. Macroscopic substances' spins are equivalent to the spins of elementary particles. In actuality, a planet's spin is equal to the sum of all of its fundamental particles' spins and orbital angular momenta.

Here, the spin of the particle is 3/2

The following are the raising and lowering operators (for spin):

$\frac{3}{2}\mathrm{h},\frac{1}{2}\mathrm{h},-\frac{1}{2}\mathrm{h},-\frac{3}{2}\mathrm{h}$

From the above equations:

$S_x=\frac{1}{2}\left(S_{+}+S_{-}\right)\left(1\right)$

As,

${\mathrm{S}}_{Â\pm }|{\mathrm{sm}}_{\mathrm{s}}>=\sout{\mathrm{h}}\sqrt{\mathrm{s}\left(\mathrm{s}+1\right)-\mathrm{m}\left(\mathrm{m}Â\pm 1\right)|{\mathrm{sm}}_{\mathrm{s}}Â\pm 1>}$

Here,the eigenstate is ${\text{|sm}}_{\mathrm{s}}>$of ${\mathrm{S}}^2$for the eigenvalue s(s+1) and of ${\mathrm{S}}_{\mathrm{z}}$with the eigenvalue m .The value of s is 1/2 , role="math" localid="1655968770915" ${\mathrm{m}}_{\mathrm{s}}=3/2,1/2,-1/2,-3/2$,So the matrix of role="math" localid="1655968803438" ${\mathrm{S}}_{-}$and ${\mathrm{S}}_{+}$will be four by four and. The elements of matrix are

$$\begin{gathered} S_{+}\overline{)\frac{3}{2}\frac{3}{2}>}=0 \\ {S}_{+}\overline{)\frac{3}{2}\frac{1}{2}>}=\sqrt{3}\sout{h}\overline{)\frac{3}{2}\frac{3}{2}>} \\ {S}_{+}\overline{)\frac{3}{2}\frac{1}{2}>}=2\sout{h}\overline{)\frac{3}{2}-\frac{1}{2}>} \\ {S}_{+}\overline{)\frac{3}{2}\frac{3}{2}>}=\sqrt{3}\sout{h}\overline{)\frac{3}{2}\frac{1}{2}>} \\ {S}_{-}\overline{)\frac{3}{2}\frac{3}{2}>}=\sqrt{3}\sout{h}\overline{)\frac{3}{2}\frac{1}{2}>} \\ {S}_{-}\overline{)\frac{3}{2}\frac{1}{2}>}=2\sout{h}\overline{)\frac{3}{2}-\frac{1}{2}>} \\ {S}_{-}\overline{)\frac{3}{2}\frac{1}{2}>}=\sqrt{3}\sout{h}\overline{)\frac{3}{2}\frac{3}{2}>} \\ {S}_{-}\overline{)\frac{3}{2}\frac{3}{2}>}=0 \end{gathered}$$

From the above elements, the matrix will be:

$S_{+}=\sout{h}\left(\begin{array}{cccc}0& \sqrt{3}& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& \sqrt{3}\\ 0& 0& 0& 0\end{array}\right)S_{-}=\sout{h}\left(\begin{array}{cccc}0& 0& 0& 0\\ \sqrt{3}& 0& 0& 0\\ 0& 2& 0& \sqrt{3}\\ 0& 0& \sqrt{3}& 0\end{array}\right)$

The value from equation (1)can be got:

$$\begin{gathered} S_x=\frac{1}{2}\left(S_{+}+S_{-}\right)=\frac{\sout{h}}{2}\left(\begin{array}{cccc}0& \sqrt{3}& 0& 0\\ \sqrt{3}& 0& 2& 0\\ 0& 2& 0& \sqrt{3}\\ 0& 0& \sqrt{3}& 0\end{array}\right) \\ {S}_x=\frac{\sout{h}}{2}\left(\begin{array}{cccc}0& \sqrt{3}& 0& 0\\ \sqrt{3}& 0& 2& 0\\ 0& 2& 0& \sqrt{3}\\ 0& 0& \sqrt{3}& 0\end{array}\right) \end{gathered}$$

To find the eigen values of $S_x$, first the characteristic equation must be solved, which is:

$$\begin{gathered} \left|S_x-\mathrm{λ}|\right|=0 \\ \left|\begin{array}{cccc}-\mathrm{λ}& \sqrt{3}& 0& 0\\ \sqrt{3}& -\mathrm{λ}& 2& 0\\ 0& 2& -\mathrm{λ}& \sqrt{3}\\ 0& 0& \sqrt{3}& -\mathrm{λ}\end{array}\right|=-\mathrm{λ}\left|\begin{array}{ccc}-\mathrm{λ}& 2& 3\\ 2& -\mathrm{λ}& \sqrt{3}\\ 0& \sqrt{3}& -\mathrm{λ}\end{array}\right|-\sqrt{3}\left|\begin{array}{ccc}\sqrt{3}& 2& 0\\ 0& -\mathrm{λ}& \sqrt{3}\\ 0& \sqrt{3}& -\mathrm{λ}\end{array}\right| \\ -\mathrm{λ}\left[-{\mathrm{λ}}^3+3\mathrm{λ}+4\mathrm{λ}\right]-\sqrt{3}\left[\sqrt{3}{\mathrm{λ}}^2-3\sqrt{3}\right] \\ ={\mathrm{λ}}^4-7{\mathrm{λ}}^2-3{\mathrm{λ}}^2+9=0 \\ \left({\mathrm{λ}}^2-9\right)\left({\mathrm{λ}}^2-1\right)=0 \\ \mathrm{λ}=Â\pm 3,Â\pm 1 \end{gathered}$$

So, ${\mathrm{S}}_{\mathrm{x}}$’s eigen values will be:

$\frac{3}{2}\mathrm{h},\frac{1}{2}\mathrm{h},-\frac{1}{2}\mathrm{h},-\frac{3}{2}\mathrm{h}$