Find the matrix representing${\mathbf{S}}_{\mathbf{x}}$for a particle of spin3/2 (using, as

always, the basis of eigenstates of${\mathit{S}}_{\mathbf{z}}$). Solve the characteristic equation to

determine the eigenvalues of${\mathbf{S}}_{\mathbf{x}}$.

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,

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

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:

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

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

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