Question

Work out the spin matrices for arbitrary spin , generalizing spin (Equations 4.145 and 4.147), spin 1 (Problem 4.31), and spin (Problem 4.52). Answer:

$$\begin{gathered} {\mathbf{S}}_{\mathbf{z}}\mathbf{=}\mathbf{\hslash }\mathbf{\left(}\begin{array}{ccccc}\mathbf{s}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}\\ \mathbf{0}& \mathbf{s}\mathbf{-}\mathbf{1}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& \mathbf{s}\mathbf{-}\mathbf{2}& \mathbf{\cdots }& \mathbf{0}\\ \mathbf{⋮}& \mathbf{⋮}& \mathbf{⋮}& \mathbf{\cdots }& \mathbf{⋮}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{-}\mathbf{s}\end{array}\mathbf{\right)} \\ {\mathbf{S}}_{\mathbf{x}}\mathbf{=}\frac{\mathbf{\hslash }}{\mathbf{2}}\mathbf{\left(}\begin{array}{ccccccc}\mathbf{0}& {\mathbf{b}}_{\mathbf{s}}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ {\mathbf{b}}_{\mathbf{s}}& \mathbf{0}& {\mathbf{b}}_{\mathbf{s}\mathbf{-}\mathbf{1}}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ \mathbf{0}& {\mathbf{b}}_{\mathbf{s}\mathbf{-}\mathbf{1}}& \mathbf{0}& {\mathbf{b}}_{\mathbf{s}\mathbf{-}\mathbf{2}}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& {\mathbf{b}}_{\mathbf{s}\mathbf{-}\mathbf{2}}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ \mathbf{⋮}& \mathbf{⋮}& \mathbf{⋮}& \mathbf{⋮}& \mathbf{\cdots }& \mathbf{⋮}& \mathbf{⋮}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& {\mathbf{b}}_{\mathbf{-}\mathbf{s}\mathbf{+}\mathbf{1}}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& {\mathbf{b}}_{\mathbf{-}\mathbf{s}\mathbf{+}\mathbf{1}}& \mathbf{0}\end{array}\mathbf{\right)} \\ {\mathbf{S}}_{\mathbf{y}}\mathbf{=}\frac{\mathbf{\hslash }}{\mathbf{2}}\mathbf{\left(}\begin{array}{ccccccc}\mathbf{0}& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ {\mathbf{ib}}_{\mathbf{s}}& \mathbf{0}& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}\mathbf{-}\mathbf{1}}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ \mathbf{0}& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}\mathbf{-}\mathbf{1}}& \mathbf{0}& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}\mathbf{-}\mathbf{2}}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}\mathbf{-}\mathbf{2}}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ \mathbf{⋮}& \mathbf{⋮}& \mathbf{⋮}& \mathbf{⋮}& \mathbf{\cdots }& \mathbf{⋮}& \mathbf{⋮}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}\mathbf{+}\mathbf{1}}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}\mathbf{+}\mathbf{1}}& \mathbf{0}\end{array}\mathbf{\right)}\mathbf{} \end{gathered}$$

where,${\mathbf{b}}_{\mathbf{j}}\mathbf{\equiv }\sqrt{\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{j}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{-}\mathbf{j}\mathbf{)}}$

Short answer

The spin matrices are, ${\mathrm{s}}_{\mathrm{y}}=\frac{\hslash }{2\mathrm{i}}\left[\begin{array}{cccccc}0& {\mathrm{b}}_{\mathrm{s}}& 0& 0& \cdots & 0\\ -{\mathrm{b}}_{\mathrm{s}}& 0& -{\mathrm{b}}_{\mathrm{s}-1}& 0& \cdots & 0\\ \cdots & -{\mathrm{b}}_{\mathrm{s}-1}& 0& -{\mathrm{b}}_{\mathrm{s}-2}& \cdots & 0\\ 0& 0& 0& 0& \cdots & -{\mathrm{b}}_{-\mathrm{s}-1}\\ 0& 0& 0& 0& -{\mathrm{b}}_{-\mathrm{s}-1}& 0\end{array}\right]$

Step-by-step solution

Definition of spin matrix

The spin related matrices are known as spin matrices. These are number of matrices. These matrices are complex that include involutory, unitary, and Hermitian.

Determination of spin matrices.

Write equation 4.135.

$s_z|sm⟩=hm|sm⟩$

Write the matrix element of $s_z$.

Write the matrix (diagonal matrix) with values of m ranging from s to -s along the diagonal.

${\mathrm{S}}_{\mathrm{z}}=\hslash \left(\begin{array}{ccccc}\mathrm{s}& 0& 0& \cdots & 0\\ 0& \mathrm{s}-1& 0& \cdots & 0\\ 0& 0& \mathrm{s}-2& 0& \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& 0& 0& 0& -\mathrm{s}\end{array}\right)$

Determine the value of ${\left(s_{+}\right)}_{nm}$.

Here,$b_{m+1}=\sqrt{(s-m)(s+m+1)}$ .

Use the property of the $\delta$ function.

$({\mathrm{s}}_{*}{)}_{\mathrm{mw}}=\hslash {\mathrm{b}}_{\mathrm{n}}{\mathrm{\delta }}_{\mathrm{nm}+1}$

Write the matrix.

${\mathrm{s}}_{+}=\hslash \left[\begin{array}{cccccc}0& {\mathrm{b}}_{\mathrm{s}}& 0& 0& \cdots & 0\\ 0& 0& {\mathrm{b}}_{\mathrm{s}-1}& 0& \cdots & 0\\ 0& 0& 0& {\mathrm{b}}_{\mathrm{s}-2}& \cdots & 0\\ \cdots & \cdots & \cdots & \cdots & \cdots & {\mathrm{b}}_{-\mathrm{s}+1}\\ 0& 0& 0& 0& \cdots & 0\end{array}\right]$

Write the value of ${\left(s\_\right)}_{nm}$.

role="math" localid="1658146052379"

Write the value of s_ .

$$\begin{gathered} \mathrm{s\_}=\sout{\mathrm{h}}000.....0{\mathrm{b}}_{\mathrm{s}}000...00{\mathrm{b}}_{\mathrm{s}-1}.......00{\mathrm{b}}_{\mathrm{s}-2}.......0000{\mathrm{s}}_{\mathrm{x}} \\ =\frac{1}{2}\left[{\mathrm{s}}_{+}+{\mathrm{s}}_{-}\right] \end{gathered}$$

Write the value ofrole="math" localid="1658143674691" $s_x$.

${\mathrm{s}}_{\mathrm{x}}=\frac{\hslash }{2}0{\mathrm{b}}_{\mathrm{s}}00\cdots 0{\mathrm{b}}_{\mathrm{s}}0{\mathrm{b}}_{\mathrm{s}-1}0\dots 00{\mathrm{b}}_{\mathrm{s}-1}0{\mathrm{b}}_{\mathrm{s}-2}\dots \cdots 00{\mathrm{b}}_{\mathrm{s}-2}00{\mathrm{b}}_{-\mathrm{s}+1}000\dots {\mathrm{b}}_{-\mathrm{s}+1}0\dots {\mathrm{b}}_{-\mathrm{s}+1}0$

Write the value of$s_y$ .

$$\begin{gathered} {\mathrm{s}}_{\mathrm{y}}=\frac{1}{2\mathrm{i}}[{\mathrm{s}}_{+}-{\mathrm{s}}_{-}] \\ {\mathrm{s}}_{\mathrm{y}}=\frac{\hslash }{2\mathrm{i}}\left[\begin{array}{cccccc}0& {\mathrm{b}}_{\mathrm{s}}& 0& 0& \cdots & 0\\ -{\mathrm{b}}_{\mathrm{s}}& 0& -{\mathrm{b}}_{\mathrm{s}-1}& 0& \cdots & 0\\ \cdots & -{\mathrm{b}}_{\mathrm{s}-1}& 0& -{\mathrm{b}}_{\mathrm{s}-2}& \cdots & 0\\ 0& 0& 0& 0& \cdots & -{\mathrm{b}}_{-\mathrm{s}-1}\\ 0& 0& 0& 0& -{\mathrm{b}}_{-\mathrm{s}-1}& 0\end{array}\right] \end{gathered}$$

Thus, the spin matrices are$\frac{\hslash }{2\mathrm{i}}\left[\begin{array}{cccccc}0& {\mathrm{b}}_{\mathrm{s}}& 0& 0& \cdots & 0\\ -{\mathrm{b}}_{\mathrm{s}}& 0& -{\mathrm{b}}_{\mathrm{s}-1}& 0& \cdots & 0\\ \cdots & -{\mathrm{b}}_{\mathrm{s}-1}& 0& -{\mathrm{b}}_{\mathrm{s}-2}& \cdots & 0\\ 0& 0& 0& 0& \cdots & -{\mathrm{b}}_{-\mathrm{s}-1}\\ 0& 0& 0& 0& -{\mathrm{b}}_{-\mathrm{s}-1}& 0\end{array}\right]$