in order to fine out the matrix elements, we just need to know 2 equations

J_z \left| j,m\right> = m \hbar \left|j,m\right>

J_\pm \left| j, m \right>=\hbar \sqrt{ j(j+1) -m ( m\pm 1) } \left |j,m \right >

The calculation is straight forward, but be careful of the sign.

i define the coefficient :

K_j(m) = \frac{\hbar}{2} \sqrt {j(j+1) -m(m+1) }

and the matrix coefficients are:

J_z ^{\mu\nu} (j) = \hbar (j -\mu +1) \delta_{\mu \nu}

J_x ^{\mu\nu} (j) = K_j(j-\mu) \delta_{\mu (\nu-1)}+K_j(j-\nu)\delta_{(\mu-1)\nu}

J_y ^{\mu\nu} (j) = -i K_j(j-\mu) \delta_{\mu (\nu-1)}+ i K_j (j-\nu)\delta_{(\mu-1)\nu}

where

\mu , \nu = 1,2,...,2j+1

the Kronecker Delta indicated that only 1st upper and lower diagonal elements are non-zero.

δμ(ν-1) means the 1st upper diagonal elements. since ν = μ+1 to make it non-zero.

For example:

J_x (1) = \frac {\hbar }{2} \begin {pmatrix} 0 & \sqrt {2} & 0 \\ \sqrt{2} & 0 &\sqrt{2} \\ 0 & \sqrt{2} & 0\end{pmatrix}

J_x (\frac {3}{2}) = \frac {\hbar }{2} \begin {pmatrix} 0 & \sqrt {3} & 0 & 0 \\ \sqrt{3} & 0 & 2 & 0 \\ 0 & 2 & 0 & \sqrt{3} \\ 0 & 0 & \sqrt{3} & 0 \end{pmatrix}

To compute J_y , we just need to multiply the upper diagonal with i and the lower diagonal with - i .

The coefficient K_j(j-\mu)  is every interesting. if we only look at the first upper diagonal. and take the square of each element.

J_x(\frac {1}{2}) = 1

J_x(1) = \begin{pmatrix} 2 & 2 \end{pmatrix}

J_x(\frac{3}{2}) = \begin{pmatrix} 3 & 4 & 3 \end{pmatrix}

and we use this to form a bigger matrix

\begin {pmatrix} ... & 5 & 4 & 3 & 2 & 1 \\ ... & 10 & 8 & 6 & 4 & 2 \\ ... & ... & 12 & 9 & 6 & 3 \\ ... & ... & ...& 12 & 8 & 4 \\ ... & ... & ... & ... & 10 & 5 \end {pmatrix}

if we read from the top right hand corner, and take the diagonal element. we can see that they fit the 1st upper diagonal element of J_x ( j) , if we take square root of each one.

and the pattern are just the multiplication table! How nice~

so, i don’t have to compute for j = 5/2.

J_x ( \frac{5}{2} ) = \frac {\hbar}{2} \begin {pmatrix} 0 & \sqrt{5} & 0 & 0& 0 & 0 \\ \sqrt{5} & 0 & \sqrt{8} & 0 & 0 & 0 \\ 0 & \sqrt{8} & 0 & \sqrt{9} & 0 & 0 \\ 0 & 0 & \sqrt{9} & 0 & \sqrt{8} & 0 \\ 0& 0 &0 & \sqrt{8} &0 & \sqrt{5} \\ 0 & 0 & 0 & 0 & \sqrt{5} & 0 \end {pmatrix}

but the physical reason for this trick , i don’t know. for the Pascal triangle, we understand the reason for each element – it is the multiplicity.

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