Compute the $\ell=1$ Rotation Operator $R_y(\theta_y)$

$$\bgroup\color{black}e^{i\theta L_y/\hbar}=\sum\limits_{n=0}^\infty{\left({i\theta L_y\over\hbar}\right)^n\over n!}\egroup$$

$$\bgroup\color{black}\left({L_y\over\hbar}\right)^0=\left(\matrix{1&0&0\cr 0&1&0\cr 0&0&1}\right)\egroup$$

$$\bgroup\color{black}\left({L_y\over\hbar}\right)^1={1\over\sqrt{2}i}\left(\matrix{0&1&0\cr -1&0&1\cr 0&-1&0}\right)\egroup$$

$$\bgroup\color{black}\left({L_y\over\hbar}\right)^2={1\over 2}\left(\matrix{1&0&-1\cr 0&2&0\cr -1&0&1}\right)\egroup$$

$$\bgroup\color{black}\left({L_y\over\hbar}\right)^3={1\over\sq... ... -2&0&2\cr 0&-2&0}\right) =\left({L_y\over\hbar}\right)\egroup$$

$$\bgroup\color{black} ... \egroup$$

All the odd powers are the same. All the nonzero even powers are the same. The $\hbar$s all cancel out. We now must look at the sums for each term in the matrix and identify the function it represents.

• The $n=0$ term contributes 1 on the diagonals.
• The $n=1,3,5,...$ terms sum to $\sin(\theta) \left({iL_y\over\hbar}\right)$.
• The $n=2,4,6,...$ terms (with a -1 in the matrix) are nearly the series for ${1\over 2}\cos(\theta)$. The $n=0$ term is is different. The middle matrix element is twice the other even terms.

Putting this all together, we get

$$\bgroup\color{black}R_y(\theta_y)=\left(\matrix{ {1\over 2}(1... ...2}}\sin(\theta_y)&{1\over 2}(1+\cos(\theta_y)) }\right).\egroup$$