Derive the Expression for Rotation Operator $R_z$ *

The laws of physics do not depend on what axes we choose for our coordinate system- There is rotational symmetry. If we make an infinitesimal rotation (through and angle $d\phi$) about the z-axis, we get the transformed coordinates

$$\begin{eqnarray*} x'&=&x-d\phi y \\ y'&=&y+d\phi x. \end{eqnarray*}$$

We can Taylor expand any function $f$,

$$\bgroup\color{black}f(x',y')=f(x,y)-{\partial f\over\partial ... ...r\partial y} d\phi x =(1+{i\over\hbar}d\phi L_z)f(x,y).\egroup$$

So the rotation operator for the function is

$$\bgroup\color{black}R_z(d\phi)=(1+{i\over\hbar}d\phi L_z)\egroup$$

A finite rotation can be made by applying the operator for an infinitesimal rotation over and over. Let $\theta_z=n d\phi$. Then

$$\bgroup\color{black}R_z(\theta_z)=\lim_{n\to \infty}(1+{i\over\hbar}{\theta_z\over n} L_z) ^n=e^{i\theta_zL_z/\hbar}.\egroup$$

The last step, converting the limit to an exponential is a known identity. We can verify it by using the log of the quantity. First we expand $ln(x)$ about $x=1$: $ln(x)=ln(1)+\left({1\over x}\right)_{x=1}(x-1)=(x-1)$.

$$\bgroup\color{black}\lim_{n\to \infty}ln(1+{i\over\hbar}{\the... ...er\hbar}{\theta_z\over n} L_z)={i\over\hbar}\theta_z L_z\egroup$$

So exponentiating, we get the identity.