Derive the Expression for Rotation Operator \bgroup\color{black}$R_z$\egroup *

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 \bgroup\color{black}$d\phi$\egroup) about the z-axis, we get the transformed coordinates

x'&=&x-d\phi y \\
y'&=&y+d\phi x.

We can Taylor expand any function \bgroup\color{black}$f$\egroup,

\begin{displaymath}\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\end{displaymath}

So the rotation operator for the function is

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

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

\begin{displaymath}\bgroup\color{black}R_z(\theta_z)=\lim_{n\to \infty}(1+{i\over\hbar}{\theta_z\over n} L_z)

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 \bgroup\color{black}$ln(x)$\egroup about \bgroup\color{black}$x=1$\egroup: \bgroup\color{black}$ln(x)=ln(1)+\left({1\over x}\right)_{x=1}(x-1)=(x-1)$\egroup.

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

So exponentiating, we get the identity.

Jim Branson 2013-04-22