Understanding Group Theory and Vector Rotation in 3D

[Phymath]

Homework Statement

I'm trying to see the relation of the rotation of a vector in a plane to the generator of rotations...

I want to see how $$e^{-i \theta J}$$ the rotation representation gives you the same result as acting on any vector with the rotation matrix say with the z direction fixed.

$$\[ \left( \begin{array}{ccc}<br /> Cos(\theta) & -Sin(\theta) & 0 \\<br /> Sin(\theta) & Cos(\theta) & 0 \\<br /> 0 & 0 & 1 \end{array} \right)\] = R_z$$

is $$R_z \textbf{v} = e^{-i \theta J_z^{(1)}} \textbf{v}$$

because a 3 dimensional vector has a spin one representation (right? because one full rotation gives the same vector back)

with $$J_z^{(1)} = \[ \left( \begin{array}{ccc}<br /> 1 & 0 & 0 \\<br /> 0 & 0 & 0 \\<br /> 0 & 0 & -1 \end{array} \right)\]$$

I get $$e^{-i \theta J_z^{(1)}} = \sum\frac{(-i \theta)^n}{n!}(J_z)^n = Cos(\theta)(J_z)^2-i J_z^{(1)} Sin(\theta)$$

$$e^{-i \theta J_z^{(1)}}= \[ \left( \begin{array}{ccc}<br /> Cos(\theta)-i Sin(\theta) & 0 & 0 \\<br /> 0 & 0 & 0 \\<br /> 0 & 0 & Cos(\theta)+i Sin(\theta) \end{array} \right)\]$$

when this matrix is applied to the vector $$\textbf{v}$$ it doesn't produce the same effect someone help finish the missing pieces thanks!

 

[Phymath]

ok I used the SO(3) matrix instead of SO(2) and that does give back the same matrix however if i still do it in 3-d i don't get the same matrix back

$$R(d\theta) = I - i d\theta J \rightarrow J = <br /> \[ \left( \begin{array}{ccc}<br /> 0 & -i & 0 \\<br /> i & 0 & 0 \\<br /> 0 & 0 & 0 \end{array} \right)\]$$

if you expand $$e^{-i \theta J} = \[ \left( \begin{array}{ccc}<br /> Cos(\theta) & -Sin(\theta) & 0 \\<br /> Sin(\theta) & Cos(\theta) & 0 \\<br /> 0 & 0 & 0 \end{array} \right)\]$$
with out the 1 in the bottom right (3,3)entry whys this?