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.

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

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)

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

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

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

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