# Lie algebra su(2)

1. Dec 14, 2008

### sunkesheng

hi ,i see from a book su(2) has the form
U$$\left(\hat{n},\omega\right)$$=1cos$$\frac{\omega}{2}$$-i$$\sigma$$sin$$\frac{\omega}{2}$$
in getting the relation with so(3),why we choose $$\frac{\omega}{2}$$,how about changing for $$\omega$$?
thank you

2. Dec 14, 2008

### Fredrik

Staff Emeritus
The ½ is there because $J_i=\frac 1 2 \sigma_i$ satisfies the commutation relations $[J_i,J_j]=i\epsilon_{ijk}J_k$.

3. Dec 14, 2008

### sunkesheng

thanks a lot,but i still cannot understand it ,because the book just gives the equation ,where can i find a detailed derivation?

4. Dec 14, 2008

### George Jones

Staff Emeritus
Note that U is an element of Lie group SU(2), not an element of the Lie algebra su(2).

Roughly, there is a factor of 1/2 because of the 2 to 1 relationship between the groups SU(2) and SO(3).

5. Dec 14, 2008

### Fredrik

Staff Emeritus
I don't think that's correct, but it's possible that I'm wrong. I think the only point of using SO(3) instead of SU(2) is that it guarantees that we can find an actual representation (with U(R')U(R)=U(R'R) for all R) instead of a projective representation. (If we take R and R' to be members of SO(3), there will sometimes be a minus sign in front of one of the U's).

I think the 1/2 appears only because a rotation operator is always $1-i\theta^iJ_i$ to first order in the parameters, with the $J_i$ satisfying the usual commutation relations.

6. Dec 14, 2008

### George Jones

Staff Emeritus
There is quite a lot of interesting stuff going on here, and I don't have time to tex it right now, but I stand by my statement. Note that what I wrote doesn't negate anything that you wrote; there are often a number of (somewhat equivalent) ways to look at the same thing.

Maybe in a couple of days I'll write a much longer post.

7. Dec 15, 2008

### sunkesheng

Last edited by a moderator: Apr 24, 2017