- #1
Splinter1
- 4
- 0
Hi. I know that the [itex]\sigma [/itex] matrices are the generators of the rotations in su(2) space. They satisfy
[tex] [\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k[/tex] It is conventional therefore to take [itex]J_i=\frac{1}{2}\sigma_i[/itex] such that [itex][J_i,J_j]=i\epsilon_{ijk}\sigma_k [/itex]. Isn't there a problem by taking these [itex] J_i [/itex] since [itex] \det J_i \neq 1[/itex]? (since we are talking about the special unitary group.)
Also, how does one arrive at the general form of the rotation matrix [itex]e^{i\bf{\sigma} \theta\cdot \bf{\hat{n}}/2}[/itex]? the factor of 1/2 obviously comes from the definition of J above. Where does the [itex] \hat n [/itex] come from?
[tex] [\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k[/tex] It is conventional therefore to take [itex]J_i=\frac{1}{2}\sigma_i[/itex] such that [itex][J_i,J_j]=i\epsilon_{ijk}\sigma_k [/itex]. Isn't there a problem by taking these [itex] J_i [/itex] since [itex] \det J_i \neq 1[/itex]? (since we are talking about the special unitary group.)
Also, how does one arrive at the general form of the rotation matrix [itex]e^{i\bf{\sigma} \theta\cdot \bf{\hat{n}}/2}[/itex]? the factor of 1/2 obviously comes from the definition of J above. Where does the [itex] \hat n [/itex] come from?
Last edited: