- #1
wdlang
- 307
- 0
it is generally known that there is a two-to-one automorphism from su(2) to so(3)
but consider the problem in this way:
all elements in so(3) are of the form (up to a unitary transform of the basis)
R(\alpha,\beta.\gamma)=e^{-i\alpha F_z} e^{-i\beta F_y} e^{-i \gamma F_z}
where F_x, F_y, F_z are the 3*3 spin operators
all elements in su(2) are of the form
R(\alpha,\beta.\gamma)=e^{-i\alpha \sigma_z/2} e^{-i\beta \sigma_y/2} e^{-i \gamma \sigma_z/2}
where \sigma_{x,y,z} are pauli matrices
F_{x,y,z} and \sigma_{x,y,z}/2 are of the same lie algebra!
so i think there should be a one-to-one correspondence between so(3) and su(2).
but consider the problem in this way:
all elements in so(3) are of the form (up to a unitary transform of the basis)
R(\alpha,\beta.\gamma)=e^{-i\alpha F_z} e^{-i\beta F_y} e^{-i \gamma F_z}
where F_x, F_y, F_z are the 3*3 spin operators
all elements in su(2) are of the form
R(\alpha,\beta.\gamma)=e^{-i\alpha \sigma_z/2} e^{-i\beta \sigma_y/2} e^{-i \gamma \sigma_z/2}
where \sigma_{x,y,z} are pauli matrices
F_{x,y,z} and \sigma_{x,y,z}/2 are of the same lie algebra!
so i think there should be a one-to-one correspondence between so(3) and su(2).