Parametrization of su(2) group

[wdlang]

all elements of su(2) can be written as

\exp(iH)

with H being a traceless hermitian matrix

thus H can be written as the sum of \sigma_x,\sigma_y,\sigma_z

H=\theta (n_x \sigma_x + n_y \sigma_y+ n_z \sigma_z).

Here (n_x,n_y,n_z) is a unit vector in R^3.

we can take \theta in the interval (-\pi,\pi]

Thus the su(2) group has the same parameter space as SO(3) group.

what is wrong?

 

[DrDu]

Not quite. A rotation about the angle phi is obtained with theta=phi/2, hence all rotations
in the interval phi in ]-pi,pi] are already obtained with theta in ]-pi/2, pi/2].
E.g. a 180 deg rotation of around x is obtained by exp(i pi/2 sigma_x)=i sigma_x which is not equal to a -180 deg rotation exp(-i pi/2 sigma_x)=-i sigma_x. Especially, a rotation of 360 deg is equal to -1 and not to 1.

 

[wdlang]

Not quite. A rotation about the angle phi is obtained with theta=phi/2, hence all rotations
in the interval phi in ]-pi,pi] are already obtained with theta in ]-pi/2, pi/2].
E.g. a 180 deg rotation of around x is obtained by exp(i pi/2 sigma_x)=i sigma_x which is not equal to a -180 deg rotation exp(-i pi/2 sigma_x)=-i sigma_x. Especially, a rotation of 360 deg is equal to -1 and not to 1.

but at which step i am wrong?

 

[wdlang]

i now see one traphole

for \theta=\pi, \exp(iH) is -1 regardless of the direction of the vector n

so in contrast to so(3), where on the surface of the pi-radius sphere, two antipodal points are identified, for su(2), the whole surface is identified.

 

[DrDu]

Exactly.