SU(3) for Spin 1 Particles?

referframe

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I am still learning about all the Groups related to the Dirac Equation for spin 1/2 particles. Apparently, the reason that the Hilbert Space for spin 1/2 particles is 2-dimensional is because when you try to map SU(2) to SO(3), the mapping is 2-to-1, i.e. SU(2) is a double cover for SO(3).

What about spin 1 particles. The spin vector space for those particles is 3-dimensional. Does that mean that SU(3) maps to SO(3) 1-to-1?

As usual, thanks in advance.
 

referframe

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Having just posted the above question, I believe I see an error in my reasoning. I am rephrasing the question:

I am still learning about all the Groups related to the Dirac Equation for spin 1/2 particles. Apparently, the reason that the spin vector for a spin 1/2 particle goes to its negative after a 2-pi rotation is because when you try to map SU(2) to SO(3), the mapping is 2-to-1, i.e. SU(2) is a double cover for SO(3).

What about spin 1 particles? The spin vector for that particle goes to itself after a 2-pi rotation. Does that mean that SU(3) maps to SO(3) 1-to-1?

As usual, thanks in advance.
 
Last edited:

Strilanc

Science Advisor
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Spin 1 means the angular momentum can be -1, 0 or +1? I think that makes it a quantum trit with operations in SU(3)? I would expect that to increase the many-to-one-ness when mapping to SO(2) instead of decreasing it. More specifically, I'd expect SU(3) to relate to SO(4) the way SU(2) relates to SO(3).

Please correct me if the above is wrong.

Related question: is the "SU(2) to SO(3) mapping is 2-to-1" statement intentionally ignoring global phase factors except for -1? Because there are four directions of rotation in SU(2) instead of three like in SO(3). There's x-wise, y-wise, z-wise, and phase-wise. At least, that's the issue when I recently tried to use quaternion spherical interpolation to gradually change between two unitary matrices: until I tweaked it to deal with the phase, the intermediate matrices ended up non-unitary.
 

dextercioby

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[...] Apparently, the reason that the spin vector for a spin 1/2 particle goes to its negative after a 2-pi rotation is because when you try to map SU(2) to SO(3), the mapping is 2-to-1, i.e. SU(2) is a double cover for SO(3).

What about spin 1 particles? The spin vector for that particle goes to itself after a 2-pi rotation. Does that mean that SU(3) maps to SO(3) 1-to-1?
.
Not the spin vector, but rather the wavefunction itself. For spin 1 particles, the symmetry group is still SU(2), no SU(3), but the <spin space> is 3 dimensional, no longer 2 dimensional, as for spin 1/2. This is from a purely Galilean perspective. .
 
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The symmetry group for any spin s particle is always SO(3) or SU(2), as they correspond to physical rotations. However the corresponding operators are different because you are representing SO(3) (SU(2)) in different spaces. For spin 1 you just have to find a representation of SO(3) on C^3.
 

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