# SU(2) and SU(3) representations to describe spin states

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• spin_100
spin_100
Spin 1/2 particles are two states system in C^2 and so it is natural for the rotations to be described by SU(2), for three states systems like spin - 1 particle, Why do we still use SU(2) and not SU(3) to describe the rotations? Is it possible to derive them without resorting to the eigenvalue conditions of J^2 and J_z, i.e. purely mathematically? I am able to derive this for the spin -1/2 case from the condition of Unitarity and det=1.

Why should we use SU(3)? Do you know why we use SU(2)? SU(2) is double cover of rotation group SO(3), and even considering higher spins we, in a sense, think about rotations in physical 3 dimensional space (namely we are considering higher dimensional representations of double cover of SO(3)). Why we use the cover instead of SO(3) is another story. You should delve into group representation theory, it really clarifies everything.

vanhees71, topsquark and dextercioby
spin_100 said:
Why do we still use SU(2) and not SU(3) to describe the rotations?
Because the N in SU(N) doesn't refer to the number of basis states. It is linked to the number of spatial dimensions of the world (for how that link works, I recommend taking @weirdoguy's advice and learning about why SU(2) is the double cover of SO(3) and why, when we include spin-1/2, that makes SU(2) the correct group for representing spatial rotations), and that doesn't change when you look at spin-1 particles instead of spin-1/2 particles.

What does change when you look at spin-1 vs. spin-1/2 particles is the representation of SU(2) that you use. Heuristically, for spin-1/2 particles you use the representation of SU(2) that uses 2x2 matrices, whereas for spin-1 particles you use the representation that uses 3x3 matrices. In other words, the size of the matrices in the representation is what refers to the number of basis states. (There is a lot more here that I am sweeping under the rug, even though you labeled this as an "A" level thread; a real "A" level discussion of this topic would take a book, and there are indeed plenty of them.)

vanhees71 and topsquark
PeterDonis said:
Because the N in SU(N) doesn't refer to the number of basis states. It is linked to the number of spatial dimensions of the world (for how that link works, I recommend taking @weirdoguy's advice and learning about why SU(2) is the double cover of SO(3) and why, when we include spin-1/2, that makes SU(2) the correct group for representing spatial rotations), and that doesn't change when you look at spin-1 particles instead of spin-1/2 particles.

What does change when you look at spin-1 vs. spin-1/2 particles is the representation of SU(2) that you use. Heuristically, for spin-1/2 particles you use the representation of SU(2) that uses 2x2 matrices, whereas for spin-1 particles you use the representation that uses 3x3 matrices. In other words, the size of the matrices in the representation is what refers to the number of basis states. (There is a lot more here that I am sweeping under the rug, even though you labeled this as an "A" level thread; a real "A" level discussion of this topic would take a book, and there are indeed plenty of them.)
Thanks. That clears a lot of things for me. So generators of SU(2) in all representations of SU(2) follow the commutation relations, i.e [J_1 , J_2 ] = ih J_3 ? Also could you recommend a beginner book for learning more about this? I have studied abstract algebra. Are there any other prerequisites?

weirdoguy said:
Why should we use SU(3)? Do you know why we use SU(2)? SU(2) is double cover of rotation group SO(3), and even considering higher spins we, in a sense, think about rotations in physical 3 dimensional space (namely we are considering higher dimensional representations of double cover of SO(3)). Why we use the cover instead of SO(3) is another story. You should delve into group representation theory, it really clarifies everything.
Also why do we choose the generators to satisfy the commutation relations? I am not able to relate it with rotation? It seems natural for 3D but not sure about Spin -1/2 particles

spin_100 said:
Also why do we choose the generators to satisfy the commutation relations?

We do not choose it, commutation relations are kind of forced by the definitions of the groups we are considering.

spin_100 said:
generators of SU(2) in all representations of SU(2) follow the commutation relations, i.e [J_1 , J_2 ] = ih J_3 ?
Yes.

I guess you can look up what Lie algebra of a given Lie group is. Generators of Lie group form a basis of this algebra. There are also instights about SU(2) written by @fresh_42, but these are more mathematical oriented.

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