Real special orthogonal group of dimension 3

Irreps of SO(3)

The Lie algebra 𝔰𝔬(3) has 2𝑗 +1 dimensional irreps 𝑑Γ𝑗 for 𝑗 ∈12ℕ0, while the group SO(3) only has irreps for 𝑗 ∈ℕ0. lie

Proof

Let 𝐽1,𝐽2,𝐽3 ∈𝔰𝔬(3) be the basis defined in Lie algebra, and 𝐽2 =⃗𝐽 ⋅⃗𝐽 be the quadratic Casimir element. Now consider a representation 𝑑Γ :𝔰𝔬(3) →𝔤𝔩(𝑉) on a finite-dimensional vector space 𝑉, which we will invoke implicitly. Let |𝑗,𝑚⟩ ∈𝑉 such that

𝐽2|𝑗,𝑚⟩=𝑗(𝑗+1)|𝑗,𝑚⟩𝐽3|𝑗,𝑚⟩=𝑚|𝑗,𝑚⟩

Then

𝐽±=𝐽1±𝑖𝐽2

are Ladder operators of 𝐽3. It follows that 𝐽±|𝑗,𝑚⟩ =(𝑚 ±1)|𝑗,𝑚±1⟩ transforms in the same irrep as |𝑗,𝑚⟩. Since 𝑉 is finite dimensional this must terminate at both ends, hence there exist 𝑝 <𝑞 such that

𝐽3|𝑗,𝑝⟩=𝑝|𝑗,𝑝⟩𝐽−|𝑗,𝑝⟩=0𝐽3|𝑗,𝑞⟩=𝑞|𝑗,𝑞⟩𝐽+|𝑗,𝑞⟩=0

In addition since

𝐽∓𝐽±=(𝐽1∓𝑖𝐽2)(𝐽1±𝑖𝐽2)=𝐽21+𝐽22±𝑖[𝐽1,𝐽2]=𝐽21+𝐽22∓𝐽3

it follows

𝐽2=𝐽23+𝐽∓𝐽±±𝐽3

and thus

𝐽2|𝑗,𝑝⟩=(𝐽23−𝐽3+𝐽+𝐽−)|𝑗,𝑝⟩=𝑝(𝑝−1)|𝑗,𝑝⟩𝐽2|𝑗,𝑞⟩=(𝐽23+𝐽3+𝐽−𝐽+)|𝑗,𝑞⟩=𝑝(𝑝+1)|𝑗,𝑞⟩

hence

𝑗(𝑗+1)=𝑝(𝑝−1)=𝑞(𝑞+1)

and since 𝑝 <𝑞 we have 𝑝 = −𝑗 and 𝑞 =𝑗. Now since 2𝑗 =𝑞 −𝑝 ∈ℕ0, we have 2𝑗 +1 dimensional irreps 𝑑Γ𝑗 of 𝔰𝔬(3) labelled by 𝑗 =0,12,1,…. Now assume 𝑑Γ has a corresponding group representation Γ𝑗. Then

𝛿𝑚𝑚′⟨𝑗,𝑚|𝑒−2𝜋𝑖𝐽3|𝑗,𝑚′⟩=⟨𝑗,𝑚|𝑒−2𝜋𝑖𝑚′|𝑗,𝑚′⟩=𝑒−2𝜋𝑖𝑚′𝛿𝑚𝑚′

which is a contradiction unless 𝑚′ ∈ℤ and thus 𝑗 ∈ℕ0.¹

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develop | en | SemBr

Footnotes

1. 2023. Groups and representations, §6.8, pp. 93–96 ↩