を持つので、これらに対応する固有状態をそれぞれ e n + {\displaystyle e_{\mathbf {n} }^{+}} 、 e n − {\displaystyle e_{\mathbf {n} }^{-}} とし、k = ?u, ?(u ? 1), …, (u ? 1), uに対し、 E n , k = c ( k ) e n + ⋅   ⋯   ⋅ e n + ⏟ u + k ⋅ e n − ⋅   ⋯   ⋅ e n − ⏟ u − k = c ( k ) S ( e n + ⊗ ⋯ ⊗ e n + ⏟ u + k ⊗ e n − ⊗ ⋯ ⊗ e n − ⏟ u − k ) {\displaystyle E_{\mathbf {n} ,k}=c(k)\underbrace {e_{\mathbf {n} }^{+}\cdot ~\cdots ~\cdot e_{\mathbf {n} }^{+}} _{u+k}\cdot \underbrace {e_{\mathbf {n} }^{-}\cdot ~\cdots ~\cdot e_{\mathbf {n} }^{-}} _{u-k}=c(k){\mathcal {S}}(\underbrace {e_{\mathbf {n} }^{+}\otimes \cdots \otimes e_{\mathbf {n} }^{+}} _{u+k}\otimes \underbrace {e_{\mathbf {n} }^{-}\otimes \cdots \otimes e_{\mathbf {n} }^{-}} _{u-k})} ∈ W 1 / 2 ⊙ 2 u = W u {\displaystyle \in W_{1/2}{}^{\odot 2u}=W_{u}} …(P1)

とする[28]。

ここでc(k)は正規化定数[28] c ( k ) = ( 2 u ) ! ( u + k ) ! ( u − k ) ! {\displaystyle c(k)={\sqrt {(2u)! \over (u+k)!(u-k)!}}} …(P2)

すると、 T ^ n ( E n , k ) = i ℏ ( D u ) ∗ ( X n ) ( E n , k ) = c ( k ) ⋅ S ( ∑ j = 1 2 u ( I ⊗ ⋯ ⊗ ( D 1 / 2 ) ∗ ( X n ) ∨ j ⊗ ⋯ ⊗ I ) ( e n + ⊗ ⋯ ⊗ e n + ⊗ e n − ⊗ ⋯ ⊗ e n − ) ) {\displaystyle {\hat {T}}_{\mathbf {n} }(E_{\mathbf {n} ,k})=i\hbar (D^{u})_{*}(X_{\mathbf {n} })(E_{\mathbf {n} ,k})=c(k)\cdot {\mathcal {S}}(\sum _{j=1}^{2u}(I\otimes \cdots \otimes {\overset {\overset {j}{\vee }}{(D^{1/2})_{*}(X_{\mathbf {n} })}}\otimes \cdots \otimes I)(e_{\mathbf {n} }^{+}\otimes \cdots \otimes e_{\mathbf {n} }^{+}\otimes e_{\mathbf {n} }^{-}\otimes \cdots \otimes e_{\mathbf {n} }^{-}))} = ℏ c ( k ) ⋅ S ( k ⋅ e n + ⊗ ⋯ ⊗ e n + ⊗ e n − ⊗ ⋯ ⊗ e n − ) = k ℏ ⋅ E n , k {\displaystyle =\hbar c(k)\cdot {\mathcal {S}}(k\cdot e_{\mathbf {n} }^{+}\otimes \cdots \otimes e_{\mathbf {n} }^{+}\otimes e_{\mathbf {n} }^{-}\otimes \cdots \otimes e_{\mathbf {n} }^{-})=k\hbar \cdot E_{\mathbf {n} ,k}}

なので、En,kは固有値 k ℏ {\displaystyle k\hbar } に対応する固有状態である。
昇降演算子詳細は「昇降演算子」を参照

nがx軸、y軸、z軸であるときの T ^ n {\displaystyle {\hat {T}}_{\mathbf {n} }} を T ^ x {\displaystyle {\hat {T}}_{x}} 、 T ^ y {\displaystyle {\hat {T}}_{y}} 、 T ^ z {\displaystyle {\hat {T}}_{z}} とし、 T ^ + := T ^ x + i T ^ y {\displaystyle {\hat {T}}_{+}:={\hat {T}}_{x}+i{\hat {T}}_{y}} T ^ − := T ^ x − i T ^ y {\displaystyle {\hat {T}}_{-}:={\hat {T}}_{x}-i{\hat {T}}_{y}}

とすると[14]、 T ^ ± ( E z , k ) {\displaystyle {\hat {T}}_{\pm }(E_{z,k})} = ℏ u ( u + 1 ) − k ( k + 1 ) ⋅ E z , k ± 1 {\displaystyle =\hbar {\sqrt {u(u+1)-k(k+1)}}\cdot E_{z,k\pm 1}}

である[14]。証明 D ^ + u := ( D ^ u ) ∗ ( X 1 ) + i ( D ^ u ) ∗ ( X 2 ) {\displaystyle {\hat {D}}_{+}^{u}:=({\hat {D}}^{u})_{*}(X_{1})+i({\hat {D}}^{u})_{*}(X_{2})} D ^ − u := ( D ^ u ) ∗ ( X 1 ) − i ( D ^ u ) ∗ ( X 2 ) {\displaystyle {\hat {D}}_{-}^{u}:=({\hat {D}}^{u})_{*}(X_{1})-i({\hat {D}}^{u})_{*}(X_{2})}