SO(3) 48 6 SO(3) t 6.1 u, v u = u 1 1 + u 2 2 + u 3 3 = u 1 e 1 + u 2 e 2 + u 3 e 3, v = v 1 1 + v 2 2 + v 3 3 = v 1 e 1 + v 2 e 2 + v 3 e 3 (6.1) i (e i ) e i e j = i j = δ ij (6.2) ( u, v ) = u v = ij u i v j i j = ij u i v j δ ij = u t v (6.3) ˆR 1 i = ˆR i = j j R ji = j j j ˆR i (6.4) ( i, j ) = ( kl k R ki, l R lj ) = kl δ kl R ki R lj = k R ki R kj = (R t R) ij (6.5) ( i, j ) = δ ij (6.6) R ki R kj = (R t R) ij = δ ij (6.7) k R t = R 1 R, u = ˆR u = i ˆR i u i = i j R ji u i (6.8)
SO(3) 49 u = Ru (6.9), i u iv i = i u iv i (C ) π π : G Hom(V, V ) : g D(g). π : R 3 V : i 1. : u u = u 1 u 2 u 3 (6.10) 6.2 i 1 0 0 R α (1) = 0 cos α sin α 0 sin α cos α R (2) β = R (3) γ = R (3) ψ 1 0 0 cos β 0 sin β 0 1 0 sin β 0 cos β cos γ sin γ 0 sin γ cos γ 0 0 0 1 = cos ψ sin ψ 0 (6.11) (6.12) R i = 1 cos ψ + 2 sin ψ e 3 y α, β, γ << 1 1 0 0 R α (1) 0 1 α 0 α 1 R (2) β R (3) γ 1 0 β 0 1 0 β 0 1 1 γ 0 γ 1 0 (6.13) 0 0 1
SO(3) 50 ( ), J i i 1 (x ) R (i) ϵ = 1 iϵj i (6.14) dr (i) θ dθ R (1) α = e iαj 1 1 iαj 1 = J 1 = = ij i (6.15) θ=0 0 0 0 0 0 i 0 i 0 0 0 0 0 0 i J 1 = 0 0 i, J 2 = 0 0 0 0 i 0 i 0 0 1 0 0 0 1 α 0 α 1 (6.16) (6.17), J 3 = 0 i 0 i 0 0 0 0 0 (6.18) [J i, J j ] = iϵ ijk J k (6.19) ϵ ijk { 1 (ijk) ϵ ijk = 1 (ijk) 0 (6.20) 6.3 i α R (i) α = e iαj i (6.21)
SO(3) 51 0 0 0 ij 1 = 0 0 1 = I 1 (6.22) 0 1 0 I1 2 = 0 0 0 0 1 0 0 0 1 = I 0 (6.23) I 2n 1 = ( 1) n I 0 I 2n+1 1 = ( 1) n I 1 (6.24) e iαj 1 1 = 1 + m! αm I1 m ( 1) n ( 1) n = 1 + (2n)! α2n I 0 + (2n + 1)! α2n+1 I 1 m=1 n=1 n=0 = 1 I 0 + cos αi 0 + sin αi 1 (6.25) e iαj 1 = 1 0 0 0 cos α sin α 0 sin α cos α (6.26) (e iαj i ) t = e iαj t i = e iαj i J t i = J i (6.27), (e iαj i ) = e iαj i = e iαj i J i = J i (6.28) Baker-Campbell-Hausdorff Baker-Campbell-Hausdorff p e X e Y = e X+Y + 1 1 [X,Y ]+ 2 12 [X Y,[X,Y ]]+ +[ [X,Y ]]]]+ (6.29) (1 + X + 1 2 X2 )(1 + Y + 1 2 Y 2 ) = 1 + X + Y + XY + 1 2 (X2 + Y 2 ) + (6.30)
SO(3) 52 1+(X +Y + 1 2 [X, Y ])+ 1 2 (X +Y )2 = 1+X +Y + 1 2 [X, Y ]+ 1 2 (X2 +Y 2 +XY +Y X)+ (6.31) q.e.d. J i J i BCH θ = θn R αβγ = e i i θ ij i (6.32) e iθn J (6.33) n J n θ = θ J i [, ] (J i, [, ]) SO(3) so(3) Lie(G) so(3) 6.4 Exp : g G (6.34) 3 S S x y x y y 1 y α y γ, β 1. z α R z α x, y, z x 1, y 1, z 1 = z 0 α < 2π 2. y 1 β S x y R y 1 β = Rz αr y β Rz α (6.35) z y z (x 1, y 1, z 1 ) (x 2, y 2 = y 1, z 2 ) 0 β π 3. z 2 γ R z 2 γ = R y 1 β Rz 1 γ R y 1 β (6.36) (x 2, y 2 = y 1, z 2 ) (X, Y, Z = z 2 ) 0 γ < 2π
SO(3) 53 4. R z 1 γ = R z αr z γr z α = R z γ (6.37) R αβγ = R z 2 γ R y 1 β Rz α = R z αr y β Rz γ (6.38) 5. (ϕ, θ, ψ) R ϕθψ = e iϕj 3 e iθj 2 e iψj 3 cos ϕ sin ϕ 0 = sin ϕ cos ϕ 0 0 0 1 cos ϕ sin ϕ 0 = sin ϕ cos ϕ 0 0 0 1 = cos θ 0 sin θ 0 1 0 sin θ 0 cos θ cos ψ sin ψ 0 sin ψ cos ψ 0 0 0 1 cos θ cos ψ cos θ sin ψ sin θ sin ψ cos ψ 0 sin θ cos ψ sin θ sin ψ cos θ cos θ cos ϕ cos ψ sin ϕ sin ψ cos θ cos ϕ sin ψ sin ϕ cos ψ sin θ cos ϕ cos θ sin ϕ cos ψ + cos ϕ sin ψ cos θ sin ϕ sin ψ + cos ϕ cos ψ sin θ sin ϕ sin θ cos ψ sin θ sin ψ cos θ 6. e 3 R ϕ,θ e 3 = r r, r = (x, y, z), r2 = r 2 (6.40) (ϕ, θ) S 2 6.5 D : G ˆR αβγ D(R αβγ ) Mat(C) (6.41) R αβγ D(R αβγ )D(R α β γ ) = D(R αβγr α β γ ) (6.42) (6.39) Hom(V,V)
SO(3) 54 6.5.1 [Ĵi, Ĵj] = iϵ ijk Ĵ k (6.43) 2 2 ( ) ( ) ( ) 0 1 0 i 1 0 σ 1 =, σ 1 0 2 =, σ i 0 3 = (6.44) 0 1 σ i σ j = δ ij 1 + iϵ ijk σ k (6.45) 1 2 2 [σ i, σ j ] = 2iϵ ijk σ k (6.46) Ĵ i = 1 2 σ i (6.47) D 1 (R θ ) = e i 1 2 θ iσ i (6.48) 2 2 2 2 j = 1 2 ( ) u1 2 u = = (u α ) D 1 (R ϕθψ ) = e i 1 2 ϕσ 3 e i 1 2 θσ 2 e i 1 2 ψσ 3 = 2 ( cos θ 2 e i 1 2 (ψ+ϕ) sin θ 2 e i 1 2 (ϕ ψ) sin θ 2 ei 1 2 (ϕ ψ) cos θ 2 ei 1 2 (ψ+ϕ) ) u 2 (6.49) Proof : g = g 3 (ϕ)g 2 (θ)g 3 (ψ) = ( e i 1 2 ϕ 0 0 e i 1 2 ϕ ) ( cos θ 2 sin θ 2 sin θ 2 cos θ 2 ) ( e i 1 2 ψ 0 0 e i 1 2 ψ ) (6.50) q.e.d.
SO(3) 55 2 1 2π 4π 2 ϕ = 0, 2π 2 SU(2) SU(2) SO(3) = SU(2)/Z 2 6.5.2 1 Z 2 SU(2) SO(3) 1 (6.51) ( hat ) 1. Ĵ 3, Ĵ ± = Ĵ1 ± iĵ2 (6.52) Ĵ± Ĵ + = Ĵ Ĵ = ϕĵ3 + ωĵ+ + ωĵ (6.53) [J 3, J ± ] = ±J ±, [J +, J ] = 2J 3 (6.54) J ± J 3 ±1 2. (Casimir ) J 2 = J 2 1 + J 2 2 + J 2 3 = J 2 3 + 1 2 (J +J + J J + ) (6.55) Proof : [J 2, J 3 ] = 0 (6.56) [J 3, J + J ] = [J 3, J + ]J + J + [J 3, J ] = J + J J + J = 0
SO(3) 56 [J 3, J J + ] = [J 3, J ]J + + J [J 3, J + ] = J J + + J J + = 0 (6.57) J 2 q.e.d. [J 2, J i ] = 0 (6.58) J 2 J 2 J 3 3. J 3 J 3 Ĵ 3 m = m m, m m = δ m m m Ĵ3 m = mδ mm (6.59) J 3 m 4. m J 3 J ± m = (m ± 1)J ± m (6.60) m J 3 J ± m = m m J ± m = (m ± 1) m J ± m (6.61) m = m ± 1 m J ± m (m m ± 1) m J ± m = 0 (6.62) 5. m (J + J J J + ) m = m J + m 1 m 1 J m m J m + 1 m + 1 J + m (6.63) f(m) = m J + m 1 m 1 J m = m J + m 1 2 = m 1 J m 2 (6.64) 12 f(m) f(m) = m J + m 1 2 f(m) 0 f(m) f(m + 1) = 2m (6.65) 12 J + m 1 = f(m) m J m = f(m) m 1
SO(3) 57 6. highest weight state J 2 f(m) m (J 2 J3 2 ) m = 1 2 m (J +J + J J + ) m = 1 (f(m) + f(m + 1)) > 0 (6.66) 2 m J 2 m m 2 > 0 (6.67) m 2 m j 0 J + j j + 1 = 0 (6.68) highest weight state f(j + 1) = 0 f(j) = 2j (6.69) f(m) = 2m + + 2j = 2 ( m + (m + 1) + + j) = (j + m)(j m + 1) (6.70) f(m) = (j + m)(j m + 1) = j(j + 1) m(m 1) (6.71). 7. (6.67) m J m 1 k m f(m) = m 1 J m 2 J j k J k+1 j = 0 (6.72) f(j k) = (2j k)(k + 1) = 0 (6.73) k > 0 k = 2j j 8. k j j = k 2 > 0 (6.74)
SO(3) 58 9. J 2 J i j J 2 = J 2 3 + 1 2 (J +J + J J + ) = J 2 3 J 3 + J + J (6.75) j, m J 2 j, m = j, m J3 2 J 3 + J + J j, m = m 2 m + j, m J + m 1 m 1 J j, m = m 2 m + f(m) = j(j + 1) (6.76) m 10. j j 2j + 1 j j, m (6.77) J 2, J 3 J 2 j, m = j(j + 1) j, m, J 3 j, m = m j, m (6.78), J ± j, m+1 J + j, m = f(m + 1) = j(j + 1) m(m + 1) = m J m+1 (6.79) j, m 13 j, m 1 = ((j + m)(j m + 1)) 1 2 J j, m (6.80) j, m + 1 = ((j m)(j + m + 1)) 1 2 J+ j, m (6.81) 13 1. D j j 1 2 Z ( )J2 j(j + 1) 2. V j j, m J 3 (weight)m weight m m = j, j 1,, j V j 2j + 1 j, j 1 = ((2j)) 1 2 J j, j j, j 2 = ((2j 1)2) 1 2 J j, j 1 = (2j(2j 1)2 1) 1 2 J 2 j, j j, j k = (2j (2j k + 1)k!) 1 2 J k (2j k)! j, j = J k (2j)!k! j, j j, m = (2j (j + m + 1)k!) 1 2 J k (j + m)! j, j = (2j)!(j m)! J j m j, j (6.82)
SO(3) 59 6.5.3 1. j = 1 2 m = ± 1 2, 1 2, ± 1 2 1 2, 1 2 J + 1 2, 1 2 = f( 1 2 ) = 1 = m J m + 1 (6.83) ( 0 1 D 1 (J + ) = 2 0 0 ) ( 0 0, D 1 (J ) = 2 1 0 1 2, m J 3 1 2, m = mδ m m, m = ± 1 2 ) (6.84) (6.85) J 1 = 1 2 (J + + J ), J 2 = i 1 2 (J + J ) 2. j = 1,m = 1, 0, 1 D 1 (J i ) = 1 2 2 σ i (6.86) 1, 1 J + 1, 0 = f(1) = 2, 1, 0 J + 1, 1 = 2 (6.87) J 3 = 1 0 0 0 0 0 0 0 1 J + J = 0 2 0, J + = 0 0 2, J = J + (6.88) 0 0 0 2 0 0 0 0 0 0 2 0, J J + = 0 2 0 (6.89) 0 0 0 0 0 2 + = x + i y, 0 = z, = x i y (6.90) J 3 x = i y, J 3 y = i x (6.91) J 1 y = i z, J 1 z = i x, J 2 x = i z, J 2 z = i x (6.92) (J 1 ± ij 2 ) + = ij 1 y ± ij 2 x = z ± z (J 1 ± ij 2 ) = ij 1 y ij 2 x = z z (6.93) D 1 (J i ) (6.18)
SO(3) 60 6.6 1. 2. ˆR θ u = i i ˆR θ j u j (6.94) i T R ϕ = T R (ϕ) r r r org 14 ϕ (r) = (T R ϕ)(r) = ϕ(r 1 r) (6.96) T R2 T R1 ϕ(r) = T R2 ϕ (r) = T R2 ϕ(r 1 1 r) = ϕ (R 2 1 r) = ϕ(r 1 1 R 2 1 r) = ϕ((r 2 R 1 ) 1 r) = T R2 R 1 ϕ(r) (6.97) T R z T R ϵ z ϕ(r) = ϕ(r ϵ 1 z r) = ϕ(x + ϵy, ϵx + y, z) (6.98) ϵ T R ϵ z ϕ(r) = (1 iϵl z )ϕ(r) (6.99) l z l z = i(x y y x ) (6.100) l k (l x, l y, l z ) l k = iϵ ijk x i j (6.101) [θ k l k, x m ] = iθ k ϵ kij x i [ j, x m ] = iθ k ϵ mki x i = i( θ x) m (6.102) 14 r e i e i = T R (e i ) = e j R ji (6.95) P x i x i = R 1 ij x j
SO(3) 61 L k = l k = x p ( ) Proof : [l k, l k ] = iϵ kk jl j (6.103) [l k, l k ] = [(ϵ ijk x i j ), (ϵ i j k x i j )] = (ϵ ijkx i ϵ jj k j ϵ i ik x i ϵ ijk j ) = (x k k x k k ) = ϵ kk iϵ ijl x j l = iϵ kk jl j (6.104) l k 15 q.e.d. T R ϵ θ ϕ(r) = (1 iϵ θ l)ϕ(r) (6.105) R(θ) = e i θ l = e i θ L/ (6.106) L D j (θ) R(θ) 6.7 6.7.1 1 L k l k l k = iϵ kij x i j (6.107) l k SO(3) l z l ± = l 1 ± il 2 = i(y z z y ) ± (z x x z ) = ( x iy) z ± z( x ± i y ) = x ± z ± 2z (6.108) l z = i(x y y x ) = 1 2 [(x + iy)( x i y ) (x iy)( x + i y )] (6.109) 15
SO(3) 62 x ± = x ± iy, ± = 1 2 ( x i y ), [ +, x + ] = [, x ] = 1 (6.110) l ± = (x ± z 2z ), l z = x + + x (6.111) highest weight state a l 6.7.2 l + Ψ ll (x, y, z) = 0 (6.112) Ψ ll = a l x l + (6.113) l k = (x z 2z + ) k (6.114) (l + m)! Ψ lm = a l (2l)!(l m)! ll m x l + (6.115) H = 1 2m p2 + V (r) = 2 2m 2 + V (r) (6.116) = 2 = 1 ( r 2 ) + 1 r 2 r r r ( l 2 ) 2 (6.117) l 2 = 1 ( sin θ ) + 2 sin θ θ θ ϕ 2 (6.118) χ(r)ψ(θ, ϕ) l 2 ψ = αψ (6.119) l 1 = i(sin ϕ θ + cot θ cos ϕ ϕ )
SO(3) 63 l 2 = i( cos ϕ θ l 3 = i ϕ + cot θ sin ϕ ϕ ) (6.120) l 2 1 = ( sin θ θ sin θ θ + 1 sin 2 θ 2 ϕ 2 ) (6.121) l + = e iϕ ( θ + i cot θ ϕ ) (6.122) 6.7.3 l = e iϕ ( θ + i cot θ ϕ ) (6.123) l 3 l 2 16 θ, ϕ Ĵi l, m = l i θ, ϕ l, m (6.124) Ĵi l θ, ϕ jm Y lm (θ, ϕ) = θ, ϕ l, m l, m Ĵi l, m = dω l, m θ, ϕ θ, ϕ Ĵi l, m = dω l, m θ, ϕ l i θ, ϕ l, m = dωy lm (θϕ)l i Y lm (θϕ) (6.125) l, m Ĵi l, m = dωy lm (θϕ)l i Y lm (θϕ) (6.126) 16 1. l 2. m
SO(3) 64 dω π 2π l i l i Y lm = θϕ Ĵi l, m = 0 0 sin θdθdϕ (6.127) l m = l Y lm lm Ĵi lm (6.128) 6.7.4 Y lm l 3 l 3 Y lm = my lm (6.129) Y lm (θ, ϕ) = ψ lm (θ)e imϕ (6.130) highest weight state l 3 Y ll = ly ll Y ll (θ, ϕ) = ψ ll (θ)e ilϕ (6.131) Y m l lowest weight state Y l l l + 1. lowest weight state lowest weight state ϕ Y l l = ψ l l (θ)e ilϕ (6.132) lowest weight state l Y l l = e i( l 1)ϕ ( θ + l cot θ)ψ l l(θ) = 0 (6.133) a ψ l l (θ) = a(sin θ) l (6.134) l + 2. l + ψ lm e imϕ = (l m)(l + m + 1)ψ lm+1 (θ)e i(m+1)ϕ (6.135) ψ l l k ψ l,k l
SO(3) 65 ψ lm = a( 1) l+m (l m)! d (2l)!(l + m)! sinm θ( d cos θ )l+m sin 2l θ (6.136) Proof : m = l m = k ψ lk+1 = l + ψ lk e ikϕ = e iϕ ( θ + i cot θ ϕ )ψ lke ikϕ 1 (l k)(l + k + 1) sin k θ θ ( 1 sin k θ ψ lk) = e i(k+1)ϕ ( θ k cot θ)ψ lk = e i(k+1)ϕ sin k θ θ ( 1 sin k θ ψ lk) (6.137) 1 = sin k θ (l + k)(l k + 1) θ ( 1 (l k)! d sin k θ a( 1)l+k (2l)!(l + k)! sink θ( d cos θ )l+k sin 2l θ) = sin k+1 θ(a( 1) l+k+1 (l k 1)! (2l)!(l + k + 1)! ( d d cos θ )l+k+1 sin 2l θ) (6.138) m = k + 1 3. a dω = sin θdθdϕ l, l l, l = 2π 0 dϕ π 0 dθ sin θψ l, lψ l, l = 2π a 2 π a = 1 (2l + 1)! 2 l l! 4π 0 sin 2l+1 θ = 2π a 2 (s l l!) 2 2 sl + 1 = 1 (6.139) (6.140) Y lm (θ, ϕ) = ( 1)l+m 2l + 1 (l m)! d 2 l l! 4π (l + m)! sinm θ( d cos θ )l+m sin 2l θe imϕ (6.141)