SOLUTIONS TO PROBLEMS IN LIE ALGEBRAS IN PARTICLE PHYSICS BY HOWARD GEORGI STEPHEN HANCOCK

Transcript

1 SOLUTIONS TO PROBLEMS IN LIE ALGEBRAS IN PARTICLE PHYSICS BY HOWARD GEORGI STEPHEN HANCOCK

2 STEPHEN HANCOCK Chpter 6 Solutions 6.A. Clerly NE α+β hs root vector α+β since H i NE α+β = NH i E α+β = N(α+β) i E α+β = (α+β) i NE α+β. Becuse the nonzero weights uniquely specify the sttes for the djoint representtion up to normliztion fctor it suffices to show tht [E α, E β ] hs root vector α + β. Indeed, since E α E β = [E α, E β ] nd H i E α E β = [H i, E α ] E β + E α H i E β = α i E α E β + E α β i E β = (α + β) i E α E β, we see tht H i [E α, E β ] = (α + β) i [E α, E β ]. Hence [E α, E β ] = NE α+β. If α + β is not root, then N = 0. 6.B. Suppose [E α, E β ] = NE α+β. By 6.A, we lso hve [E α, E α β ] = N E β nd [E β, E α β ] = N E α for some N nd N. Using the Jcoi identity nd [E γ, E γ ] = γ H, we otin 0 = [E α, [E β, E α β ]] + [E β, [E α β, E α ]] + [E α β, [E α, E β ]] = [E α, N E α ] + [E β, N E β ] + [E α β, NE α+β ] = N α H N β H N(α + β) H = ((N N)α (N + N)β) H. Since α nd β re linerly independent, nd the genertors H i (components of H) re linerly independent, it follows tht N = N nd N = N. Therefore, [E β, E α β ] = NE α nd [E α, E α β ] = NE β. 6.C. Tke H = σ 3 = dig(,,, ) nd H = σ 3 τ 3 = dig(,,, ). For the four dimensionl representtion, the sttes nd ssocited weights vectors re clerly (, 0, 0, 0) T with weight (, ) (0,, 0, 0) T with weight (, ) (0, 0,, 0) T with weight (, ) (0, 0, 0, ) T with weight (, ). The weights of the djoint representtion re the differences of these weights, long with the two elements of the Crtn sulger. The distinct differences, up to sign, re (, ) (, ) = (0, ), (, ) (, ) = (, ), (, ) (, ) = (, 0), nd (, ) (, ) = (, ), so the roots re (0, 0), (0, 0), (0, ), (, ), (, 0), (, ), (0, ), (, ), (, 0), (, ). H H

3 SOLUTIONS TO PROBLEMS IN LIE ALGEBRAS IN PARTICLE PHYSICS BY HOWARD GEORGI 3 Chpter 8 Solutions 8.A. The simple roots re the positive roots tht cnnot e written s the sum of other positive roots, nmely α = (0, ) nd α = (, ). Roots (, 0) nd (, ) re not simple ecuse (, 0) = (, ) + (0, ) nd (, ) = (, ) + (0, ). The fundmentl weights µ j re defined y The fundmentl weights re therefore α j µ k α j = δ jk. µ = (, ) nd µ = (, 0) since (0, ) µ (, ) µ (0, ) = (, ) = nd The ngle etween α nd α is (, ) µ (0, ) µ (, ) = (0, ) = 0. cos θ α α = α α (0, ) (, ) α α = (0, ) (, ) = = θ α α = 35. The Dynkin digrm for the lger is thus the following. 8.B. We my chnge sis nd consider the following six genertors for the lger: 0 0 J = σ + σ η = K = σ σ η = 0 0 J = σ + σ η J 3 = σ 3 + σ 3 η = = We esily clculte tht 0 0 i i 0 0 i i i i 0 0 i i K = σ σ η K 3 = σ 3 σ 3 η = = [J i, J j ] = iε ijk J k, [K i, K j ] = iε ijk K k, [J i, K j ] = 0. Now tke J 3 nd K 3 s the Crtn genertors nd 0 0 J + = J + ij = 0 0 J = J ij = K + = K + ik = K = K ik = i i 0 0 i i i i 0 0 i i

4 STEPHEN HANCOCK s the other four genertors for the djoint representtion. We hve [J 3, J ± ] = ±J ±, [K 3, J ± ] = 0, [J 3, K ± ] = 0, [K 3, K ± ] = ±K ±. The six root vectors re therefore J 3 with root (0, 0) J + with root (, 0) J with root (, 0) K 3 with root (0, 0) K + with root (0, ) K with root (0, ). We should lso hve [J +, J ] = (, 0) (J 3, K 3 ) = J 3 nd [K +, K ] = (0, ) (J 3, K 3 ) = K 3, nd it is esily checked tht these re stisfied. K 3 J 3 Both {J, J, J 3 } nd {K, K, K 3 } re nontrivil invrint sulgers, nd thus the lger is not simple. The commuttion reltions show tht the group generted is in fct SU() SU(), nd thus there re no nontrivil elin sugroups nd the group is semisimple. The simple roots re α = (, 0) nd α = (0, ). These re orthogonl, i.e., θ α α = 90, nd so the Dynkin digrm is s follows. 8.C. We re given α =, α =, α 3 =, nd the Dynkin digrm elow. Using Georgi s convention, the Crtn mtrix is defined y A ji = αj α i. α i It follows tht α i A ji = α j A ij nd A ji A ij = cos θ α i α j. Note tht for i j, cos θ α i α j is the numer of lines etween αi nd α j in the Dynkin digrm. Clerly A = A = A 33 = nd A 3 = A 3 = 0. The ove reltions give A /A = A A = nd A 3 /A 3 = A 3 A 3 =, with nonpositive solution A =, A =, A 3 =, A 3 =. The Crtn mtrix is 0. 0

5 SOLUTIONS TO PROBLEMS IN LIE ALGEBRAS IN PARTICLE PHYSICS BY HOWARD GEORGI 5 Chpter Solutions.A. Using the method discussed in the text, we clculte Cnceling columns with 3 oxes (fctors of ε), we hve shown = or (, ) (, ) = (, ) (5, 0) (, 3) (3, ) (3, ) (0, ) (, ) (, ) (, 0) (0, ). In terms of the corresponding dimensionlities, we cn write this s 5 5 = , where we hve defined 5 (, ) nd 5 (, 0). We use the following procedure to determine which representtions pper symmetriclly in the product nd which pper ntisymmtriclly. Trnspositions in rows contriute fctor of +,

6 6 STEPHEN HANCOCK while trnspositions in columns contriute fctor of. It follows tht or c c c d c c c d c c c d c c c d c c c d 3.E. We clculte + c c c d c c c d c c c d + c c c d c c d c c c c d c c c d c c c d c c c d c c c d + c c c d + + [(, ) (, )] S = (, ) (3, ) (0, ) (, ) (0, ) [(, ) (, )] AS = (5, 0) (, 3) (3, ) (, ) (, 0) [5 5] S = 60 S 5 S 5 3 [5 5] AS = AS 5 AS 6. Chpter 3 Solutions = so tht [] [, ] = [,, ] [3, ]. Using the fctors over hooks rule, we hve nd D([] [, ]) = D[] D[, ] = D([,, ] [3, ]) = D[,, ] + D[3, ] = = N(N )(N + + N ) 8 N(N ) N(N + ) N(N + )(N + )(N ) = N(N )(N) 8 c c c d c c d c c c c d c c c d = N (N ) N(N + )(N )(N ) + = N (N ). Thus D([] [, ]) = D([,, ] [3, ]), nd the dimensions check out for ritrry N. Chpter 9 Solutions 9.A. We must check tht the elements σ, τ, η, nd σ τ η c close under commuttion. Clerly nd we know tht [σ, τ ] = [τ, η ] = [η, σ ] = 0, [σ, σ ] = iε c σ c, [τ, τ ] = iε c τ c, [η, η ] = iε c η c.

7 SOLUTIONS TO PROBLEMS IN LIE ALGEBRAS IN PARTICLE PHYSICS BY HOWARD GEORGI 7 Next, we hve Finlly, [σ, σ τ c η d ] = [σ, σ ]τ c η d = iε e σ e τ c η d [τ, σ τ c η d ] = [τ, τ c ]η d σ = iε ce σ τ e η d [η, σ τ c η d ] = [η, η d ]σ τ c = iε de σ τ c η e. [σ τ η c, σ d τ e η f ] = [σ, σ d ]τ e τ η f η c + [τ, τ e ]η f η c σ σ d + [η c, η f ]σ σ d τ τ e = iε dg σ g (δ e iε eh τ h )(δ cf iε cfi η i ) + iε eh τ h (δ cf iε cfi η i )(δ d + iε dg σ g ) + iε cfi η i (δ d + iε dg σ g )(δ e + iε eh τ h ) = iδ e δ cf ε dg σ g + δ e ε cfi ε dg η i σ g + δ cf ε dg ε eh σ g τ h iε dg ε eh ε cfi σ g τ h η i + iδ cf δ d ε eh τ h δ cf ε dg ε eh σ g τ h + δ d ε eh ε cfi τ h η i + iε dg ε eh ε cfi σ g τ h η i + iδ d δ e ε cfi η i δ d ε eh ε cfi τ h η i δ e ε cfi ε dg η i σ g iε dg ε eh ε cfi σ g τ h η i = iδ e δ cf ε dg σ g + iδ cf δ d ε eh τ h + iδ d δ e ε cfi η i iε dg ε eh ε cfi σ g τ h η i. The terms lying outside the given set of mtrices cnceled! So we see tht the lger is closed, i.e., the 36 mtrices form Lie lger. Tke the Crtn genertors to e σ 3 = H = dig(,,,,,,, ) τ 3 = H = dig(,,,,,,, ) η 3 = H 3 = dig(,,,,,,, ) σ 3 τ 3 η 3 = H = dig(,,,,,,, ). The weights of the defining representtion re therefore ν = (,,, ), ν = (,,, ), ν 3 = (,,, ), ν = (,,, ), ν 5 = (,,, ), ν 6 = (,,, ), ν 7 = (,,, ), ν 8 = (,,, ). Noting tht ν 5 = ν, ν 6 = ν 3, ν 7 = ν, nd ν 8 = ν, ll differences of weights long with the four elements of the Crtn sulger re 0, 0, 0, 0, ±ν, ±ν, ±ν 3, ±ν, ±(ν + ν ), ±(ν ν ), ±(ν + ν 3 ), ±(ν ν 3 ), ±(ν + ν ), ±(ν ν ), ±(ν + ν 3 ), ±(ν ν 3 ), ±(ν + ν ), ±(ν ν ), ±(ν 3 + ν ), ±(ν 3 ν ). These re the 36 roots. Explicitly, they re (0, 0, 0, 0), (0, 0, 0, 0) (0, 0, 0, 0) (0, 0, 0, 0), ±(,,, ), ±(,,, ) ±(,,, ) ±(,,, ), ±(,, 0, 0), ±(0, 0,, ), ±(, 0,, 0), ±(0,, 0, ), ±(, 0, 0, ), ±(0,,, 0), ±(, 0, 0, ) ±(0,,, 0), ±(, 0,, 0), ±(0,, 0, ), ±(,, 0, 0), ±(0, 0,, ). We cn check this y finding eigen-genertors of the Crtn genertors under commuttion. Let σ ± = (σ ±iσ )/, τ ± = (τ ±iτ )/, nd η ± = (η ±iη )/. We chnge sis nd consider

8 8 STEPHEN HANCOCK the 3 genertors σ ± ± σ ± τ 3 η 3, τ ± ± σ 3 τ ± η 3, η ± ± σ 3 τ 3 η ±, σ ± τ ± η 3, σ ± τ 3 η ±, σ 3 τ ± η ±, σ ± τ ± η ±, long with our originl Crtn genertors. We find tht [H i, H j ] = α i H j where α = (0, 0, 0, 0) [H i, σ ± ± σ ± τ 3 η 3 ] = α i (σ ± ± σ ± τ 3 η 3 ) where α = (±, 0, 0, ±± ) [H i, τ ± ± σ 3 τ ± η 3 ] = α i (τ ± ± σ 3 τ ± η 3 ) where α = (0, ±, 0, ±± ) [H i, η ± ± σ 3 τ 3 η ± ] = α i (η ± ± σ 3 τ 3 η ± ) where α = (0, 0, ±, ±± ) [H i, σ ± τ ± η 3 ] = α i σ ± τ ± η 3 where α = (±, ±, 0, 0) [H i, σ ± τ 3 η ± ] = α i σ ± τ 3 η ± where α = (±, 0, ±, 0) [H i, σ 3 τ ± η ± ] = α i σ 3 τ ± η ± where α = (0, ±, ±, 0) [H i, σ ± τ ± η ± ] = α i σ ± τ ± η ± where α = (±, ±, ±, ±± ± ), which is in greement with the roots listed previously. The simple roots re seen to e α = (0, 0,, ), α = (0,,, 0), α 3 = (0, 0,, ), α = (,,, ) ecuse ll other positive roots cn e written s sums of these. Using cos θ αβ = α β/( α β ), the ngles etween them re θ α α = 0, θ α α 3 = 0, θ α 3 α = 35, with ll other ngles 90. Also note tht α = α = α 3 = nd α = so tht the first three simple roots re of equl length, while α is longer. The Dynkin digrm is therefore This corresponds to the lger Sp(8) = C. Chpter Solutions.A. Georgi constructs the spinor rep genertors in Chpter. We need simply permute indices 3 3 in his definitions to mke them liner comintions of the 0 mtrices given. We therefore let σ ± = (σ 3 ± iσ )/ nd τ ± = (τ 3 ± iτ )/. The Crtn genertors re σ = H nd τ = H, or explicitly σ = H = 0 0 i i i i 0 0 τ = H = 0 i 0 0 i i 0 0 i 0.

9 SOLUTIONS TO PROBLEMS IN LIE ALGEBRAS IN PARTICLE PHYSICS BY HOWARD GEORGI 9 The genertors corresponding to the roots re σ ±, σ τ ±, nd σ ±τ ±, or explicitly 0 i 0 0 i 0 σ + = 0 0 i i 0 0 σ = 0 0 i i i 0 0 i 0 σ τ + = σ +τ + = 8 σ +τ = 8 It is esily checked tht Tht is, the roots re 0 0 i 0 0 i i 0 0 i 0 0 i i i i i i i i i i i i i i i i σ τ = σ τ + = 8 σ τ = 8 [H i, H j ] = α i H j where α = (0, 0) [H i, σ ±] = α i ( σ ±) where α = (±, 0) [H i, σ τ ± ] = α i ( σ τ ± ) where α = (0, ±) [H i, σ ±τ ± ] = α i ( σ ±τ ± ) where α = (±, ± ). 0 0 i 0 0 i i 0 0 i 0 0 i i i i i i i i i i i i i i i i (0, 0), (0, 0), (, 0), (0, ), (, 0), (0, ), (, ), (, ), (, ), (, ), which defines the spinor rep. We now check tht the genertors ct on the sttes s expected. Tke = ( ) ( ) = i i i i = ( ) ( ) = i i = ( ) ( ) = i i i i These re simultneous eigenvectors of H nd H. We find tht σ + = σ + = σ = σ = σ τ + = σ τ + = σ τ = σ τ = = ( ) ( ) = i i. i i i i. σ +τ + = σ +τ = σ τ + = σ τ =,

10 0 STEPHEN HANCOCK while ny other comintion gives 0, s expected for the spinor rep. The simple roots re α = (0, ) nd α = (, ). Note tht θ α α = 35 nd α > α, yielding the following Dynkin digrm corresponding to SO(5). Following Georgi, the mtrix R is R = σ τ 3 = It is trivil to check tht R = R nd tht T = RT R for ech genertor T..