Solved Problems groups

BY

BY HOWHOWARD ARD GEORGEORGIGI

STEPHEN

STEPHEN HANCOCHANCOCKK

1 1

6.A.  _Clearly

_|

_NE α+β

_

_{has root vector α +β  since H i}

_|

_NE α+β

_

_{= N H i}

_|

_E α+β

_

_{= N (α+β )i}

_|

_E α+β

_

₌

(α+β )i

|

NE α+β



. Because the nonzero weights uniquely specify the states for the adjoint representation— up to a normalization factor—it suffices to show that

 _|

[E α, E β]



  has root vector α + β . Indeed, since E α

|

E β



=

|

[E α, E β]



and

H iE α

|

E β



= [H i, E α]

|

E β



+ E αH i

|

E β



= αiE α

|

E β



+ E αβ i

|

E β



= (α + β )iE α

|

E β



,

we see that H i

|

[E α, E β]



= (α + β )i

|

[E α, E β]



. Hence [E α, E β] = N E α+β. If  α + β   is not a root, then N  = 0.

6.B. _{Suppose [E α, E β] = N E α+β. By 6.A, we also have [E α, E −α−β] = N} _{E −β and [E β, E −α−β] =}

N 

E −_α for some N  and N . Using the Jacobi identity and [E _γ , E −_γ ] = γ 

·

H , we obtain

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 and β   are linearly independent, and the generators H i  (components of  H ) are linearly independent, it follows that N 

= N  and N 

=

₋

N . Therefore, [E β, E −_α−_β] = N E −_α and

[E α, E −_α−_β] =

−

NE −_β.

6.C.  _{Take H 1 = σ3 = diag(1, 1,}

₋

_1,

₋

_{1) and H 2 = σ3τ 3 = diag(1,}

₋

_1,

₋

_{1, 1). For the four}

dimen-sional representation, the states and associated weights vectors are clearly (1, 0, 0, 0)T  with weight (1, 1)

(0, 1, 0, 0)T  with weight (1,

₋

1) (0, 0, 1, 0)T  with weight (

₋

1,

₋

1) (0, 0, 0, 1)T  with weight (

₋

1, 1).

The weights of the adjoint representation are the differences of these weights, along with the two elements of the Cartan subalgebra. The distinct differences, up to sign, are (1, 1)

₋

(1,

₋

1) = (0, 2), (1,

₋

1)

₋

(

₋

1, 1) = (2,

₋

2), (1, 1)

₋

(

₋

1, 1) = (2, 0), and (1, 1)

₋

(

₋

1,

₋

1) = (2, 2), so the roots are

(0, 0), (0, 0), (0, 2), (2,

₋

2), (2, 0), (2, 2), (0,

₋

2), (

₋

2, 2), (

₋

2, 0), (

₋

2,

₋

2).

H 1

Chapter 8 Solutions

8.A. _{The simple roots are the positive roots that cannot be written as the sum of other positive}

roots, namely

α1 = (0, 2) and α2 = (2,

₋

2).

Roots (2, 0) and (2, 2) are not simple because (2, 0) = (2,

₋

2)+(0, 2) and (2, 2) = (2,

₋

2)+2(0, 2). The fundamental weights µ j are defined by

2α j

_·

µk

α j2 = δ  jk. The fundamental weights are therefore

µ1 = (1, 1) and µ2 = (2, 0) since 2(0, 2)

_·

µ1 (0, 2)2 = 2(2,

₋

2)

_·

µ2 (2,

₋

2)2   = 1 and 2(2,

₋

2)

_·

µ1 (2,

₋

2)2 = 2(0, 2)

_·

µ2 (0, 2)2 = 0.

The angle between α1 and α2 is

cos θ_α1 α2 = α 1

·

α2

|

α1

||

α2

|

= (0, 2)

_·

(2,

₋

2)

|

(0, 2)

_||

(2,

₋

2)

_|

=

−

1

√ 

₂ =

_⇒

θ_α1 α2 = 135◦. The Dynkin diagram for the algebra is thus the following.

8.B. _{We may change basis and consider the following six generators for the algebra:}

J 1 = σ 1 + σ1η1 4 = 1 4









0 0 1 1 0 0 1 1 1 1 0 0 1 1 0 0









J 2 = σ 2 + σ2η1 4 = 1 4









0 0

₋

i

₋

i 0 0

₋

i

₋

i i i 0 0 i i 0 0









J 3 = σ 3 + σ3η1 4 = 1 4









1 1 0 0 1 1 0 0 0 0

₋

1

₋

1 0 0

₋

1

₋

1









K 1 = σ 1

−

σ1η1 4 = 1 4









0 0 1

₋

1 0 0

₋

1 1 1

₋

1 0 0

−

1 1 0 0









K 2 = σ 2

−

σ2η1 4 = 1 4









0 0

₋

i i 0 0 i

₋

i i

₋

i 0 0

−

i i 0 0









K 3 = σ 3

−

σ3η1 4 = 1 4









1

₋

1 0 0

−

1 1 0 0 0 0

₋

1 1 0 0 1

₋

1









. We easily calculate that

[J i, J  j] = iεijkJ k, [K i, K  j] = iεijkK k, [J i, K  j] = 0. Now take J 3 and K 3 as the Cartan generators and

J + = J 1 + iJ 

_√ 

2 2 = 1 2

√ 

2









0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0









J − = J 1

_√ 

−

iJ 2 2 = 1 2

√ 

2









0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0









K + = K 1 + iK 

_√ 

2 2 = 1 2

√ 

2









0 0 1

₋

1 0 0

₋

1 1 0 0 0 0 0 0 0 0









K − = K 1

_√ 

−

iK 2 2 = 1 2

√ 

2









0 0 0 0 0 0 0 0 1

₋

1 0 0

−

1 1 0 0









[J 3, J ±] =

±

J ±, [K 3, J ±] = 0, [J 3, K ±] = 0, [K 3, K ±] =

±

K ±.

The six root vectors are therefore

|

J 3



 with root (0, 0)

|

J +

_

 with root (1, 0)

|

J −

_

 with root (

₋

1, 0)

|

K 3



 with root (0, 0)

|

K +

_

 with root (0, 1)

|

K −

_

 with root (0,

₋

1). We should also have [J +, J −

] = (1, 0)

_·

(J 3, K 3) = J 3 and [K +, K −] = (0, 1)

·

(J 3, K 3) = K 3, and it

is easily checked that these are satisfied.

J 3

K 3

Both

_{

J 1, J 2, J 3

}

and

{

K 1, K 2, K 3

}

are nontrivial invariant subalgebras, and thus the algebra is not

simple. The commutation relations show that the group generated is in fact SU (2)

_⊕

SU (2), and thus there are no nontrivial abelian subgroups and the group is semisimple. The simple roots are

α1 = (1, 0) and α2 = (0, 1).

These are orthogonal, i.e., θα1_α2 = 90◦, and so the Dynkin diagram is as follows.

8.C. _{We are given α}12_{= 2, α}22 _{= 2, α}32 _{= 1, and the Dynkin diagram below.}

Using Georgi’s convention, the Cartan matrix is defined by A ji = 2α  j

·

αi αi2 . It follows that αi2A ji = α j 2

Aij and A jiAij = 4cos2θαi_αj.

Note that for i

_

= j , 4cos2θαi_αj is the number of lines between αi and α j in the Dynkin diagram.

Clearly A11 = A22 = A33 = 2 and A13 = A31 = 0. The above relations give

A12/A21 = 1

A12A21 = 1 and

A23/A32 = 2

A23A32 = 2,

with nonpositive solution A12 =

−

1, A21 =

−

1, A23 =

−

2, A32 =

−

1. The Cartan matrix is





2

₋

1 0

−

1 2

₋

2 0

₋

1 2





.

Chapter 12 Solutions 12.A. _{Using the method discussed in the text, we calculate}

⊗

a a a_b

→

a a a

→

_b a a a

⊕

a a a b a a a

→

a a a b

⊕

a a a b a a a

→

a a b a a a a

→

a a a b

⊕

a a a b a a a

→

a a b a

⊕

a a a b a a a

→

a a a b

Canceling columns with 3 boxes (factors of  ε), we have shown

⊗

=

_⊕

_⊕

_⊕

⊕

⊕

⊕

⊕

⊕

⊕

or

(2, 1)

_⊗

(2, 1) = (4, 2)

_⊕

(5, 0)

_⊕

(2, 3)

_⊕

(3, 1)

_⊕

(3, 1)

_⊕

(0, 4)

_⊕

(1, 2)

_⊕

(1, 2)

_⊕

(2, 0)

_⊕

(0, 1).

In terms of the corresponding dimensionalities, we can write this as

15

_⊗

15 = 60

_⊕

42

_⊕

24

_⊕

24

_⊕

21

_⊕

15

_⊕

15

_⊕

15

⊕

6

_⊕

3, where we have defined 15

_≡

(2, 1) and 15

≡

(4, 0).

We use the following procedure to determine which representations appear symmetrically in the product and which appear antisymmatrically. Transpositions in rows contribute a factor of +1,

c c c a a a d b

→

+   a a a c c cb d c c c a a a d b

→ −

a a a c c c b d c c c a a d a b

→ −

  a a a c c b c d c c c a a d a b

→

+ a a a c cb c d c c c a a d b a

→ −

a a a c c b d c c c c a d a a b

→

+   a a a cb c c d c c c a d a a b

→ −

a a a c b c c d c c c a d a b a

→

+ a a a cb c d c c c c a d a a b

→ −

a a a c b c c d c c c d a a a b

→

+ a a ab c c c d It follows that [(2, 1)

_⊗

(2, 1)]S = (4, 2)

⊕

(3, 1)

⊕

(0, 4)

⊕

(1, 2)

⊕

(0, 1) [(2, 1)

_⊗

(2, 1)]AS = (5, 0)

⊕

(2, 3)

⊕

(3, 1)

⊕

(1, 2)

⊕

(2, 0) or [15

_⊗

15]S = 60

⊕

24S

⊕

15S

⊕

15

⊕

3 [15

_⊗

15]AS = 42

⊕

24AS

⊕

21

⊕

15AS

⊕

6. Chapter 13 Solutions 13.E. _{We calculate}

⊗

a a = a a

_⊕

a a

so that [2]

_⊗

[1, 1] = [2, 1, 1]

_⊕

[3, 1]. Using the factors over hooks rule, we have D([2]

_⊗

[1, 1]) = D[2]

_×

D[1, 1] = N (N 

 −

1) 2

_·

1

×

N (N  + 1) 2

_·

1 = N 2(N 2

₋

1) 4 and D([2, 1, 1]

_⊕

[3, 1]) = D[2, 1, 1] + D[3, 1] = N (N  + 1)(N  + 2)(N 

 −

1) 4

_·

2

_·

1

_·

1 + N (N  + 1)(N 

 ₋

1)(N 

 ₋

2) 4

_·

1

_·

2

_·

1 = N (N  2

−

1)(N  + 2 + N 

 ₋

2) 8 = N (N 2

₋

1)(2N ) 8 = N 2(N 2

₋

1) 4 .

Thus D([2]

_⊗

[1, 1]) = D([2, 1, 1]

_⊕

[3, 1]), and the dimensions check out for arbitrary N .

Chapter 19 Solutions

19.A. _{We must check that the elements σa, τ a, ηa, and σaτ bηc close under commutation. Clearly}

[σa, τ b] = [τ a, ηb] = [ηa, σb] = 0, and we know that

Next, we have [σa, σbτ cηd] = [σa, σb]τ cηd = 2iεabeσeτ cηd [τ a, σbτ cηd] = [τ a, τ c]ηdσb = 2iεaceσbτ eηd [ηa, σbτ cηd] = [ηa, ηd]σbτ c = 2iεadeσbτ cηe. Finally, [σaτ bηc, σdτ eηf ] = [σa, σd]τ eτ bηf ηc + [τ b, τ e]ηf ηcσaσd + [ηc, ηf ]σaσdτ bτ e

= 2iεadgσg(δ be

−

iεbehτ h)(δ cf 

 −

iεcf iηi) + 2iεbehτ h(δ cf 

 −

iεcf iηi)(δ ad + iεadgσg) + 2iεcf iηi(δ ad + iεadgσg)(δ be + iεbehτ h)

= 2iδ beδ cf εadgσg + 2δ beεcfiεadgηiσg + 2δ cf εadgεbehσgτ h

−

2iεadgεbehεcfiσgτ hηi + 2iδ cf δ adεbehτ h

−

2δ cf εadgεbehσgτ h + 2δ adεbehεcfiτ hηi + 2iεadgεbehεcf iσgτ hηi + 2iδ adδ beεcfiηi

−

2δ adεbehεcf iτ hηi

−

2δ beεcf iεadgηiσg

−

2iεadgεbehεcfiσgτ hηi = 2iδ beδ cf εadgσg + 2iδ cf δ adεbehτ h + 2iδ adδ beεcf iηi

−

2iεadgεbehεcf iσgτ hηi.

The terms lying outside the given set of matrices canceled! So we see that the algebra is closed, i.e., the 36 matrices form a Lie algebra.

Take the Cartan generators to be

σ3 = H 1 = diag(1, 1, 1, 1,

−

1,

−

1,

−

1,

−

1)

τ 3 = H 2 = diag(1, 1,

−

1,

−

1, 1, 1,

−

1,

−

1)

η3 = H 3 = diag(1,

−

1, 1,

−

1, 1,

−

1, 1,

−

1)

σ3τ 3η3 = H 4 = diag(1,

−

1,

−

1, 1,

−

1, 1, 1,

−

1).

The weights of the defining representation are therefore

ν 1 = (1, 1, 1, 1), ν 2 = (1, 1,

₋

1,

₋

1), ν 3 = (1,

₋

1, 1,

₋

1), ν 4 = (1,

₋

1,

₋

1, 1), ν 5 = (

₋

1, 1, 1,

₋

1), ν 6 = (

₋

1, 1,

₋

1, 1), ν 7 = (

₋

1,

₋

1, 1, 1), ν 8 = (

₋

1,

₋

1,

₋

1,

₋

1). Noting that ν 5 ₌

 −

ν 4_{, ν} 6 ₌

 −

ν 3_{, ν} 7 ₌

 −

ν 2_{, and ν} 8 ₌

 −

ν 1_{, all differences of weights along with}

the four elements of the Cartan subalgebra are

0, 0, 0, 0,

±

2ν 1,

_±

2ν 2,

_±

2ν 3,

_±

2ν 4,

±

(ν 1+ ν 2),

_±

(ν 1

₋

ν 2),

_±

(ν 1+ ν 3),

_±

(ν 1

₋

ν 3),

±

(ν 1+ ν 4),

_±

(ν 1

₋

ν 4),

_±

(ν 2+ ν 3),

_±

(ν 2

₋

ν 3),

±

(ν 2+ ν 4),

_±

(ν 2

₋

ν 4),

_±

(ν 3+ ν 4),

_±

(ν 3

₋

ν 4). These are the 36 roots. Explicitly, they are

(0, 0, 0, 0), (0, 0, 0, 0) (0, 0, 0, 0) (0, 0, 0, 0),

±

(2, 2, 2, 2),

_±

(2, 2,

₋

2,

₋

2)

_±

(2,

₋

2, 2,

₋

2)

_±

(2,

₋

2,

₋

2, 2),

±

(2, 2, 0, 0),

_±

(0, 0, 2, 2),

_±

(2, 0, 2, 0),

_±

(0, 2, 0, 2),

±

(2, 0, 0, 2),

_±

(0, 2, 2, 0),

_±

(2, 0, 0,

₋

2)

_±

(0, 2,

₋

2, 0),

±

(2, 0,

₋

2, 0),

_±

(0, 2, 0,

₋

2),

_±

(2,

₋

2, 0, 0),

_±

(0, 0, 2,

₋

2).

We can check this by finding “eigen-generators” of the Cartan generators under commutation. Let σ± = (σ1

±

iσ2)/

√ 

2, τ ± = (τ 1

±

iτ 2)/

√ 

2, and η± = (η1

±

iη2)/

√ 

2. We change basis and consider

σ±

±

 σ±τ 3η3, τ ±

±

 σ3τ ±η3, η±

±

 σ3τ 3η±,

σ±τ ±η₃, σ±τ 3η±, σ₃τ ±η±,

σ±τ ±η_±,

along with our original 4 Cartan generators. We find that

[H i, H  j] = αiH  j   where α = (0, 0, 0, 0) [H i, σ±

±

σ±τ 3η3] = αi(σ±

±

 σ±τ 3η3) where α = (

±

2, 0, 0,

±±

2) [H i, τ ±

±

σ3τ ±η3] = αi(τ ±

±

σ3τ ±η3) where α = (0,

±

2, 0,

±±

2) [H i, η±

±

σ3τ 3η±] = α_i(η±

±

 σ3τ 3η±) where α = (0, 0,

±

2,

±±

2) [H i, σ±τ ±η₃] = α_iσ±τ ±η₃   where α = (

±

2,

±

 2, 0, 0) [H i, σ±τ 3η±] = α_iσ±τ 3η±   where α = (

±

2, 0,

±

 2, 0) [H i, σ3τ ±η±] = α_iσ₃τ _±η_±   where α = (0,

±

2,

±

 2, 0) [H i, σ±τ ±η±] = α_iσ±τ ±η±   where α = (

±

2,

±

 2,

_±

2,

_±±



_±

2), which is in agreement with the roots listed previously.

The simple roots are seen to be

α1 = (0, 0, 2, 2), α2 = (0, 2,

₋

2, 0), α3 = (0, 0, 2,

₋

2), α4 = (2,

₋

2,

₋

2, 2)

because all other positive roots can be written as sums of these. Using cos θαβ = α

·

β/(

|

α

||

β 

|

), the angles between them are

θα1_α2 = 120◦, θ_α2_α3 = 120◦, θ_α3_α4 = 135◦, with all other angles 90◦

. Also note that

 _|

α1

_|

=

 _|

α2

_|

=

 _|

α3

_|

= 2

√ 

2 and

 _|

α4

_|

 = 4 so that the first three simple roots are of equal length, while α4 is

√ 

2 longer. The Dynkin diagram is therefore

This corresponds to the algebra S p(8) = C 4.

Chapter 21 Solutions

21.A. _{Georgi constructs the spinor rep generators in Chapter 21. We need simply permute indices}

3

_→

2

_→

1

_→

3 in his definitions to make them linear combinations of the 10 matrices given. We therefore let σ± = (σ3

±

iσ1)/

√ 

2 and τ ± = (τ 3

±

iτ 1)/

√ 

2. The Cartan generators are 1₂σ2 = H 1

and 1₂τ 2 = H 2, or explicitly 1 2σ2 = H 1 = 1 2









0 0

₋

i 0 0 0 0

₋

i i 0 0 0 0 i 0 0









1 2τ 2 = H 2 = 1 2









0

₋

i 0 0 i 0 0 0 0 0 0

₋

i 0 0 i 0









.

The generators corresponding to the roots are 1₂σ±, 1₂σ2τ ±, and 1₄σ±τ ±, or explicitly 1 2σ+ = 1 2

√ 

2









1 0 i 0 0 1 0 i i 0

₋

1 0 0 i 0

₋

1









1 2σ2τ + = 1 2

√ 

2









0 0

₋

i 1 0 0 1 i i

₋

1 0 0

−

1

₋

i 0 0









1 4σ+τ + = 1 8









1 i i

₋

1 i

₋

1

₋

1

₋

i i

₋

1

₋

1

₋

i

−

1

₋

i

₋

i 1









1 4σ+τ − = 1 8









1

₋

i i 1

−

i

₋

1 1

₋

i i 1

₋

1 i 1

₋

i i 1









1 2σ− = 1 2

√ 

2









1 0

₋

i 0 0 1 0

₋

i

−

i 0

₋

1 0 0

₋

i 0

₋

1









1 2σ2τ − = 1 2

√ 

2









0 0

₋

i

₋

1 0 0

₋

1 i i 1 0 0 1

₋

i 0 0









1 4σ−τ + = 1 8









1 i

₋

i 1 i

₋

1 1 i

−

i 1

₋

1

₋

i 1 i

₋

i 1









1 4σ−τ − = 1 8









1

₋

i

₋

i

₋

1

−

i

₋

1

₋

1 i

−

i

₋

1

₋

1 i

−

1 i i 1









.

It is easily checked that

[H i, H  j] = αiH  j   where α = (0, 0) [H i,1₂σ±] = α_i(1₂σ±) where α = (

±

1, 0) [H i,1₂σ2τ ±] = αi(1₂σ2τ ±) where α = (0,

±

1) [H i, 14σ±τ ±] = α_i(1₄σ±τ ±) where α = (

±

1,

±

 1). That is, the roots are

(0, 0), (0, 0), (1, 0), (0, 1), (

₋

1, 0), (0,

₋

1), (1, 1), (1,

₋

1), (

₋

1, 1), (

₋

1,

₋

1), which defines the spinor rep.

We now check that the generators act on the states as expected. Take

|↑↑

=

_√ 

1 2



1 i



⊗

√ 

1₂



1_i



= 1 2









1 i i

−

1









|↑↓

=

_√ 

1 2



1 i



⊗

√ 

1₂



₋

1_i



= 1 2









1

−

i i 1









|↓↑

=

_√ 

1 2



1

−

i



⊗

√ 

1₂



1_i



= 1 2









1 i

−

i 1









|↓↓

=

_√ 

1 2



1

−

i



⊗

√ 

1₂



₋

1_i



= 1 2









1

−

i

−

i

−

1









.

These are simultaneous eigenvectors of  H 1 and H 2. We find that

1 2σ+

|↓↑

= 1

√ 

₂

|↑↑

1 2σ+

|↓↓

= 1

√ 

₂

|↑↓

1 2σ−

|↑↑

= 1

√ 

₂

|↓↑

1 2σ−

|↑↓

= 1

√ 

₂

|↓↓

1 2σ2τ +

|↑↓

= 1

√ 

₂

|↑↑

1 2σ2τ +

|↓↓

=

−

1

√ 

₂

|↓↑

1 2σ2τ −

|↑↑

= 1

√ 

₂

|↑↓

1 2σ2τ −

|↓↑

=

−

1

√ 

₂

|↓↓

1 4σ+τ +

|↓↓

= 1 2

|↑↑

1 4σ+τ −

|↓↑

= 1 2

|↑↓

1 4σ−τ +

|↑↓

= 1 2

|↓↑

1 4σ−τ −

|↑↑

= 1 2

|↓↓

,

The simple roots are

α1 = (0, 1) and α2 = (1,

₋

1).

Note that θα1_α2 = 135◦ and

 |

α2

|

>

 |

α1

|

, yielding the following Dynkin diagram corresponding to SO(5).

Following Georgi, the matrix R is

R = σ1τ 3 =









0 0 1 0 0 0 0

₋

1 1 0 0 0 0

₋

1 0 0









.