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ALGEBRAS IN PARTICLE ARTICLE PHYSICS PHYSICS SOLUTI SOLUTIONS ONS TO PROBL PROBLEMS EMS IN LIE ALGEBRAS BY HOWARD GEORGI GEORGI
STEPHEN HANCOCK HANCOCK
1
2
STEPHEN HANCOCK
Chapter 6 Solutions 6.A. Clearly
|NE has root vector α +β since H |NE = N H |E = N (α+β ) |E = (α+β ) |NE . Because the nonzero weights uniquely specify the states for the adjoint representation— up to a normalization factor—it suffices to show that | [E , E ] has root vector α + β . Indeed, since E |E = |[E , E ] and H E |E = [H , E ]|E + E H |E = α E |E + E β |E = (α + β ) E |E , we see that H |[E , E ] = (α + β ) |[E , E ]. Hence [E , E ] = N E . If α + β is not a root, i
i
α+β
i
α+β
i
α+β
α+β
α+β
α
α
α
β
i
α
β
i
β
i
α
β
α
α
β
i
β
i
α
i
β
α
α i
β
α
β
β
i
β
α
β
α+β
then N = 0.
[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 6.B. Suppose
·
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 ] = 6.C. Take
−NE
−β
.
H 1 = σ 3 = diag(1, 1, 1, 1) and H 2 = σ 3 τ 3 = diag(1, 1, 1, 1). For the four dimensional 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 (0, 0, 0, 1)T
− with weight (−1, −1) 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 2
H 1
SOLUTIONS TO PROBLEMS IN LIE ALGEBRAS IN PARTICLE PHYSICS BY HOWARD GEORGI
3
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 j 2
−
= δ jk .
The fundamental weights are therefore µ1 = (1, 1) since
2(0, 2) µ1 2(2, 2) µ2 = =1 (0, 2)2 (2, 2)2 The angle between α1 and α 2 is
·
− · −
and
µ2 = (2, 0)
and
2(2, 2) µ1 2(0, 2) µ2 = = 0. (2, 2)2 (0, 2)2
− · −
α1 α 2 (0, 2) (2, 2) 1 = = = α α1 α2 (0, 2) (2, 2) 2 The Dynkin diagram for the algebra is thus the following. cos θα
1
8.B. We
J 1 =
J 2 =
J 3 =
2
=
· | || | |
·
· − − √ ⇒ || − |
θα
1
α2 =
135◦ .
may change basis and consider the following six generators for the algebra:
σ1 + σ1 η1 4
σ2 + σ2 η1 4
σ3 + σ3 η1 4
0 1 0 = 4 1 10 1 0 = 4 i 1i 1 1 = 4 0
1 1 0 0
0 0 i i
−i −i −i −i
1 1 0 0 0
We easily calculate that
1 1 0 0
0 0 1 1
0 0
K 1 =
0 0 0 0 −1
0 0 1 1
− − −1
[J i , J j ] = iεijk J k ,
K 2 =
K 3 =
σ1
−σ η
1 1
4
σ2
−σ η
2 1
4
σ3
−σ η
3 1
4
0 1 0 = 4 1 −01 1 0 = 4 i −i1 1 −1 = 4 0
0 0 1 1
−1
1
−1
1 0 0
− 0 −i i 0 i −i −i 0 0 i 0 0 −1 0 0 1 0 0 . 0 −1 1
0
[K i , K j ] = iεijk K k ,
−
1 1 0 0
0
[J i , K j ] = 0.
Now take J 3 and K 3 as the Cartan generators and J +
J −
0 J + iJ 1 0 = √ = √ 2 2 2 0 00 J − iJ 1 0 = √ = √ 2 2 2 1 1
1
2
2
0 0 0 0
0 0 1 1 1
1 1 0 0
0 0 0 1 1 0 0
K +
0 0 0 0 0
K −
0 0 1 −1 K + iK 1 0 0 −1 1 √ 2 = 2√ = 0 0 0 0 2 0 0 0 0 0 0 00 K − iK 1 0 0 0 0 √ 2 = 2√ = 2 1 −1 0 0 1
2
1
2
−1
1 0 0
4
STEPHEN HANCOCK
as the other four generators for the adjoint representation. We have [J 3 , J ± ] =
±
±J
[K 3 , J ± ] = 0,
,
[J 3 , K ± ] = 0,
The six root vectors are therefore J 3 with root (0, 0)
[K 3, K ± ] =
±
±K
.
| |K with root (0, 0) |J with root (1, 0) |K with root (0, 1) |J with root (−1, 0) |K with root (0, −1). We should also have [J , J ] = (1, 0) · (J , K ) = J and [K , K ] = (0, 1) · (J , K ) = K , and it +
3
+
+
−
−
−
3
3
+
3
−
3
3
3
is easily checked that these are satisfied.
K 3
J 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) These are orthogonal, i.e., θ α
1
8.C. We
2
α2
α2 = (0, 1).
and
= 90◦ , and so the Dynkin diagram is as follows.
2
2
are given α 1 = 2, α 2 = 2, α 3 = 1, and the Dynkin diagram below.
Using Georgi’s convention, the Cartan matrix is defined by A ji = It follows that
2
2α j αi
·
αi 2
.
2
αi A ji = α j Aij and A ji Aij = 4cos2 θαi αj . Note that for i = j , 4cos2 θαi αj is the number of lines between αi and α j in the Dynkin diagram. Clearly A 11 = A22 = A 33 = 2 and A 13 = A31 = 0. The above relations give A12 /A21 = 1 A23 /A32 = 2 and A12 A21 = 1 A23 A32 = 2,
with nonpositive solution A 12 =
−1, A = −1, A = −2, A 2 −1 0 −1 2 − 2 . 0 −1 2 21
23
32
=
−1. The Cartan matrix is
SOLUTIONS TO PROBLEMS IN LIE ALGEBRAS IN PARTICLE PHYSICS BY HOWARD GEORGI
5
Chapter 12 Solutions 12.A. Using
the method discussed in the text, we calculate
a a a
a a a
a a a b
→
a a
a b
→
b a a a b
→
a a a
→
a b
a
⊕
a a b
a
→
a
⊕
a
a a
a a
a a
→
b
a a
a a
⊕
b
a a
⊗
a a a
a a a
→
a b a
a
⊕
a a b
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,
≡
≡
6
STEPHEN HANCOCK
while transpositions in columns contribute a factor of 1.
−
c c c a a a d b c c c a a a d b c d c d b
c c a a a b c c a a a
c c c a a d b a
→
+ a a a c c c b d a a a c c c b d
→−
a b a + b d
a a c c c d a a c c c
→− →
→−
a a a c c b d c
c d c d b
c a c a
c a a b c a a
c d a c d a
c c a a b c c a a b
c c c d a a a b
→
→
+ a b a b d
a c a c
a + b c a b c
a a c c d
→− →
a c c d a c c
a a c c d
→−
a a a + b c c c d
It follows that [(2, 1)
⊗ (2, 1)] = (4, 2) ⊕ (3, 1) ⊕ (0, 4) ⊕ (1, 2) ⊕ (0, 1) [(2, 1) ⊗ (2, 1)] = (5, 0) ⊕ (2, 3) ⊕ (3, 1) ⊕ (1, 2) ⊕ (2, 0) S
AS
or [15
⊗ 15] = 60 ⊕ 24 ⊕ 15 ⊕ 15 ⊕ 3 [15 ⊗ 15] = 42 ⊕ 24 ⊕ 21 ⊕ 15 ⊕ 6. S
S
AS
S
AS
AS
Chapter 13 Solutions 13.E. We
calculate
⊗
so that [2]
a a =
a a
a
⊕
a [3, 1]. Using the factors over hooks rule, we have
⊗ [1, 1] = [2, 1, 1] ⊕ N (N − 1) N (N + 1) N (N − 1) = × D([2] ⊗ [1, 1]) = D[2] × D[1, 1] = 2·1 2·1 4 2
and D([2, 1, 1]
2
+ 2)(N − 1) N (N + 1)(N − 1)(N − 2) + ⊕ [3, 1]) = D[2, 1, 1] + D[3, 1] = N (N + 1)(N 4·2·1·1 4·1·2·1 N (N − 1)(N + 2 + N − 2) N (N − 1)(2N ) N (N − 1) = = = . 2
Thus D([2]
2
⊗ [1, 1]) = D([2, 1, 1] ⊕
2
2
8 8 4 [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 σa τ b ηc close under commutation. Clearly [σa , τ b ] = [τ a , ηb ] = [ηa , σb ] = 0,
and we know that [σa, σb ] = 2iεabc σc ,
[τ a , τ b ] = 2iεabc τ c ,
[ηa, ηb ] = 2iεabc ηc .
SOLUTIONS TO PROBLEMS IN LIE ALGEBRAS IN PARTICLE PHYSICS BY HOWARD GEORGI
7
Next, we have [σa , σb τ c ηd ] = [σa, σb ]τ c ηd = 2iεabeσe τ c ηd [τ a , σb τ c ηd ] = [τ a , τ c ]ηd σb = 2iεace σb τ e ηd [ηa , σb τ c ηd ] = [ηa, ηd ]σb τ c = 2iεadeσb τ c ηe . Finally, [σaτ b ηc , σd τ e ηf ] = [σa, σd ]τ e τ b ηf ηc + [τ b , τ e ]ηf ηc σaσd + [ηc , ηf ]σa σd τ b τ e = 2iεadg σg (δ be
− iε
beh τ h )(δ cf
− iε
cf i ηi ) + 2iεbeh τ h (δ cf
− iε
cf i ηi )(δ ad + iεadg σg )
+ 2iεcf i ηi (δ ad + iεadg σg )(δ be + iεbehτ h ) = 2iδ be δ cf εadg σg + 2δ be εcfi εadg ηi σg + 2δ cf εadg εbehσg τ h 2iεadg εbehεcfi σg τ hηi + 2iδ cf δ ad εbehτ h 2δ cf εadg εbehσg τ h + 2δ ad εbehεcfi τ h ηi + 2iεadg εbeh εcf i σg τ h ηi + 2iδ ad δ be εcfi ηi 2δ ad εbehεcf i τ h ηi 2δ be εcf i εadg ηi σg 2iεadg εbehεcfi σg τ h ηi = 2iδ be δ cf εadg σg + 2iδ cf δ adεbehτ h + 2iδ ad δ be εcf i ηi 2iεadg εbeh εcf i σg τ h ηi .
−
− −
−
−
−
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 = H = diag(1, 1, −1, −1, 1, 1, −1, −1) η = H = diag(1, −1, 1, −1, 1, −1, 1, −1) σ τ η = H = diag(1, −1, −1, 1, −1, 1, 1, −1). 3
2
3
3
3 3 3
4
The weights of the defining representation are therefore ν 1 = (1, 1, 1, 1),
ν 2 = (1, 1, 1, 1),
3
4
7
8
ν = (1, −1, 1, −1), ν = (1, −1, −1, 1), − − ν = (−1, 1, 1, −1), ν = (−1, 1, −1, 1), ν = (−1, −1, 1, 1), ν = (−1, −1, −1, −1). Noting that ν = −ν , ν = −ν , ν = −ν , and ν = −ν , all differences of weights along with 5
6
5
4
6
3
7
2
8
1
the four elements of the Cartan subalgebra are 0,
0, 1
2
±2ν , ±(ν + ν ), ±(ν + ν ), ±(ν + ν ), 1
2
1
4
2
4
0, 3
±2ν , ±(ν − ν ), ±(ν − ν ), ±(ν − ν ), 1
2
1
4
2
4
0,
±2ν , ±(ν + ν ), ±(ν + ν ), ±(ν + ν ), 1
3
2
3
3
4
4
±2ν , ±(ν − ν ), ±(ν − ν ), ±(ν − ν ). 1
3
2
3
3
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 = (σ ± iσ )/ 2, τ = (τ ± iτ )/ 2, and η = (η ± iη )/ 2. We change basis and consider ±
1
2
±
1
2
±
1
2
8
STEPHEN HANCOCK
the 32 generators σ± σ± τ 3 η3 , σ± τ ± η3 , σ± τ ± η± ,
τ ± σ3 τ ± η3 , σ± τ 3 η± ,
±
η± σ3 τ 3 η± , σ3 τ ± η± ,
±
±
along with our original 4 Cartan generators. We find that [H i , H j ] = αi H j [H i , σ± [H i , τ ± [H i , η±
where α = (0, 0, 0, 0)
where α = ( 2, 0, 0,
where
3 ± 3
3 3 ±
where
± σ τ η ] = α (σ ± σ τ η ) ± σ τ η ] = α (τ ± σ τ η ) ± σ τ η ] = α (η ± σ τ η ) 3
i
±
3 ± 3
i
±
3 3 ±
i
±
± 3
± 3
3
[H i , σ± τ ± η3] = αi σ± τ ± η3
where
[H i , σ± τ 3η± ] = αi σ± τ 3η±
where
[H i , σ3 τ ± η± ] = αi σ3 τ ± η±
where
[H i , σ± τ ± η± ] = αi σ± τ ± η±
where
±± 2) α = (0, ±2, 0, ±± 2) α = (0, 0, ±2, ±± 2) α = (±2, ± 2, 0, 0) α = (±2, 0, ± 2, 0) α = (0, ±2, ± 2, 0) 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 angles between them are θα
1
α2 =
120◦ ,
θα
2
α3 =
120◦ ,
θα
3
α4 =
− − = α · β/(|α||β |), the
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 α 4 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 12 σ2 = H 1 and 12 τ 2 = H 2 , or explicitly
→ → →
±
√
0 1 1 0 σ = H = 2 2 i 2
1
0 0 0 0 i
√
±
−i
0 0 0
0 i 0 0
−
0 1 1 i τ = H = 2 2 0 2
2
0
−i
0 0 0 0 0 0 i
0 0 i 0
−
.
SOLUTIONS TO PROBLEMS IN LIE ALGEBRAS IN PARTICLE PHYSICS BY HOWARD GEORGI
The generators corresponding to the roots are 12 σ± , 12 σ2 τ ± , and 14 σ± τ ± , or explicitly
1 0 i 0 1 1 0 1 0 i σ = √ i 0 1 0 − 2 2 2 0 0 i 0 0 −−i 1 1 1 1 0 0 1 i σ τ = √ 0 0 2 2 2 i −1 1−1 i −i i 0−10 1 1 i −1 −1 −i σ τ = 4 8 i −1 −1 −i −11 −−ii −ii 11 1 1 −i −1 1 −i σ τ = − i 1 1 i 4 8
1 0 −i 0 1 1 0 1 0 −i σ = √ i 0 1 0 − − 2 2 2 0 0 −0 i −i 0 −−1 1 1 1 0 0 −1 i σ τ = √ 1 0 0 2 2 2 i 1 1 −i i −i0 1 0 1 1 i −1 1 i σ τ = 1 −1 −i 4 8 −i 11 −ii −−ii −11 1 1 −i −1 −1 i σ τ = . − − − i 1 1 i 4 8
+
−
2 +
2 −
+ +
− +
+ −
1
It is easily checked that
−i
i
− −
1
[H i , H j ] = α i H j
−1
i
1
where α = (0, 0)
[H i , 12 σ± ] = α i ( 12 σ± )
where α = ( 1, 0)
[H i , 12 σ2 τ ± ] = α i ( 12 σ2 τ ± )
where α = (0, 1)
±
[H i , 14 σ± τ ± ] = α i ( 14 σ± τ ± )
i
±
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 1 1 1 1 i 1 |↑↑ = √ 2 i ⊗ √ 2 i = 2 i −1 1 −ii |↑↓ = √ 12 1i ⊗ √ 12 −1i = 21 1
These are simultaneous eigenvectors of H 1 and 1 1 1 σ+ = σ2 τ + 2 2 2 1 1 1 σ+ = σ2 τ + 2 2 2 1 1 1 = σ− σ2 τ − 2 2 2 1 1 1 σ− = σ2 τ − 2 2 2
− −
1 1 1 1 1 i 1 |↓↑ = √ 2 −i ⊗ √ 2 i = 2 −i 1 1 1 1 1 1 −i 1 |↓↓ = √ 2 −i ⊗ √ 2 −i = 2 −i .
H 2 . We find that 1 = 2 1 = 2 1 = 2 1 = 2
|↓↑ √ |↑↑
|↑↓ √ |↑↑
|↓↓ √ |↑↓
|↓↓ − √ |↓↑
|↑↑ √ |↓↑
|↑↑ √ |↑↓
|↑↓ √ |↓↓
|↓↑ − √ |↓↓
−1
1 σ+ τ + 4 1 σ+ τ − 4 1 σ− τ + 4 1 σ− τ − 4
|↓↓ = 21 |↑↑ |↓↑ = 21 |↑↓ |↑↓ = 21 |↓↑ |↑↑ = 21 |↓↓,
9
10
STEPHEN HANCOCK
while any other combination gives 0, as expected for the spinor rep. The simple roots are α1 = (0, 1) and α2 = (1, 1). Note that θα α = 135◦ and α2 > α1 , yielding the following Dynkin diagram corresponding to SO(5). 1
2
−
| | | |
Following Georgi, the matrix R is
0 0 R = σ τ = 1 1 3
0
It is trivial to check that R = R −1
0 0 0 1
1 0 0 0
−
0 1 0 0
.
− and that T = −RT R for each generator T . a
∗
a
a
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