Spinor Tech Summary

Spinor Tech Summary

Kiyoshi Shiraishi June 15, 2014

Abstract

We summarize some definitions and identities on the spinorial tech- nique in writing the scattering amplitudes.

1 Two-spinors and Basic definitions

Here

k ² ≡ k ^µ k µ ≡ η µν k ^µ k ^ν ≡ (k ⁰ ) ² − (k ¹ ) ² − (k ² ) ² − (k ³ ) ² = 0 (1) k ₀ = k ⁰ , k _i = − k ⁱ (i = 1, 2, 3) (2)

|j ^± ≡ u _± (k j ) , j ^± | ≡ u _± (k j ) (3) ij ≡ i ⁻ |j ⁺ = u ₋ (k i )u + (k j ) (4) [ij] ≡ i ⁺ | j ⁻ = u ₊ (k _i )u ₋ (k _j ) (5) u + (k) = k +

k ₋ e ^iφ

, u ₋ (k) = k ₋ e ^{− iφ}

− k +

(6) where

e ^±iφ

= k ¹ ± ik ²

(k ¹ ) ² + (k ² ) ² = k ¹ ± ik ²

k + k ₋ , k _± = k ⁰ ± k ³ (7) u ₊ (k) = u ^† ₊ (k) =

k ₊ ,

k ₋ e ^{− iφ}

, u ₋ (k) = u ^† ₋ (k) =

k ₋ e ^iφ

, − k ₊

(8) ij =

k _i− k _j+ e ^iφ

−

k _i+ k _j− e ^iφ

(9)

[ij] =

k i+ k j − e ^{− iφ}

−

k i − k j+ e ^−iφ

(10) ij[ji] = 2k i · k j (11)

|ij| = |[ij]| =

2k i · k j (12)

ji = −ij , [ji] = −[ij] (13) especially,

ii = i ⁻ |i ⁺ = u ₋ (k i )u + (k i ) = 0 (14) [ii] = i ⁺ | i ⁻ = u ₊ (k _i )u ₋ (k _i ) = 0 (15)

u ₊ (k)u ₊ (k) = k +

k ₋ e ^iφ

k ₊ ,

k ₋ e ^{− iφ}

=

k ₊

k ₊ k ₋ e ^−iφ

k ₊ k ₋ e ^iφ

k ₋

=

k ⁰ + k ³ k ¹ − ik ² k ¹ + ik ² k ⁰ − k ³

= k ⁰ σ ⁰ + k ⁱ σ ⁱ = k ₀ σ ⁰ − k _i σ ⁱ (16)

u ₋ (k)u ₋ (k) = k ₋ e ^{− iφ}

−

k ₊ k ₋ e ^iφ

, − k ₊

=

k ₋ −

k ₊ k ₋ e ^{− iφ}

−

k ₊ k ₋ e ^iφ

k ₊

=

k ⁰ − k ³ − k ¹ + ik ²

− k ¹ − ik ² k ⁰ + k ³

= k ⁰ σ ⁰ − k ⁱ σ ⁱ = k 0 σ ⁰ + k i σ ⁱ (17) where

σ ⁰ =

1 0 0 1

, σ ¹ =

0 1 1 0

, σ ² =

0 −i i 0

, σ ³ =

1 0 0 −1

(18) (k ₀ σ ⁰ ± k _i σ ⁱ )u _± (k) = 0 (‘Dirac equation ) (19) u ₋ (k) = iσ ² u ^∗ ₊ (k) , u ₋ (k) = u ^T ₊ (k)( − i)σ ² (20) u ₋ (k i )u ₋ (k j ) = u ^T ₊ (k i )(−i)σ ² iσ ² u ^∗ ₊ (k j ) = u ^T ₊ (k i )u ^∗ ₊ (k j ) = u + (k j )u + (k i ) (21)

u ₋ (k i )σ ⁱ u ₋ (k j ) = u ^T ₊ (k i )(−i)σ ² σ ⁱ iσ ² u ^∗ ₊ (k j ) = −u ^T ₊ (k i )(σ ⁱ ) ^T u ^∗ ₊ (k j ) = −u + (k j )σ ⁱ u + (k i ) (22)

Fierz identity

σ ^µ = (1, σ ⁱ ) (23)

(σ ^µ ) ab (σ µ ) cd = δ ab δ cd − (σ ⁱ ) ab (σ ⁱ ) cd = −2(σ ² ) ac (σ ² ) bd (24) i ⁺ | σ ^µ | j ⁺ p ⁺ | σ µ | q ⁺ = u ^† ₊ (k i )σ ^µ u + (k j )u ^† ₊ (k p )σ µ u + (k q )

= − 2u ^† ₊ (k _i )u ₋ (k _p )u ^† ₋ (k _j )u ₊ (k _q )

= −2jq[ip] (25)

Schouten identity

ab cd = δ ac δ bd − δ ad δ bc (26) ac db = δ ad δ cb − δ ab δ cd (27) _{ad bc} = δ _ab δ _dc − δ _ac δ _db (28) Thus

_{ab cd} + _{ac db} + _{ad bc} = 0 (29) Then

ijk + ijk + ikj = 0 (30)

“eikonal” identity

k − 1

i=j

i (i + 1)

iqq (i + 1) = jk

jqqk (31)

which can be proved by jk

jqqk + k (k + 1)

kqq (k + 1) = jk q (k + 1) − jq k (k + 1) jqqkq (k + 1)

= − kj q (k + 1) − jq k (k + 1) jqqkq (k + 1)

= qkj (k + 1) jq qk q (k + 1)

= j (k + 1)

jq q (k + 1) (32)

where Schouten identity is used.

2 Four-component spinors

gamma matrices

Bjorken-Drell rep.

γ ⁰ =

1 0 0 − 1

, γ ⁱ =

0 σ ⁱ

− σ ⁱ 0

, γ ⁵ =

0 1 1 0

(33) {γ µ , γ ν } = 2η µν , γ ⁵ γ ^µ + γ ^µ γ ⁵ = 0 , (γ ⁵ ) ² = 1 (34) four component spinor

U + (k) = 1

√ 2

u ₊ (k) u ₊ (k)

, U − (k) = 1

√ 2

u ₋ (k)

− u ₋ (k)

(35)

γ ⁵ U _± (k) = ±U _± (k) (36)

U + (k) = U ₊ ^† (k)γ ⁰ = 1

√ 2 (u + (k) , −u + (k)) (37) U ₋ (k) = U ₋ ^† (k)γ ⁰ = 1

√ 2 (u ₋ (k) , u ₋ (k)) (38) U _± (k)γ ⁵ = ∓U _± (k) (39) U ± (k _i )γ ⁰ U ± (k _j ) = u _± (k _i )u _± (k _j ) (40) U _± (k i )γ ⁱ U _± (k j ) = ±u _± (k i )σ ⁱ u _± (k j ) (41) U _∓ (k i )γ ^µ U _± (k j ) = ±U _∓ (k i )γ ^µ γ ⁵ U _± (k j ) = ∓U _∓ (k i )γ ⁵ γ ^µ U _± (k j )

= −U ∓ (k _i )γ ^µ U ± (k _j ) = 0 (42)

|j ^± ≡ U _± (k j ) , j ^± | ≡ U _± (k j ) N ote! the same notation as 2spinors (43) i ^± | j ^± = U ± (k _i ) U ± (k _j ) = 0 (44) i ⁻ |j ⁺ = U ₋ (k i )U + (k j ) = ij , i ⁺ |j ⁻ = U + (k i )U ₋ (k j ) = [ij] (45) i ^± | γ ^µ | i ^± = U ± (k _i )γ ^µ U ± (k _i ) = 2k ^µ _i (46) i ⁻ |γ ^µ |j ⁻ = U ₋ (k i )γ ^µ U ₋ (k j ) = U + (k j )γ ^µ U + (k i ) = j ⁺ |γ ^µ |i ⁺ (47) i ^∓ |γ ^µ |j ^± = U _∓ (k i )γ ^µ U _± (k j ) = 0 (48) Fierz identity

i ⁻ |γ ^µ |j ⁻ p ⁺ |γ µ |q ⁺ = j ⁺ |γ ^µ |i ⁺ p ⁺ |γ µ |q ⁺ = −2iq[jp] (49)

i ⁻ |γ ^µ |j ⁻ γ µ = 2(|j ⁻ i ⁻ | + |i ⁺ j ⁺ |) (50) i ⁺ | γ ^µ | j ⁺ γ _µ = 2( | j ⁺ i ⁺ | + | i ⁻ j ⁻ | ) (51)

‘Dirac equation’

k/ U ± (k) = k/ | k ^± = 0 (52) Projector

U _± (k)U _± (k) = 1 2

u _± (k)u _± (k) ∓u _± (k)u _± (k)

±u _± (k)u _± (k) −u _± (k)u _± (k)

= 1

2 (1 ± γ ⁵ )k/ (53)

re-discover:

ij[ji] = i ⁻ |j ⁺ j ⁺ |i ⁻ = tr U ₋ (k i )U + (k j )U + (k j )U ₋ (k i )

= tr

1 − γ ⁵ 2 k/ _i k/ _j

= 2k _i · k _j (54)

3 QED (QCD) Examples (I)

First of all, remember that

s _ij = (k _i + k _j ) ² = 2k _i · k _j = ij [ji] (55)

3.1 e ⁻ e ⁺ → µ ⁻ µ ⁺

(Of course, the amplitude for e ⁻ e ⁺ → q q ¯ can be obtained similarly) A 4 = ie ² A ₄ δ(

i

k _i ) (56)

3.1.1 e ⁻ _L (1)e ⁺ _R (2) → µ ⁻ _R (3)µ ⁺ _L (4)

A 4 = 2 ⁻ | γ ^µ | 1 ⁻ 3 ⁺ | γ _µ | 4 ⁺ s 12

= −2 24 [13]

s 12

(57) here we used (49).

| A ₄ | = 2 |s 13 |

s ₁₂ ∝ 1 − cos θ (CMF ) (58) helicity suppressed as 1 3 or 2 4

A ₄ = − 2 24 [13]

s 12

= − 2 24 [13] 13

12[21]13 = 2 24 [24] 24

12[24]43 = − 2 24 ²

1234 (59) here we used

4 j=1

ij [jk] = 0 (derived f rom (16)) , s ₁₂ = s ₃₄ , s ₁₃ = s ₂₄ (60) Exercise: Show

A 4 = −2 [13] ²

[12][34] (61)

3.1.2 e ⁻ _R (1)e ⁺ _L (2) → µ ⁻ _R (3)µ ⁺ _L (4) Using (47), we obtain

A 4 = +2 14 ²

1234 (62)

3.1.3 e ⁻ _L (1)e ⁺ _R (2) → µ ⁻ _L (3)µ ⁺ _R (4)

A 4 = +2 23 ²

12 34 (63)

3.1.4 e ⁻ _R (1)e ⁺ _L (2) → µ ⁻ _L (3)µ ⁺ _R (4)

A 4 = −2 13 ²

1234 (64)

3.1.5 Unpolarized, helicity-summed cross sections

dσ(e

e

→µ

µ

)

d cos θ ∝ 1

2

helicity

|A 4 | ² = 4 24 ² 1234

² + 14 ²

1234 ²

= 4 s ² ₁₃ + s ² ₁₄

s ² ₁₂ ∝ (1 − cos θ) ² + (1 + cos θ) ² ∝ 1 + cos ² θ(65) e ⁻ e ⁺ → µ ⁻ µ ⁺

s 12 = s, s 13 = t, s 14 = u dσ dΩ = α ²

2s t ² + u ²

s ² (66)

(In the limit of t → 0, _dΩ ^dσ → ^α _2s

) e ⁻ µ ⁻ → e ⁻ µ ⁻

s 12 = t, s 13 = s, s 14 = u dσ dΩ = α ²

2s

s ² + u ²

t ² (67)

(In the limit of t → 0, _dΩ ^dσ → ^α _t

^s )

3.2 e ⁻ e ⁻ , e ⁻ e ⁺ scattering

e(1)e(2) → e(3)e(4)

A + −− + = 3 ⁻ |γ ^µ |1 ⁻ 4 ⁺ |γ µ |2 ⁺ s 13

= +2 23[14]

s 13

(68)

A + − + − = − 4 ⁻ |γ ^µ |1 ⁻ 3 ⁺ |γ µ |2 ⁺ s 14

= −2 24[13]

s 14

(note the sign f rom exchange 3 ↔ 4) (69)

A ₊₊₋₋ = 3 ⁻ | γ ^µ | 1 ⁻ 4 ⁻ | γ _µ | 2 ⁻

s ₁₃ − 4 ⁻ | γ ^µ | 1 ⁻ 3 ⁻ | γ _µ | 2 ⁻ s ₁₄

= − 2 34 [12]

1 s 13

+ 1 s 14

= +2 34 [12]

s ₁₂ s 13 s 14

(70)

since s 12 + s 13 + s 14 = 0.

Unpolarized, helicity-summed cross sections

|A| ² ∝ s ² ₁₄ s ² ₁₃ + s ² ₁₃

s ² ₁₄ + s ⁴ ₁₂

s ² ₁₃ s ² ₁₄ = s ⁴ ₁₂ + s ⁴ ₁₃ + s ⁴ ₁₄

s ² ₁₃ s ² ₁₄ (71) For e ⁻ e ⁻ → e ⁻ e ⁻ or e ⁺ e ⁺ → e ⁺ e ⁺ (Møller scattering),

s 12 = s, s 13 = t, s 14 = u

|A| ² ∝ s ⁴ + t ⁴ + u ⁴

t ² u ² (72)

dσ

dΩ = α ² 2s

s ⁴ + t ⁴ + u ⁴

t ² u ² (73)

= α ² 2s

s ² + u ²

t ² + s ² + t ² u ² + 2 s ²

ut

(74)

= α ² 4s

s − u

t + s − t u

2

(75) (In the limit of t → 0, _dΩ ^dσ → ^α _t

^s )

For e ⁻ e ⁺ → e ⁻ e ⁺ (Bhabha scattering), s 13 = s, s 14 = t, s 12 = u

| A | ² ∝ s ⁴ + t ⁴ + u ⁴

s ² t ² (76)

dσ

dΩ = α ² 2s

s ⁴ + t ⁴ + u ⁴

s ² t ² (77)

= α ² 2s

s ² + u ²

t ² + u ² + t ² s ² + 2 u ²

st

(78)

= α ² 4s

s − u

t + t − u s

2

(79)

(In the limit of t → 0, _dΩ ^dσ → ^α _t

^s )

4 Spinor-helicity rep. for polarization

4.1 Polarization vectors

⁺ _µ (k, q) = q ⁻ √ |γ µ |k ⁻

2qk , ⁻ _µ (k, q) = q ⁺ √ |γ µ |k ⁺

2[kq] (80)

where q · k does not vanish.

k · ^± (k, q) = 0 (T ransverse) (f rom Dirac eq.) (81) ⁺ · ⁻ = q ⁻ |γ ^µ |k ⁻ q ⁺ |γ µ |k ⁺

2 qk [kq] = −2qk[kq]

2 qk [kq] = −1 (82) ⁺ · ⁺ = q ⁻ |γ ^µ |k ⁻ q ⁻ |γ µ |k ⁻

2 qk ² = q ⁻ |γ ^µ |k ⁻ k ⁺ |γ µ |q ⁺

2 qk ² = 0 (83) ⁺ _µ ⁻ _ν + ⁻ _µ ⁺ _ν = − η _µν + k _µ q _ν + k _ν q _µ

k · q (84)

where we used (47), (53), (46) and γ µ q/γ ν + γ ν q/γ µ = 2(q ν γ µ + q µ γ ν − q/η µν ). It is easy to confirm k · ^{+ −} _ν + k · ^{− +} _ν = q · ^{+ −} _ν + q · ^{− +} _ν = 0.

From (53),

U ± (p)U ± (p) = 1

2 (1 ± γ ⁵ )p / (85)

then for arbitrary p p · ⁺ (k, q) = q √ ⁻ |p /|k ⁻

2qk = qp[pk] √

2qk , p · ⁻ (k, q) = q √ ⁺ |p /|k ⁺

2[kq] = [qp]pk √ 2[kq] (86) Thus we find

q · ^± = 0 (and rediscover k · ^± = 0) (87) Exercise. Show

⁺ (k 1 , q) · ⁺ (k 2 , q) = ⁻ (k 1 , q) · ⁻ (k 2 , q) = 0 (88) ⁺ (k ₁ , q) · ⁻ (k ₂ , k ₁ ) = 0 (89) Actually, you must show

⁺ (k, q) · ⁺ (k , q ) = − qq [kk ]

qk q k (90) ⁺ (k, q) · ⁻ (k , q ) = − qk [kq ]

qk [k q ] (91) Using (51), we find

/ ⁺ (k, q) =

√ 2

qk (|k ⁻ q ⁻ | + |q ⁺ k ⁺ |) , / ⁻ (k, q) =

√ 2

[kq] (|k ⁺ q ⁺ | + |q ⁻ k ⁻ |) (92) Then

/ ^± (k, q) | q ^± = 0 , q ^∓ | / ^± (k, q) = 0 (93)

4.2 Significance of q

⁺ _µ (k, q ) = q √ ⁻ | γ _µ | k ⁻

2 q k = q ⁻ √ | γ _µ | k ⁻ k ⁻ | q ⁺

2 q k kq = √ q ⁻ | γ _µ k/ | q ⁺ 2 q k kq

= − q √ ⁻ |k/γ µ |q ⁺ 2q kkq +

√ 2q q q k kq k µ

= ⁺ _µ (k, q) +

√ 2q q

q k kq k _µ (94)

It’s a gauge degree of freedom!

5 QED (QCD) example (II)

5.1 two fermions and two photons

A _+−λ

_λ

= 2 ⁻ | / ^λ

(k ₄ , q ₄ )(p / ₁ + k/ ₃ ) / ^λ

(k ₃ , q ₃ ) | 1 ⁻ s ₁₃

+ 2 ⁻ | / ^λ

(k ₃ , q ₃ )(p / ₁ + k/ ₄ ) / ^λ

(k ₄ , q ₄ ) | 1 ⁻ s 14

(95) For λ 3 = λ 4 = −, this vanishes if we take q 3 = q 4 = p 1 .

For λ 3 = λ 4 = +, this vanishes if we take q 3 = q 4 = p 2 .

A ₊₋₊₋ = 2 ⁻ | / ⁻ (k 4 , q 4 )(p / 1 + k/ 3 ) / ⁺ (k 3 , p 2 )|1 ⁻ s ₁₃

=

√ 2

[4q ₄ ] 24 q ₄ ⁺ | (p / ₁ + k/ ₃ ) | 2 ⁺ [31]

√ 2 23

1

s ₁₃ (96)

The choice q 4 = k 3 leads to A + − + − =

√ 2

[43] 24 3 ⁺ | p / 1 | 2 ⁺ [31]

√ 2 23

1

s ₁₃ = 2 24[31]12[31]

[43] 23 s ₁₃

= 2 24 [31] 12 [31]

[43]2313[31] = 2 24 [34] 42

[34]2313 = − 2 24 ²

1323 (97) or

A + − + − = 2 24[31]12[31]

[43] 23 s ₁₃ = −2 [13] ² 2412 23 [34]s ₂₄

= +2 [13] ² 2412

21 [14] 24 [42] = 2 [13] ²

[14][24] (98)

By exchanging 3 ↔ 4, we obtain

A ₊₋₋₊ = − 2 23 ²

1424 (99)

Unpolarized, helicity-summed cross sections

|A| ² ∝ |s 14 |

| s ₁₃ | + |s 13 |

| s ₁₄ | (100)

For e ⁻ γ → e ⁻ γ or e ⁺ γ → e ⁺ γ,

s 13 = s, s 12 = t, s 14 = u

|A| ² ∝ s ² + u ²

|su| (101)

For e ⁻ e ⁺ → γγ,

s 12 = s, s 13 = t, s 14 = u

| A | ² ∝ t ² + u ²

|tu| (102)

5.2 e ⁻ e ⁺ → qg¯ q

Using Fierz identity, we find A ₅ = 25

s 12

1 ⁺ | (k/ ₃ + k/ √ ₄ )/ ⁺ (k ₄ , q) | 3 ⁻

2s ₃₄ + [13]

s 12

2 ⁻ | (k/ ₄ + k/ √ ₅ )/ ⁺ (k ₄ , q) | 5 ⁺

2s ₄₅ (103)

Using (92), we can rewrite A ₅ as A ₅ = 25

s 12

1 ⁺ | (k/ ₃ + k/ ₄ ) | q ⁺ [43]

s 34 q4 + [13]

s 12

2 ⁻ | (k/ ₄ + k/ ₅ ) | 4 ⁻ q5

s 45 q4 (104)

Choosing q = k ₅ , we remove the second term and get A 5 = − 25

s 12

1 ⁺ | (k/ ₃ + k/ ₄ ) | 5 ⁺ [43]

s 34 45

= 25 1 ⁺ | (k/ ₁ + k/ ₂ + k/ ₅ ) | 5 ⁺ [43]

12[21]34[43]45

= 25[12]25[43]

12 [21] 34 [43] 45

= − 25 ²

12 34 45 (105)

where we used 5

i=1 k i = 0, Dirac eqs. and (53).

Note that

25 ²

123445 = 35 3445

25 ²

1235 (106)

6 Gluon amplitudes

• “Color” factors are separated (and ordered).

• Gluon vertices include · .

• Since ⁺ (k ₁ , q) · ⁺ (k ₂ , q) = 0, A(+ + · · · +) = 0 can be shown.

• Using ⁻ (k ₁ , q) · ⁺ (k ₂ , k ₁ ) = 0, A( − + · · · +) = 0 is also shown.

6.1 4-point amplitude

6.1.1 an amplitude

• Calculate A ₄ (1 ⁻ 2 ⁻ 3 ⁺ 4 ⁺ ).

• Choose q ₁ = q ₂ = k ₃ , q ₃ = q ₄ = k ₂ , only nonvanishing · is ⁻ (k ₁ , k ₃ ) · ⁺ (k 4 , k 2 ) = _{24 [31]} ^{21 [43]} .

• only s ₁₂ channel contributes

A 4 (1 ⁻ 2 ⁻ 3 ⁺ 4 ⁺ )

= i

√ 2 2

[ − 2k ₁ · ⁻ (k ₂ , k ₃ ) ⁻ _µ (k ₁ , k ₃ )] − ig ^µν s 12

[2k ₄ · ⁺ (k ₃ , k ₂ ) ⁺ _ν (k ₄ , k ₂ )]

= − 2i s 12

k ₁ · ⁻ (k ₂ , k ₃ ) k ₄ · ⁺ (k ₃ , k ₂ ) ⁻ (k ₁ , k ₃ ) · ⁺ (k ₄ , k ₂ )

= − 2i s ₁₂

[31] √ 12 2[23]

24 [43]

√ 2 23

21 [43]

24 [31] = − i 12 ² [34] ² s ₁₂ s ₂₃

= i 12 ⁴

12 23 34 41 (107)

Using cyclic symmetry, we get

A ₄ (1 ⁺ 2 ⁻ 3 ⁻ 4 ⁺ ) = i 23 ⁴

12233441 (108) 6.1.2 decoupling identity

We consider SU (N) gauge theory, not U (N) gauge theory. The vertices come

from the commutater in the field strength, thus if one of the generator T ^a is

replaced by unity, the amplitude (possibly thought as in U (N) theory) must

vanish.

The full amplitude looks like

A = g ²

noncyclic perms

Tr(T ^a

T ^a

T ^a

T ^a

)A(1234) (109) (which can be generalized to n-point amplitudes)

Noncyclic means that, roughly speaking, 1 is fixed (or permutation σ ∈ S _n /Z _n ), because of the cyclicity A(1234) = A(4123).

Turning to the discussion, the following holds:

0 = T r(T ^a

T ^a

T ^a

)[A(1234) + A(1243) + A(1423)]

+ T r(T ^a

T ^a

T ^a

)[A(1324) + A(1342) + A(1432)] (110) Therefore we find

A(1234) + A(1243) + A(1423) = A(1324) + A(1342) + A(1432) = 0 (111) and then

A(1 ⁻ 2 ⁺ 3 ⁻ 4 ⁺ ) = − A(1 ⁻ 2 ⁺ 4 ⁺ 3 ⁻ ) − A(1 ⁻ 4 ⁺ 2 ⁺ 3 ⁻ ) (112)

A(1 ⁻ 2 ⁺ 3 ⁻ 4 ⁺ ) = −A(1 ⁻ 2 ⁺ 4 ⁺ 3 ⁻ ) − A(1 ⁻ 4 ⁺ 2 ⁺ 3 ⁻ )

= − A(3 ⁻ 1 ⁻ 2 ⁺ 4 ⁺ ) − A(3 ⁻ 1 ⁻ 4 ⁺ 2 ⁺ )

= − i

31 ⁴

31122443 + 31 ⁴ 31144223

= −i 13 ³ 24

1

12 34 + 1 14 23

= i 13 ³ 24

14 23 + 12 34 12233441

= i 13 ³ 24

− 13 42

12233441 = i 13 ⁴

12233441 (113) where we used Schouten identity (30).

6.1.3 cross section

colors

|A| ² = 2N ² (N ² − 1)g ⁴ (|A(1234)| ² + |A(1342)| ² + |A(1423)| ² ) (114) because of (6.1.2) and the reflection symmetry ¹ A(1234) = A(4321). Then

colors

|A| ² (1 ⁻ 2 ⁻ 3 ⁺ 4 ⁺ ) = 2N ² (N ² − 1)g ⁴ s ⁴ 1

s ² u ² + 1 t ² s ² + 1

u ² t ²

(115)

1

A(12 · · · n) = ( − 1)

A(n · · · 21) in general

and summing over helicities, we find

colors

helicities

|A| ² = 4N ² (N ² − 1)g ⁴ (s ⁴ + t ⁴ + u ⁴ ) 1

s ² u ² + 1 t ² s ² + 1

u ² t ²

(116)

(for averaging over the initial colors we divide this by 4(N ² − 1) ² ) Column

Fun with Mandelstam!

s + t + u = 0, okey?

0 = (s + t + u) ² = s ² + t ² + u ² + 2(st + tu + us)

(s ² + t ² + u ² ) ² = s ⁴ + t ⁴ + u ⁴ + 2(s ² t ² + t ² u ² + u ² s ² ) = 4(s ² t ² + t ² u ² + u ² s ² ) + 8(s ² tu + t ² us + u ² st) = 4(s ² t ² + t ² u ² + u ² s ² ) (s ⁴ + t ⁴ + u ⁴ )(s ² + t ² + u ² ) = − 4(st + tu + us)(s ² t ² + t ² u ² + u ² s ² ) =

−4(st +tu+us)(s ² t ² +t ² u ² +u ² s ² −(s ² tu+t ² us+u ² st)) = −4(s ³ t ³ + t ³ u ³ + u ³ s ³ − 3s ² t ² u ² )

then

(s ⁴ + t ⁴ + u ⁴ ) ₁

s

u

+ _t

¹ s

+ _u

¹ t

= 4

3 − ^tu _s

− ^us _t

− _u ^st

7 Five-parton amplitude

• Calculate A 5 (1 ⁻ _{q ¯} 2 ⁺ _q 3 ⁻ 4 ⁺ 5 ⁺ )

• Use ⁻ (k ₃ , k ₂ ), ⁺ (k ₄ , k ₁ ), ⁺ (k ₅ , k ₁ ) to apply (93)

A 5 (1 ⁻ _{q ¯} 2 ⁺ _q 3 ⁻ 4 ⁺ 5 ⁺ ) = A ^(a) + A ^(b) (117) A ^(a) = − i

√ 2

2 ⁺ | (k/ ₃ − k/ ₄ − k/ ₅ ) | 1 ⁺ s 12 s 45 × [ ⁻ (k 3 , k 2 ) · ⁺ (k 5 , k 1 ) ⁺ (k 4 , k 1 ) · k 5

− ⁻ (k ₃ , k ₂ ) · ⁺ (k ₄ , k ₁ ) ⁺ (k ₅ , k ₁ ) · k ₄ ] (118)

A ^(b) = − i

√ 2s 12 s 34

{2 ⁺ |(k/ 3 + k/ 4 − k/ 5 )|1 ⁺ × [ 1

2 ⁻ (k 3 , k 2 ) · ⁺ (k 4 , k 1 ) ⁺ (k 5 , k 1 ) · (k 3 − k 4 )

− ⁻ (k 3 , k 2 ) · ⁺ (k 5 , k 1 ) ⁺ (k 4 , k 1 ) · k 3 ]

− 2 ⁺ | (k/ ₃ − k/ ₄ ) | 1 ^{+ −} (k ₃ , k ₂ ) · ⁺ (k ₄ , k ₁ ) ⁺ (k ₅ , k ₁ ) · (k ₃ + k ₄ ) (119) }

A ^(a) = −i [23]31 s 12 s 45

− [25]13 [23]15

15[54]

14 + [24]13 [23]14

14[45]

15

= i [23] 13 ² [45]

s ₁₂ s ₄₅ [23] 14 15 [ − 15 [52] − 14 [42]]

= − i [23] 13 ³ [45]

s 12 s 45 1415 = i [23] 13 ³

s 12 141545 (120) A ^(b) = i [25] 13 ³ [34]

s 12 s 34 1415 = − i [25] 13 ³

s 12 141534 (exercise!) (121) A 5 (1 ⁻ _{¯ q} 2 ⁺ _q 3 ⁻ 4 ⁺ 5 ⁺ ) = A ^(a) + A ^(b)

= −i 13 ³ (−[23]34 − [25]54) s ₁₂ 14 15 34 45

= − i 13 ³ [21]14 s ₁₂ 14 15 34 45

= i 13 ³ 23

1223344551 (122)

Here we used momentum conservation, (86) and (91).

8 General gluon amplitudes

8.1 Recursive relations

Berends-Giele current J ^µ (1 · · · n) J ^µ (1 · · · n) = − i

K _1,n ² [

n−1

j=1

V ₃ ^µνλ (K _1,j , K _j+1,n )J _ν (1, . . . , j)J _λ (j + 1, . . . , n)

+

n−2

j=1 n−1

!=j+1

V ₄ ^µνλρ J _ν (1, . . . , j)

× J _λ (j + 1, . . . , )J _ρ ( + 1, . . . , n)] (123) where V are the color-ordered gluon self-interactions

V ₃ ^µνλ (P, Q) = i

√ 2 [η ^νλ (P − Q) ^µ + 2η ^λµ Q ^ν − 2η ^µν Q ^λ ] (124) V ₄ ^µνλρ = i

2 [2η ^µλ η ^νρ − η ^µν η ^λρ − η ^µρ η ^νλ ] (125) and

K _i,j ≡ k _i + k _i+1 + · · · + k _j (126)

• decoupling identity

J ^µ (123 · · · n) + J ^µ (213 · · · n) + · · · + J ^µ (23 · · · n1) = 0

• reflection identity

J ^µ (123 · · · n) = (−1) ⁿ⁺¹ J ^µ (n · · · 321)

• conservation K _1,n ^µ J µ (12 · · · n) = 0

8.2 example: J ^µ (+ · · · +)

By explicit calculation, one can get

J ^µ (1 ⁺ 2 ⁺ 3 ⁺ 4 ⁺ ) = √ q ⁻ | γ ^µ K / _1,4 | q ⁺

2 q1 1 · · · 4 4q (127) where 1 · · · n ≡ 1223 · · · (n − 1) n.

Ansatz

J ^µ (1 ⁺ · · · n ⁺ ) = √ q ⁻ |γ ^µ K / 1,n |q ⁺

2q11 · · · nnq (128)

(Reflection and Conservation are trivially satisfied; ?Decoupling?)

· Note that

J ^µ (1 ⁺ ) = q √ ⁻ | γ ^µ k/ ₁ | q ⁺

2 q1 1q = q √ ⁻ | γ _µ | 1 ⁻

2 q1 = ^+µ (k ₁ , q) (129)

In the right-hand side of (123), only the second and third term in V 3 con- tribute.

J ^µ (1 ⁺ 2 ⁺ · · · n ⁺ ) = 1

√ 2K _1,n ² q11 · · · nnq

n − 1

j=1

j (j + 1) jq q (j + 1)

×(q ⁻ |γ ^µ K / j+1,n |q ⁺ K j+1,n ^ν q ⁻ |γ ν K / 1,j |q ⁺

−q ⁻ |γ ^µ K / 1,j |q ⁺ K _1,j ^ν q ⁻ |γ ν K / j+1,n |q ⁺ ) (130) Note that

q ⁻ |γ µ γ ν |q ⁺ = −q ⁻ |γ ν γ µ |q ⁺ + 2η µν q ⁻ |q ⁺ = −q ⁻ |γ ν γ µ |q ⁺ , (131) and K 1,j + K j+1,n = K 1,n .

We find

J ^µ (1 ⁺ 2 ⁺ · · · n ⁺ ) = √ q ⁻ |γ ^µ K / 1,n |q ⁺ 2K _1,n ² q11 · · · nnq

×

n − 1

j=1

j (j + 1)

jqq (j + 1) q ⁻ |K / j+1,n K / 1,j |q ⁺

= √ q ⁻ |γ ^µ K / 1,n |q ⁺ 2K _1,n ² q11 · · · nnq

×

n − 1 j=1

j (j + 1)

jqq (j + 1) q ⁻ |K / j+1,n K / 1,n |q ⁺

= − √ q ⁻ |γ ^µ K / 1,n |q ⁺ 2K _1,n ² q11 · · · nnq

×

n − 1 j=1

j (j + 1)

jqq (j + 1) q ⁻ |K / 1,j K / 1,n |q ⁺ (132) for

q ⁻ |K /K / |q ⁺ = K ² q ⁻ |q ⁺ = 0 . (133)

To proceed, we use

n − 1 j=1

j (j + 1)

jqq (j + 1) q ⁻ |K / 1,j =

n − 1 j=1

j

!=1

j (j + 1)

jqq (j + 1) q ⁻ |k/ !

= n

!=1 n − 1

j=!

j (j + 1)

jqq (j + 1) q ⁻ |k/ ! = n

!=1

n

qq n q ⁻ |k/ !

= n

!=1

1

q n n ⁻ |k/ ! = 1

q n n ⁻ |K / 1,n (134)

where “eikonal” identity (31), (‘open’) Schouten identity

nq ⁻ | + qn ⁻ | + nq ⁻ | = 0 (135)

and Dirac equations ⁻ |k/ ! = 0 are used.

Then (128) is shown.

Show A(+ · · · +) = A(− + · · · +) = 0 by using J (+ · · · +).

8.3 J ^µ (− + · · · +) and MHV amplitude

J ^µ (1 ⁻ 2 ⁺ · · · n ⁺ ) = √ 1 ⁻ |γ ^µ K / 2,n |1 ⁺ 21 · · · nn1

n m=3

1 ⁻ |k/ m K / 1,m |1 ⁺

K _1,m− ² ₁ K _1,m ² (136) where q 1 = k 2 , q 2 = · · · = q n = k 1 . Show this!!

A(1 ⁻ 2 ⁺ · · · n ⁺ (n+1) ⁻ ) = − i n √ ⁺ | γ _µ | (n + 1) ⁺ 2[n (n + 1)]

1 ⁻ | γ ^µ K / _1,n | 1 ⁺

√ 2 1 · · · n n1

1 ⁻ | k/ _n K / _1,n | 1 ⁺ K _1,n−1 ²

(137) where we choose q _n+1 = k _n , and note that K / _2,n | 1 ⁺ = K / _1,n | 1 ⁺ . Moreover we note that k _n+1 = − K _1,n is null.

Fierz identity leads to

n ⁺ | γ _µ | (n + 1) ⁺ 1 ⁻ | γ ^µ K / _1,n | 1 ⁺ = 2 1 (n + 1) n ⁺ | K / _1,n | 1 ⁺ (138) while

n ⁺ |K / 1,n |1 ⁺ = −n ⁺ |k/ n+1 |1 ⁺ = −[n (n + 1)](n + 1) 1 (139) and

1 ⁻ |k/ n K / 1,n |1 ⁺ = −1 ⁻ |k/ n k/ n+1 |1 ⁺ = −1n[n (n + 1)](n + 1) 1 (140) Therefore

A(1 ⁻ 2 ⁺ · · · n ⁺ (n + 1) ⁻ ) = i 1 (n + 1) [n (n + 1)] (n + 1) 1 (n + 1) 1 1 · · · n s _n,n+1

= i 1 (n + 1) ⁴

1 · · · (n + 1) (n + 1) 1 (141)

9 ‘Supersymmetry’

9.1 ‘susy’ generator

Q(η) = ¯ η ^α Q _α is bosonic.

[Q(η), G ^± (k)] = ± Γ ^± (k, η)Λ ^± (k) , [Q(η), Λ ^± (k)] = ∓ Γ ^∓ (k, η)G ^± (k) (142) where Γ(k, η) is linear in η and fermionic.

The Jacobi identity

0 = [[Q(η), Q(ζ)], Φ(k)] + [[Q(ζ), Φ(k)], Q(η)] + [[Φ(k), Q(η)], Q(ζ)] (143) leads to a possible choice Γ ⁺ (k, q) = θ[qk] and Γ ⁻ (k, q) = θqk where θ is a fermionic parameter (¯ η(q) ∼ θU + (q)).

9.2 ‘Supersymmetry’ Ward identity

0 = 0|[Q, Φ 1 Φ 2 · · · Φ n ]|0 = n

i=1

0|Φ 1 · · · [Q, Φ i ] · · · Φ n |0 (144)

9.2.1 all helicities positive 0 = 0 | [Q(η(q)), Λ ⁺ ₁ G ⁺ ₂ · · · G ⁺ _n ] | 0

= −Γ ⁻ (k 1 , q)A n (G ⁺ ₁ , G ⁺ ₂ , · · · , G ⁺ _n ) + Γ ⁺ (k 2 , q)A n (Λ ⁺ ₁ , Λ ⁺ ₂ , G ⁺ ₃ , · · · , G ⁺ _n ) + · · · + Γ ⁺ (k n , q)A n (Λ ⁺ ₁ , G ⁺ ₂ , · · · , Λ ⁺ _n ) (145) Helicity conservation of ‘gluino’ implies that the fermionic amplitudes vanish, thus A n (G ⁺ ₁ , G ⁺ ₂ , · · · , G ⁺ _n ) = 0.

9.2.2 only one negative helicity 0 = 0 | [Q(η(q)), Λ ⁺ ₁ G ⁻ ₂ G ⁺ ₃ · · · G ⁺ _n ] | 0

= −Γ ⁻ (k 1 , q)A n (G ⁺ ₁ , G ⁻ ₂ , G ⁺ ₃ , · · · , G ⁺ _n ) − Γ ⁻ (k 2 , q)A n (Λ ⁺ ₁ , Λ ⁻ ₂ , G ⁺ ₃ , · · · , G ⁺ _n )

+0 + · · · + 0 (146)

If we choose q = k 1 , we get A n (Λ ⁺ ₁ , Λ ⁻ ₂ , G ⁺ ₃ , · · · , G ⁺ _n ) = 0. If we choose

q = k 2 , we get A n (G ⁺ ₁ , G ⁻ ₂ , G ⁺ ₃ , · · · , G ⁺ _n ) = 0.

9.2.3 two negative helicity

0 = 0 | [Q(η(q)), G ⁻ ₁ G ⁻ ₂ Λ ⁺ ₃ G ⁺ ₄ · · · G ⁺ _n ] | 0

= Γ ⁻ (k 1 , q)A n (Λ ⁻ ₁ , G ⁻ ₂ , Λ ⁺ ₃ , · · · , G ⁺ _n ) + Γ ⁻ (k 2 , q)A n (G ⁻ ₁ , Λ ⁻ ₂ , Λ ⁺ ₃ , · · · , G ⁺ _n )

− Γ ⁻ (k ₃ , q)A _n (G ⁻ ₁ , G ⁻ ₂ , G ⁺ ₃ , · · · , G ⁺ _n ) (147) If we choose q = k ₁ , we get

A n (G ⁻ ₁ , G ⁻ ₂ , G ⁺ ₃ , · · · , G ⁺ _n ) = 12

13 A n (G ⁻ ₁ , Λ ⁻ ₂ , Λ ⁺ ₃ , G ⁺ ₄ , · · · , G ⁺ _n ) (148)

References

[1] L. Dixon, hep-ph/9601359, and his recent talks which can be found in various conferences.

[2] M. Mangano and S. Parke, Phys. Rep. 200 (1991) 301. The preprint of this can be obtained from Parke’s page.

[3] D. A. Kosower, “N=4 Supersymmetric Gauge Theory, Twistor Space, and

Dualities”, Saclay Lectures (Fall 2004).