positivity of scattering amplitudes in the …positivity of scattering amplitudes in the...
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Positivity of Scattering Amplitudes in the Amplituhedron
Jaroslav Trnka California Institute of Technology
Montreal, July 30, 2015
In collaboration with Nima Arkani-Hamed and Andrew Hodges, 1412.8478
You already heard about Amplituhedron todayfrom Nima and Hugh
Scattering Amplitudes
✤ Basic objects in Quantum Field Theory (QFT)
✤ Predictions for colliders: cross-sections
✤ My motivation: new ideas in QFT
Perturbative QFT
✤ Loop expansion
✤ Integrand: rational function before integration
sum of Feynman diagrams
⌦ = d4`1 . . . d4`L I(`j , ki, si)
I(`j , ki, si)
A =
Z
`j2R⌦
Integrand form
Special case: Planar N=4 SYM
✤ “Simplest Quantum Field Theory”
✤ Yangian symmetry of the integrand
✤ Toy model for QCD: tree-level amplitudes identical
✤ Formulation using on-shell diagrams, positive Grassmannian and the Amplituhedron
The Amplituhedron
(Arkani-Hamed, JT 2013)
Definition of the Amplituhedron
✤ The full definition that Nima and Hugh described fits on one slide:
Definition of the Amplituhedron
✤ The full definition that Nima and Hugh described fits on one slide:
✤ The full definition that Nima and Hugh described fits on one slide:
Definition of the Amplituhedron
⌦n,k,`
Form with logarithmic singularities on the boundaries of Amplituhedron
Integrand of amplitudes in planar N=4 SYM can be extracted from it
`nk
- number of loops- number of external legs- SU(4) R-charge
Amplituhedron form
✤ Geometry set-up for
Projective space
points in this space
-dimensional projective plane
lines
⌦n,k,`,m
Zj
Y
n
k
` Lk = (AB)k
Pm+k�1
⌦n,k,`,m(Y, Zj , ABk)
Amplituhedron form
✤ Geometry set-up for
Projective space
points in this space
-dimensional projective plane
lines
⌦n,k,`,m
Zj
Y
n
k
` Lk = (AB)k
Simple case:m = 2 k = 1 ` = 0
points
Point
Pm+k�1 n Z1, . . . , Zn 2 P2
Y 2 P2
2
1
54
3
Y
⌦n,k,`,m(Y, Zj , ABk)
Simplest example: triangle
✤ Polygon for n=3:
✤ We can expand:
✤ First boundaries:
✤ Differential form:
Y, Z1, Z2, Z3
Y = Z1 + c2Z2 + c3Z3
Y 2 (23) : c2, c3 ! 1,c2c3
fixed
Y 2 (12) : c3 = 0Y 2 (13) : c2 = 0
⌦ =dc2c2
dc3c3
Y
1
2
3
Simplest example: triangle
✤ Polygon for n=3:
✤ We can expand:
✤ We can solve in terms of SL(3) invariants
and we get
Y, Z1, Z2, Z3
Y = Z1 + c2Z2 + c3Z3
c2 =hY 13ihY 23i , c3 =
hY 12ihY 23i
⌦ =hY d2Y ih123i2
hY 12ihY 23ihY 13i
hX1X2X3i = ✏abcXa1X
b2X
c3
1
2
3
Y
Polygon
✤ General polygon
✤ We can again expand
✤ The result is not
✤ It is some non-trivial two-form
Y = Z1 + c2Z2 + c3Z3 + · · ·+ cnZn
⌦ =dc2c2
dc3c3
. . .dcncn
1
2 3
45
Y
Triangulation of polygon
✤ We can triangulate
✤ In each triangle
and we have
✤ The form for polygon:
Y = Z1 + ciZi + ci+1Zi+1
(Z1, Zi, Zi+1)
⌦i =dcici
dci+1
ci+1=
hY d2Y ih1 i i+ 1i2
hY 1 iihY 1 i+ 1ihY i i+ 1i
⌦ =n�1X
i=2
⌦i
1
i
i+ 1
Y
Triangulation of polygon
✤ For n=4 we have
⌦ =hY d2Y ih123i2
hY 12ihY 23ihY 13i +hY d2Y ih134i2
hY 13ihY 14ihY 34i
hY 13i is spurious
1
23
4
Triangulation of polygon
✤ For n=4 we have
⌦ =hY d2Y i (hY 23ih134ih124i � hY 41ih123ih234i)
hY 12ihY 23ihY 34ihY 41i
1
23
4
Triangulation of polygon
✤ For n=4 we have
⌦ =hY d2Y i N (Y, Z)
hY 12ihY 23ihY 34ihY 41i
We want to fix the numerator
from singularities1
23
4
Positivity in the Amplituhedron
(Arkani-Hamed, Hodges, JT 2014)
Numerator of the form
✤ First singularities OK
✤ Second singularities: some are bad
⌦ =hY d2Y i N (Y, Z)
hY 12ihY 23ihY 34ihY 41i
hY 12i = hY 34i = 0
hY 23i = hY 14i = 0Y = X24 = (23) \ (41) = Z2h341i � Z3h241i
Y = X13 = (12) \ (34) = Z1h234i � Z2h134ihY 12i = hY 23i = 0
Y = Z2
GOOD BAD
etc.
The numerator should vanish on X13, X24
Numerator of the form
⌦ =hY d2Y ihY X13X24i
hY 12ihY 23ihY 34ihY 41i
1
2
3
4
X13
X24
Pentagon
✤ For n=5 the form is
✤ The numerator N (Y, Z) = CIJYIY J
1
2 3
45
Y
⌦ =hY d2Y iN (Y, Z)
hY 12ihY 23ihY 34ihY 45ihY 51i
Pentagon
1
2 3
4
5
X13
X24
X35X41
X52
✤ Five bad points:
✤ Numerator vanishes
Pentagon
X13 = (12) \ (34) X24 = (23) \ (45)
X35 = (34) \ (51) X41 = (45) \ (12)
X52 = (51) \ (23)
N (Y = X) = CIJXIXJ = 0
Pentagon
1
2 3
4
5
X13
X24
X35X41
X52
Pentagon
Numerator vanishes if these six points are on the conic
N = ✏I1J1,I2J2,I3J3,I4J4,I5J5,I6J6YI1Y J1XI2
13XJ213X
I324X
J324X
I435X
J435X
I541X
J541X
I652X
J652
General polygon
✤ General polygon:
✤ We have forbidden points
✤ Points lie on algebraic curve of degree n-3
⌦ =hY d2Y iN (Y, Z)
hY 12ihY 23ihY 34i . . . hY n1i
n(n� 3)
2Xij = (i i+ 1) \ (j j + 1)
Y,Xij
Lesson from polygon
✤ The numerator is fully specified by a set of its zeroes
✤ These zeroes: illegal singularities from denominator
✤ Illegal: points outside the polygon
✤ Outside = they can not be written as
Trivial: it is a polynomial in Y
Y = c1Z1 + c2Z2 + · · ·+ cnZn ci > 0
Positivity
✤ Interesting property: positive if Y inside polygon
✤ Denominator manifestly positive
✤ Numerator also positive (zeroes outside polygon)
✤ This simple case: area of polygon = positive
⌦ =hY d2Y iN (Y, Z)
hY 12ihY 23ihY 34i . . . hY n1i
Other example
✤ We consider but
✤ We have a line in and n points
✤ These n points are positive
✤ Line where
✤ Boundaries of this space
m = 2, ` = 0 k = 2
P3Y ↵� Zj
hijkli > 0
Y = C · Z C 2 G+(2, n) Z 2 M+(4, n)
hY i i+ 1i
Four point example
✤ For n=4 we have
✤ The logarithmic form on the boundaries is just
✤ Rewrite using SL(4) invariants
Y = (C · Z) 2 G+(2, 4)
C =
✓1 c1 0 �c20 c3 1 c4
◆
⌦ =dc1c1
dc2c2
dc3c3
dc4c4
Four point example
✤ We can choose two points on
✤ Then we can solve for
C =
✓1 c1 0 �c20 c3 1 c4
◆
Y = Y1Y2
Y1 = Z1 + c1Z2 � c2Z4
Y2 = c3Z2 + Z3 + c4Z4
ci
c2 =hY 12ihY 24i c3 =
hY 34ihY 24ic1 =
hY 41ihY 24i c4 =
hY 23ihY 24i
dc1dc2 =hY d2Y1ih1234i
hY 24i2 dc3dc4 =hY d2Y2ih1234i
hY 24i2
Four point example
✤ The form is then
✤ Manifestly positive
✤ No numerator in this case
⌦ =dc1c1
dc2c2
dc3c3
dc4c4
=hY d2Y1ihY d2Y2ih1234i2
hY 12ihY 23ihY 34ihY 41i
General case
✤ In general,
✤ The differential form is
✤ The numerator
Y = (C · Z) : G+(2, n) ! G(2, 4)
⌦ =hY d2Y1ihY d2Y2i N (Y, Z)
hY 12ihY 23ihY 34i . . . hY n1i
N (Y ) = Ca1b1...an�4bn�4Ya1b1 . . . Y an�4bn�4 ⌘ (C · Y Y . . . Y )
General case
✤ Four-dimensional object
✤ Legal boundaries (consistent with positivity):
✤ Illegal: all others
hY i i+ 1i = 0Level 1:
Level 2: hY i i+ 1i = hY j j + 1i = 0
Level 3: hY i� 1 ii = hY i i+ 1i = hY j j + 1i = 0
hY i� 1 ii = hY i i+ 1i = hY j � 1 ji = hY j j + 1i = 0Level 4:
hY 12i = hY 34i = hY 56i = 0
E.g.
but also Level 3:
hY 12i = hY 34i = hY 56i = hY 67i = 0
General case
✤ In general we have many illegal Level 3 boundaries
✤ The line Y is then given by
✤ Vanishing of the numerator
hY i i+ 1i = hY j j + 1i = hY k k + 1i = 0
X = Zi + ↵Zi+1whereY = (X j j + 1) \ (X k k + 1) = X1 + ↵X2 + ↵2X3
N (Y ) =2n�8X
k=0
C(k)↵k = 0
2n� 7conditions
Fixing the numerator
✤ Number of all independent conditions
✤ All illegal points/lines outside the Amplituhedron
✤ Numerator completely fixed by vanishing there
2
✓n4
◆�
✓n3
◆� 1
Number of degrees of freedomin the numerator
Overall constant
More cases
✤ Next example:
✤ The form is
✤ Double poles in the denominator
m = 3, k = 1, ` = 0 Point Y in P3
⌦ =hY d3Y iN (Y, Z)
hY 123ihY 125ihY 145ihY 235ihY 345ihY 134i
hY 123i = hY 125i = 0 Y = Z1 + ↵Z2We have:
hY 145i = ↵h2145iOn that configuration: hY 134i = ↵h2134i
⌦ ⇠ d↵N (↵)
↵2Legal boundary but double pole
N (↵) ⇠ ↵
More cases
✤ Numerator specified by:
✤ We also checked another case
✤ This construction makes manifest:
Vanishing on all illegal boundariesPreserve logarithmic singularities on all legal boundaries
m = 4, k = 1, ` = 0
“The form” is positive in the Amplituhedron
Numerical checks
✤ Other cases: too complicated to construct
✤ Check: positivity of the form (without measure)
✤ Evidence for such construction for all cases
✤ Form = Amplitude: No physics interpretation
N
Evidence for dual picture
✤ Definition: is a differential form with special properties
✤ Positivity: evidence of (dual) volume interpretation
…… Nima might say more on Monday……
⌦
Beyond the Amplituhedron
✤ Differential form on the Amplituhedron
✤ Final amplitude: integrate over loop momenta
✤ This translates into some complicated complex contour in the Amplituhedron
⌦
= Integrand of scattering amplitudes
A =
Z
`2R3,1
d4` I(`)
Beyond the Amplituhedron
✤ We do not know what it means geometrically
✤ Also IR divergencies, we take ratios of amplitudes
✤ Ratio function
✤ The result depends on Y, Z.
R6 =A(k=1)
6
A(k=0)6
n = 6, ` = 1,m = 4
Beyond the Amplituhedron
✤ The result is
✤ Rational prefactors
✤ Transcendental functions
R6 = H1 · [(2)� (3) + (4)] +H2 · [(3)� (4) + (5)] +H3 · [(4)� (5) + (6)]
(1) =hY d4Y ih12345i4
hY 1234ihY 2345ihY 3451ihY 4512ihY 5123i
H1 = Li2(1� u1) + Li2(1� u2) + Li2(1� u3) + log(u3) log(u1)� 2⇣2
u1 =hY 1234ihY 4561ihY 1245ihY 3461i , u2 =
hY 2345ihY 5612ihY 2356ihY 4512i , u3 =
hY 3456ihY 6123ihY 3461ihY 5623i
Beyond the Amplituhedron
✤ The result is positive
✤ Rational prefactors
✤ Transcendental functions
R6 = H1 · [(2)� (3) + (4)] +H2 · [(3)� (4) + (5)] +H3 · [(4)� (5) + (6)]
(1) =hY d4Y ih12345i4
hY 1234ihY 2345ihY 3451ihY 4512ihY 5123i
H1 = Li2(1� u1) + Li2(1� u2) + Li2(1� u3) + log(u3) log(u1)� 2⇣2
u1 =hY 1234ihY 4561ihY 1245ihY 3461i , u2 =
hY 2345ihY 5612ihY 2356ihY 4512i , u3 =
hY 3456ihY 6123ihY 3461ihY 5623i
Conclusion
✤ The Amplituhedron: generalization of
✤ Differential forms -> amplitudes in planar N=4 SYM
✤ Interesting properties of this form
✤ Extends beyond the integrand form
No triangulation, fixing the form from geometryPositivity of the form
G+(k, n)
Thank you!