Analytical Amplitudes from Numerical Exploration
arXiv:1904.04067 (JHEP) G. De Laurentis, D. Maître </font size>
arXiv:1910.11355 (JHEP) G. De Laurentis </font size>
arXiv:2002.04018 (JHEP) L. Budge, J. M. Campbell, G. De Laurentis, R. K. Ellis, S. Seth </font size>
Giuseppe De Laurentis
PhD supervisor: Daniel Maître
Summer Meeting - Freiburg\
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Table of Contents
A. Reconstruction of analytical spinor-helicity amplitudes
1. Motivation 2. Singular limits 3. Partial fractions 4. Ansatze fitting
B. Some applications
1. Six-gluon amplitude at one-loop 2. Two-loop five-parton finite remainders
3. Higgs + 4-parton (w/ finite top-mass) 4. Tree amplitudes from CHY formalism
A 1.1 Motivation
Cross sections at hadron colliders
$$d\hat{σ}_{n}=\frac{1}{2\hat{s}}dΠ_{n-2};(2π)^4δ^4\big(∑_{i=1}^n p_i\big);|\overline{\mathcal{A}(p_i,μ_F, μ_R)}|^2$$
Better predictions require both more loops and higher multiplicity.
IR singularities have to be cancelled between real and virtual corrections.
More loops $\rightarrow$ analytical complexity; more legs $\rightarrow$ algebraic complexity
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Chromo-dynamics and kinematics
Color ordering at tree level and one loop
F.A.Berends, W.Giele , Z.Bern, D.A.Kosower , … </font size>
$\mathcal{A}^{tree}_{n}({p_i, λ_i, a_i}) = ; g^{n-2} ∑_{σ\in S_n/Z_n} \text{Tr}(T^{a_σ(1)}\dots T^{a_σ(n)}) A^{tree}_n(σ(1^{λ_1}),\dots ,σ(n^{λ_n}))$</font size>
$A^{tree},, d_i,, c_i,, b_i,$ and $R$ are rational functions of kinematic invariants only.
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Intermediate and final expressions
Brute force calculations are a mess:
<img src="Five_gluons_mess.png”; style="max-width:500px;float:center;border:none;margin-top:-5px;“>
Results are often much easier
S.J.Parke, T.R.Taylor ,
F.A.Berends, W.T.Giele </font size>
$A^{tree}(1^{+}_{g}2^{+}_{g}3^{+}_{g}4^{-}_{g}5^{-}_{g}) = \frac{i,⟨45⟩^{4}}{⟨12⟩⟨23⟩⟨34⟩⟨45⟩⟨51⟩}$
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Numerical and Analytical: pros and cons
Numerical calculations efficiently bypass algebraic complexity
BlackHat [C.F.Berger et al. ], CutTools [G.Ossola et al. ], MadGraph [J.Alwall et al. ], Rocket [W.T.Giele, G.Zanderighi ], Samurai [P.Mastrolia et al. ], NGluon [S.Badger et al. ], OpenLoops [F.Cascioli et al. ], … </font size>
Lowest-laying representations of the Lorentz group
$\boldsymbol{(j_{-},j_{+})}$
dim.
name
quantum field
kinematic variable
(0,0)
1
scalar
$h$
m
(0,1/2)
2
right-handed Weyl spinor
$\chi_{R,\alpha}$
$\lambda_\alpha$
(1/2,0)
2
left-handed Weyl spinor
$\chi_L^{,\dot\alpha}$
$\bar{\lambda}^{\dot\alpha}$
(1/2,1/2)
4
rank-two spinor/four vector
$A^\mu/A^{\dot\alpha\alpha}$
$P^\mu/P^{\dot\alpha\alpha}$
(1/2,0)$\oplus$(0,1/2)
4
bispinor (Dirac spinor)
$\Psi$
$u, v$
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Spinor Covariants
Weyl spinors are sufficient for massless particles:
$\text{det}(P^{\dot\alpha\alpha})=m^2 \rightarrow 0 \quad \Longrightarrow \quad P^{\dot\alpha\alpha} = \bar\lambda^{\dot\alpha}\lambda^\alpha$.</font size>
$$
\lambda_\alpha=\frac{1}{\sqrt{p^0+p^3}}\begin{pmatrix}p^0+p^3 \\ p^1+ip^2\end{pmatrix} , , ;;; \lambda^\alpha=\epsilon^{\alpha\beta} \lambda_\beta =\frac{1}{\sqrt{p^0+p^3}}\begin{pmatrix}p^1+ip^2 \\ -p^0+p^3\end{pmatrix}
$$</font size>
$\bar\lambda_{\dot\alpha}=\frac{1}{\sqrt{p^0+p^3}}\begin{pmatrix}p^0+p^3 \\ p^1-ip^2\end{pmatrix} , , ;;; \bar\lambda^{\dot\alpha}=\epsilon^{\dot\alpha\dot\beta}\bar\lambda_{\dot\beta}=\frac{1}{\sqrt{p^0+p^3}}\begin{pmatrix}p^1-ip^2 \\ -p^0+p^3\end{pmatrix}$</font size>
$$
\bar\lambda_{\dot\alpha} = (\lambda_\alpha)^* \quad if \quad p^i \in \mathbb{R}; \quad \quad \bar\lambda_{\dot\alpha} \neq (\lambda_\alpha)^* \quad if \quad p^i \in \mathbb{C}
$$</font size>
Spinor Helicity Invariants
$$
s_{ij} = ⟨ij⟩[ji]
$$
$$
⟨i;|;(j+k);|;l] = (λ_i)^α (\not P_j + \not P_k )_{α\dotα} \barλ_l^\dotα
$$
$$
⟨i;|;(j+k);|;(l+m);|;n⟩ = (λ_i)^α (\not P_j + \not P_k )_{α \dot α} (\bar{\not P_l} + \bar{\not P_m} )^{\dot α α} (λ_n)_α
$$
$$
tr_5(ijkl) = tr(\gamma^5 \not P_i \not P_j \not P_k \not P_l) = [i,|,j,|,k,|,l,|,i⟩ - ⟨i,|,j,|,k,|,l,|,i]
$$
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Why use complex momenta?
We want to be able to distinguish between different poles.
For example, we want to tell apart:
$$
\frac{1}{⟨ij⟩} \quad and \quad \frac{1}{[ji]}
$$
but with real momenta we would have:
$$
⟨ij⟩ \sim [ji]^* \sim \sqrt{s_{ij}}
$$
Singular limits $\Rightarrow$ poles of the amplitude
Let $\{r\}$ be a set of Lorentz invariants
$r_i \in \{ ⟨12⟩, ⟨13⟩, \dots, ⟨1|2+3|4], \dots, s_{123}, \dots \}$,</font size>
and let $, \mathbb{f} ,$ be the function we want to reconstruct.
We want to build phase space points where a single invariant from $\{r\}$ vanishes:
$r_i \rightarrow ε \ll 1, \quad r_{j \neq i} \sim \mathcal{O}(1), \quad \mathbb{f} \rightarrow ε^α ; \Rightarrow ; log(\mathbb{f}) \rightarrow α\cdot log(ε)$</font size>
$\Rightarrow$ The slope of the log-log plot gives us the type of singularity, if it exists.
Tree-level example
$\mathbb{f} = A^{tree}(1^{+}_{g}2^{+}_{g}3^{+}_{g}4^{-}_{g}5^{-}_{g}6^{-}_{g})$
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Note: the invariant on the x-axis gets smaller from left to right.</font size>
The least common denominator
Studying all the limits yields the least common denominator for $\mathbb{f}$:
$\mathbb{f} = \frac{\mathcal{N_{LCD}}}{\mathcal{D_{LCD}}} = \frac{\mathcal{N_{LCD}}}{⟨12⟩⟨16⟩[16]⟨23⟩⟨34⟩[34][45][56]s_{234}s_{345}}$.
The complexity of the numerator depends on two parameters:
mass dimension; $\quad \quad \quad$ 2. little group scalings.
In this case $\mathcal{N_{LCD}}$ has mass dimension 10, and phase weights [-1, 0, -1, 1, 0, 1].
The ansatz has 1326 independent terms.
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How big is the ansatz?
Easiest to count at constant null phase weights; the size of the ansatz is a function of:1. its mass dimension ($d$) $\quad\quad$ 2. multiplicity of phase space ($m$).
$|s_{ij}| = \frac{m(m-3)}{2}$ $\quad\quad$ $|tr_5| = {m-1 \choose 4}$
$\left(\mkern -9mu \begin{pmatrix}, |s_{ij}| , \\ , d/2 , \end{pmatrix} \mkern -9mu \right) \leq$ ansatz size $\leq \left(\mkern -9mu \begin{pmatrix} , |s_{ij}| , \\ , d/2 , \end{pmatrix} \mkern -9mu \right) + |tr_5| \left(\mkern -9mu \begin{pmatrix} , |s_{ij}| , \\ , (d-4)/2 , \end{pmatrix} \mkern -9mu \right)$
The upper bound is saturated for $(\forall m \wedge d \leq 4)$ and for $(\forall d \wedge m \leq 5)$.
Otherwise it is an overcounting due to Schouten identity for 4-momenta:
$tr_5(2345)1_\mu - tr_5(1345)2_\mu + tr_5(1245)3_\mu - tr_5(1235)4_\mu + tr_5(1234)5_\mu = 0$</font size>
Partial fractioning
Except for the easiest cases, we should really think about $\mathbb{f}$ as:
$\mathbb{f} = \sum_i \frac{\mathcal{N}_i}{\mathcal{D}_i} = \sum_i \frac{\mathcal{N}_i}{\mathcal{R}_i\mathcal{S}_i}$,
where $\mathcal{R}_i$ are products of subsets of $\mathcal{D_{LCD}}$ (i.e. real poles), and $\mathcal{S}_i$ are products of factors not in $\mathcal{D_{LCD}}$ (i.e. spurious poles).
Doubly singular limits
The required information can be accessed from doubly singular limits.
We now want phase space points where two invariants vanish:
$r_i \rightarrow ε \ll 1, \quad r_j \rightarrow ε \ll 1, \quad \mathbb{f} \rightarrow ε^α ; \Rightarrow ; log(\mathbb{f}) \rightarrow α\cdot log(ε)$
In general we cannot guarantee uniqueness anymore, even with complex momenta. $\exists ;r_{k \neq i, j} \sim \epsilon$.
Information from doubly singular limits
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The first number is the slope of the log-log plot in the limit, the second number is the degeneracy of the phase space in the limit.
Which poles share the same denominator?
The following is single line of the table in the previous slide:
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<img src="Graph.gv.svg”; style="max-width:720px;float:center;border:none;margin-bottom:-30px;“>
Diagramatic representation of relation between poles</font size>
Partial fraction decomposition #1
(Spoiler: $ \quad \small \mathcal{N_1} = 1i⟨4|2+3|1]^3, \quad \mathcal{N_2} = -1i⟨6|1+2|3]^3$)</font size>
Where does $⟨2|1+6|5]$ </font size> come from?
$\{⟨12⟩, [56]\}_{\epsilon} , \cap , \{⟨16⟩, s_{234}\}_{\epsilon} , \cap , \{⟨23⟩, [45]\}_{\epsilon} , \cap , \{[34], s_{234}\}_{\epsilon} , \dots$ </font size>
Partial fraction decomposition #2
(Spoiler: $ \quad \small \mathcal{N_1} = 1i[23]^3⟨56⟩^3, \quad \mathcal{N_2} = -1i[12]^3⟨45⟩^3, \quad \mathcal{N_3} = 1is_{123}^3 $)</font size>
$⟨1|2+3|4] = \{⟨12⟩, [34]\}_{\epsilon} , \cap , \{⟨23⟩, s_{234}\}_{\epsilon} , \cap , \{⟨16⟩, [45]\}_{\epsilon} , \cap , \dots$</font size>
$⟨3|1+2|6] = \{⟨23⟩, [16]\}_{\epsilon} , \cap , \{⟨12⟩, s_{345}\}_{\epsilon} , \cap , \{⟨34⟩, [56]\}_{\epsilon} , \cap , \dots$</font size>
Both of these partial fractions correspond to some BCFW shift, but we are not limited to these representations.
($\small \mathcal{N_1} = 1i[12]⟨45⟩⟨4|2+3|1]^2, , \mathcal{N_2} = 1i[23]⟨56⟩⟨6|1+2|3]^2, , \mathcal{N_3} = 1i⟨4|2+3|1]⟨6|1+2|3]s_{123}$) </font size>
We now have no spurious poles, but $s_{234}$ and $s_{345}$ appear in the same denominator, although the doubly singular limit suggests they shouldn’t.
This means that in the limit of $\{s_{234}, s_{345}\} \rightarrow \epsilon$ the individual terms will behave like $\epsilon^{-2}$ but the sum as $\epsilon^{-1}$.
Different representations can be exploited to ensure numerical stability.
Spinor ansatz
Let’s consider the first representation:
$\mathbb{f}=$
$\frac{\mathcal{N_1}}{[16]⟨23⟩⟨34⟩[56]⟨2|1+6|5]s_{234}}+$
$\frac{\mathcal{N_2}}{⟨12⟩⟨16⟩[34][45]⟨2|1+6|5]s_{345}}$
$[⟨24⟩⟨24⟩⟨24⟩[12][12][12],$
$⟨24⟩⟨24⟩⟨34⟩[12][12][13],$
$⟨24⟩⟨24⟩⟨45⟩[12][12][15],$
$⟨24⟩⟨34⟩⟨34⟩[12][13][13],$
$⟨24⟩⟨34⟩⟨45⟩[12][13][15],$
$⟨24⟩⟨45⟩⟨45⟩[12][15][15],$
$⟨34⟩⟨34⟩⟨34⟩[13][13][13],$
$⟨34⟩⟨34⟩⟨45⟩[13][13][15],$
$⟨34⟩⟨45⟩⟨45⟩[13][15][15],$
$⟨45⟩⟨45⟩⟨45⟩[15][15][15]]$,
much smaller than the one for $\mathcal{N_{LCD}}$ which had more than 1000 terms!
Gaussian elimination
$\mathbb{f}=$
$\frac{\mathcal{N_1}}{[16]⟨23⟩⟨34⟩[56]⟨2|1+6|5]s_{234}}+$
$\frac{\mathcal{N_2}}{⟨12⟩⟨16⟩[34][45]⟨2|1+6|5]s_{345}}$
Isolate the first term by generating phase space points in the limit of, say, $s_{234} \rightarrow \epsilon$.
Then we can reconstruct the numerical coefficient of the ansatz entries by Gaussian elimination.
$$
M_{ij}c_j = \mathbb{f}(P_i) \quad \text{i.e.} \quad \begin{pmatrix} \leftarrow ansatz(P_1) \rightarrow \\ \leftarrow ansatz(P_2) \rightarrow \\ \dots \end{pmatrix} \cdot \begin{pmatrix} c_1 \\ c_2 \\ \dots \end{pmatrix} = \begin{pmatrix} \mathbb{f}(P_1) \\ \mathbb{f}(P_2) \\ \dots \end{pmatrix}
$$
We get:
Brief litterature review
First complete numerical calculation
R.K.Ellis, W.T.Giele, G.Zanderighi </font size>
Pheno applications to 4-jet production
Z.Bern, G.Diana, L.J.Dixon, F.FebresCordero, S.Hoeche, D.A.Kosower, H.Ita, D.Maitre, K.Ozeren </font size>
S.Badger, B.Biedermann, P.Uwer, V.Yundin </font size>
NMHV analytical cut constructible part
R.Britto, B.Feng, P.Mastrolia </font size>
NMHV analytical rational part
Z.Xiao, G.Yang, C.Zhu </font size>
More litterature info
David C. Dunbar </font size>
Our results
First complete analytical results presented with a single notation.
All expressions are manifestely gauge invariant and rational (no square roots).
Best speed up is a factor of ~75 in the split NMHV configuration (compared to numerical computations in BlackHat).
Worst one is a factor of ~2 in the alternated NMHV configuration.
Bottleneck: numerators of subleading poles.
Three mass triangle
<img src="three_mass_triangle.svg”; width=25%; height=25%; style="float:center;border:none;margin-bottom:0px;“>
$\mathcal{D_{LCD}} = ⟨12⟩[12]⟨34⟩[34]⟨56⟩[56]⟨1|3+4|2]^4⟨3|1+2|4]^4⟨5|1+2|6]^4Δ_{135}^3$</font size>
We obtain an explicitly rational expression, like for $\mathcal{N}=1$ SUSY by N.E.J.Bjerrum-Bohr, D.C.Dunbar, W.B.Perkins . </font size>
Pole structure: are there radicals?
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How to avoid square roots
All branch cuts should have been taken care of by unitarity cuts.
We should be able to explain this without using square roots.
$\Delta_{135} = (K_1 \cdot K_2)^2 - K_1^2 K_2^2$</font size>
$(\Omega_{351})^2 \equiv (2s_{12}s_{56}-(s_{12}+s_{56}-s_{45})s_{123})^2 = 4s_{123}^2\Delta_{135}-4s_{12}s_{56}\langle 4|1+2|3]\langle 3|1+2|4]$</font size>
$(\Pi_{351})^2 \equiv (s_{123}-s_{124})^2 = 4\Delta_{135}-4\langle 4|1+2|3]\langle 3|1+2|4]$</font size>
The three-mass triangle rational coefficient
(Showing ⟨1|3+4|2], ⟨3|1+2|4] and ⟨5|1+2|6] quadruple and triple poles) </font size>
A six-gluon one-loop rational part
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The pole structure
is a bit of a mess:
<img src="GraphRational.gv.svg”; style="maxwidth:1000px; float:center; border:none;“>
The rational part
(Showing ⟨1|2+3|1], ⟨1|2+3|6] and symmetric poles) </font size>
Preview of refitted results into spinor helicity of [S. Abreu, J. Dormans, F. Febres Cordero, H. Ita, B. Page, V. Sotnikov ] </font size>
Expression complexity
Using the Mandelstam expressions as numerical input, we reconstruct them in spinor helicity.
<img src="LeafCountRatios.png”; style="max-width:560px;float:center;border:none;“>
Example: uubggg pmpmp Nf1 #3
<img src="uubggg_pmpmp_nf1_nb3.png”; style="max-width:1024px;float:center;border:none;“>
and
are equivalent.
Example: ggggg mpmpp Nf1 # 9
<img src="ggggg_mpmpp_nf1_9.png”; style="max-width:512px;float:center;border:none;“>
and </font size>
$-1\frac{[12]³[15][23]⟨25⟩³[35]³}{[13]⁴[25]⟨5|1+2|5]³}+\frac{97}{12}\frac{[12]⁴⟨25⟩[35]⁴}{[13]⁴[25]³⟨5|1+2|5]}$
$+\frac{13}{3}\frac{[12]⁴⟨15⟩[15][35]⁴}{[13]⁴[25]⁴⟨5|1+2|5]}+\frac{1}{4}\frac{[12]⁴⟨15⟩[15]⟨25⟩[35]⁴}{[13]⁴[25]³⟨5|1+2|5]²}$
$-\frac{3}{2}\frac{[12]²⟨25⟩²[25][35]²}{[13]²[25]⟨5|1+2|5]²}+\frac{7}{4}\frac{[12]³⟨25⟩²[35]³}{[13]³[25]⟨5|1+2|5]²}$
$-\frac{43}{3}\frac{[12]³⟨25⟩[35]³}{[13]³[25]²⟨5|1+2|5]}$
$-\frac{25}{3}\frac{[12]³⟨15⟩[15][35]³}{[13]³[25]³⟨5|1+2|5]}$
$-\frac{3}{2}\frac{[12]⟨25⟩[25][35]}{[13][25]⟨5|1+2|5]}$
$+4\frac{[12]²⟨25⟩[35]²}{[13]²[25]⟨5|1+2|5]}$
$-\frac{15}{2}\frac{[12]²[35]²}{[13]²[25]²}$
$+\frac{7}{2}\frac{[12][35]}{[13][25]}$
$-\frac{2}{3}$
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are equivalent.</font size>
Biggest linear systems to solve
Recall that size of the linear system is directly related to the mass dimension of the numerators by this formula
<img src="MaximumMassDimensionComparison.png”; style="max-width:560px;float:center;border:none;“>
By performing the partial fraction decomposition before fitting the numerators the size of the biggest systems to solve is significantly reduced.
Numerical stability
<img src="RelativeError.png”; style="max-width:700px;float:center;border:none;“>
<img src="RelativeErrorTailLogY.png”; style="max-width:700px;float:center;border:none;“>
Errors made by double-precision evaluations over $1k$ $collider$ phase space points.
Example of cut diagram
<img src="HiggsBox.png”; style="max-width:300px;float:center;border:none;“>
Only singularity involving $m_{top}$ (from pentagon contributions)
$16 |S_{1×2×3×4}| = −s_{12} , s_{23} , s_{34} , \langle 1 |2 + 3|4] , \langle 4|2 + 3|1] + m^2_{top} , tr_5(1234)^2$
We can generate point near this singularity in a similar fashion.
Structure of the coefficients
The massive external leg (the Higgs) is easily accomodated by considering it as a pair of massless particles (think decay products).
In the end all dependance on $P_{Higgs}$ is removed by using momentum conservation.
The coefficients are Taylor expasions in $m_{top}$:
$C^{(0)} + m^2_{top} C^{(2)}$.
with $C^{(0)}$ and $C^{(2)}$ resabling the six-gluon coefficients.
Brief overview
Massless $n-$point tree amplitudes are given by
$\textit{A}_n , = , z_1^4 \cdot i \sum_{j = 1}^{(n-3)!} \frac{I_{\scriptscriptstyle CHY}(z^{(j)}(k); k; ϵ)}{\det(ϕ_{rst}^{ijk})(z^{(j)}(k); k)}$
The $z’s$ are related to the momenta through the scattering equations.
The only process dependant part is $I_{CHY}$, the rest depends only on multiplicity.
For instance:
(See backup slides for specific definitions of functions)
Numerical amplitudes
Developed two python packages:
>>> pip install seampy
>>> pip install lips
seampy from Scatterin Equations and AMplitudes with PYthon
lips from Lorentz Invarinat Phase Space
Arbitrary-precision floating-point amplitudes with a couple of lines of code
>>> from seampy import theories, NumericalAmplitude
>>> from lips import Particles
>>> theories
[YM, EG, BS, BI, NLSM, Galileon, CG, DF2]
>>> oDF2Amp = NumericalAmplitude(theory= "DF2" , helconf= "+++++" )
>>> oParticles = Particles(5 )
>>> oDF2Amp(oParticles)
mpc(real= '#nbr' , imag= '#nbr' )