with Daniel Maitre
IPPP - Durham University
<img src="IP3_logo_blue.png”; style="max-width:280px;float:center;border:none;“> <img src="DurhamLogo.svg”; style="max-width:280px;float:center;border:none;“>
1. Motivation and introduction
2. Singularity structure
3. Ansatz and amplitude reconstruction
4. Some results (Yang-Mills in the Standard Model)
Cross sections at hadron colliders are given by:
$d\hat{σ}_{n}=\frac{1}{2\hat{s}}dΠ_{n-2};(2π)^4δ^4\big(∑_{i=1}^n p_i\big);|\overline{\mathcal{A}(p_i,μ_F, μ_R)}|^2$ </font size>
Better predictions require both more loops and higher multiplicity.
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Powers of coupling in pure gluon scattering.</font size>
We can split dynamics from kinematics with the following identities.
$\mathcal{A}^{tree}_{n}({p_i, λ_i, a_i}) = ; g^{n-2} ∑_{σ\in S_n/Z_n} \text{Tr}(T^{a_σ(1)}\dots T^{a_σ(n)}) A^{tree}_n(σ(1^{λ_1}),\dots ,σ(n^{λ_n}))$</font size>
$A^{tree},, d_i,, c_i,, b_i,$ and $R$ are rational functions of kinematic invariants only.
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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 [6, 7]:
$A^{tree}(1^{+}_{g}2^{+}_{g}3^{+}_{g}4^{-}_{g}5^{-}_{g}) = \frac{i,⟨45⟩^{4}}{⟨12⟩⟨23⟩⟨34⟩⟨45⟩⟨51⟩}$
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Numerical calculations efficiently bypass algebraic complexity, see: BlackHat [8], CutTools [9], MadGraph [10], Rocket [11], Samurai [12], NGluon [13], OpenLoops [14]… </font size>
The lowest-laying representations of the Lorentz group are:
| $\boldsymbol{(j_{-},j_{+})}$ | 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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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>
Some definitions:
$$ s_{ij} = ⟨ij⟩[ji] $$
$$ ⟨i;|;(j+k);|;l] = (λ_i)^α (\not P_j + \not P_k )_{α\dotα} \barλ_l^\dotα $$
$$ ⟨i;|;(j+k);|;(l+m);|;n⟩ = (λ_i)^α (\not P_j + \not P_k )_{α \dot α} (\bar{\not P_l} + \bar{\not P_m} )^{\dot α α} (λ_n)_α $$
$$ tr_5(ijkl) = tr(\gamma^5 \not P_i \not P_j \not P_k \not P_l) = [i,|,j,|,k,|,l,|,i⟩ - ⟨i,|,j,|,k,|,l,|,i] $$
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Singular limits give us information about the poles of the amplitude.
We need a set of possible poles of the amplitudes:
$r_i \in \{ ⟨12⟩, ⟨13⟩, \dots, ⟨1|2+3|4], \dots, s_{123}, \dots \}$,</font size>
and let $, \mathbb{f} ,$ be the function we want to reconstruct.
We want to build phase space points where a single invariant 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.
As an example, let us consider the following amplitude:
$\mathbb{f} = A^{tree}(1^{+}_{g}2^{+}_{g}3^{+}_{g}4^{-}_{g}5^{-}_{g}6^{-}_{g})$
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Note: the invariant on the x-axis gets smaller from left to right.</font size>
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:
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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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).
This information can be accessed by studying 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 taking the 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.
The slope in the doubly singular limit tells us whether two poles should be in the same denominator and the degeneracy how to separate them.
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>
Let us briefly consider the coefficient of a three mass triangle:
<img src="three_mass_triangle.svg”; width=25%; height=25%; style="float:center;border:none;margin-bottom:0px;“>
$\mathcal{D_{LCD}} = ⟨12⟩[12]⟨34⟩[34]⟨56⟩[56]⟨1|3+4|2]^4⟨3|1+2|4]^4⟨5|1+2|6]^4Δ_{135}^3$
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Do we need square roots of momentum invariants?
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>
Going back to our tree level example, let’s see how we can group together the poles of the amplitude.
(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>
(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.
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!
We want to isolate the first term in
$\mathbb{f}=$ $\frac{\mathcal{N_1}}{[16]⟨23⟩⟨34⟩[56]⟨2|1+6|5]s_{234}}+$ $\frac{\mathcal{N_2}}{⟨12⟩⟨16⟩[34][45]⟨2|1+6|5]s_{345}}$
this can be done 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:
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>
As an application we obtained analytical expressions for all parts of the six-gluon amplitude with a gluon in the loop.
For the three mass triangle mentioned in an earlier section, which was previously obtained in [16], the poles are:
$\mathcal{D_{LCD}} = ⟨12⟩[12]⟨34⟩[34]⟨56⟩[56]⟨1|3+4|2]^4⟨3|1+2|4]^4⟨5|1+2|6]^4Δ_{135}^3$</font size>
We obtain an explicitly rational expression, like for $\mathcal{N}=1$ SUSY in [17].
Another interesting piece is the rational part:
An analytical expression for this component was previously obtained in [18] from Feynman diagrams. </font size>
The pole structure is a bit of a mess:
<img src="GraphRational.gv.svg”; style="maxwidth:1000px; float:center; border:none;“>
The aim of this talk was to show that it is possible to reconstruct an analytical expression from numerical evaluations only.
Thank you very much!