Alfred Švarc Rudjer Bošković Institute, Zagreb, Croatia The importance of energy and angle dependent phase rotations in partial wave analysis Alfred Švarc Rudjer Bošković Institute, Zagreb, Croatia Yannick Wunderlich, Helmholtz-Institut für Strahlen- und Kernphysik, Bonn Mainz 2016

𝑭 𝒊 𝒓𝒐𝒕 𝑾,𝒙 =𝒆 𝒊 𝝓(𝑾,𝒙) 𝑭 𝒊 𝑾,𝒙 𝑭 𝒊 𝑾,𝒙 Continuum ambiguity If Fi (W, x) are invariant functions for a physical process then the observable quantities are invariant with respect to an simultaneous energy and angle dependent phase rotation of all invariant functions: 𝑭 𝒊 𝑾,𝒙 𝑭 𝒊 𝒓𝒐𝒕 𝑾,𝒙 =𝒆 𝒊 𝝓(𝑾,𝒙) 𝑭 𝒊 𝑾,𝒙 Mainz 2016

𝑭 𝑾,𝒙 = 𝒌=𝟎 𝑵 𝑴 𝒌 𝑾 𝑷 𝒌 𝒙 ; 𝒙= 𝐜𝐨𝐬 𝜽 The problem: 𝑭 𝑾,𝒙 = 𝒌=𝟎 𝑵 𝑴 𝒌 𝑾 𝑷 𝒌 𝒙 ; 𝒙= 𝐜𝐨𝐬 𝜽 The question: What is the analytic structure of the rotated function? 𝑭 𝒓𝒐𝒕 𝑾,𝒙 =𝒆 𝒊 𝝓(𝑾,𝒙) 𝑭 𝑾,𝒙 Mainz 2016

Angle independent phase rotation 𝒆 𝒊 𝝓(𝑾, 𝒙) 𝒆 𝒊 𝝓(𝑾) Trivial: 𝒆 𝒊 𝝓(𝑾) 𝑭 𝑾,𝒙 = 𝒌=𝟎 𝑵 𝒆 𝒊 𝝓(𝑾) 𝑴 𝒌 (𝑾) 𝑷 𝒌 𝒙 Energy dependent phase rotation of multipoles Mainz 2016

𝒆 𝒊 𝝓(𝑾) 𝑴 𝒌 (𝑾) = 𝒙+𝒊 𝒚 𝒆 𝒊 𝝓(𝑾) 𝑴−𝑾 + 𝒆 𝒊 𝝓(𝑾) ℘ Further simplification: If poles and background are separated using L + P expansion 𝑴 𝒌 𝑾 = 𝒙+𝒊 𝒚 𝑴−𝑾 + ℘ Then again the result is trivial! 𝒆 𝒊 𝝓(𝑾) 𝑴 𝒌 (𝑾) = 𝒙+𝒊 𝒚 𝒆 𝒊 𝝓(𝑾) 𝑴−𝑾 + 𝒆 𝒊 𝝓(𝑾) ℘ The same pole Rotated residue Rotated Pietarinen Mainz 2016

We still do not know, we are working on it. Question! Is rotated L + P expansion indeed the L + P expansion of rotated function? Answer! We still do not know, we are working on it. (Presumably the answer will be given in answering the question whether the rotated Laurent decomposition is Laurent decomposition of the rotated function.) Present attempt: Test it on solvable model. Mainz 2016

Instead of using general model we use a Toy model from paper PRC C 88, 035206 (2013) because we know what we should get Mainz 2016

𝑇 𝑇𝑜𝑦 𝑟𝑜𝑡 𝑊 = 𝑇 𝑇𝑜𝑦 𝑊 𝑒 − 𝑖 𝜆 𝜙(𝑊) We introduce the rotation in the following way: (inspired by reduced multipoles used by MAID group) We calculate the phase of the Toy model 𝑇 𝑇𝑜𝑦 𝑊 = 𝑇 𝑇𝑜𝑦 𝑊 𝑒 𝑖 𝜙(𝑊) We rotate the Toy model „back” for the phase 𝑇 𝑇𝑜𝑦 𝑟𝑜𝑡 𝑊 = 𝑇 𝑇𝑜𝑦 𝑊 𝑒 − 𝑖 𝜆 𝜙(𝑊) Mainz 2016

We obtain: Mainz 2016

𝒆 𝒊 𝝓(𝑾) ℘ Problem: Fitting function 𝒙+𝒊 𝒚 𝒆 𝒊 𝝓(𝑾) 𝑴−𝑾 is no problem. Problem is how can Pietarinen expansion fit the function 𝒆 𝒊 𝝓(𝑾) ℘ Why? Rotation is mixing real and imaginary part, so new cut structure needs MORE Pietarinen terms to describe them! Empirical facts: L + P fitting Mainz 2016

Rotated, λ = 0.00 N = 30 Fixed Free Mainz 2016

Dependence on the number of Pietarinen terms Mainz 2016

Dependence on the number of Pietarinen terms Mainz 2016

Dependence on the number of Pietarinen terms Mainz 2016

I have tried with a COMPLEX BRANCH POINT! Is it a resonant effect? I have tried with a COMPLEX BRANCH POINT! Mainz 2016

Rotated, λ = 0.50 N = 15 Real bp Complex bp Mainz 2016

Final fit after a lot of fitting skill Mainz 2016

Rotated, λ = 0.25 N = 50 Fixed Free Mainz 2016

Rotated, λ = 0.50 N = 80 Fixed Free Mainz 2016

Rotated, λ = 0.75 N = 80 Fixed Free Mainz 2016

Rotated, λ = 0.99 N = 100 Fixed Free Mainz 2016

𝑭 𝒛 ≡ −𝒛 → 𝒆 𝒊 𝒛 −𝟓 𝒛+𝟓 −𝒛 Question: How phase rotation influences the analytic structure? To answer this question we have to know analytic continuation of the phase rotation into the complex energy plane, and THIS IS TOUGH! Instead, I give just an illustration ! Function is square root with the branch-point to the right, and I use a simple rotation: 𝑭 𝒛 ≡ −𝒛 → 𝒆 𝒊 𝒛 −𝟓 𝒛+𝟓 −𝒛 Mainz 2016

The result: Mainz 2016

We have been advised to pay attention to absolute value and phase: Mainz 2014

Mainz 2014

𝒆 𝒊 𝝓(𝑾) ℘ I conclude with a question: Is energy dependent phase rotation defined as before allowed from the aspect of analyticity? Namely, we want that every background, even the rotated one Maintains the good analytic structure, namely that it contains only real and complex branch-points generating corresponding cuts. Is that indeed so always or we need some SELECTION RULES? 𝒆 𝒊 𝝓(𝑾) ℘ Mainz 2016

Angle independent phase rotation 𝒆 𝒊 𝝓(𝑾, 𝒙) ***************** 𝑭 𝑾,𝒙 = 𝒌=𝟎 𝑵 𝑴 𝒌 𝑾 𝑷 𝒌 𝒙 ; 𝒙= 𝐜𝐨𝐬 𝜽 𝑭 𝒓𝒐𝒕 𝑾,𝒙 =𝒆 𝒊 𝝓(𝑾,𝒙) 𝑭 𝑾,𝒙 Now we decompose 𝒆 𝒊 𝝓(𝑾,𝒙) over Legendre polynomials too 𝒆 𝒊 𝝓(𝑾,𝒙) = 𝒍=𝟎 𝑵 𝒍 𝑳 𝒍 𝑾 𝑷 𝒍 (𝒙) Mainz 2016

We obtain: 𝒆 𝒊 𝝓 𝑾,𝒙 𝑭 𝑾,𝒙 = 𝒌=𝟎 𝑵 𝒍=𝟎 𝑵 𝒍 𝑳 𝒍 𝑾 𝑷 𝒍 𝒙 𝑴 𝒌 𝑾 𝑷 𝒌 𝒙 == 𝒌=𝟎 𝑵 𝒍=𝟎 𝑵 𝒍 𝑳 𝒍 𝑾 𝑴 𝒌 𝑾 𝑷 𝒍 𝒙 𝑷 𝒌 𝒙   and use 𝑷 𝒍 𝒙 𝑷 𝒌 𝒙 = 𝒋=𝟎 𝑴 𝑨 𝟐𝒋 𝑷 𝒍+𝒌−𝟐𝒋 𝒙 Mainz 2016

𝑭 𝒓𝒐𝒕 (𝑾,𝒙)= 𝒌=𝟎 𝑵 𝑴 𝒌 𝑾 𝒍=𝟎 𝑵 𝒍 𝑳 𝒍 𝑾 𝒋=𝟎 𝑴 𝑨 𝟐𝒋 𝑷 𝒍+𝒌−𝟐𝒋 𝒙 Final result: 𝑭 𝒓𝒐𝒕 (𝑾,𝒙)= 𝒌=𝟎 𝑵 𝑴 𝒌 𝑾 𝒍=𝟎 𝑵 𝒍 𝑳 𝒍 𝑾 𝒋=𝟎 𝑴 𝑨 𝟐𝒋 𝑷 𝒍+𝒌−𝟐𝒋 𝒙 Or after some book-keeping: Let us separate individual terms 𝑭 𝒓𝒐𝒕 𝒌,𝒍 𝑾 and 𝑭 𝒓𝒐𝒕 𝒌 (𝑾) : k=0:   l=0 𝐹 𝑟𝑜𝑡 0,0 (𝑊,𝑥)= 𝑀 0 𝑊 𝐿 0 𝑊 𝐴 0 𝑃 0 (𝑥) l=1 𝐹 𝑟𝑜𝑡 0,1 (𝑊,𝑥)= 𝑀 0 𝑊 𝐿 1 𝑊 𝐴 0 𝑃 1 (𝑥) l=2 𝐹 𝑟𝑜𝑡 0,2 𝑊,𝑥 = 0 𝐹 𝑟𝑜𝑡 0 𝑊,𝑥 = 𝑀 0 𝑊 { 𝐿 0 𝑊 𝐴 0 𝑃 0 𝑥 + 𝐿 1 𝑊 𝐴 0 𝑃 1 𝑊 } Mainz 2016

l=0 𝐹 𝑟𝑜𝑡 1,0 (𝑊,𝑥)= 𝑀 1 (𝑊) 𝐿 0 (𝑊) 𝐴 0 𝑃 1 (𝑥) k=1:   l=0 𝐹 𝑟𝑜𝑡 1,0 (𝑊,𝑥)= 𝑀 1 (𝑊) 𝐿 0 (𝑊) 𝐴 0 𝑃 1 (𝑥) l=1 𝐹 𝑟𝑜𝑡 1,1 𝑊,𝑥 = 𝑀 1 𝑊 𝐿 1 𝑊 𝐴 0 𝑃 2 𝑥 + 𝐿 1 𝑊 𝐴 2 𝑃 0 (𝑥) l=2 𝐹 𝑟𝑜𝑡 1,2 𝑊,𝑥 = 𝑀 1 𝑊 𝐿 2 𝑊 𝐴 0 𝑃 3 𝑥 + 𝐿 2 𝑊 𝐴 2 𝑃 1 (𝑥) 𝐹 𝑟𝑜𝑡 1 𝑊,𝑥 = 𝑀 1 𝑊 { 𝐿 1 𝑊 𝐴 2 𝑃 0 𝑥 + + 𝐿 0 𝑊 𝐴 0 + 𝐿 2 𝑊 𝐴 2 𝑃 1 𝑥 + + 𝐿 1 𝑊 𝐴 0 𝑃 2 𝑥 + 𝐿 2 𝑊 𝐴 0 𝑃 3 𝑥 } Mainz 2016

𝑴 𝟏 (𝑾) = 𝑴 𝟎 𝑾 𝑳 𝟏 𝑾 𝑨 𝟎 + 𝑴 𝟏 𝑾 𝑳 𝟎 𝑾 𝑨 𝟎 + 𝑳 𝟐 𝑾 𝑨 𝟐 +⋯⋯ And finally: 𝑴 𝟎 (𝑾) = 𝑴 𝟎 𝑾 𝑳 𝟎 𝑾 𝑨 𝟎 + 𝑴 𝟏 𝑾 𝑳 𝟏 𝑾 𝑨 𝟐 +⋯⋯   𝑴 𝟏 (𝑾) = 𝑴 𝟎 𝑾 𝑳 𝟏 𝑾 𝑨 𝟎 + 𝑴 𝟏 𝑾 𝑳 𝟎 𝑾 𝑨 𝟎 + 𝑳 𝟐 𝑾 𝑨 𝟐 +⋯⋯ Angle dependent phase rotation is mixing multipoles in the sense that in the k-th multipole can get contribution from any other multipole ! Question: How does poles behave? Mainz 2016

Let us show it in a different way! Let us assume that the invariant amplitude F(W, x) is an operator. Let us represent it as a matrix in a „Legendre polynomial” orthonormal base. 𝒆 𝟎 ∙ ∙ ⋅ 𝒆 𝒏 ≡ 𝑷 𝟎 (𝒙) ∙ ∙ ⋅ 𝑷 𝒏 (𝒙) Mainz 2016

Then the matrix of the operator F(W, x) in Legendre polynomial base is: 𝑭 𝑾, 𝒙 = 𝑭 𝟏 (𝑾) ⋯ 𝟎 ⋮ ⋱ ⋮ 𝟎 ⋯ 𝑭 𝒏 (𝑾) The situation will be much cleaner if we assume that there exists a set of invariant functions for we can uniquely define the arbitrary phase! This is usually achieved in coupled-channel formalism where unitarity is restored, and is used to fix the unknown phase. Mainz 2016

𝑭 𝑾, 𝒙 = 1 𝑊− 𝑊 1 0 ∙ 0 0 1 𝑊− 𝑊 2 ∙ 0 ⋅ ∙ ∙ ∙ 0 0 ∙ 1 𝑊− 𝑊 𝑛 Then for such a solution we may write the matrix of invariant function operator in Legendre polynomial basis: 𝑭 𝑾, 𝒙 = 1 𝑊− 𝑊 1 0 ∙ 0 0 1 𝑊− 𝑊 2 ∙ 0 ⋅ ∙ ∙ ∙ 0 0 ∙ 1 𝑊− 𝑊 𝑛 Matrix is diagonal, and we assume only one pole per angular momentum. Mainz 2016

𝒆 𝒊 𝝓 𝑾,𝒙 1 𝑊− 𝑊 1 0 ∙ 0 0 1 𝑊− 𝑊 2 ∙ 0 ⋅ ∙ ∙ ∙ 0 0 ∙ 1 𝑊− 𝑊 𝑛 = Now, each possible invariant function can be obtained from this matrix using a phase rotation! 𝒆 𝒊 𝝓 𝑾,𝒙 1 𝑊− 𝑊 1 0 ∙ 0 0 1 𝑊− 𝑊 2 ∙ 0 ⋅ ∙ ∙ ∙ 0 0 ∙ 1 𝑊− 𝑊 𝑛 = = 𝑎 1 𝑊− 𝑊 1 + 𝑎 2 𝑊− 𝑊 2 +…+ 𝑎 𝑛 𝑊− 𝑊 𝑛 ℳ 12 ∙ ℳ 1𝑛 ℳ 21 𝑏 1 𝑊− 𝑊 1 + 𝑏 2 𝑊− 𝑊 2 +…+ 𝑏 𝑛 𝑊− 𝑊 𝑛 ∙ ℳ 2𝑛 ⋅ ∙ ∙ ∙ ℳ 𝑛1 ℳ 𝑛2 ∙ 𝑧 1 𝑊− 𝑊 1 + 𝑧 2 𝑊− 𝑊 2 +…+ 𝑧 𝑛 𝑊− 𝑊 𝑛 Mainz 2016

Mainz 2016