1. 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

2. 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

3. 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

4. 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

5. 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

6. 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

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

8. 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

9. We obtain:
Mainz 2016

10. 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

11. Rotated, λ = N = 30
Fixed
Free
Mainz 2016

12. Dependence on the number of Pietarinen terms
Mainz 2016

13. Dependence on the number of Pietarinen terms
Mainz 2016

14. Dependence on the number of Pietarinen terms
Mainz 2016

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

16. Rotated, λ = N = 15
Real bp
Complex bp
Mainz 2016

17. Final fit after a lot of fitting skill
Mainz 2016

18. Rotated, λ = N = 50
Fixed
Free
Mainz 2016

19. Rotated, λ = N = 80
Fixed
Free
Mainz 2016

20. Rotated, λ = N = 80
Fixed
Free
Mainz 2016

21. Rotated, λ = N = 100
Fixed
Free
Mainz 2016

22. 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

23. The result:
Mainz 2016

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

25. Mainz 2014

26. 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

27. 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

28. 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

29. 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= 𝐹 𝑟𝑜𝑡 0,0 (𝑊,𝑥)= 𝑀 0 𝑊 𝐿 0 𝑊 𝐴 0 𝑃 0 (𝑥)
l= 𝐹 𝑟𝑜𝑡 0,1 (𝑊,𝑥)= 𝑀 0 𝑊 𝐿 1 𝑊 𝐴 0 𝑃 1 (𝑥)
l= 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

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

31. 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

32. 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

33. 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

34. 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 ∙ 𝑊− 𝑊 2 ∙ 0 ⋅ ∙ ∙ ∙ 0 0 ∙ 1 𝑊− 𝑊 𝑛
Matrix is diagonal, and we assume only one pole per angular momentum.
Mainz 2016

35. 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 𝑊− 𝑊 ∙ 𝑊− 𝑊 2 ∙ 0 ⋅ ∙ ∙ ∙ 0 0 ∙ 1 𝑊− 𝑊 𝑛 =
= 𝑎 1 𝑊− 𝑊 𝑎 2 𝑊− 𝑊 2 +…+ 𝑎 𝑛 𝑊− 𝑊 𝑛 ℳ 12 ∙ ℳ 1𝑛 ℳ 𝑏 1 𝑊− 𝑊 𝑏 2 𝑊− 𝑊 2 +…+ 𝑏 𝑛 𝑊− 𝑊 𝑛 ∙ ℳ 2𝑛 ⋅ ∙ ∙ ∙ ℳ 𝑛1 ℳ 𝑛2 ∙ 𝑧 1 𝑊− 𝑊 𝑧 2 𝑊− 𝑊 2 +…+ 𝑧 𝑛 𝑊− 𝑊 𝑛
Mainz 2016

36. Mainz 2016