Renormalons in Quantum Mechanics Cihan Pazarba s Bo gazi ci - PowerPoint PPT Presentation

Renormalons in Quantum Mechanics Cihan Pazarba¸ sı Bo˘ gazi¸ ci University based on 1906.07198 collobration with Dieter Van den Bleeken 06.09.2019, ICTP

Introduction • Perturbative Series are generically divergent f ( λ ) = � f n λ n ( f n ∼ A − n n ! ) • Borel summation − → Tame the divergence

Introduction • Perturbative Series are generically divergent f ( λ ) = � f n λ n ( f n ∼ A − n n ! ) • Borel summation − → Tame the divergence � ∞ � ∞ d s e − s ˆ d s s n e − s n ! = = ⇒ f ( λ ) = f ( sλ ) 0 0 n ! ( sλ ) n = f ( sλ ) = � ∞ ˆ f n A n =0 A − sλ Im( s ) Im f ( λ ) = Res ( e − s ˆ f ( sλ )) | s = A/λ C + Re( s ) C − Choosing C + or C − results in complex conjugate results! − → Borel Ambiguity

Introduction • Perturbative Series are generically divergent f ( λ ) = � f n λ n ( f n ∼ A − n n ! ) • Borel summation − → Tame the divergence (Ambigious) • Types of Divergences: Proliferation of Diagrams • Proliferation of number diagrams − → Tunneling B-ZJ • Borel Ambiguity ← cancellation Instanton/WKB Action − − − − − →

Introduction • Perturbative Series are generically divergent f ( λ ) = � f n λ n ( f n ∼ A − n n ! ) • Borel summation − → Tame the divergence (Ambigious) • Types of Divergences: Proliferation of Diagrams • Proliferation of number diagrams − → Tunneling B-ZJ • Borel Ambiguity ← cancellation Instanton/WKB Action − − − − − → Loop Expansion • Remnant of renormalization process of loop • Borel Ambiguity − → ? � n ∼ d te − t t n ∼ n ! ln µ d k 2 k 2 � � � IR-Region: k 2 � n � d k 2 � d te − t t n ∼ n ! ln k 2 � UV Region: ∼ k 2 µ

Introduction Renormalon Problems: • No resolution to Borel ambiguity! • No proof/conjecture that the divergence will survive or cancel out! Solution: • Simplify the problem → Find the simplest toy model • Try NR Quantum Mechanics?

Introduction Renormalon Problems: • No resolution to Borel ambiguity! • No proof/conjecture that the divergence will survive or cancel out! Solution: • Simplify the problem → Find the simplest toy model • Try NR Quantum Mechanics? • Delta potentials for D ≥ 2 have UV divergence → Candidate for renormalons in QM!

Introduction Renormalon Problems: • No resolution to Borel ambiguity! • No proof/conjecture that the divergence will survive or cancel out! Solution: • Simplify the problem → Find the simplest toy model • Try NR Quantum Mechanics? • Delta potentials for D ≥ 2 have UV divergence → Candidate for renormalons in QM! Outline: • Review of δ (2) ( x ) scattering. • Construction of Renormalons in QM. • Resolution of Renormalon ambiguity.

Renormalization in QM Perturbative Scattering: S = S (0) − iT where T = V � ( GV ) n Consider 2-body scattering with V = δ (2) ( x, y ) � ∞ T (1) = λ 2 du 2 0 1 Loop: 4 π u 2 0 f − u 2 ren = λ 2 l ( E f ) 4 π log e iπ z T (1) 1 Renormalization: , l ( E f ) = µ

Renormalization in QM Perturbative Scattering: S = S (0) − iT where T = V � ( GV ) n Consider 2-body scattering with V = δ (2) ( x, y ) � ∞ T (1) = λ 2 du 2 0 1 Loop: 4 π u 2 0 f − u 2 ren = λ 2 l ( E f ) 4 π log e iπ z T (1) 1 Renormalization: , l ( E f ) = µ Generalization to higher loops is immediate! . . . . . . . . . N-1 Loop: (Only diagram) ren = λ n ( l ( E f )) n − 1 Easy to sum! T ( n ) Renormalization:

Renormalization in QM Perturbative Scattering: S = S (0) − iT where T = V � ( GV ) n Consider 2-body scattering with V = δ (2) ( x, y ) � ∞ T (1) = λ 2 du 2 0 1 Loop: 4 π u 2 0 f − u 2 ren = λ 2 l ( E f ) 4 π log e iπ z T (1) 1 Renormalization: , l ( E f ) = µ Generalization to higher loops is immediate! . . . . . . . . . N-1 Loop: (Only diagram) ren = λ n ( l ( E f )) n − 1 Easy to sum! T ( n ) Renormalization: • NP Bound State: E = − µe 4 π/λ λ T = • This mathces with exact solution 4 π (log E f 1 − λ µ + iπ ) • No Renormalon Integral

Renormalons In Quantum Mechanics 4 Point Interaction . . . . . . • Renormalon Loop Exists • Particle number conserves • But still complicated

Renormalons In Quantum Mechanics 4 Point Interaction . . . . . . • Renormalon Loop Exists • Particle number conserves • But still complicated Write the problem with a combinations of background potentials 3D Model: ∗− ⋆ − ⋆ − . . . − ⋆ −∗ H = P 2 2 + λ 0 δ 2 ( x, y ) + V ( x, y, z )

Renormalons In Quantum Mechanics 4 Point Interaction . . . . . . • Renormalon Loop Exists • Particle number conserves • But still complicated Write the problem with a combinations of background potentials 3D Model: ∗− ⋆ − ⋆ − . . . − ⋆ −∗ H = P 2 2 + λ 0 δ 2 ( x, y ) + V ( x, y, z ) � E f − u 2 � � � � � � n � d 2 Q 2 T ( n +3) = λ ∗ d 2 Q n +3 � λ ∗ � du 2 log 2 − iπ (2 π ) 2 F ( P 2 , P n +3 ) � µ 2 (2 π ) 2 4 π u 2 = u n +3 ∗ renormalon integral ∼ n ! V ( P f − P n +3 ) ˜ ˜ V ( P 2 − P i ) • ( x, y, z ) → ( Q, u ) F ( P 2 , P n +3 ) = ( E f − E n +3 )( E f − E 2 )

Renormalons In Quantum Mechanics 4 Point Interaction . . . . . . • Renormalon Loop Exists • Particle number conserves • But still complicated Write the problem with a combinations of background potentials 3D Model: ∗− ⋆ − ⋆ − . . . − ⋆ −∗ H = P 2 2 + λ 0 δ 2 ( x, y ) + V ( x, y, z ) � E f − u 2 � � � � � � n � d 2 Q 2 T ( n +3) = λ ∗ d 2 Q n +3 � λ ∗ � du 2 log 2 − iπ (2 π ) 2 F ( P 2 , P n +3 ) � µ 2 (2 π ) 2 4 π u 2 = u n +3 ∗ renormalon integral ∼ n ! • ( x, y, z ) → ( Q, u ) • This implies 3 rd direction leads to the renormalon integral

Renormalons In Quantum Mechanics 4 Point Interaction . . . . . . • Renormalon Loop Exists • Particle number conserves • But still complicated Write the problem with a combinations of background potentials 3D Model: ∗− ⋆ − ⋆ − . . . − ⋆ −∗ H = P 2 2 + λ 0 δ 2 ( x, y ) + V ( x, y, z ) � E f − u 2 � � � � � � n � d 2 Q 2 T ( n +3) = λ ∗ d 2 Q n +3 � λ ∗ � du 2 log 2 − iπ (2 π ) 2 F ( P 2 , P n +3 ) � µ 2 (2 π ) 2 4 π u 2 = u n +3 ∗ renormalon integral ∼ n ! Questions • Does it survive in the full expansion? • Physical implication?

Renormalons In Quantum Mechanics H = P 2 2 + λ 0 δ 2 ( x, y ) + V ( x, y, z ) V ( x, y, z ) = κδ ( z cos θ − y sin θ ) θ → 0 = ⇒ S θ =0 = S 2d S 1d − → No renormalon

Renormalons In Quantum Mechanics H = P 2 2 + λ 0 δ 2 ( x, y ) + V ( x, y, z ) V ( x, y, z ) = κδ ( z cos θ − y sin θ ) θ → 0 = ⇒ S θ =0 = S 2d S 1d − → No renormalon ∗ − ⋆ − . . . − ⋆ − ∗ All Diagrams at Order κ 2 ⋆ − . . . − ⋆ −∗ − ⋆ − . . . − ⋆ −∗ � �� � � �� � a n − a ∗ − ⋆ − . . . − ⋆ −∗ − ⋆ − . . . − ⋆ � �� � � �� � n − a a ⋆ − . . . − ⋆ −∗ − ⋆ − . . . − ⋆ −∗ − ⋆ − . . . − ⋆ � �� � � �� � � �� � a n − a − b b

Renormalons In Quantum Mechanics H = P 2 2 + λ 0 δ 2 ( x, y ) + V ( x, y, z ) V ( x, y, z ) = κδ ( z cos θ − y sin θ ) θ → 0 = ⇒ S θ =0 = S 2d S 1d − → No renormalon ∗ − ⋆ − . . . − ⋆ − ∗ All Diagrams at Order κ 2 ⋆ − . . . − ⋆ −∗ − ⋆ − . . . − ⋆ −∗ � �� � � �� � a n − a ∗ − ⋆ − . . . − ⋆ −∗ − ⋆ − . . . − ⋆ � �� � � �� � n − a a ⋆ − . . . − ⋆ −∗ − ⋆ − . . . − ⋆ −∗ − ⋆ − . . . − ⋆ + � �� � � �� � � �� � a n − a − b b � λ � n T n, 2 = 9 2(cos θ log cos 2 θ ) 2 κ 2 µ − 3 ( n − 3)! 2 6 π Standart Borel summation λ � λ � 2 θ = 0 2 (cos θ log cos 2 θ ) 2 κ 2 µ − 3 2 e − 6 π Im T 2 = ∓ 9 πi Ambiguity: − − − − → 0 6 π

Resolution to Renormalon Ambiguity λ � λ � 2 2 (cos θ log cos 2 θ ) 2 κ 2 µ − 3 2 e − 6 π Im T 2 = ∓ 9 πi Ambiguity: 6 π • No need to cancel imaginary part! ( T ∈ C ) • Is there a “natural” sign?

Resolution to Renormalon Ambiguity λ � λ � 2 2 (cos θ log cos 2 θ ) 2 κ 2 µ − 3 2 e − 6 π Im T 2 = ∓ 9 πi Ambiguity: 6 π • No need to cancel imaginary part! ( T ∈ C ) • Is there a “natural” sign? (YES!) dq f ( q ) l ( p 2 f − q 2 ) n − 1 λ n � → T = � Take a step back − n Here there are two options: 1. First integrate, then sum (what we’ve done) 2. First sum, then integrate (Observe we have a geometric series!) λ � T = dq f ( q ) f − q 2 ) − → Pole requires an analytical continuation 1 − λ l ( p 2