Path integrals on Riemannian Manifolds Bruce Driver Department of - PowerPoint PPT Presentation

Path integrals on Riemannian Manifolds Bruce Driver Department of Mathematics, 0112 University of California at San Diego, USA http://math.ucsd.edu/ ∼ bdriver Nelder Talk 1. 1pm-2:30pm, Wednesday 29th October, Room 340, Huxley Imperial College, London

Newtonian Mechanics on R d Given a potential energy function V : R d → R we look to solve q ( t ) = −∇ V ( q ( t )) for q ( t ) ∈ R d m ¨ that is Force = mass · acceleration Recall that p = m ˙ q and H ( q, p ) = 1 2 mp · p + V ( q ) = Conserved Energy q ) := 1 q | 2 + V ( q ) = E ( q, ˙ 2 m | ˙ Bruce Driver 2

Q.M. and Canonical Quantization on R d We want to find � � ih ˆ t H ψ 0 ψ ( t, x ) = e ( x ) i.e. solve the Schr¨ odinger equation i � ∂ψ ∂t = ˆ Hψ ( t ) for ψ ( t ) ∈ L 2 � R d � with ψ (0 , x ) = ψ 0 ( x ) where by “Canonical Quantization,” ∂ p = � i ∇ = � q � ˆ q = M q , p � ˆ ∂q and i p ) = − � 2 2 m ∇ 2 + M V ( q ) . H ( q, p ) � H (ˆ q, ˆ Bruce Driver 3

Feynman Path Integral Feynman explained that the solution to the Schr¨ odinger equation should be given by � 1 � T � � i � ˆ T i H ψ 0 0 ( K.E. - P .E. )( t ) dt ψ 0 ( ω ( T )) d vol ( ω ) e ( x ) = e (1) h Z ( T ) W x,T ( R 3 ) where ψ 0 ( x ) is the initial wave function, .E. ) ( t ) = m ω ( t ) | 2 − V ( ω ( t )) , 2 | ˙ ( K.E. - P and � � T i 0 ( K.E. )( t ) dt d vol ( ω ) . Z ( T ) = e h W x 0 ,T ( R 3 ) ω ( T ) ω x = the paths in R d starting at x which are parametrized by [0 , T ] . � R d � Figure 1: W x,T Bruce Driver 4

The Path Integral Prescription on R d Theorem 1 (Meta-Theorem – Feynman (Kac) Quantization) . Let V : R d → R be a nice function and W � R d ; x, T � [0 , T ] → R d � := � ω ∈ C � : ω (0) = x � . Then � ( x ) = “ 1 e − � T � � e − T ˆ H f 0 E ( ω ( t ) , ˙ ω ( t )) dt f ( ω ( T )) D ω ” (2) Z T W ( R d ; x,T ) 2 m | v | 2 + V ( x ) is the classical energy and where E ( x, v ) = 1 � � T ω ( t ) | 2 dt D ω ”. e − 1 0 | ˙ “ Z T := 2 W ( R d ; x,T ) Bruce Driver 5

Proof of the Path Integral Prescription Theorem 2 (Trotter Product Formula) . Let A and B be n × n matrices. Then � n e ( A + B ) = lim � A B e n e . n n →∞ Proof: Since d dε | 0 log( e εA e εB ) = A + B, log( e εA e εB ) = ε ( A + B ) + O � ε 2 � , i.e. e εA e εB = e ε ( A + B )+ O ( ε 2 ) and therefore e n − 1 A + n − 1 B + O ( n − 2 ) � n � ( e n − 1 A e n − 1 B ) n = = e A + B + O ( n − 1 ) → e ( A + B ) as n → ∞ . Bruce Driver 6

• Let A := 1 2 ∆ ; � e t ∆ / 2 f � � ( x ) = R d p t ( x, y ) f ( y ) dy where � d/ 2 � � � 1 1 2 t | x − y | 2 p t ( x, y ) = exp 2 πt • Let B = − M V – multiplication by V ; e − tM V = M e − tV • By Trotter ( x 0 := x ) , n V � n �� � T n ∆ / 2 e − T e f ( x ) � n V ( x 1 ) . . . p T n ( x 0 , x 1 ) e − T n ( x n − 1 , x n ) e − T n V ( x n ) f ( x n ) dx 1 . . . dx n = p T n ( R d ) n n | x i − x i − 1 | 2 − T − n � 1 � � V ( x i ) 2 T n = e f ( x n ) dx 1 . . . dx n i =1 i =1 Z n ( T ) ( R d ) n � 1 e − � T 2 | ω ′ ( s ) | 2 + V ( ω ( s + )) ] ds f ( ω ( T )) dm H n ( ω ) 0 [ 1 = (3) Z n ( T ) H n Bruce Driver 7

n T � n where Z n ( T ) := (2 πT/n ) dn/ 2 , P n = � k k =0 , and : ω ′′ ( s ) = 0 for s / ω ∈ W � R d ; x, T � H n = � � ∈ P n . Q.E.D. Bruce Driver 8

Euclidean Path Integral Quantization on R d Theorem 3 (Meta-Theorem – Path integral quantization) . We can define ˆ H by; � ( x ) “ = ” 1 e − � T � � e − T ˆ H ψ 0 0 E ( ω ( t ) , ˙ ω ( t )) dt ψ 0 ( ω ( T )) D ω (4) Z T ω (0)= x where � � T ω ( t ) | 2 dt D ω ”. e − 1 0 | ˙ “ Z T := 2 ω (0)=0 and D ω = “Infinite Dimensional Lebesgue Measure.” • Question: what does this formula really mean? 1. Problems, Z T = lim n →∞ Z n ( T ) = 0 . 2. There is not Lebesgue measure in infinite dimensions. 3. The paths ω appearing in Eq. (4) are very rough and in fact nowhere differentiable. Bruce Driver 9

Summary of Flat Results • Let P := { 0 = t 0 < t 1 < · · · < t n = T } be a partition of [0 , T ] . ω : [0 , T ] → R d : ω (0) = 0 and ¨ � R d � := � ∈ P � • Let H P ω ( t ) = 0 ∀ t / � R d � • λ P be Lebesgue measure on H P � T � ω ( t ) | 2 dt � − 1 • Z P := � 0 | ˙ H P ( R d ) exp dλ P ( ω ) 2 � T ω ( t ) | 2 dt � � 1 − 1 • dµ P := 0 | ˙ Z P exp dλ P ( ω ) 2 Theorem 4 (Wiener 1923) . There exist a measure µ on W � [0 , T ] , R d � such that ⇒ µ as |P| → 0 . µ P = Bruce Driver 10

2 | v | 2 + V ( x ) where V is a nice potential, then Theorem 5 (Feynman Kac) . If E ( x, v ) = 1 � T � � 1 ⇒ e − � T 0 V ( ω ( s )) ds dµ ( ω ) − exp E ( ω ( t ) , ˙ ω ( t )) dt dλ P ( ω ) = Z P 0 and morever, � T � � � 1 � � e − t ˆ H f − (0) = lim exp E ( ω ( t ) , ˙ ω ( t )) dt f ( ω ( T )) dλ P ( ω ) Z P |P|→ 0 H P ( R d ) 0 � e − � T 0 V ( ω ( s )) ds f ( ω ( T )) dµ ( ω ) . = W ( [0 ,T ] , R d ) Bruce Driver 11

Norbert Wiener Figure 2: Norbert Wiener (November 26, 1894 – March 18, 1964). Graduated High School at 11, BA at Tufts College at the age of 14, and got his Ph.D. from Harvard at 18. Bruce Driver 12

Classical Mechanics on a Manifold • Let ( M, g ) be a Riemannian manifold. • Newton’s Equations of motion m ∇ ˙ σ ( t ) = −∇ V ( q ( t )) , (5) dt i.e. Force = mass · tangential acceleration • In local coordinates ( q 1 , . . . , q d ); H ( q, p ) = 1 2 mg ij ( q ) p i p j + V ( q ) where ds 2 = g ij ( q ) dq i dq j Bruce Driver 13

(Not) Canonical Quantization on M H ( q, p ) = 1 2 g ij ( q ) p i p j + V ( q ) √ gg ij ( q ) p j + V ( q ) . = 1 1 √ gp i 2 • To quantize H ( q, p ) , let p i := 1 ∂ ? q i � ˆ q i := M q i , p i � ˆ ∂q i , and H ( q, p ) � H (ˆ q, ˆ p ) . i • Is ∂ 2 H = − 1 ˆ 2 g ij ( q ) ∂q i ∂q j + V ( q ) • or is it √ gg ij ( q ) ∂ H = − 1 1 ∂ ∂q j + V ( q ) = − 1 ˆ √ g 2∆ M + M V , ∂q i 2 • or something else? Bruce Driver 14

Path Integral Quantization of ˆ H The previous formulas on R d suggest we can define ˆ H in the manifold setting by; � ( x 0 ) = 1 e − � T � � e − T ˆ H ψ 0 0 E ( σ ( t ) , ˙ σ ( t )) dt ψ 0 ( σ ( T )) D σ (6) Z T σ (0)= x 0 where E ( x, v ) = 1 2 g ( v, v ) + V ( x ) is the classical energy. • Formally, there no longer seems to be any ambiguity as there was with canonical quantization. • On the other hand what does Eq. (6) actually mean? Bruce Driver 15

Back to Curved Space Path Integrals • Recall we now wish to mathematically interpret the expression; Z ( T ) e − � T 1 σ ( t ) | 2 + V ( σ ( t )) ] dt D σ. 0 [ 1 2 | ˙ dν ( σ ) “ = ” o σ ( T ) σ Figure 3: A path in W o,T ( M ) . • To simplify life (and w.o.l.o.g.) set V = 0 , T = 1 so that we will now consider, � 1 � 1 σ ( t ) | 2 dt ψ 0 ( σ (1)) D σ. e − 1 0 | ˙ 2 Z W o ( M ) • We need introduce (recall) six geometric ingredients. Bruce Driver 16

I. Geometric Wiener Measure ( ν ) over M Fact (Cartan’s Rolling Map) . Relying on Itˆ o to handle the technical (non-differentiability) � R d � difficulties, we may transfer Wiener’s measure, µ, on W 0 ,T to a measure, ν, on W o,T ( M ) . � R d � Figure 4: Cartan’s rolling map gives a one to one correspondance between, W 0 ,T and W o,T ( M ) . Bruce Driver 17

II. Riemannian Volume Measures • On any finite dimensional Riemannian manifold ( M, g ) there is an associated volume measure , � � � �� ∂ Σ , ∂ Σ d Vol g = det g dt 1 . . . dt n (7) ∂t i ∂t j where R n ∋ ( t 1 , . . . , t n ) → Σ ( t 1 , . . . , t n ) ∈ M is a (local) parametrization of M. Example 1. Suppose M is 2 dimensional surface, then we teach, dS = � ∂ t 1 Σ ( t 1 , t 2 ) × ∂ t 2 Σ ( t 1 , t 2 ) � dt 1 dt 2 . (8) Combining this with the identity, � a × b � 2 = � a � 2 � b � 2 − ( a · b ) 2 � � a · a a · b = det a · b b · b shows, � � � ∂ t 1 Σ · ∂ t 1 Σ ∂ t 1 Σ · ∂ t 2 Σ dS = det dt 1 dt 2 ∂ t 1 Σ · ∂ t 2 Σ ∂ t 2 Σ · ∂ t 2 Σ that is Eq. (7) reduces to Eq. (8) for surfaces in R 3 . Bruce Driver 18

III. Scalar Curvature • On any finite dimensional Riemannian manifold ( M, g ) there is an associated function called scalar curvatue , Scal : M → R such that at a point m ∈ M, ε 2 � � � � � B R d 6( d + 2)Scal( m ) + O ( ε 3 ) 1 − for ε ∼ 0 , Vol g ( B ε ( m )) = ε (0) � � � � � � B R d � is the volume of a ε – Euclidean ball in R d . ε (0) where � � Bruce Driver 19

IV. Tangent Vectors in Path Spaces • The space � 1 � � σ ( t ) | 2 dt < ∞ H ( M ) = σ ∈ W o ( M ) : E ( σ ) := | ˙ 0 is an infinite dimensional Hilbert manifold. • The tangent space to σ ∈ H ( M ) is � � X : [0 , 1] → TM : X ( t ) ∈ T σ ( t ) M and T σ H ( M ) = G 1 ( X, X ) := � 1 . � � ∇ X ( t ) dt , ∇ X ( t ) 0 g dt < ∞ dt Bruce Driver 20

o X ( t ) σ ( t ) σ Figure 5: A tangent vector at σ ∈ H ( M ) . Bruce Driver 21