Wiener Measure Heuristics and the Feynman-Kac formula Theorem 1 - PowerPoint PPT Presentation

Wiener Measure Heuristics and the Feynman-Kac formula Theorem 1 (Trotter Product Formula) . Let A and B be d × d matrices. Then � n � e ( A + B ) = lim A B e n e . n n →∞ Path integrals over Euclidean spaces Proof: By the chain rule, d dε | 0 log( e εA e εB ) = A + B. Hence by Taylor’s theorem with remainder, Bruce Driver log( e εA e εB ) = ε ( A + B ) + O ε 2 � Visiting Miller Professor � (Permanent Address) Department of Mathematics, 0112 University of California at San Diego, USA which is equivalent to e εA e εB = e ε ( A + B )+ O ( ε 2 ) . http://math.ucsd.edu/ ∼ driver Taking ε = 1 /n and raising the result to the n th – power gives Student Topology Seminar e n − 1 ( A + B )+ O ( n − 2 ) � n ( e n − 1 A e n − 1 B ) n = � University of California, Berkeley, August 29, 2007 = e A + B + O ( n − 1 ) → e ( A + B ) as n → ∞ . Q.E.D. Bruce Driver 2 University of California, Berkeley, August 29, 2007 Fact (Trotter product formula) . For “nice enough” V, e T (∆ / 2 − V ) = strong– lim n V ] n . 2 n ∆ e − T T n →∞ [ e (1) See [1] for a rigorous statement of this type. Lemma 2. Let V : R d → R be a continuous function which is bounded from below, then Figure 1: A typical path in W m,T . n V � n �� � T n ∆ / 2 e − T e f ( x 0 ) � Theorem 4. Let T > 0 and n ∈ N be given. For τ ∈ [0 , T ] , let τ + = i n V ( x 1 ) . . . p T n T if 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 τ ∈ ( i − 1 n T, i n T ] . Then Eq. (2) may be written as, R dn n V � n �� � n ∆ / 2 e − T T dn � e f ( x 0 ) ⎛ ⎞ n n 1 − n � | x i − x i − 1 | 2 − T � V ( x i ) 2 T n 1 � � T = e f ( x n ) dx 1 . . . dx n . ⎜ ⎟ (2) 2 | ω ′ ( τ ) | 2 + V ( x 0 + ω ( τ + )) ] dτ f ( x 0 + ω ( T )) dm n ( ω ) i =1 i =1 0 [ 1 e − = � ⎝ ⎠ 2 π T Z n ( T ) W n,T n ( R d ) n where Notation 3. Given T > 0 , and n ∈ N , let W n,T denote the set of piecewise C 1 – paths, � � T e − 1 0 | ω ′ ( τ ) | 2 dτ dm n ( ω ) . � i ω : [0 , T ] → R d such that ω (0) = 0 and ω ′′ ( τ ) = 0 if τ / � n Z n ( T ) := 2 ∈ n T i =0 =: P n ( T ) – see W n,T Figure 1. Further let dm n denote the unique translation invariant measure on W n,T which Moreover, by Trotter’s product formula, is well defined up to a multiplicative constant. e T (∆ / 2 − V ) f ( x 0 ) 1 � With this notation we may rewrite Lemma 2 as follows. � T 0 [ 1 2 | ω ′ ( τ ) | 2 + V ( x 0 + ω ( τ + )) ] dτ f ( x 0 + ω ( T )) dm n ( ω ) . e − = lim (3) Z n ( T ) n →∞ W n,T Bruce Driver 3 University of California, Berkeley, August 29, 2007 Bruce Driver 4 University of California, Berkeley, August 29, 2007

Following Feynman, at an informal level (see Figure 2), W n,T → W T as n → ∞ , where “normalization” constant (or partition function) given by � � T 0 | ω ′ ( τ ) | 2 dτ D ω. ” [0 , T ] → R d � e − 1 � � � W T := ω ∈ C : ω (0) = 0 . Z ( T ) = “ 2 W T Moreover, formally passing to the limit in Eq. (3) leads us to the following heuristic This expression may also be written in the Feynman – Kac form as � � T e T (∆ / 2 − V ) f ( x 0 ) = e − 0 V ( x 0 + ω ( τ )) dτ f ( x 0 + ω ( T )) dµ ( ω ) , W T where 1 � T 0 | ω ′ ( τ ) | 2 dτ D ω ” Z ( T ) e − 1 dµ ( ω ) = “ (5) 2 is the informal expression for Wiener measure on W T . Thus our immediate goal is to make sense out of Eq. (5). Let � T � � | h ′ ( τ ) | 2 dτ < ∞ Figure 2: A typical path in W T may be approximated better and better by paths in W m,T H T := h ∈ W T : as m → ∞ . 0 � T 0 | h ′ ( τ ) | 2 dτ := ∞ if h is not absolutely continuous. Further let with the convention that e T (∆ / 2 − V ) f � � ( x 0 ) ; expression for � T h ′ ( τ ) · k ′ ( τ ) dτ for all h, k ∈ H T � h, k � T := 1 � � T � � 0 [ 1 2 | ω ′ ( τ ) | 2 + V ( x 0 + ω ( τ )) ] dτ f ( x 0 + ω ( T )) D ω ” e T (∆ / 2 − V ) f e − ( x 0 ) = “ (4) 0 Z ( T ) W T and X h ( ω ) := � h, ω � T for h ∈ H T . Since where D ω is the non-existent Lebesgue measure on W T , and Z ( T ) is the 1 Z ( T ) e − 1 2 � ω � 2 dµ ( ω ) = “ HT D ω, ” (6) Bruce Driver 5 University of California, Berkeley, August 29, 2007 Bruce Driver 6 University of California, Berkeley, August 29, 2007 dµ ( ω ) should be a Gaussian measure on H T and hence we expect, Gaussian Measures “on” Hilbert spaces � − 1 � � e i � h,ω � T dµ ( ω ) = exp 2 � h � 2 . (7) H T Goal Given a Hilbert space H , we would ideally like to define a probability measure µ on H T B ( H ) such that � 2 � λ � 2 for all λ ∈ H e i ( λ,x ) dµ ( x ) = e − 1 µ ( h ) := ˆ (8) H so that, informally, dµ ( x ) = 1 Ze − 1 2 | x | 2 H D x. (9) The next proposition shows that this is impossible when dim( H ) = ∞ . Proposition 5. Suppose that H is an infinite dimensional Hilbert space. Then there is no probability measure µ on B = B ( H ) such that Eq. (8) holds. Proof: Suppose such a Gaussian measure were to exist. Let { λ i } ∞ i =1 be an orthonormal set in H and for M > 0 and n ∈ N let W M n = { x ∈ H : | ( λ i , x ) | ≤ M for i = 1 , 2 , . . . , n } . Let µ n be the standard Gaussian measure on R n , dµ n ( y ) = (2 π ) − n/ 2 e − y · y/ 2 dy. Bruce Driver 7 University of California, Berkeley, August 29, 2007 Bruce Driver 8 University of California, Berkeley, August 29, 2007

Then for all bounded measurable functions f : R n → R , we have Guassian Measure for ℓ 2 � � f (( λ 1 , x ) , . . . , ( λ n , x )) dµ ( x ) = f ( y ) dµ n ( y ) R n H Remark 6. Suppose that H = ℓ 2 the space of square summable sequences { x n } ∞ n =1 . In and therefore, this case the Gaussian measure that we are trying to construct is formally given by the n ) = µ n ( { y ∈ R n : | y i | ≤ M for i = 1 , . . . , n } ) µ ( W M expression � M � n ∞ � � � 1 − 1 e − 1 2 y 2 dy � (2 π ) − 1 / 2 2Σ ∞ i =1 x 2 √ = → 0 as n → ∞ . dµ ( x ) = 2 π ) ∞ exp dx i i ( − M i =1 � 1 ∞ ∞ Because � � 2 πe − 1 2 x 2 � i dx i √ = =: p 1 ( dx i ) , W M n ↓ H M := { x ∈ H : | ( λ i , x ) | ≤ M ∀ i = 1 , 2 , . . . } , i =1 i =1 µ ( H M ) = lim n →∞ µ ( W M n ) = 0 for all M > 0 . Since H M ↑ H as M ↑ ∞ we learn that where p 1 ( dx ) is the heat kernel defined by µ ( H ) = lim M →∞ µ ( H M ) = 0 , i.e. µ ≡ 0 . Q.E.D. � � 1 − 1 Moral : The measure µ must be defined on a larger space. This is somewhat analogous 2 tx 2 √ p t ( x ) = 2 πt exp . to trying to define Lebesgue measure on the rational numbers. In each case the measure can only be defined on a certain completion of the naive initial space. This suggests that we define µ = p ⊗ N 1 , the infinite product measure on R N . Theorem 7. Let µ = p ⊗ N � R N , F � where µ and F . be the infinite product measure on 1 Also let a = ( a 1 , a 2 , . . . ) ∈ (0 , ∞ ) N be a sequence and define �� X a = ℓ 2 ( a ) = { x ∈ R N : ∞ i =1 a i x 2 i := � x � a < ∞} . Bruce Driver 9 University of California, Berkeley, August 29, 2007 Bruce Driver 10 University of California, Berkeley, August 29, 2007 So X = L 2 ( N , a ) where a now denotes the measure on N determined by a ( { i } ) = a i which shows that B ( X a ) ⊂ F and hence B ( X a ) ⊂ F X a . To prove the reverse inclusion, let i : X a → R N be the inclusion map and recall that for all i ∈ N . Then X a ∈ F , F X a := { A ∩ X a : A ∈ F} = B ( X a ) ( B ( X a ) is the Borel σ – field on X a ) and F X a = i − 1 ( F ) = i − 1 � π − 1 � �� σ j ( B ) : j ∈ N and B ∈ B R ⎧ ∞ � i − 1 π − 1 1 a i < ∞ � � if = σ j ( B ) : j ∈ N and B ∈ B R ⎨ µ ( X a ) = (10) i =1 � ( π j ◦ i ) − 1 ( B ) : j ∈ N and B ∈ B R � 0 otherwise. = σ = σ ( π j ◦ i : j ∈ N ) . ⎩ ∞ Since π j ◦ i ∈ X ∗ � a for all j, we see from this expression that a i < ∞ , µ a := µ | B ( X a ) is a the unique probability measure on Assuming that i =1 F X a ⊂ σ ( X ∗ a ) = B ( X a ) . ( X a , B ( X a )) which satisfies � � Let us now prove Eq. (10). Letting q and q N be as defined above, for any ε > 0 , f ( x 1 , . . . , x n ) dµ a ( x ) = f ( x 1 , . . . , x n ) p 1 ( dx i ) . . . p 1 ( dx n ) (11) � � � M.C.T. X a R n R N e − εq/ 2 dµ = N →∞ e − εq N / 2 dµ R N e − εq N / 2 dµ R N lim = lim N →∞ for all f ∈ ( B ( R n )) b and n = 1 , 2 , 3 , . . . N N − ε � � a i x 2 Proof: For N ∈ N , let q N : R N → R be defined by q N ( x ) = � N i � i =1 a i x 2 2 = lim R N e p 1 ( dx i ) i . Then it is easily 1 N →∞ seen that q N is F – measurable. Therefore, q := sup N ∈ N q N (also notice that q N ↑ q as i =1 N N → ∞ ) is F – measurable as well and hence � e − ε 2 a i x 2 p 1 ( dx ) . � = lim (12) X a = { x ∈ R N : q ( x ) < ∞} ∈ F . N →∞ R i =1 Similarly, if x 0 ∈ X a , then q ( · − x 0 ) = sup N ∈ N q N ( · − x 0 ) is F – measurable and Using therefore for r > 0 , � � 1 � 1 2 π 1 e − λ e − λ +1 2 x 2 p 1 ( x ) dx = 2 x 2 dx = √ √ √ λ + 1 = B ( x 0 , r ) = { x ∈ X a : � x − x 0 � a < r } = { x ∈ R N : q ( · − x 0 ) < r 2 } ∈ F λ + 1 2 π 2 π Bruce Driver 11 University of California, Berkeley, August 29, 2007 Bruce Driver 12 University of California, Berkeley, August 29, 2007