Lecture Notes on Guassian Measures -- I.E. Gaussian Noises

Home , Gaussian integral

PATH INTEGRAL NOTES 1. Finite Dimensional Gaussian Integrals Let A be a positive deﬁnite symmetric N × N matrix , then by a standard BRUCE K. DRIVER calculation, Z s − 1 Ax·x 2π These are some notes on heuristic and rigorous aspects of (1.1) Z := e 2 dx = det . Abstract. A A path integrals. N R Indeed we know that  N Z Z √ − 1 |y|2 − 1 y2 N e 2 dy =  e 2 dy = ( 2π)

N R R √ from which Eq. (1.1) follows after making the change of variables: y = Ax. N Contents Notation 1.1. For A > 0, let µA be the probability measure on R defined by 1. Finite Dimensional Gaussian Integrals 1 r 2. Infinite Dimensional Gaussian Integrals 4 1 − 1 Ax·x A − 1 Ax·x (1.2) dµA(x) := e 2 dx = det e 2 dx. 3. Semi-Group Property and the Operator Connection 9 ZA 2π 3.1. Heuristics 9 Further let 3.2. Rigorous Interpretation of the above Heuristics 10 N X ∂ 4. Harmonic Oscillator Examples 11 (1.3) L := LA := A−1∂ ∂ where ∂ := . i,j i j i ∂x 4.1. Finite Dimensional Case 11 i,j=1 i 4.2. An Infinite Dimensional Example 12 More invariantly, if we let (v, w) := (Av, w) N , then we may write L = A R 5. Path Integral Quantization 12 PN 2 N N ∂ where {uj} is any orthonormal basis for , (·, ·) . 6. Gross’ QFT Notes Introduction 13 j=1 uj j=1 R A 6.1. The harmonic oscillator 13 Proposition 1.2 (Gaussian Integral formulas). The following Gaussian inte- 6.2. A quantized field; informalities 14 gral formulas hold; 6.3. Ground state transformation. 14 Z λ·x 1 (A−1λ·λ) N 6.4. Back to the quantized fields 16 (1.4) e dµA(x) = e 2 ∀ λ ∈ C ,

7. Appendix – Some Wiener Space Results 17 N R 8. After thoughts? 18 Z √ t L 8.1. Extras on `2 – Gaussian Measures. 18 (1.5) f x − ty dµA(y) = e 2 f (x) ,

N 9. Classical Wiener Measure 19 R References 22 Z L (1.6) f (x) dµA(x) = e 2 f (0) ,

N R and for v ∈ RN , Z Z Date: January 7, 2013 File:topclass.tex. (1.7) (∂vf)(x) dµA(x) = (Av, x) f (x) dµA(x)

Key words and phrases. Path integrals, Wiener measure. N N R R 1 2 DRIVER or equivalently, and therefore, tL/2 λ·x t A−1λ·λ λ·x Z Z e e = e 2 e . (1.8) (v, x) f (x) dµ (x) = (∂ −1 f)(x) dµ (x). A A v A Q.E.D. N N R R Remark 1.3 (Feynman diagrams interpretation 1). Let {v }n ⊂ N . Using Applying this formula with f replaced by fg gives the integration by parts for- j j=1 R mula; the integration by parts formula we find, n n−1 Z Z Z Y Z Y (1.9) (∂A−1vf)(x) g (x) dµA(x) = f (x)(−∂A−1v + (v, x)) g (x) dµA(x) (vj, x) dµA (x) = (vj, x)(vn, x) dµA (x) N N N N R j=1 R j=1 R R Z n−1 which we may abbreviate as; Y ∗ = (v , x) ∂ −1 1dµ (x) j (A vn) A N ∗ R (1.10) ∂A−1v = −∂A−1v + M(v,x). j=1 n−1 We also have the formula, Z Y −1 = ∂(A vn) (vj, x) dµA (x) N Z Z Z R j=1 λ·x λ·x p (x) dµA(x) = p (Dλ) e |λ=0dµA(x) = p (Dλ) e dµA(x) Z n−1 X −1 Y N N N = A vn, vk (vj, x) dµA (x) . R R R λ=0 N R k=1 1 (A−1λ·λ) ,j∈{ / k,n} (1.11) = p (Dλ) e 2 . Continuing this way inductively one easily shows that to compute the integral, R Qn Proof: Using the identity N (vj, x) dµA (x) , we should: R j=1 −1 −1 −1 n Ax · x − 2λ · x = A(x − A λ) · (x − A λ) − A λ · λ, (1) put down n dots labeled by the {vj}j=1 , (2) to each perfect pairing (n must be even else we get zero) of the vj assign we find −1 Z Z a weight which is the product vi,A vj over all pairs vi and vj which λ·x 1 − 1 (Ax·x−2λ·x) e dµA(x) = e 2 dx are paired, and N Z N (3) sum the result over all perfect pairings. R R Z 1 A−1λ·λ 1 − 1 A(x−A−1λ)·(x−A−1λ) = e 2 e 2 dx Definition 1.4. Given a polynomial, p (x) , on N , the new polynomial, Z N R ZR e−L/2p, often denoted by : p : is called the Wick ordered version of p. 1 A−1λ·λ 1 − 1 Ax·x 1 A−1λ·λ = e 2 e 2 dx = e 2 Z N Before going further let us point out that in much of what we do for a while, R where in the third equality we used the translation invariance of Lebesgue mea- we may simplify the notation and take A = I without any loss of generality. sure. Eq. (1.4) now follows from the previous equation by analytic continuation. The following lemma explains why this is the case. Let us finish by showing that Eq. (1.5) and 1.4) are consistent. To this end, Lemma 1.5. Let (v, w) := Av · w – an inner product on N . Then for any λ·x A R suppose that f (x) = e . Then on one hand; N N orthonormal basis, {uj}j=1 of R , (·, ·)A we have Z √ Z √ Z √ λ·(x− ty) λ·x − tλ·y f x − ty dµA(y) = e dµA(y) = e e dµA(y) N N X −1 X N N N (1.12) Aij ei ⊗ ej = uk ⊗ uk R R R i,j=1 k=1 λ·x t A−1λ·λ t −1 = e e 2 = exp λ · x + A λ · λ 2 and N N X X while on the other hand, (1.13) A−1∂ ∂ = ∂2 . ij i j uk λ·x −1 λ·x −1 λ·x i,j=1 k=1 Le = Aij λiλj · e = A λ · λ e PATH INTEGRAL NOTES 3

N Proof: Let v, w ∈ R , then by contracting Eq. (1.12) with (v, ·)A ⊗ (w, ·)A Then working inductively it follows that we must show, ∞ X X h(n) (t) = A−1 ...A−1 (∂ . . . ∂ P (t)) (∂ . . . ∂ Q (t)) N N ii,j1 in,jn i1 in j1 jn X −1 X i1,j1,...,in,jn i1,j1,...,in,jn=1 (1.14) Aij (v, ei)A (w, ej)A = (v, uk)A (w, uk)A = (v, w)A . i,j=1 k=1 and in particular that, ∞ However, X X h(n) (0) = A−1 ...A−1 (∂ . . . ∂ p)(∂ . . . ∂ q) . ii,j1 in,jn i1 in j1 jn N X i1,j1,...,in,jn i1,j1,...,in,jn=1 (v, e ) = v · Ae = v · e A and i A i m im Since h is a polynomial in t (hence analytic), we have m=1 ∞ N X 1 X eL/2 e−L/2p · e−L/2q = eL/2 (P (1) · Q (1)) = h (1) = h(n) (0) (w, ej)A = w · Aej = w · enAjn n! n=0 n=1 ∞ ∞ X 1 X −1 −1 so the left side of Eq. (1.14) becomes, (1.17) = A ...A (∂i . . . ∂i p)(∂j . . . ∂j q) . n! ii,j1 in,jn 1 n 1 n n=0 i1,j1,...,in,jn=1 N X −1 −1 Evaluating this expression at x = 0 gives Eq. (1.15). Replacing p and q by Aij (v, ei)A (w, ej) = Aij AimAjn (v · em)(w · en) = δjmAjn (v · em)(w · en) A L/2 L/2 i,j=1 e p and e q respectively in Eq. (1.17) and then evaluating at x = 0 gives Eq. (1.16). Q.E.D. = Amn (v · em)(w · en) = Av · w = (v, w)A Corollary 1.7. Let S N denote the symmetric tensor algebra over N which as desired. Eq. (1.13) follows from Eq. (1.12) by mapping N ⊗ N to the R R R R we identify with the space of polynomial functions on N . On this space, let space of constant coefficient purely second order differential operators in such R ∞ ∞ as way that v ⊗ w → ∂v∂w. Q.E.D. X 1 X −1 −1 −L/2 −L/2 hp, qi = A ...A ∂i . . . ∂i e p (0) ∂j . . . ∂j e q (0) A n! ii,j1 in,jn 1 n 1 n Proposition 1.6. For all polynomials, p, and q, n=0 i1,j1,...,in,jn=1 (1.15) ∞ ∞ ∞ X 1 X −L/2 −L/2 = ∂u . . . ∂u e p ∂u . . . ∂u e q |x=0 X 1 X −1 −1 n! 1 n 1 n µA (: p : · : q :) = A ...A (∂i . . . ∂i p) (0) (∂j . . . ∂j q) (0) n! ii,j1 in,jn 1 n 1 n n=0 u1,...,un∈S n=0 i ,j ,...,i ,j =1 1 1 n n N where S is an orthonormal basis for R , (·, ·)A . Then the (Fock-Kakutani-Ito and – Segal-Bargmann) map, (1.16) ∞ ∞ N −L/2 N S , (·, ·) 2 3 p → e p ∈ S , h·, ·i X 1 X −1 −1 R L (µA) R A µA (p · q) = A ...A µA (∂i . . . ∂i p)·µA (∂j . . . ∂j q) . n! ii,j1 in,jn 1 n 1 n n=0 i1,j1,...,in,jn=1 is an isometric isomorphism of vector spaces. By completing the right side of this equation and using continuity, we get a unitary map between Hilbert spaces; Proof: Let 2 −L/2 N L (µA) 3 p → “e p” ∈ S (R ), h·, ·iA . P (t, x) := e−tL/2p (x) ,Q (t, x) = e−tL/2q (x) , and The right hand side is called the Bosonic Fock space over RN . h (t) := etL/2 (P (t) · Q (t)) – where x has been suppressed. Remark 1.8. Let us consider the free Euclidean field, i.e. the Gaussian mea- Then by a simple exercise in the chain and product rule we find sure is informally given by; ˙ X −1 tL/2 1 1 Z h i h (t) = Aij e (∂iP (t) · ∂jQ (t)) . (1.18) dµ (ϕ) = exp − |∇ϕ (x)|2 + m2ϕ2 (x) dx Dϕ i,j Z 2 d R 4 DRIVER 1 1 2 Therefore, = exp − −∆ + m ϕ, ϕ L2( d,dx) Dϕ, Z 2 R ∗ :(v, x) p (x) := (v, x): p (x): −∂A−1v : p (x) := ∂A−1v : p (x): which we view as a measure on L2 d, dx . Working as above if f, g ∈ L2 d , R R and hence then we should have ∗ Z −1 µA (: (v, x) p (x): ·q (x)) = µA (∂A−1v : p (x): ·q (x)) = µA (: p (x): ∂A−1vq (x)) . (1.19) (ϕ, f)(ϕ, g) dµ (ϕ) = −∆ + m2 f, g . 2 d L (R ,dx) Q.E.D. Taking f and g, formally, to be δx and δy respectively, the above formula leads Remark 1.10 (Feynman Interpretation 2). Let us now work out the Feynman to rules for computing an integral of the form, Z 2−1 Z n ϕ (x) ϕ (y) dµ (ϕ) = −∆ + m δx, δy =: ∆m (x − y) , Y mj 2 d L (R ,dx) J ((v1, m1) ,..., (vn, mn)) := :(vj, x) : dµA (x) . N R j=1 where ∆m is called the Euclidean propagator. This propagator satisfies: (1) If d = 1 then Using the integration by parts formula in Eq. (1.20) we have, n 1 −m|x| Z ∆m (x) = e . Y mj mn−1 2m :(vj, x) : · :(vn, x) (vn, x): dµA (x) N (2) For d ≥ 2, R j=1   − ln |x| if d = 2 Z n Y mj mn−1 ∆m (x) d−2 for |x| 1 −1 v = ∂A vn :(vj, x) : :(vn, x) : dµA (x) , |x| if d ≥ 3 N R j=1 and ∆ (x) e−m|x| for |x| 1. In particular for d ≥ 2, we have m v ≫ where Z ϕ2 (x) dµ (ϕ) = ∞ n n−1 Y mj X Y mj mn−1 mk 2 d −1 −1 L (R ,dx) ∂A vn :(vj, x) = :(vj, x) : · :(vn, x) : ·∂A vn :(vk, x) : from which we might expect that there is no Gaussian measure, µ, on j=1 k=1 j6=k,n L2 d, dx (or any class of functions for that matter) such that Eq. n−1 R X Y mj mn−1 −1 mk−1 (1.19) holds. = :(vj, x) : · :(vn, x) : ·mk A vn, vk :(vk, x) : . k=1 j6=k,n Lemma 1.9. For any polynomials, p (x) and q (x) , on N we have R Using these formula inductively and a little thought leads to the following Feyn- (1.20) µA (: (v, x) p (x): ·q (x)) = µA (: p (x): ∂A−1vq (x)) man rules for computing J ((v1, m1) ,..., (vn, mn)) : n th where (1) Draw n dots labeled by {vj} and to the j – dot, attach mj legs. Z j=1 (2) Consider all possible pairings of the legs (iff possible) with no self con- µA (f) = µA (f (x)) := f (x) dµA (x) . N R nections, i.e. no pairings of a leg of some vi with another leg of vi. Proof: Since (3) Given a pairing graph as above associate the weight which is the product −1 over terms of the form A vi, vj – one for each leg which connects vi d −tL/2 tL/2 −tL/2 L tL/2 e M(v,x)e = −e ,M(v,x) e to v . dt 2 j (4) Sum the above weights over all allowed pairing graphs. −tL/2 −1 tL/2 −1 = −e Aij (v, ei) ∂ei e = −∂A v, it follows that 2. Infinite Dimensional Gaussian Integrals −tL/2 tL/2 For general references on Gaussian measures in infinite dimensions, see [1, 2, (1.21) e M(v,x)e = M(v,x) − t∂A−1v 6]. and hence It will now be convenient to get rid of the matrix A in the above formula. −L/2 −L/2 (1.22) e M(v,x) = M(v,x) − t∂A−1v e . We do this by letting (v, w) := (v, w)A := v · Aw. With this notation, a small PATH INTEGRAL NOTES 5

2 exercise in linear algebra shows that Eq. (1.4) may be written in this notation i.e. Xa = L (N, a) where a now denotes the measure on N determined by as a ({i}) = ai for all i ∈ N. Then Xa ∈ B, BXa := {A ∩ Xa : A ∈ B} (BXa is the Z 1 ∗ λ(x) (λ,λ)∗ N Borel σ – field on X ) and e dµA(x) = e 2 ∀ λ ∈ R , a  ∞ N R  1 if P a < ∞ ∗ i where (·, ·) is the inner product on N dual to (·, ·) . With this as moti- (2.5) µ(Xa) = i=1 ∗ R A  vation we now proceed to the infinite dimensional setting. 0 otherwise. Given a real separable Hilbert space H, we would like to understand, ∞ Assuming that P a < ∞, µ := µ| is a the unique probability measure on i a BXa 1 − 1 (x,x) i=1 (2.1) dµ (x) := e 2 H Dx Z (Xa, BXa ) which satisfies as a Gaussian measure on H. A formal definition would be that µ is to be the Z Z unique measure on H such that (2.6) f(x1, . . . , xn)dµa(x) = f(x1, . . . , xn)p1(dxi) . . . p1(dxn) n Xa Z 1 R λ(x) (λ,λ) ∗ ∗ (2.2) e dµ (x) = e 2 H for all λ ∈ H . for all bounded measurable function f : n → and n = 1, 2, 3,.... Moreover, H R R ∞ if λ = (λ1, λ2, . . . , λn, 0, 0,... ) , then Let us now suppose that we have chosen an orthonormal basis {ej}j=1 for ∞ Z 1 2 (λ,x) 2 (λ,λ) 2 H and let us identify H with ` via, x → {xj = (x, ej)}j=1 . In this notation, (2.7) e ` dµ (x) = e 2 ` . Eqs. (2.1) and (2.2) become, Xa N P N 2 1 1 − 1 (x,x) Proof: For N ∈ N, let qN : R → R be defined by qN (x) = i=1aixi . (2.3) dµ (x) := e 2 `2 Dx, Z Then for any ε > 0, ∞ Z Z Z P∞ 2 Q −εq/2 −εqN /2 M.C.T. −εqN /2 where (x, x)`2 := j=1 xj , Dx = dxi, and e dµ = lim e dµ = lim e dµ N→∞ N→∞ i=1 RN RN RN Z N N 1 Z ε P 2 (λ,x) 2 (λ,λ)`2 − 2 aixi (2.4) e dµ (x) = e for all λ ∈ F 1 Y 2 = lim e p1(dxi) ` N→∞ N R i=1 where F ⊂ `2 is the collection of sequences with only finitely many non-zero terms. 1Technicalities: It is easily seen that q is B – measurable. Therefore, q := sup q N N∈N N In this case, the Gaussian measure that we are trying to construct is formally (also notice that qN ↑ q as N → ∞) is B – measurable as well and hence given by the expression, Xa = {x ∈ RN : q(x) < ∞} ∈ B. ∞ 1 1 X ∞ 2 Y Similarly, if x0 ∈ Xa, then q(· − x0) = supN∈ qN (· − x0) is B – measurable and therefore for dµ(x) = √ exp − x dxi N ( 2π)∞ 2 i=1 i r > 0, i=1 2 B(x0, r) = {x ∈ Xa : kx − x0ka < r} = {x ∈ RN : q(· − x0) < r } ∈ B ∞ ∞ 1 1 2 Y − x Y which shows that BXa ⊂ B and hence BXa ⊂ {A ∩ Xa : A ∈ B} . To prove the reverse inclu- = √ e 2 i dxi =: p1(dxi), sion, let i : Xa → RN be the inclusion map and recall that i=1 2π i=1 −1 −1 −1 {A ∩ Xa : A ∈ B} = i (B) = i σ π (B): j ∈ and B ∈ B 1 − 1 y2 j N R where p1(dy) := √ e 2 dy . 2π −1 −1 = σ i π (B): j ∈ N and B ∈ B ⊗N N j R This suggests that we define µ = p1 , the infinite product measure on R . −1 = σ (πj ◦ i) (B): j ∈ N and B ∈ BR = σ (πj ◦ i : j ∈ N) . ⊗N N Theorem 2.1. Let µ = p1 be the infinite product measure on R , B := B N R Since π ◦ i ∈ X∗ for all j, we see from this expression that ⊗N N j a where µ = p1 as described above. For a = (a1, a2,... ) ∈ (0, ∞) , define {A ∩ X : A ∈ B} ⊂ σ (X∗) ⊂ B ⊂ {A ∩ X : A ∈ B} . q a a Xa a 2 N X ∞ 2 Xa = ` (a) = {x ∈ R : i=1aixi =: kxka < ∞}, 6 DRIVER

N Z ε 2 Finally Eq. (2.6) follows from the definition of µ and the fact that Y − aix (2.8) = lim e 2 p1(dx). N→∞ Z Z i=1 R f(x1, . . . , xn)dµa(x) = f(x1, . . . , xn)dµ(x). ∗ RN {A ∩ Xa : A ∈ B} ⊂ σ (Xa ) ⊂ BXa ⊂ {A ∩ Xa : A ∈ B} . Xa Using By the definition of infinite product measure, the integral on the left side Z Z r of Eq. (2.7) may be reduced to a finite dimensional Gaussian integral. The − λ x2 1 − λ+1 x2 1 2π 1 e 2 p1(x)dx = √ e 2 dx = √ = √ resulting integral is easily computed to give the right side Eq. (2.7). Q.E.D. 2π 2π λ + 1 λ + 1 in Eq. (2.8) we learn that Theorem 2.2. Suppose that H and K are separable Hilbert spaces, H is a dense subspace of K, and the inclusion map, i : H → K is continuous. Then v Z N u N there exists a Gaussian measure, ν, on K such that −εq/2 Y 1 u Y −1 e dµ = lim √ = t lim (1 + εai) Z N→∞ N→∞ λ(x) 1 ∗ ∗ RN 1 + εai 1 1 (2.11) e dν (x) = exp (λ, λ)H∗ for all λ ∈ K ⊂ H K 2 or equivalently that iff i : H → K is Hilbert Schmidt. Recalling the Hilbert Schmidt norm of i and ∞ Z 1 X its adjoint, i∗, are the same, the following conditions are equivalent; (2.9) − log e−εq/2dµ = ln(1 + εa ). 2 i RN i=1 (1) i : H → K is Hilbert Schmidt, (2) i∗ : K → H is Hilbert Schmidt, Notice that there is δ > 0 such that (3) tr (i i∗) < ∞ (2.10) ln(1 + x) ≤ x ∀x ≥ 0 and ln(1 + x) ≥ x/2 for x ∈ [0, δ]. (4) tr (i∗i) < ∞. ∞ P Proof: Let us start with the direction that we are most interested in. If lim supi→∞ ai 6= 0, then ln(1 + εai) = ∞ for all ε > 0. If limi→∞ ai = 0 i=1 Namely, if i : H → K is Hilbert Schmidt, then there exists a measure ν on P∞ ∗ but i=1 ai = ∞, then using Eq. (2.10) ln(1 + εai) ≥ εai/2 for all i large and K such that Eq. (2.11) holds. In this case, the operator, A := i i : H → H, ∞ P P∞ is a self-adjoint trace class operator and hence by the spectral theorem, there hence again ln(1 + εai) = ∞. If i=1 ai < ∞ then by Eq. (2.10), ∞ i=1 exists an orthonormal basis, {ej}j=1 for H such that Aej = ajej with aj > 0 ∞ ∞ and P∞ a < ∞. Observe that X X j=1 j ln(1 + εai) ≤ ε ai → 0 as ε ↓ 0. ∗ i=1 i=1 (ej, ek)K = (iej, iek)K = (i iej, ek)H = (Aej, ek)H = ajδjk ∞ In summary, n −1/2 o and therefore, aj ej is an orthonormal basis for K. Hence the map, ∞ j=1 Z P ∞ −εq/2 ∞ if i=1 ai = ∞ P 2 − lim log e dµ = P∞ U : X (a) → K defined by Ux = j=1 xjej is unitary and so is U|`2 : ` → H. ε↓0 0 if ai < ∞ RN i=1 (Notice that or equivalently, ∞ ∞ 2 X 2 2 X 2 Z P∞ kUxkK = xj kejkK = xj aj.) −εq/2 0 if i=1 ai = ∞ lim e dµ = P∞ j=1 j=1 ε↓0 1 if ai < ∞. RN i=1 −1 We may now define v := U∗µ := µ◦U – a probability measure on K. Moreover, −εq/2 −εq/2 ∗ Since e ≤ 1 and limε↓0 e = 1Xa , the previous equation along with the if λ ∈ K , then dominated convergence theorem shows that Z Z λ(k) λ(Ux) 1 1 Z P∞ e dν (k) = e dµ (x) = exp (λ ◦ U, λ ◦ U)(`2)∗ = exp (λ, λ)H∗ . 0 if i=1 ai = ∞ K K 2 2 µ(Xa) = 1Xa dµ = P∞ 1 if ai < ∞. RN i=1 We will sketch two proofs of the converse direction – each of which makes use proving Eq. (2.5). of a result (see [1, 2, 6]) which will not be proved here. In each case we start PATH INTEGRAL NOTES 7 with H densely contained and continuous embedded in K and assume there i : H → K will be Hilbert Schmidt and hence K will support a Gaussian measure exists a measure ν on K satisfying Eq. (2.11). with variance determined by H. 1. For the first proof, we use the fact that under the above assumptions, ∗ Remark 2.4. Suppose that H ⊂ K and ν is a Gaussian measure on K such i : H → K, is compact. Therefore A = i i is a compact self-adjoint operator. ∞ ∞ that Eq. (2.11) holds and {ej}j=1 is any orthonormal basis for H. Then for Therefore there exists an orthonormal basis, {ej}j=1 for H such that Aej = ajej k1, k2 ∈ K, we have with aj > 0 and limj→∞ aj = 0. Let PN : K → span (e1, . . . , eN ) be defined by PN Z ∞ PN (k) = j=1 (k, ej)H ej. Then, working as above, (ej, ek)K = ajδjk so that X (k1, k)K (k2, k)K dν (k) = ((k1, ·)K (k2, ·)K )H∗ = (k1, ej)K (k2, ej)K N K 2 X 2 2 j=1 kPN (k)kK = aj (k, ej)H ↑ kkkK . ∞ ∞ j=1 X X ∗ ∗ = (k1, iej)K (k2, iej)K = (i k1, ej)H (i k2, ej)H Therefore by the monotone convergence theorem, j=1 j=1 ∗ ∗ ∗ ∗ N Z Z = (i k1, i k2)H = (k1, ii k2)K = (k1, i k2)K 2 X 2 kkk dν (k) = lim aj (k, ej) dν (k) K N→∞ H K j=1 K Example 2.5 (Wiener measure). Let H denote the set of functions h : d N ∞ ∞ [0,T ] → R which are absolutely continuous and satisfy h (0) = 0 and X X X 2 R 1 0 2 = lim aj = aj = kejk . (h, h)H = 0 |h (s)| ds < ∞. The informal expression for Wiener measure is N→∞ K j=1 j=1 j=1 then given by R 2 Z T ! However, by Fernique’s or Skorohod’s theorem, we know that K kkkK dν (k) < 1 1 2 ∞. dµ (ω) = exp − |ω˙ (t)| dt Dω for ω ∈ H. ∞ Z 2 0 2. It is known that for any orthonormal basis {ej}j=1 of H which is contained 2 d in We are going to show that the measure µ may be constructed on L [0,T ] , R . ∗ ∗ 2 H∗ = {h ∈ H :(h, ·)H extends to an element of K } Before doing this let us compute i where i : H → L is the inclusion operator. PN p To this end, let h ∈ H and f ∈ L2, then has the property that PN (k) = j=1 (k, ej)H ej → k in L (ν) for all p < ∞. Therefore, Z T Z T Z t Z Z (ih, f) = h (t) · f (t) dt = h0 (s) ds · f (t) dt 2 2 L2 kkk dν (k) = lim kPN (k)k dν (k) 0 0 0 K N→∞ K K K Z Z T Z T ! 2 0 0 Z N = 10≤s≤th (s) · f (t) dtds = h (s) · f (t) dt ds X [0,T ]2 0 s = lim (k, ej) ej dν (k) N→∞ H ∗ K j=1 = (h, i f) , K H N Z X where = lim (k, ej)H (k, el)H (ej, ek)K dν (k) N→∞ K Z τ Z T ! Z j,l=1 ∗ (i f)(τ) = f (t) dt ds = 10≤s≤τ · 1s≤t≤T f (t) dtds N ∞ 0 s [0,T ]2 X X 2 = lim δjk (ej, ek) = kejk . T N→∞ K K Z j,l=1 j=1 = min (t, τ) f (τ) dτ. So again we may now apply Fernique’s theorem to finish the proof. Q.E.D. 0 From this we expect that Lemma 2.3. If H is a Hilbert space and A : H → H is a positive trace class operator. We may define (x, y) := (Ax, y) for all x, y ∈ H. Then let K denote Z T A p tr (i i∗) = d · min (t, t) dt = d · T 2/2 the completion of H in the norm, k·kA := (·, ·)A. Then the inclusion map, 0 8 DRIVER which is finite. To see this is correct use Corollary 7.2 below to conclude {f ≥ ε} as a disjoint union of {Aj} where ∞ > m (Aj) > 0. Then the functions, ∞ ∞ T T 1 2 Z Z √ 1A forms an orthonormal subset of L (m) and therefore, X 2 2 m(A ) j |hn (t)| dt = d tdt = d · T /2. j j=1 n=1 0 0 ∞ ! ∞ X 1 1 X 1 Z It now follows from Remark 2.4 that tr (M ) ≥ 1 , f 1 = fdm f p Aj p Aj m (A ) Z Z j=1 m (Aj) m (Aj) j=1 j Aj ∗ (f, ω)L2 (g, ω)L2 dµ (ω) = (f, i g)L2 = f (s) g (t) min (s, t) dsdt. ∞ Z 2 2 X 1 L [0,T ] ≥ εdm = ε · ∞ = ∞. m (Aj) A In particular taking f = δs and g = δt leads us to expect that j=1 j Z d Example 2.7. Part of the problem above was the non-compactness of Rd. X d d ω (s) ⊗ ω (t) dµ (ω) = min (s, t) ei ⊗ ei To avoid this issue, let us replace R be a T – the d – dimensional torus L2 d i=1 which we identify with [0, 2π] / v where v is the usual identification of the is a meaningful computation. In fact, Wiener [8, 7] proved in 1923 that endpoints. We will denote points in [0, 2π]d by θ and let dθ denote normalized µ (W ) = 1 where in·θ T Haar measure on d. In this case √ e is an orthonormal basis for T 2 2 knk +m d d n∈Z WT := ω ∈ C([0,T ] → R ): ω(0) = 0 . d H T equipped with the inner product; Example 2.6. Now consider the measure dµ (ϕ) of Remark 1.8. In this case Z 2 d 2 one can show that the measure never lives on L no matter the dimension. (f, g)H( d) := ∇f (θ) · ∇g (θ) + m f (θ) g (θ) dθ. R T d To see this let H be the Sobolev space of one derivative in L2. In this case we T −1 d 2 d have i∗ = −∆ + m2 . Indeed, if u := i∗g, then Hence if i : H T → L T is the inclusion map, then 2 Z in·θ 2 2 X e X 1 (f, g)L2 = (if, g)L2 = (f, u)H = ∇f · ∇u + m f · u dx ∀ f ∈ H, kik = = d H.S. q 2 R 2 knk + m2 n∈ d knk + m2 n∈ d Z 2 d Z which, by Elliptic regularity or by the Fourier transform, implies the distribu- L (T ) tion, ∆u, is an L2 – function and which is finite iff d = 1. When d = 2 the sum is logarithmically divergent 2 and it is worse when d = 3. This can also be understood by noting that (f, g) 2 = f, −∆ + m u . −1 L −∆ + m2 (θ − α) has the same singularity structure as Example 2.6 above. −1 2 d Hence we have −∆ + m2 u = g or i∗g = u = −∆ + m2 g as claimed. Hence we need to take K even bigger than L T – however only just barely ∞ 2 Informally, now, when d = 2. For example for any s ∈ R and f ∈ C T , Z s 2 ∗ 2−1 2 X 2 2 ˆ tr (i i ) = −∆ + m (x, x) dx = ∞. kfks := knk + m f (n) d 2 R n∈Z −1 −1 because when d ≥ 2, −∆ + m2 (x, x) = ∞ and or d = 1 −∆ + m2 (x, x) = ˆ R −in·θ 2 2 2 where f (n) := 2 f (θ) e dθ. So kfk0 = kfkL2( 2) , kfk1 = kfkH and for c > 0 for all x ∈ . To give a rigorous proof, notice that i i∗ is unitarily equiv- T T R any s < 0, kfk2 ≤ kfk2 . Let K denote the completion of C∞ 2 in the s – 2 2−1 s 0 s T alent to the the multiplication operator, k + m which has continuous norm. This is the Sobolev space of s – derivatives in L2. For any s < 0, the 2−1 spectrum, it follows that −∆ + m is not compact let alone trace class. In inclusion map, i : H = K1 → Ks is Hilbert Schmidt. Indeed, we now have general on non-atomic spaces with no infinite atoms any non-zero multiplication 2 operator is not trace class. Indeed, suppose M is the multiplication operator in·θ s f 2 X e X 1 2 2 2 on L (X, m) and observe that we may assume f ≥ 0 since |M | = M . Then kikH.S. = = 2 knk + m f |f| q 2 2 2 2 2 knk + m for some ε > 0 we will have µ (f ≥ ε) > 0. Since m is non-atomic, we may write n∈Z knk + m n∈Z Ks PATH INTEGRAL NOTES 9

X 1 d = < ∞. For x ∈ R and T > 0 let us define Wiener “measure,” on H (x, T ) by 1+|s| 2 2 2 ! n∈Z knk + m Z T 0 1 1 2 dµx,T (ω) := 0 exp − |ω˙ (t)| Dω. ZT 0 2 3. Semi-Group Property and the Operator Connection Lemma 3.1 (Heuristic). If 0 < t < T and f : H (x, t) → R, then Let V : d→ be a “nice” potential function and for x, y ∈ d and T > 0, R R R Z Z let 0 0 f ω|[0,t] dµx,T (ω) = f (ω) dµx,t (ω) . ( Z T ) H(x,T ) H(x,t) d 2 H (x, y, T ) := ω : [0,T ] → R : ω (0) = x, ω (T ) = y and |ω˙ (t)| dt < ∞ . More generally if g (ω) = G ω| is another function on paths, then 0 [0,T −t] Z 0 3.1. Heuristics. We begin with a heuristic discussion. We now define (infor- f ω|[0,t] g (ω (t + ·) − ω (t)) dµx,T (ω) H(x,T ) mally) the partition function by Z Z 0 0 = f (ω) dµx,t (ω) · g (ω (t + ·) − ω (t)) dµ0,T (ω) . Z Z T ! H(x,t) H(0,T −t) V 1 2 ZT (x, y) := exp − |ω˙ (t)| + V (ω (t)) dt Dω, Proof: Working informally and making use of Eq. (3.2) we have, H(x,y,T ) 0 2 Z Q 0 where we are thinking of Dω := t∈(0,T ) dm (ω (t)) where m is Lebesgue mea- f ω|[0,t] dµx,T (ω) H(x,T ) sure on Rd. We will also let Z Z t Z T ! ( Z T ) 1 1 2 1 2 d 2 = 0 f ω|[0,t] exp − |ω˙ (t)| exp − |ω˙ (t)| Dω H (x, T ) := ω : [0,T ] → R : ω (0) = x, and |ω˙ (t)| dt < ∞ ZT H(x,T ) 0 2 t 2 0 Z Z t 1 1 2 0 and = 0 f ω|[0,t] exp − |ω˙ (t)| ZT −tDω ZT H(x,t) 0 2 Z Z Z T ! V V 1 2 Z Z t ZT (x, y) := ZT (x, y) dy = exp − |ω˙ (t)| + V (ω (t)) dt Dω 1 1 2 d 2 = f ω|[0,t] exp − |ω˙ (t)| Dω R H(x,T ) 0 0 Zt H(x,t) 0 2 Q Z where now Dω := t∈(0,T ] dm (ω (t)) . If S > 0, then (informally) 0 = f (ω) dµx,t (ω) . Z Z S+T ! H(x,t) V 1 2 ZS+T (x, y) = exp − |ω˙ (t)| + V (ω (t)) dt Dω The “proof” of the second equation is left to the reader. Q.E.D. H(x,y,S+T ) 0 2 Z Definition 3.2 (Heuristic). For t > 0, let − R S ( 1 |ω˙ (t)|2+V (ω(t)))dt − R S+T ( 1 |ω˙ (t)|2+V (ω(t)))dt = e 0 2 e S 2 Dω Z H(x,y,S+T ) V 1 V Rt f (x) = 0 Zt (x, y) f (y) dy Z Z d V V t R = ZS (x, z) ZT (z, y) dz. Z d − R t V (ω(s))ds 0 R = f (ω (t)) e 0 dµx,t (ω) and H(x,t) Z Z − R t V (ω(s))ds 0 V V V = f (ω (t)) e 0 dµ (ω) , (3.1) ZS+T (x) = ZS (x, z) ZT (z) dz. x,T d H(x,T ) R 0 where the last expression is valid for any T > t. In the special case where V = 0,ZT (z) is independent of z and hence we will 0 0 simply write ZT for ZT (z) . It then follows from Eq. (3.1) that V 2 d 2 d Theorem 3.3 (Heuristic). Rt : L R , m → L R , m is a semi-group with 0 0 0 ˆ 1 (3.2) ZS+T = ZSZT . infinitesimal generator, H := − 2 ∆ + MV . 10 DRIVER

Proof: We have Therefore, Z V 1 V Z R f (x) = Z (x, y) f (y) dy − R t V (ω(s))ds 0 t+s 0 t+s e 0 f (ω (t + h)) dµ (ω) Zt+s d x,T R H(x,T ) 1 Z V V ∞ Z = 0 0 Zt (x, z) Zs (z, y) f (y) dydz X − R t V (ω(s))ds (n) n 0 Z Z d d 0 s t R ×R = e f (ω (t)) (∆ω ) dµx,T (ω) Z n=0 H(x,T ) 1 V V V V = 0 Zt (x, z) Rs f (z) dz = Rt Rs f (x) . ∞ Z d Z R t Z t R X − V (ω(s))ds (n) 0 n 0 = e 0 f (ω (t)) dµx,T (ω) · ω (h) dµ0,T (ω) Also, n=0 H(x,T ) H(0,T ) Z ∞ V − R t V (ω(s))ds 0 Z 0 X n lim Rt f (x) = lim f (ω (t)) e dµx,T (ω) = RV f (n) (x) · ω (h) dµ0 (ω) . t↓0 t↓0 H(x,T ) t 0,T H(0,T ) Z n=0 − R 0 V (ω(s))ds 0 = f (ω (0)) e 0 dµx,T (ω) = f (x) . R n 0 Informal Fact: If we let cn (h) := H(0,T ) ω (h) dµ0,T (ω) , then c0 (h) = 1, H(x,T ) P d V codd (h) = 0, c2 (h) = h j ej ⊗ ej and Now to compute Rt f (x) , we have dt Z d d n 0 n/2 RV f (x) = RV f (x) | |ω (h)| dµ0,T (ω) = Cn |h| . dt t ds t s=t H(0,T ) Z d − R t V (ω(s))ds 0 Therefore, = |s=t f (ω (s)) e 0 dµx,T (ω) ds H(x,T ) Z − R t V (ω(s))ds 0 Z 0 d − R s V (ω(s))ds 0 e f (ω (t + h)) dµx,T (ω) + |s=t f (ω (t)) e 0 dµx,T (ω) . H(x,T ) ds H(x,T ) h X = RV f (x) + RV f (2) (x) e ⊗ e + O h3/2 The second term gives, t 2 t j j j d Z R s − 0 V (ω(s))ds 0 |s=t f (ω (t)) e dµx,T (ω) V h V 3/2 ds H(x,T ) = Rt f (x) + Rt ∆f (x) + O h . Z 2 − R t V (ω(s))ds 0 = − f (ω (t)) V (ω (t)) e 0 dµx,T (ω) Diﬀerentiating this equation in h shows, H(x,T ) Z V d − R t V (ω(s))ds 0 1 V (3.3) = −Rt (V f)(x) . (3.4) |s=t f (ω (s)) e 0 dµx,T (ω) = Rt ∆f (x) . ds H(x,T ) 2 For the ﬁrst term, we will suppose that s = t + h with h > 0 and write ∆ω := ω (t + h) − ω (t) and Combining Eqs. (3.3) and (3.4) shows

0 1 00 2 3 d 1 f (ω (t + h)) = f (ω (t)) + f (ω (t)) ∆ω + f (ω (t)) ∆ω + O ∆ω . RV f (x) = −RV (V f)(x) + RV (∆f)(x) = RV Hfˆ (x) . 2 dt t t 2 t t Then using Lemma 3.1 we ﬁnd, Q.E.D. Z − R t V (ω(s))ds 0 e 0 g (ω (t)) u (∆ω) dµx,T (ω) H(x,T ) 3.2. Rigorous Interpretation of the above Heuristics. Z R t 0 − 0 V (ω(s))ds 0 Remark 3.4. If µ were concentrated on diﬀerentiable paths, we would have = e g (ω (t)) u (∆ω) dµx,T (ω) x,T H(x,T ) |ω (t + h) − ω (t)| ≤ C (ω) h and therefore would expect that Z Z − R t V (ω(s))ds 0 0 Z Z e 0 g (ω (t)) dµ0,T (ω) · u (ω (h)) dµ0,T (ω) . 2 0 2 2 0 H(x,T ) H(0,T ) |ω (t + h) − ω (t)| dµx,T ≤ h C (ω) dµx,T h. PATH INTEGRAL NOTES 11

d V If this were the case the above computation would have lead to dt Rt f (x) = transformation which takes, H0 to ωq·∇ which an Euler type operator and hence V d 0 easy to understand. We now carry this out in detail. (In fact what we are about −Rt (V f)(x) and in particular when V = 0 that dt Rt f (x) = 0. Thus we 0 to do was already done in Eq. (1.21) above and we could just use the results would have Rt f = f which would be absurd. derived there.) The rigorous interpretation of the following considerations is the following Let L = Bij∂i∂j with B as symmetric matrix to be chosen explicitly shortly. Feynman - Kac formula. Observe that L is the quantum mechanical Hamiltonian for a free particle. Then d Theorem 3.5 (Feynman - Kac Formula). Let V : R → R be a reasonable d tL/2 −tL/2 1 tL/2 −tL/2 ˆ 1 e (ωq · ∇) e = e [L, ωq · ∇] e potential bounded from below. Then, H := − 2 ∆ + V is essentially self-adjoint dt 2 and where Z Z t 1 ˆ [L, ωq · ∇] = B ωe · ∇∂ = ωBe · ∇∂ = ∂ ∂ . e−tH f (x) = f (x + ω (t)) exp − V (ω (s)) ds dµ (ω) 2 ij j i i i ωBei ei W 0 Therefore, where µ is Wiener measure on tL/2 −tL/2 e (ωq · ∇) e = ωq · ∇ + t∂ωBei ∂ei d W := ω ∈ C [0, ∞) → R : ω (0) = 0 . and hence it we choose B = ω−1, then etL/2 (ωq · ∇) e−tL/2 = ωq · ∇ + t∂ ∂ = ωq · ∇ + t∆ 4. Harmonic Oscillator Examples ei ei and therefore, 4.1. Finite Dimensional Case. Consider the following Newton’s equations tL/2 −tL/2 1 for a harmonic oscillator; e (H0) e = − ∆ + ωq · ∇ + t∆. 2 q¨(t) + Aq (t) = 0 Taking t = 2 then implies where A > 0 is a N × N matrix. The quantum Hamiltonian and its ground L −L e H0e = ωq · ∇ state and ground state energy, λ0, are given by: 1 1 or equivalently put Hˆ := − ∆ + (Aq, q) acting on L2 N , and −L L 2 2 R H0 = e ωq · ∇e −1 N r where L := ω ∂i∂j. Let {ξj} be an orthonormal basis of Eigenvectors of ω 4 π 1 ij j=1 Ω0 (q) := det exp − (ωq, q) −L α ω 2 (or equivalently A) with ωξj = ωjξj. Now suppose that hα (x) = e ξ where α N α 1 we are writing ξ to be the polynomial function on R defined by ξ (q) := with Hˆ Ω0 = λ0Ω0 where λ0 = tr (ω) . QN αj 2 j=1 (ξj, q) . Then √ where ω := A. We now pass to the ground state representation by introducing ωq · ∇ = (q, ξj) ωξj · ∇ = (q, ξj) ωjξj · ∇ = ωj (q, ξj) ∂ξj the Hilbert space so that r 2 N 2 2 N π α α α−ej α K = K := L , Ω (q) dq = L , det exp (− (ωq, q)) dq ωq · ∇ξ = ωj (q, ξj) ∂ξj ξ = ωj (q, ξj) αjξ (q) = ωjαjξ (q) ω R 0 R ω = (ω · α) ξα. which is unitarily equivalent to L2 N via, U : L2 N → K defined by R R Hence it follows that Uf := Ω−1f. Let 0 −L L −L α −L α H0hα = e ωq · ∇e hα = e ωq · ∇ξ = e (ω · α) ξ = (ω · α) hα. ˆ −1 1 H0 = U H − λ0 U = − ∆ + ∇ ln Ω0 · Ω0 Thus we have now found a total orthonormal subset of eigenvectors for H , 2 0 1 {hα} with corresponding eigenvalues, σ Hˆ0 = {ω · α} . The spec- = − ∆ + ωq · ∇. a∈N0 α∈N0 2 trum of Hˆ is thus given by This last expression is the quantization of the function of the form p2/2±iωq·p. 1 Under the action of the free Newtonian dynamics, q → q + tp and p → p and σ Hˆ = ω · α + tr (ω) . 2 therefore the term ωq·p → ω (q + tp)·p. Hence we may expect to find a similarity α∈N0 12 DRIVER

4.2. An Infinite Dimensional Example. Let us now go back to the path The goal now is to make sense out of the following expression integral expressions like that in Eq. (1.18). Here we will consider the related 1 − R ∞ 1 |ω0(t)|2+V (ω(t)) dt (5.1) dµ(ω) = e −∞( 2 ) Dω, “measure,” Z Z ! where Dω is “Lebesgue measure”, Z is a normalization constant chosen so that 1 1 2 dµ (ϕ) = exp − |∇ϕ| + m2ϕ2 dx Dϕ. µ becomes probability measure. Notice that 1 |ω0(t)|2 + V (ω(t)) is the classical Z 2 1 2 [0,T ]×S energy of a system with mass m = 1 in subject to a the force −∇V. By viewing ϕ (t, θ) as a function of t with values in L2 S1 , the above expression To make sense of Eq. (5.1), let us begin by truncating the time interval. To may be written as this end let T > 0 and α and β be two probability densities (or more generally ! ! measures) on RN . Let begin by considering the informal expression 1 Z kϕ˙ (t)k2 (4.1) dµ (ϕ) = exp − + V (ϕ (t)) dt Dϕ 1 R T 1 0 2 − −T ( 2 |ω (t)| +V (ω(t)))dt Z [0,T ] 2 (5.2) dµT (ω) = e α(ω(−T ))β(ω(T ))Dω, ZT where kϕk2 := R ϕ2 (θ) dθ and where now S1 Z − R T 1 |ω0(t)|2+V (ω(t)) dt 2 2 2 2 2 Z = e −T ( 2 ) α(ω(−T ))β(ω(T ))Dω. V (f) := k∂θfk + m kfk = −∂ + m f, f . T θ N C([−T,T ]→R ) Thus it is reasonable to associate µ to the the elliptic differential operator, The first observation we should make is that (at least formally) in Eq. (5.1) ˆ 1 and (5.2) may replace V by V − λ0 without changing the measure µT and µ. (4.2) H := − ∆L2(S1) + MV 2 So we now assume this has been done and we will rename V − λ0 by V. Let T N which is the Hamiltonian associated to an infinite dimensional quantum me- W := C([−T,T ] → R ), P = {−T = t0 < t1 < ··· < tn = T } be a partition chanical oscillator. Using the considerations developed above, we have of [−T,T ] and T T 00 2 2 1 WP = {ω ∈ W : ω (t) = 0 for all t∈ / P}. A = −∂θ + m , and λ0 = tr (A) = ∞. 2 T Given a function F : W → R, let Hence we conclude, formally, that Hˆ ≥ ∞I which is of course nonsense. We (5.3) ˆ 1 Z R T 1 0 2 must renormalize H to take care off this by subtracting of this ground state P − −T ( 2 |ω (t)| +V (ω(t)))dt µT (F ) := P e F (ω)α(ω(−T ))β(ω(T ))dλP (ω) Z T energy. (The idea here is that the potential energy is only defined up to an T WP additive constant and hence subtracting λ0 (albeit ∞) from Hˆ does not change P P where as usual ZT is a normalization constant so as to make µT a probability the physics being described.) It is also to our benefit to go to the ground T measure. Let νT denote Wiener measure on W with initial distribution α and state representation where we take the quantum mechanical Hilbert space to T end point distribution given by β. For ω ∈ W , let ωP denote the unique path be, informally, T P in W such that ωP = ω on P. Then we may write µ as ! P T q − R T V (ω (t))dt 2 1 2 R −T P L H (S) , dν (f) := “ exp (−∂ + m2)f, f Df” . T e F (ωP )dνT (ω) 1/2 0 θ (5.4) µP (F ) := W . Z L2(S) T R T R − −T V (ωP (t))dt W T e dνT (ω) Again this has to be made sense of using the considerations introduced above. To get a rigorous interpretation of Eq. (5.2), let us pass to the limit |P| → 0 in There is of course much more to be said, but it will not be done here and now. Eq. (5.4) to find for continuous and bounded F that − R T V (ω(t))dt 5. R −T Path Integral Quantization P W T e F (ω)dνT (ω) (5.5) µT (F ) = lim µT (F ) = R T . N |P|→0 R − −T V (ω(t))dt Suppose that V : R → R is a smooth potential such that V (x) → ∞ as W T e dνT (ω) 1 |x| → ∞. Let H0 := − ∆ + V and λ0 denoted the lowest eigenvalue of H0 2 Suppose that F (ω) = f(ω(s1), ω(s2), . . . , ω(sk)) where −T < s1 < s2 < and Ω0 be the corresponding eigenfunction which we may assume is strictly ··· < sk < T. Making use of the Feynman Kac formula, we have that positive. The renormalized Hamiltonian is now defined to be H = H0 − λ0I, so that H ≥ 0, 0 = inf σ(H) and HΩ0 = 0 with Ω0 > 0. µT (F ) = PATH INTEGRAL NOTES 13

R Qk+1 H N k+2 f(x1, x2, . . . , xk)α(x0)β(xk+1) i=1 p∆ s(xi−1, xi) dx0dx1 . . . dxk+1 6. Gross’ QFT Notes Introduction = (R ) i R Qk+1 H N k+2 α(x0)β(xk+1) p (xi−1, xi) dx0dx1 . . . dxk+1 (These are notes of Leonard Gross which were originally written sometime (R ) i=1 ∆is R Qk+1 H around 1975.) The purpose of these notes is to show how the quantization of a N k+2 f(x1, x2, . . . , xk)α(x0)β(xk+1) i=1 p∆ s(xi−1, xi) dx0dx1 . . . dxk+1 = (R ) i , classical field leads to the study of an elliptic differential operator in infinitely (α, e−2TH β) many variables. We will see here, in a simple instance, how such a differential H operator arises as the Hamiltonian operator for the quantized field. The fol- where s0 = −T, sk+1 = T, ∆is := si − si−1, and pt (x, y) is the integral kernel −tH −τH lowing exposition is distilled from the many similar discussions in the physics of e . Making use of the fact that e α → (α, Ω0)Ω0 as τ → ∞, we find using the previous equation that literature. See e.g [vN,p. ] or [D, p].

µ(F ) = lim µT (F ) T →∞ 6.1. The harmonic oscillator. Recall that Newton’s equation, F = ma, re- " R # duces, in the case of a harmonic oscillator, to (α, Ω0)(β, Ω0) N k f(x1, x2, . . . , xk) (R ) Qk H (6.1) − kx = mx¨ ×Ω0(x1)Ω0(xk) i=2 p∆ s(xi−1, xi) dx1dx2 . . . dxk = i (α, Ω0)(β, Ω0) where x is the oscillator position,x ˙ = dx/dt, and k is the spring constant. The k 2 k force F is −grad V , with V = x . The energy of the oscillator is therefore Z Y 2 = f(x , x , . . . , x )Ω (x )Ω (x ) pH (x , x ) dx dx . . . dx . 1 2 k 0 1 0 k ∆is i−1 i 1 2 k N k 1 2 k 2 (R ) i=2 E = mx˙ + x 2 2 The conclusion of the above computations may be summarized as follows. or Theorem 5.1. The informal expression in Eq. (5.1) should satisfy 1 k Z (6.2) E = p2 + x2, 1 − R ∞ 1 |ω0(t)|2+V (ω(t)) dt F (ω)e −∞( 2 ) Dω 2m 2 Z W where p = mv = mx˙ is the momentum. k Z Y Quantization of this system yields the Hamiltonian = f(x , x , . . . , x )Ω (x )Ω (x ) pH (x , x ) dx dx . . . dx 1 2 k 0 1 0 k ∆is i−1 i 1 2 k N k+2 2 (R ) i=2 1 d k 2 (6.3) H = − 2 + x , where F (ω) = f(ω(s1), ω(s2), . . . , ω(sk)) and s1 < s2 < ··· < sk. In particular 2m dx 2 we have that which is to be interpreted as a self–adjoint operator on L2((−∞, ∞); dx) (and Z Z 1 ∂ 1 − R ∞ 1 |ω0(t)|2+V (ω(t)) dt which is obtained from (6.2) by the usual substitution p → , x → mult. −∞( 2 ) 2 i ∂x f(ω(0))e Dω = f(x)Ω0(x)dx Z N by x). W R so that the time 0 distribution of µ has a density equal to the square of the ground state of H. Moreover, 6.1.1. n harmonic oscillators. Similarly the Hamiltonian for a system consisting of n independent harmonic oscillators of masses m and spring constants k may Z j j 1 − R ∞ 1 |ω0(t)|2+V (ω(t)) dt f(ω(0))g(ω(t))e −∞( 2 ) Dω be derived from the corresponding Newton’s equations, Z W 2 Z d xj H (6.4) mj 2 = −kjxj j = 1, . . . , n. = f(x)g(y)Ω0(x)Ω0(y)pt (x, y)dxdy dt N N R ×R −tH The Hamiltonian is given by = (fΩ0, e (gΩ0)). n 2 ! Diﬀerentiating this equation at t = 0 gives, X 1 ∂ kj 2 (6.5) H = − 2 + xj . Z 2mj ∂x 2 d 1 − R ∞ ( 1 |ω0(t)|2+V (ω(t)))dt j=1 j (fΩ0,H(gΩ0)) = − |0 f(ω(0))g(ω(t))e −∞ 2 Dω dt Z W H is to be interpreted as a self–adjoint operator on a suitable domain in from which we recover the Hamiltonian of the system. 2 n n ∞ n L (R ; d x) (e.g., it may be deﬁned as the closure of its restriction to Cc (R )). 14 DRIVER

6.2. A quantized field; informalities. We will show now how the quanti- Of course u(x, t) is determined by the two sequences {qj(0)} and {q˙j(0)} zation of a classical (non-interacting) field may be regarded as just an infinite which, in view of (6.8), are determined by the initial values u(x, 0) andu ˙(x, 0). dimensional version of the preceding procedure. For simplicity we consider a Comparison of equations (6.9) with (6.4) shows that the boundary value prob- single component u of the electromagnetic field. We will assume there are no lem (6.6), (6.7) is equivalent to a mechanical system consisting of infinitely charges or currents present. I.e., it is not interacting with anything. Then many harmonic oscillators with masses mj all equal to one and spring constants ∂2u k = j2. The canonical coordinates are {q }∞ and {p = m q˙ =q ˙ }∞ . Ac- the field u(x, t) satisfies the homogeneous wave equation ∆u = ∂t2 . Whereas j j j=1 j j j j j=1 quantum mechanics arises from quantizing an ordinary differential equation, cording to the principles of quantum mechanics one should quantize this system namely Newton’s equation, quantum field theory arises from quantizing a par- by taking the Hilbert space to be tial differential equation (the field equation). In order to keep the exposition 2 ∞ (6.10) K = L (R , dq1dq2 ··· ) very elementary at this stage we will reduce the number of space dimensions to one in the preceding field equation, thereby obtaining the equation and the Hamiltonian to be (in analogy to (6.5)) ∞ ! 2 2 2 ∂ u ∂ u X 1 ∂ 2 2 (6.6) = , (6.11) H = − + ω q 2 2 2 ∂q2 j ∂t ∂x j=1 j which is the equation for a vibrating string. Moreover the main ideas will not where ωj = j. be altered if we consider the string only on the interval (0, π) and assume it is This definition of K is meaningless, however, because of the appearance of fixed at the endpoints. I.e., infinite dimensional Lebesgue measure. Further, the expression for H contains (6.7) u(0, t) = 0 u(π, t) = 0 for all t. problems of its own, as we will see. If Q denotes the quantized version of q (informally Q is multiplication by In accordance with the method of separation of variables a general solution of j j j q on K) the principles of quantum mechanics assert, in view of (6.8), that the (6.6) and (6.7) can be written in the form j quantized field ϕ at time zero is given by ∞ X ∞ (6.8) u(x, t) = qj(t)uj(x) X ϕ(x, 0) = Qjuj(x). j=1 j=1 where uj(x) and qj(t) each satisfy ordinary differential equations and uj is zero The field is therefore an operator valued function. Actually this series of oper- at the endpoints. Specifically we may take ators does not converge for each x. However, as is known, the series converges r 2 in the sense of distributions to an operator valued distribution ϕ(f, 0) which uj(x) = sin jx j = 1, 2,... corresponds informally to π Z π which is a normalized eigenfunction for d2/dx2 with Dirichlet boundary condi- ϕ(x, 0)f(x)dx tions and with eigenvalue −j2. That is, one has ∂2u /∂x2 = −j2u . So the 0 j j where f is a test function. functions q satisfy j In order to rescue the preceding discussion of the quantized field we return d2q briefly to the harmonic oscillator. (6.9) j = −j2q j = 1, 2,.... dt2 j 6.3. Ground state transformation. All of this follows from (6.8), the orthonormality of the uj and the equations 6.3.1. One oscillator. Consider a single harmonic oscillator with m = 1 and 2 ∞ ∂ u X 2 k = ω2. Its Hamiltonian is = −j qj(t)uj(x), ∂x2 2 j=1 0 1 d 1 2 2 2 (6.12) H = − 2 + ω x on L ((−∞, ∞); dx). ∞ 2 dx 2 ∂2u X = q¨j(t)uj(x). Let ∂t2 1/4 −ωx2/2 j=1 ψ(x) = (ω/π) e . in accordance with the usual procedure of the method of separation of variables. It is straightforward to verify that PATH INTEGRAL NOTES 15

R ∞ 2 ˆ 2 ˆ (1) −∞ ψ(x) dx = 1 and Clearly (Hf, f) ≥ 0. Moreover the constant functions are in L (µ) and H1 = 0. (2) H0ψ = (ω/2)ψ So inf(spectrum Hˆ ) = 0. Of course this just reflects the fact that inf(spectrum 0 So ψ is a normalized eigenfunction for H0 and, since ψ is positive, ω/2 is (H − ω/2)) = 0, which we already knew. 0 0 the lowest eigenvalue of H and ψ is the ground state of H . We now apply 6.3.2. n oscillators. Let us return now to a system of n independent harmonic JACOBI’S GROUND STATE TRANSFORMATION. Let 2 oscillators all of mass one and spring constants ωj . Specializing (6.5) to this (6.13) dµ(x) = ψ(x)2dx. case we find the Hamiltonian to be n Then µ is a probability measure on . Define X 2 2 2 2 2 n R (6.18) Hn = (1/2) (−∂ /∂xj + ωj xj ) acting in L (R , dx) 2 n 2 n U : L (R , dx) → L (R , µ) j=1 Let by 1/4 −ω2x2/2 ψj(xj) = (ωj/π) e j j (Uf)(x) = f(x)/ψ(x). and U is clearly unitary. Moreover one can compute that n ψ(x) = Πj=1ψj(xj), x = (x1, . . . , xn) (6.14) U(H0 − ω/2)U −1 = (1/2)(−d2/dx2 + ωxd/dx). Then clearly   The computation can be carried out easily by observing that (U −1g)(x) = n 1 X g(x)ψ(x) so that (6.19) H ψ = ω ψ n 2 j j=1 (H0U −1g)(x) = (1/2)(−d2/dx2 + ω2x2)(g(x)ψ(x)) because H is a sum of independent harmonic oscillator Hamiltonians. So ψ is = (ω/2)g(x)ψ(x) − (1/2)g0(x)ψ0(x) − (1/2)g00(x)ψ(x) n the ground state of Hn and the “zero point energy” (i.e. inf spectrum H) is 0 00 = [(ω/2)g + (1/2)ωxg (x) − (1/2)g (x)]ψ(x), n X which proves (6.14). (6.20) En ≡ (1/2) ωj. Define j=1 Define (6.15) Hˆ = (1/2) −d2/dx2 + ωxd/dx on L2( , µ) R n 2 n (6.21) dµn(x) = Πj=1[ψj(x) dxj] on R So n Then µn is a probability measure on R and the map f → Uf = f/ψ is again (6.16) Hˆ = U(H0 − ω/2)U −1 2 n 2 n unitary from L (R , dx) onto L (R , µn). Just as for a single oscillator one sees The virtue of unitarily transforming H0 − ω/2 to Hˆ is that Hˆ happens to be that n the operator associated to the Dirichlet form for µ. To be somewhat precise X (6.22) U(H − E )U −1 = (1/2) (−∂2/∂x2 + ω x ∂/∂x ). about this note that an integration by parts gives n n j j j j j=1 Z Z f 0(x)g0(x)dµ(x) = f 0(x)g0(x)ψ(x)2dx So define n R R X Z (6.23) Hˆ = (1/2) (−∂2/∂x2 + ω x ∂/∂x ) acting in L2( n, µ ). = (−f 00(x) + ωxf 0(x))g(x)ψ(x)2dx n j j j j R n j=1 R ˆ = 2(Hf, g)L2(µ). Then −1 ˆ So we have the identity (6.24) UHnU = Hn + En. Z ˆ ˆ 0 0 Hn differs from the sum of the first n terms in (6.11) in two ways. First, it (6.17) (Hf, g)L2(µ) = (1/2) f (x)g (x)dµ(x) incorporates the subtraction of the zero point energy En, which does not affect R 16 DRIVER the physics. And second, it incorporates the unitary transform U, which also and easy mechanism to do this because Dirichlet forms in finite and infinite does not affect the physics. dimensions are well understood. Let us try to understand better, but at an informal level, the infinite product 6.4. Back to the quantized fields. We can see now that the operator H measure µ defined in (6.25). Write Dq = Π∞ dq for Lebesgue “measure” on ∞ j=1 j “defined” in (6.10) is bounded below by (1/2) P ω , which diverges because ∞ j=1 j R . Then (6.25) gives ωj = j in that field theory. So we can say, informally , that H ≥ +∞, which P∞ 2 ∞ 1/2 − j=1 ωj qj shows that H is at best meaningless. The customary resolution of this problem (6.29) dµ(x) = [Πj=1(ωj/π) ][e ]Dq. goes like this. Any potential is defined only up to an additive constant because In the case of primary interest to us we have ω = j. So the first of the three only grad V has direct physical meaning. Thus there is no change in the physics j factors in (6.29) is infinite. The second factor happens to be zero a.e. with if we subtract the infinite constant (1/2) P∞ ω from the operator H. j=1 j respect to µ. And the third factor is meaningless (though there have been some We may therefore attempt to give the infinite sum in (6.11) a mathematical attempts to give a useful interpretation to Dq.) It is therefore particularly interpretation by passing to the limit n → ∞ in the equivalent system (6.23) edifying that the product of all three factors makes perfectly good sense as a To this end note first that the measure µ in (6.21) is just a product of n measure on R∞ probability measures, an infinite product of which is a perfectly respectable At time t = 0 the equation (6.8) reads measure. Define ∞ ∞ 2 ∞ X (6.25) dµ(q) = Πj=1[ψj(qj) dqj] on R (6.30) v(x) ≡ u(x, 0) = qjuj(x). Let j=1 ∞ Now denote by B the nonnegative self-adjoint operator on L2((0, 2π)) which is ˆ X 2 2 2 2 2 (6.26) H = (1/2) (−∂ /∂qj + ωjxj∂/∂qj) the square root of the Dirichlet Laplacian. Thus B = −d /dx with Dirichlet j=1 boundary conditions. Then ˆ Since there are no zeroth order terms in H (unlike (6.18)) the infinite sum makes (6.31) Buj = juj. perfectly good sense when applied to many functions on the infinite product ∞ Hence space. For example if f : R → C depends on only finitely many coordinates, say f(q) = g(q , . . . , q ), with g ∈ C∞( ∞), then there are only finitely many ∞ 1 n c R X 2 (6.32) (Bv, v) 2 = jq . nonzero terms in the sum for Hfˆ . Such functions are dense in L2 (Rn, µ) . So L ((0,2π)) j Hˆ is densely defined by the formula (6.26). Thus by subtracting the infinite j=1 zero point energy from (6.11) and unitarily transforming to the ground state Moreover, since the qj are Cartesian coordinates in the Hilbert space measure µ we have given a meaning to (6.11) as a densely defined operator in Re L2((0, 2π)), (6.29) may be written in terms of the initial data v instead 2 n L (R , µ). of the qjs. So, writing Dv for Lebesgue “measure” instead of Dq, we may write The operator Hˆ is particularly nicely related to the measure µ. Just as in −(Bv,v) (6.33) dµ(v) = Z−1e L2((0,2π)) Dv, (6.17) the operator Hˆn in (6.23) clearly satisfies Z where Z−1 is the normalization constant (which is actually +∞.) Let us trans- ˆ 2 (6.27) (Hnf, g)L (µn) = (1/2) (∇f · ∇g)dµ form this informal notation for the well defined measure µ once more, taking ∞ R into account the fact that (Bv, v) = kB1/2vk2, which is exactly the Sobolev Similarly the field theoretic Hamiltonian satisfies norm for the Sobolev space H1/2((0, 2π)) (with Dirichlet boundary conditions.) ∞ Thus we may write Z X ˆ 2 n (6.28) (Hf, g)L (R ,µ) = (1/2) (∂f/∂qj)(∂g/∂qj)dµ(x) 2 ∞ −1 −kvkH R j=1 (6.34) dµ(v) = Z e 1/2 Dv If one wished to pursue the functional analytic questions left untouched in this The informal extension of the preceding discussion to higher dimensions, with discussion, such as whether Hˆ actually has a self-adjoint version in L2(µ), the or without a box, is straight forward. The important thing to keep in mind is description of Hˆ as the operator associated to a Dirichlet form offers a quick that the quadratic form in the exponent in (6.29) comes from the square root PATH INTEGRAL NOTES 17

2 1/2 of the operator whose eigenvalues were ωj . For example if we begin with the Similarly the change of coordinates xj → sj = ωj xj changes the measure in Klein-Gordon equation n (6.21) into γ ≡ γ × · · · × γ (n times) and the Hamiltonian Hˆn in (6.23) into 2 2 2 3 (6.35) ∂ u/∂t = −(m − ∆)u on × n ( 2 ) R R X ∂ ∂ (6.40) H˜ = (1/2) ω − + s . n j ∂s2 j ∂s as our classical field equation then the preceding informal discussion suggests j=1 j j that the ground state measure for the quantized field should be (informally) There is a virtue in using the form (6.38) because it has the convenient expression 2 1/2 n −1 −((m −∆) v,v)L2( 3) (6.36) dν(v) = Z e R Dv X ∗ ωjaj aj where Dv is Lebesgue “measure” on Re L2(R3). Further, the preceding discus- j=1 sion suggests that the Hamiltonian should be determined by 2 n n in terms of the annihilation operators aj in L (R , γ ). However when this Z procedure is used on the coordinates qj in (6.30) the new coordinates are no 2 2 3 2 (6.37) (Hf, g)L (ν) = (1/2) (∇f(v), ∇g(v))L (R )dν(v). longer orthonormal coordinates in L ((0, 2π)) and the Sobolev space H1/2 gets 2 3 Re L (R ) lost. Nevertheless this change of coordinates is convenient for other purposes Gaussian measures such as (6.36) are well understood. One knows that in and is frequently made in the name of convenience. See e.g. [Von Neumann’s actuality the measure ν lives as a countably additive probability measure on book, page 265], [Dirac’s book p. 136], [Bjorken and Drell, Vol.2 page 8], [Jauch “large” sets such as Re S0(R3), but not on “small” sets such as Re L2(R3). In and Rohrlich, page 35 Equation 2-62], [Boguliubov and Shirkov, page 32] any case let us note that the determining norm for the Gauss measure is the We have developed now the Hamiltonian for a field which does not interact Sobolev H1/2 norm since with charges, currents or other sources. When such sources or self–interactions are present the Hamiltonian H0 will be modified by the presence of an addi- 2 1/2 2 1/4 2 2 ((m − ∆) v, v) 2 3 = k(m − ∆) vk 2 3 = kuk . tional term V which is usually a multiplication operator. Thus the theory of L (R ) L (R ) H1/2 interacting fields is the theory of elliptic differential operators H in infinitely The appearance of the H1/2 norm in this heuristic discussion is consistent with many variables, of which a Schrödinger operator the (rigorous) appearance of its dual norm H−1/2 in Euclidean quantum field theory for the space of test functions [Book by Glimm and Jaffe], [Book Eu- H = H0 + V clidean QFT, by Barry Simon]. See also e.g. L. Gross’s paper in “Functional in infinitely many dimensions is typical. integration and its applications” Ed. A.M. Arthurs (1974) Oxford U. Press (Cumberland Lodge Conference) for further discussion of the connection at a rigorous level. 7. Appendix – Some Wiener Space Results Let W = {ω ∈ C([0, 1] → R): ω(0) = 0} and let H denote the set of functions 6.4.1. Comparison with physics literature. In comparing the previous discussion R 1 0 2 h ∈ W which are absolutely continuous and satisfy (h, h) = 0 |h (s)| ds < ∞. with the standard treatments in physics texts one should be aware of a change The space H is called the Cameron-Martin space and is a Hilbert space when of coordinates that is very often made that may obscure the comparison. Let equipped with the inner product

1 2 Z 1 (6.38) dγ(s) = √ e−s ds. 0 0 π (h, k)H = h (s)k (s)ds for all h, k ∈ H. 0 Then γ((−∞, ∞)) = 1. Referring back to the measure µ in (6.13) one sees that The space W is a Banach space when equipped with the sup-norm, the change of coordinates x → s = ω1/2x transforms the measure µ into γ. kfk = max |f(x)| . Moreover the Hamiltonian Hˆ of (6.15) is transformed into x∈[0,1] d2 d Proposition 7.1. Let G(s, t) = min(s, t). Then G is the reproducing kernel (6.39) H˜ = (1/2)ω − + s . ds2 ds for H, i.e. (G(s, ·), h) = h(s) for all s ∈ [0, 1]. 18 DRIVER

Proof: The proof follows from the fundamental theorem of calculus for so µ is a Gaussian and q(`ξ, `n) = (ξ, n)`2 . absolutely continuous functions, 8.1. Extras on `2 – Gaussian Measures. Z 1 0 (G(s, ·), h) = 1t≤sh (t)dt = h(s). Proposition 8.2. Deﬁne for ` ∈ X∗, 0 Z Q.E.D. x` ≡ `(x)x dµ(x). ∞ Corollary 7.2. Let {hn}n=1 be any Orthonormal. basis for H. Then ∞ R P R ∞ If ` = `ξ, then x` = ξixix dµ(x) and hence (x`)i = Σξixixj dµ(x) = ξi. X X i=1 hn(s)hn(t) = G(s, t). ∗ J 2 2 n=1 So x`ξ = ξ, and X −→ X, J`ξ := ξ ∈ ` and (ξ, η) = q(`ξ, `n) = (ξ, n)` . (·,·) Proof: The proof is simply Bessel’s equality, Therefore J(X∗) = `2 ∞ ∞ X X Question: What happened to the Hilbert space `2 that µ was modeled on. hn(s)hn(t) = (G(s, ·), hn)(G(t, ·), hn) ∞ ∞ P 2 n=1 n=1 Some answers: Let a ∈ ` such that ai < ∞, ai > 0. For w ∈ ` (a) we i=1 = (G(s, ·),G(t, ·)) = G(s, t). 2 have `(x) = (w, x)a is deﬁned on ` (a). Now

Q.E.D. N N ∞ i P w x a − 1 P a2w2 − 1 P a2w2 i`(x) i i i 2 i i 2 i i Proposition 7.3 (A Sobolev Theorem). The inclusion map i : H → W is Ee = lim Ee i=1 = lim e 1 = e 1 N→∞ N→∞ continuous and in fact ∞ ∞ Notice P `(e )2 = P a2w2 khkW ≤ khkH for all h ∈ H. i i i i=1 i=1 1 i` − (`,`) ∗ 2 Proof: By Proposition 7.1, for s ∈ [0, 1], (1) Therefore E[e ] = e 2 H and H = ` has reappeared. 2 2 (2) Let again `(x) = (w, x) , w ∈ ` (a), then E(`(x) ) = (`, `) ∗ . |h(s)| = |(G(s, ·), h)H | ≤ kG(s, ·)kH khkH . a H (3) Consider properties of µ under translations by a ∈ RN. This proves the Proposition since dµ(x+a) − x·a+ 1 |a|2 Formally: = e ( 2 ). Z 1 dµ(x) 2 2 2 P 2 kG(s, ·)kH = (1t≤s) dt = s ≤ 1. (4) `(x) = (w, x) with w ∈ ` , let `N (x) = wnxn. Then E`N = 0 n≤N P 2 Q.E.D. wn, more generally n≤N

8. After thoughts? 2 X 2 E(`M − `N ) = wn → 0 as M,N → ∞. ∞ P N

X when f = 1A then f(x − a) = 1A(x − a) = 1A+a(x). PATH INTEGRAL NOTES 19

So (8.1) is equivalent to Ψ(x) = x Heaviside(x)−(2x−1) Heaviside(x−1/2)+(x−1) Heaviside(x−1)

na(x) Z Z k −zx}|· a{− 1 |a|2 Proof: Let β := {1} ∪ ψkj : 0 ≤ k and 0 ≤ j < 2 be the Haar basis for (8.2) f(x − a)dµ(x) = f(x)e 2 dµ(x) 2 R t L ([0, 1], dλ) introduced in Exercise ?? above and let Ψ0(t) := 0 1ds = t for all bounded measurable f. It suﬃce to check (8.2) for f(x) = eα·x, α ∈ ` . R t k ˜ fin and Ψkj(t) := 0 ψkj(s)ds for 0 ≤ k and 0 ≤ j < 2 . Then β = {Ψ0} ∪ Now k Z Ψkj : 0 ≤ k and 0 ≤ j < 2 is an orthonormal basis for H which satisﬁes, α·(x−a) −α·a + 1 |α|2 e dµ(x) = e e 2 , 1 while k/2 −(k+1) −k/2 Z kΨ0k∞ = 1 and kΨkjk∞ = 2 2 = 2 . α·x −x·a− 1 |a|2 1 |α−a|2− 1 |a|2 −α·a+ 1 |α|2 2 e e 2 dµ(x) = e 2 2 = e 2

Q.E.D. The following pictures shows the graphs of Ψ00, Ψ1,0, Ψ1,1, Ψ2,1, Ψ2,2 and Ψ2,3 respectively. 2 Theorem 8.4. If a ∈ RN \ ` , then µa ⊥ µ.

Proof: First notice that: µa ⊥ µ iﬀ kµa − µk = 2. Let us compute kµa − µk 2 making use of the notation, z := e−x·a/2+|a| /4 : Z Z −x·a+ 1 |a|2 2 kµa − µk = e − 1 dµ = |z − 1| |z + 1|dµ Z Z Z ≥ |z − 1|2dµ = (z2 − 2z + 1)dµ = 2(1 − zdµ).

Now Plot of Ψ0, 0. Z Z − 1 (x·a+ 1 |a|2) zdµ = e 2 2 dµ Z −x· a − 1 |a|2 = e 2 dµ e 4

1 | a |2− 1 |a|2 − 1 |a|2 = e 2 2 4 = e 8 . Therefore for a ∈ `2, − 1 |a|2 2 kµa − µk ≥ 2(1 − e 8 ) = 2 if a∈ / ` .

Q.E.D. Plot of Ψ10. Plot of Ψ11. 9. Classical Wiener Measure Because of Proposition 7.3, the transpose itr : W ∗ → H∗ is continuous as ∗ well. Hence by the Riesz theorem, for each φ ∈ W there exists a unique hφ ∈ H ∗ such that (hφ, ·)H = φ|H . Hence we may deﬁne an inner product q on W by ∗ q(φ, ψ) = (hφ, hψ) for all φ, ψ ∈ W . Theorem 9.1. There exists a unique (Gaussian) measure µ on W such that Z (9.1) eiφ(x)dµ(x) = e−q(φ)/2, W Plot of Ψ20. Plot of Ψ21. where q(φ, ψ) = (hφ, hψ) as described above. 20 DRIVER

Proof: To prove this reﬁnement let α ∈ [0, 1] and consider X Γk(t) := xjΨkj(t) j<2k

where {xj}j<2k are a ﬁxed sequence of real numbers. By the mean value theo- k/2 rem and using the disjoint supports of {Ψkj} and the fact that Ψ˙ kj = 2 , ∞ we ﬁnd k/2 Plot of Ψ22. Plot of Ψ23. |Γk(t) − Γk(s)| ≤ M2 |t − s| k k Because Ψkj : 0 ≤ k and 0 ≤ j < 2 have disjoint supports, where M := max xj : j < 2 . Hence

1 |Γ (t) − Γ (s)|   p k k k/2 1−α α ≤ 2 M |t − s| . X 1 − k k 1 − k X p |t − s| max xjΨkj(t) ≤ 2 2 max |xj| : j < 2 ≤ 2 2  |xj|  t 2 2 −k j<2k j<2k If |t − s| ≤ 2 this gives for any p ∈ [1, ∞). Integrating this inequality, using H¨older’sinequality, shows |Γk(t) − Γk(s)| k/2 −k(1−α) −k( 1 −α) ≤ 2 2 M = 2 2 M that |t − s|α

1 k   p −k 1 − 2 Z Z If |t − s| ≥ 2 , recalling kΓkk∞ ≤ 2 2 M,we have X 1 − k X p xkjΨkj(t) dµ(x) ≤ 2 2  |xkj| dµ(x) 2 |Γk(t) − Γk(s)| kα 1 − k −k( 1 −α) j<2k j<2k α ≤ 22 2 2 M = 2 2 M. ∞ |t − s| 2 1 k k − 2 p 1 = 2 2 Cp −k( 2 −α) 2 That is to say, kΓkkα ≤ 2 M and working as above this implies 1 where   p 2 − 1 x Z 1 1 Z Z e 2 X X X −k( −α) X p p xkjΨkj(t) dµ(x) ≤ 2 2  |xkj| dµ(x) Cp = |x| √ dx < ∞. 2 2π k k k k j<2 α j<2 R X 1 −k( 1 −α) k Therefore = C 2 2 2 p < ∞ 2 p k Z ∞ X X 1 X −( 1 − 1 )k xkjΨkj(·) dµ(x) ≤ Cp 2 2 p < ∞ provided that Z k k k=1 1 1 j<2 ∞ − α − > 0 2 p providing p > 2. Letting or α < 1/2 − 1/p. Since we may take p as large as we like it follows that   X X lim kS(x) − SN (x)kα = 0 for µ – a.e. x SN =  xkjΨkj(·) N→∞ k≤N j<2k and therefore S(x) ∈ C0,α([0, 1], R) for µ – a.e. x. Q.E.D. we have it follows that S(x) := k·k –lim S (x) exists for µ – a.e. x. This ∞ N→∞ N Lemma 9.3. Let {b } be a one dimensional Brownian motion. Then for completes the proof. Q.E.D. s s≥0 any T > 0 and λ ∈ [0, π2/ 4T 2); Remark 9.2. We can use the above argument to prove more. Namely that " Z T !# √ Wiener measure lives on the space of paths which are α – H¨oldercontinuous λ 2 −1/2 (9.2) E exp bsds = cos λT < ∞. provided α < 1/2. 2 0 PATH INTEGRAL NOTES 21

∞ (All we really need for the qualitative result is to observe that, using Fernique’s Rigorous Proof. Let {Nk}k=1 be a sequence of i.i.d. normal random ∞ theorem, variables and {hk}k=1 be an orthonormal basis for H = H ([0, 1] , R) . Then " !# (see Proposition ?? below) λ Z T λT 2 Z 1 λT 2 Z 1 exp b2ds = exp b2ds = 1 + b2ds + O λ2T 4 ∞ E 2 s E 2 s 2 E s X 0 0 0 bs := Nkhk (s) 2 Z 1 2 k=1 λT 2 4 λT 2 4 = 1 + sds + O λ T = 1 + + O λ T 1 2 0 4 R 2 is a standard one dimensional Brownian motion. Letting ak := 0 hk (s) ds, we for sufficiently small λT 2.) then have Z 1 ∞ 2 X 2 Proof: When T = 1, simply follow the proof of [5, Eq. (6.9) on p. 472] with bsds = Nk ak 0 λ replaced by −λ. For general T > 0 observe that k=1 and therefore that Z T Z 1 Z 1 b2ds = T b2 dt =d T 2 b2dt Z 1 " ∞ !# s tT t λ 2 λ X 2 0 0 0 E exp bsds = E exp Nk ak 2 0 2 and therefore, k=1 ∞ ∞ " !# Y λ 2 Y 1 λ Z T λT 2 Z 1 √ (9.3) = E exp akN = √ . 2 2 −1/2 2 2 1 − λa E exp bsds = E exp bsds = cos λT k=1 k=1 k 2 0 2 0 provided that λ < 1/ak for all k. Assuming this restriction on λ, the latter provided that λ ∈ [0, π2/ 4T 2 . P∞ product is finite iff k=1 ak < ∞. But Heuristic Proof. ∞ ∞ Z 1 Z 1 ∞ Z 1 1 1 X X 2 X 2 λ Z 1 Z 1 Z ak = h (s) ds = h (s) ds = sds = 1/2 < ∞. exp b2ds = exp − x˙ 2 (t) − λx2 (t) dt Dx k k E 2 s Z 2 k=1 k=1 0 0 k=1 0 0 H([0,1],R) 0 1 Z 1 Z 1 A calculus of variation exercises show = exp − −D2x (t) − λx (t) x (t) dt Dx Z 2 R 1 h2 (s) ds H([0,1],R) 0 0 4 sup 1 = 2 2 h6=0 R ˙ 2 π 2 d 0 h (s) ds where D = dt2 with Dirichlet boundary conditions at 0 and Neumann 2 2 2 boundary conditions at 1. The eigenfunctions for D are proportional to and therefore ak ≤ 4/π for all k and the condition on λ becomes λ < π /4. 2 2 h 1 i {sin (lπx/2)}l∈2 −1 with eigenvalues being −l π /4. Therefore, λ R 2 N To get the more explicit expression for E exp 2 0 bsds let us observe 1 dim H/2 that Z 1 Z (2π) √ exp − x˙ 2 (t) − λx2 (t) dt Dx = ( ) 2 p 2 2 2 π H([0,1],R) 0 det (−D − λ) hl (t) = sin l t s πl 2 Y 2π l odd = n √ o l2π2/4 − λ ˙ π is an orthonormal basis for H. Indeed, hl (t) = 2 cos l 2 t is the or- l odd l odd 2 2 and Z is this expression with λ = 0. Hence we find, thonormal basis of eigenfunctions associated to d /dt with Neumann boundary conditions at 0 and Dirichlet boundary conditions at 1. With this choice of s !−1/2 λ Z 1 l2π2/4 4λ basis we find 2 Y Y Z 1 E exp bsds = = 1 − 8 π 4 2 l2π2/4 − λ l2π2 a = sin2 l t dt = 0 l odd l odd l 2 2 2 2 π l 0 2 π l provided that 4λ < π2. which combined with Eq. (9.3) gives Eq. (??). Q.E.D. 22 DRIVER

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Department of Mathematics, 0112, University of California, San Diego, La Jolla, CA 92093-0112 E-mail address: [email protected]