Hermite functions. Jordan Bell Department of Mathematics, University of Toronto. September 9, 2015

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1 Hermite functions Jordan Bell Department of Mathematics, University of Toronto September 9, 25 Locally convex spaces If V is a vector space and {p α : α A} is a separating family of seminorms on V, then there is a unique topology with which V is a locally convex space and such that the collection of finite intersections of sets of the form {v V : p α (v < ɛ}, α A, ɛ > is a local base at. We call this the topology induced by the family of seminorms. If {p n : n } is a separating family of seminorms, then d(v, w = 2 n p n (v w, v, w V, + p n (v w is a metric on V that induces the same topology as the family of seminorms. If d is a complete metric, then V is called a Fréchet space. 2 Schwartz functions For φ C (, C and n, let p n (φ = sup k n u sup( + u 2 n/2 φ (k (u. We define S to be the set of those φ C (, C such that p n (φ < for all n. S is a complex vector space and each p n is a norm, and because each p n is a norm, a fortiori {p n : n } is a separating family of seminorms. With the topology induced by this family of seminorms, S is a Fréchet space. 2 As well, D : S S defined by (Dφ(x = φ (x, x Theorem and Theorem 4. 2 Walter udin, Functional Analysis, second ed., p. 84, Theorem 7.4.

2 and M : S S defined by (Mφ(x = xφ(x, x are continuous linear maps. 3 Hermite functions Let λ be Lebesgue measure on and let (f, g L 2 = fgdλ. With this inner product, L 2 (λ is a separable Hilbert space. We write f 2 L = (f, f 2 L 2 = f 2 dλ. For n, define H n : by H n (x = ( n e x2 D n e x2, which is a polynomial of degree n. H n are called Hermite polynomials. It can be shown that For m, n, exp(2zx z 2 = For n, define h n : by n! H n(xz n, z C. ( H m (xh n (xe x2 dλ(x = 2 n n! πδ m,n. h n (x = (2 n n! π /2 e x2 /2 H n (x = ( n (2 n n! π /2 e x2 /2 D n e x2. h n are called Hermite functions. Then for m, n, (h m, h n L 2 = h m (xh n (xdλ(x = δ m,n. One proves that {h n : n } is an orthonormal basis for (L 2 (λ, (, L 2. 3 We remind ourselves that for x, 4 5. e x2 = 2 π /2 e y2 /4 e ixy dy, 3 Theorem Lemma 2

3 and by the dominated convergence theorem this yields D n e x2 = 2 π /2 ( iy n e y2 /4 e ixy dy, and so h n (x = (2 n n! π /2 e x2 /2 2 π /2 4 Mehler s formula We now prove Mehler s formula for the Hermite functions. 5 (iy n e y2 /4 e ixy dy. (2 Theorem (Mehler s formula. For z C with z < and for x, y, h n (xh n (yz n = π /2 ( z 2 /2 2 + z2 exp ( z 2 (x2 + y 2 + 2z z 2 xy. Proof. Using (2, = h n (xh n (yz n ( ( π 2 2 n n! e(x +y 2 /2 z n (2πiξ n e π2 ξ 2 e 2πixξ dξ (2πiζ n e π2 ζ 2 e 2πiyζ dζ = πe (x2 +y 2 /2 e π2 ξ 2 π 2 ζ 2 2πixξ 2πiζy ( 2π 2 ξζz n dξdζ n! = πe (x2 +y 2 /2 e π2 ξ 2 π 2 ζ 2 2πixξ 2πiζy e 2π2ξζz dξdζ. Now, writing a = iy π + ξz, we calculate e π2 ζ 2 2πiζy 2π 2 ξζz dζ = e π2 (ζ+a 2 +π 2 a 2 dζ = π e π2 a 2 = π exp ( y 2 + 2πiyξz + π 2 ξ 2 z 2. 5 Sundaram Thangavelu, An Introduction to the Uncertainty Principle: Hardy s Theorem on Lie Groups, p. 8, Proposition.2.. 3

4 Then, for α = ( z 2 π 2, h n (xh n (yz n =e (x2 +y 2 /2 =e (x2 y 2 /2 =e (x2 y 2 /2 e π2 ξ 2 2πixξ y 2 +2πiyξz+π 2 ξ 2 z 2 dξ e αξ2 2πi(x yzξ dξ π α exp ( π2 (x yz2 α ( =π /2 e (x2 y 2 /2 ( z 2 /2 exp =π /2 ( z 2 /2 exp ( x2 ( =π /2 ( z 2 /2 exp (x yz2 z 2 z 2 + 2xyz z 2 y2 z 2 z 2 + x2 2 y2 2 + z 2 2 z 2 (x2 + y 2 + 2z z 2 xy. 5 The Hermite operator We define A : S S by, (Aφ(x = φ (x + (x 2 + φ(x, x, A = D 2 + M 2 +, which is a continuous linear map S S, which we call the Hermite operator. S is a dense linear subspace of the Hilbert space L 2 (λ, and A : S S is a linear map, so A is a densely defined operator in L 2 (λ. For φ, ψ S, integrating by parts, (Aφ, ψ L 2 = ( φ (x + (x 2 + φ(xψ(xdλ(x = φ (xψ(xdλ(x + (x 2 + φ(xψ(xdλ(x = φ(xψ (xdλ(x + (x 2 + φ(xψ(xdλ(x = (φ, Aψ L 2, showing that A : S S is symmetric. Furthermore, also integrating by parts, (Aφ, φ L 2 = (φ (xφ (x + (x 2 + φ(xφ(xdλ(x, 4

5 so A is a positive operator. It is straightforward to check that each h n belongs to S. For n, we calculate that h n(x + (2n + x 2 h n (x =, and hence (Ah n (x = (2n + x 2 h n (x + x 2 h n (x + h n (x = (2n + 2h n (x, Ah n = (2n + 2h n. Therefore, for each h n, A h n = 2n+2 h n, and it follows that there is a unique bounded linear operator T : L 2 (λ L 2 (λ such that 6 The operator norm of T is T h n = A h n = (2n + 2 h n, n. (3 T = sup n 2n + 2 = 2. The Hermite functions are an orthonormal basis for L 2 (λ, so for f L 2 (λ, f = (f, h n L 2h n. For f, g L 2 (λ, ( (T f, g L 2 = (f, h n L 2T h n, (g, h n L 2h n ( = (f, h n L 2(2n + 2 h n, (g, h n L 2h n = (2n + 2 (f, h n L 2(g, h n L 2, from which it is immediate that T is self-adjoint. For p, T p h n 2 L 2 = (2n + 2 p h n 2 L 2 = (2n + 2 2p h n 2 L 2 = (2n + 2 2p. Therefore for p, T p h n 2 L = (2n + 2 2p = 2 2p m 2p = 2 2p ζ(2p. 2 This means that for p, T p is a Hilbert-Schmidt operator with Hilbert- Schmidt norm 7 T p HS = 2 p ζ(2p. m= 6 Theorem L 2 L 2 5

6 6 Creation and annihilation operators Taking the derivative of ( with respect to x gives 2 n! H n(xz n+ = n! H n(xz n, so H = and for n, n! H n(x = (n! 2H n (x, and so H n = 2nH n, h n(x = (2n /2 h n (x xh n (x, Dh n = (2n /2 h n Mh n. Furthermore, from its definition we calculate h n(x = xh n (x (2n + 2 /2 h n+ (x, Dh n = Mh n (2n + 2 /2 h n+. We define B : S S, called the annihilation operator, by (Bφ(x = φ (x + xφ(x, x, B = D + M, which is a continuous linear map S S. For n, we calculate Bh n = (2n /2 h n, and h (x = π /4 e x2 /2, so Bh =. We define C : S S, called the creation operator, by (Cφ(x = φ (x + xφ(x, x, C = D + M, which is a continuous linear map S S. For n, we calculate Thus, For φ S, Furthermore, and Ch n = (2n + 2 /2 h n+. h n = (2 n n! /2 C n h = π /4 (2 n n! /2 C n (e x2 /2. (4 B C = 2D. BC = D 2 + M 2 + = A CB = D 2 + M 2 = A 2. 6

7 7 The Fourier transform Define F : S S, for φ S, by (F φ(ξ = ˆφ(ξ = φ(xe iξx dx, (2π /2 ξ. For ξ, by the dominated convergence theorem we have in other words, ˆφ(ξ + h lim ˆφ(ξ = ( ixφ(xe iξx dx h h (2π,, /2 xφ(x(ξ = i D ˆφ(ξ = id ˆφ(ξ, F (Mφ = id(f φ. (5 Also, by the dominated convergence theorem we obtain in other words, For φ S, φ(x = Dφ(ξ = iξ ˆφ(ξ, F (Dφ = im(f φ. (6 ˆφ(ξe ixξ dξ, x. (7 (2π /2 φ ˆφ is an isomorphism of locally convex spaces S S. 8 Using (7 and the Cauchy-Schwarz inequality φ ( + ξ 2 /2 ( + ξ 2 /2 ˆφ(ξ dξ (2π /2 ( /2 ( /2 (2π /2 ( + ξ 2 dξ ( + ξ 2 ˆφ(ξ 2 dξ ( /2 = 2 /2 ( + ξ 2 dξ ˆφ(ξ 2, and using (6 and the fact that ˆφ L 2 = φ L 2, φ 2 2 ˆφ(ξ 2 dξ + 2 = 2 ˆφ(ξ 2 dξ + 2 = 2 φ 2 L φ 2 L 2, ξ 2 ˆφ(ξ 2 dξ (F φ (ξ 2 dξ 8 Walter udin, Functional Analysis, second ed., p. 86, Theorem

8 and therefore φ 2 /2 ( φ L 2 + φ L 2. (8 We remind ourselves that A = D 2 + M 2 +, B = D + M, C = D + M. Using F D = imf, DF = i F M, we get F A = F ( D 2 + M 2 + = (imf D + (idf M + F = im(imf + id(idf + F = M 2 F D 2 F + F = AF, and and F B = F (D + M = imf + idf = ibf F C = F ( D + M = imf + idf = icf. We now determine the Fourier transform of the Hermite functions. Theorem 2. For n, F h n = ( i n h n. Proof. For n, by induction, from F C = icf we get F C n = ( ic n F. From (4, h n = π /4 (2 n n! /2 C n (e x2 /2. Writing g(x = e x2 /2, it is a fact that F g = g, and using this with the above yields F h n = π /4 (2 n n! /2 F C n g = π /4 (2 n n! /2 ( ic n F g = π /4 (2 n n! /2 ( ic n g = π /4 (2 n n! /2 ( i n π /4 (2 n n! /2 h n = ( i n h n. 8

9 There is a unique Hilbert space isomorphism F : L 2 (λ L 2 (λ such that F f = ˆf for all f S. 9 For f L 2 (λ, f = (f, h n L 2h n, and then F f = (f, h n L 2F h n = (f, h n L 2( i n h n. 8 Asymptotics For x =, ( reads thus n! H n(z n = exp( z 2 = ( z 2 n, n! H 2n ( = ( n (2n!, H 2n+ ( =. n! Similarly, taking the derivative of ( with respect to x yields H 2n( =, H 2n+( n (2n +! = 2(. n! For u(x = e x2 /2 H n (x, u (x = xu+e x2 /2 H n(x, u (x = u xu xe x2 /2 H n(x+e x2 /2 H n(x. Using we get and thence H n(x = 2xH n (x H n+ (x, H n(x = 2nH n (x H n(x 2xH n(x + 2nH n (x =, u = u + x 2 u 2nu. Thus, writing f(x = x 2 u(x, u satisfies the initial value problem v + (2n + v = f, v( = H n (, v ( = H n(. (9 Now, for λ >, two linearly independent solutions of v +λv = are v (x = cos(λ /2 x and v 2 (x = sin(λ /2 x. The Wronskian of (v, v 2 is W = λ /2, and 9 Walter udin, Functional Analysis, second ed., p. 88, Theorem 7.9. N. N. Lebedev, Special Functions and Their Applications, p. 66,

10 using variation of parameters, if v satisfies v +λv = g then there are c, c 2 such that v(x = c v + c 2 v 2 + Av + Bv 2, where x A(x = x W v 2(tg(tdt, B(x = W v (tg(tdt. We calculate that the unique solution of the initial value problem v + λv = g, v( = a, v ( = b, is v(x = av (x + bλ /2 v 2 (x λ /2 v (x x v 2 (tg(tdt + λ /2 v 2 (x x v (tg(tdt = a cos(λ /2 x + bλ /2 sin(λ /2 x x + λ /2 (cos(λ /2 t sin(λ /2 x sin(λ /2 t cos(λ /2 xg(tdt x = a cos(λ /2 x + bλ /2 sin(λ /2 x + λ /2 sin(λ /2 (x tg(tdt. Therefore the unique solution of the initial value problem (9 is v(x = H n ( cos((2n + /2 x + H n((2n + /2 sin((2n + /2 x x + (2n + /2 sin((2n + /2 (x t t 2 u(tdt, where u(x = e x2 /2 H n (x. If n = 2k then We calculate v(x = ( k (2k! cos((4k + /2 x k! x + (4k + /2 sin((4k + /2 (x t t 2 u(tdt = ( k (2k! k! cos((4k + /2 x + (4k + /2 r 2k (x. ( ( x x r 2k (x 2 t 4 dt u(t 2 dt x 5 e t2 H 2k (t 2 dt = x 5 22k (2k! π, /4 x 5/2 r 2k (x π 2 k (2k!.

11 By Stirling s approximation, 2 k (2k! (2k! k! Thus for α 2k = (2k! k!, = 2k k! 2 k (2πk /2 k k e k (2k! ((4πk /2 (2k 2k e 2k = /2 π/4 k /4. r 2k (x α 2k = O( x 5/2 k /4 k /2 = O( x 5/2 k /4. Thangavelu states the following inequality and asymptotics without proof, and refers to Szegő and Muckenhoupt. Lemma 3. There are γ, C, ɛ > such that for N = 2n +, and h n (x C(N /3 + x 2 N /4, x 2 2N Ce γx2, x 2 > 2N, h n (x N /8 (x N /2 /4 e ɛn /4 (x N /2 3/2 for N /2 + N /6 x (2N /2. Lemma 4. For N = 2n +, x N 2 N 6, and θ = arccos(xn 2, h n (x = ( /2 2 (N x 2 /4 cos π ( N(2θ sin θ π Theorem 5.. h n p n 2p 4 for p < h n p n 8 log n for p = h n p n 6p 2 for 4 < p. 4 +O(N /2 (N x 2 7/4. ather than taking the pth power of h n, one can instead take the pth power of H n and integrate this with respect to Gaussian measure. Writing dγ(x = (2π /2 e x2 /2 dx and taking H n to be the Hermite polynomial that is monic, now write H n p p = H n p dγ. Larsson-Cohn 2 proves that for < p < 2 there is an explicit c(p such that H n p = c(p n /4 n!( + O(n, Sundaram Thangavelu, Lectures on Hermite and Laguerre Expansions, pp , Lemma.5. and Lemma.5.2; Gábor Szegő, Orthogonal Polynomials; Benjamin Muckenhoupt, Mean convergence of Hermite and Laguerre series. II, Trans. Amer. Math. Soc. 47 (97, , Lemma 5. 2 Lars Larsson-Cohn, L p -norms of Hermite polynomials and an extremal problem on Wiener chaos, Ark. Mat. 4 (22,

12 and for 2 < p < there is an explicit c(p such that H n p = c(p n /4 n!(p n/2 ( + O(n. This uses the asymptotic expansion of Plancherel and otach. 3 3 M. Plancherel and W. otach, Sur les valeurs asymptotiques des polynomes d Hermite, Commentarii mathematici Helvetici (929,