Math 118, Handout 4: Hermite functions and the Fourier transform. n! where D = d/dx, are a basis of eigenfunctions for the Fourier transform

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1 The Hermite functions defined by h n (x) = ( )n e x2 /2 D n e x2 where D = d/dx, are a basis of eigenfunctions for the Fourier transform f(k) = f(x)e ikx dx on L 2 (R). Since h 0 (x) = e x2 /2 h (x) = 2xe x2 /2 h 2 (x) = 2 (4x2 2)e x2 /2 h 3 (x) = 6 (8x3 24x)e x2 /2 we see that h n (x) = H n(x)e x2 /2 where H n (x) is the classical Hermite polynomial of degree n Theorem Proof: By the product rule, h n(x) = ( )n Theorem 2 H n (x) = ( ) n e x2 D n e x2 h n(x) = xh n (x) (n + )h n+ (x) xe x2 /2 D n e x2 + ( )n e x2 /2 D n+ e x2 = xh n (x) (n+)h n+ (x) h n(x) = xh n (x) + 2h n (x) Proof: First observe that D n satisfies a commutation relation xd n f(x) D n (xf(x)) = nd n f(x)

2 because of Leibniz Rule for the nth derivative of a product fg. Hence h n(x) = xh n (x) + ( )n e x2 /2 D n ( 2xe x2 ) = xh n (x) + 2h n (x). (Subtraction gives (n + )h n+ = 2xh n 2h n.) These two recurrence relations are called ladder relations in quantum mechanics, and constitute a factorization of the operator since Thus we have proved Theorem 3: D 2 x 2 + I = (D x)(d + x) (D x)(d + x)h n (x) = 2(D x)h n = 2nh n (x). h n(x) x 2 h n (x) = (2n + )h n (x). Exercise H4.: Show that K = D 2 x 2 is a symmetric operator on L 2 (R): for nice smooth functions f, g L 2 (R) we have f(x)kg(x) dx =< f, Kg >=< Kf, g >. Hence h n are eigenfunctions of the symmetric operator K with distinct eigenvalues (2n + ). As a standard consequence, they are orthogonal: (2n + ) < h n, h m >=< Kh n, h m >= (2m + ) < h n, h m > so < h n, h m >= 0 for n m. We will show that they are eigenfunctions of the Fourier transform as well in Theorem 4 ĥ n (k) = h n (k). Proof: For n = 0, this says the Fourier transform of e x2 /2 is e k2 /2 which we proved by direct calculation. For n =, we have h (x) = 2h 0(x). Since the 2

3 Fourier transform of f is ik ˆf(k), we have ĥ(k) = 2ikĥ0(k) = 2ikh 0 (k) = ih (k). For n >, we compute by differentiating under the integral that ĥ n(k) = and by integration by parts that Thus addition gives and kĥn(k) = ( ixh n (x))e ikx dx ( ih n(x))e ikx dx ĥ n(k) + kĥn(k) = 2iĥn (k) ĥ n(k) kĥn(k) = i(n + )ĥn+(k) Subtracting to eliminate the derivatives and solving for ĥn+, i n+ (n + )ĥn+(k) = 2xi n ĥ n (k) 2i n ĥ n (k) so that i n ĥ n (k) satisfies the same two-term recurrence relation as h n (x). Since the initial values also match, the theorem is proved by induction. Thus h n are eigenfunctions of the Fourier transform. They span L 2 (R) by Theorem 5: Any smooth function f L 2 (R) which is orthogonal to h n for every n must be 0. Proof: If < f, h n >= 0 for every n then y n < f, h n >= 0 for all y R. The Taylor series for e (x y)2 is so ( y) n e (x y)2 = D n e x2 = e x2 /2 y n h n (x), 0 = y n < f, h n >= e x2 /2 e (x y)2 f(x)dx = e y2 h 0 f(2y). 3

4 Hence h 0 f = 0 and taking Fourier transforms gives ĥ0(k) ˆf(k) = h 0 (k) ˆf(k) = 0. Since h 0 never vanishes we must have ˆf = 0 and by Parseval s identity we must have f = 0. Thus we have a basis of eigenfunctions for the Fourier transform. Once we evaluate their norms h n, we can define and compute the Fourier transform by the standard eigen-expansions and f(x) = ˆf(k) = h n < f, h 2 n > h n (x) = h n < f, h 2 n > h n (k) = At least formally we thus have e ikx = h n 2 h n(k)h n (x). h n 2 h n(x)h n (y)f(y)dy h n 2 h n(k)h n (x)f()dx Exercise H4.2: Show that (Hint: Square the expansion h n 2 = π 2n. y n h n (x) = e x2 /2 e (x y)2 and integrate.) Exercise H4.3: them to compute Exercise H3.4: Calculate the first three Hermite polynomials and use (a) Show that x 2 e x2 dx. < Kf, f >= f (x) 2 + x 2 f(x) 2 dx = (2n + ) < f, h n > 2 h n 2 4

5 for real-valued f L 2 (R). (b) Prove the weak Heisenberg inequality for such f. f (x) 2 + x 2 f(x) 2 dx f(x) 2 dx Exercise H3.5: Show that e 2its = e t2 (Hint: Seek an expansion of the form and use orthogonality of the H n s.) (it) n H n (s) e 2its = f n (t)h n (s) Exercise H3.6: Use Cramer s inequality H n (s).09 2 n/2 e s2 /2 and Stirling s approximation to show that the error in N terms of the approximation in H3.5 is bounded by e 2its N f n (t)h n (s) 0 ( ) 2e N/2 N for N > 0, t, and s 2. How many terms are required to get 0-digit accuracy? 5