MATH 105B EXTRA/REVIEW TOPIC: HERMITE POLYNOMIALS PROF. MICHAEL VANVALKENBURGH Schr¨odinger'sequation for a quantum harmonic oscillator is 1 d2 − + x2 u = u. 2 dx2 This says u is an eigenfunction with eigenvalue . In physics, represents an energy value. Let 1 d a = p x + the \annihilation" operator, 2 dx and 1 d ay = p x − the \creation" operator. 2 dx Note that, as an operator, d2 aya + aay = − + x2; dx2 which we can use to rewrite Schr¨odinger'sequation. Fact. The function −1=4 −x2=2 u0(x) = π e represents a quantum particle in its \ground state." (Its square is the probability density function of a particle in its lowest energy state.) Note that au0(x) = 0; which justifies calling the operator a the \annihilation" operator! On the other hand, y u1(x) : = a u0(x) 2 = π−1=42−1=2(2x)e−x =2 represents a quantum particle in its next highest energy state. So the operator ay \creates" one unit of energy! 1 MATH 105B EXTRA/REVIEW TOPIC: HERMITE POLYNOMIALS 2 Each application of ay creates a unit of energy, and we get the \Hermite functions" −1=2 y n un(x) : = (n!) (a ) u0(x) −1=4 −n=2 −1=2 −x2=2 = π 2 (n!) Hn(x)e ; where Hn(x) is a polynomial of order n called \the nth Hermite polynomial." In particular, H0(x) = 1 and H1(x) = 2x. Fact: Hn satisfies the ODE d2 d − 2x + 2n H = 0: dx2 dx n For review, let's use the power series method: We look for Hn of the form 1 X k Hn(x) = Akx : k=0 We plug in and compare coefficients to find: A2 = −nA0 2(k − n) A = A for k ≥ 1: k+2 (k + 2)(k + 1) k Note how there is a natural split into odd and even terms, similar to what happened for Legendre's equation in x5.2. We can use this recurrence relation to find formulas for the Hn, and compare with wikipedia. [Note: Our Hn are sometimes called \physicists' Hermite polynomials." Physicists use different conventions than probabilists.] Now consider the Sturm-Liouville problem d 2 dH 2 e−x + λe−x H = 0: dx dx with the boundary condition that H not grow faster than a polynomial as x ! ±∞. Fact. When λ = n is a non-negative integer, the Hermite polynomial Hn(x) is a solution. Then Sturm-Liouville theory, with p = e−x2 , q = 0, and r = e−x2 , gives orthogonality: Z 1 ( −x2 0 when n 6= m e Hn(x)Hm(x) dx = −∞ 6= 0 when n = m. This can be rewritten in terms of the un: ( Z 1 0 when n 6= m un(x)um(x) dx = −∞ 1 when n = m. We chose the constants above (π−1=4, for example) so that we get = 1. MATH 105B EXTRA/REVIEW TOPIC: HERMITE POLYNOMIALS 3 We can use this orthogonality to find the Fourier-Hermite representation of a given function f: 1 X f(x) = Anun(x): n=0 This is useful because the Fourier transform acts very simply on Hermite functions: n F[un](w) = (−i) un(w): That is, the Hermite function un is an eigenfunction of the Fourier transform, with eigen- value (−i)n..