Introduction to Quantum Mechanics · Problem 2.16

Time-Independent Schrödinger Equation — David J. Griffiths

Physics — Quantum Mechanics · hermite_polynomial_properties

Problem

In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials. (a) The Rodrigues formula says that Hn(ξ)=(−1)neξ2(ddξ)ne−ξ2.(2.87)H_n(\xi) = (-1)^n e^{\xi^2} \left(\frac{d}{d\xi}\right)^n e^{-\xi^2}. \tag{2.87} Use it to derive H3H_3 and H4H_4. (b) The following recursion relation gives you Hn+1H_{n+1} in terms of the two preceding Hermite polynomials: Hn+1(ξ)=2ξHn(ξ)−2nHn−1(ξ).(2.88)H_{n+1}(\xi) = 2\xi H_n(\xi) - 2n H_{n-1}(\xi). \tag{2.88} Use it, together with your answer in (a), to obtain H5H_5 and H6H_6. (c) If you differentiate an nnth-order polynomial, you get a polynomial of order (n−1)(n-1). For the Hermite polynomials, in fact, dHndξ=2nHn−1(ξ).(2.89)\frac{dH_n}{d\xi} = 2n H_{n-1}(\xi). \tag{2.89} Check this, by differentiating H5H_5 and H6H_6. (d) Hn(ξ)H_n(\xi) is the nnth zz-derivative, at z=0z = 0, of the generating function exp⁡(−z2+2zξ)\exp(-z^2 + 2z\xi); or, to put it another way, it is the coefficient of zn/n!z^n/n! in the Taylor series expansion for this function: e−z2+2zξ=∑n=0∞znn!Hn(ξ).(2.90)e^{-z^2+2z\xi} = \sum_{n=0}^{\infty} \frac{z^n}{n!} H_n(\xi). \tag{2.90} Use this to obtain H1H_1, H2H_2, and H3H_3.

Browse the problem library