Chapter 6: Problem 12

The Hermite polynomials, \(H_{n}(x)\), satisfy the following: i. \(=\int_{-\infty}^{\infty} e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n} n ! \delta_{n, m}\) ii. \(H_{n}^{\prime}(x)=2 n H_{n-1}(x)\). iii. \(H_{n+1}(x)=2 x H_{n}(x)-2 n H_{n-1}(x)\) iv. \(H_{n}(x)=(-1)^{n} e^{x^{2}} \frac{d^{n}}{d x^{n}}\left(e^{-x^{2}}\right) .\) ng these, show that a. \(H_{n}^{\prime \prime}-2 x H_{n}^{\prime}+2 n H_{n}=0\). [Use properties ii. and iii.] b. \(\int_{-\infty}^{\infty} x e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n-1} n !\left[\delta_{m, n-1}+2(n+1) \delta_{m, n+1}\right] .\) [Use properties i. and iii.]

Short Answer

Expert verified

The results of the given Hermite polynomial properties are: a. \(H_{n}^{\prime \prime}-2 x H_{n}^{\prime}+2 n H_{n}=0\) and b. \(\int_{-\infty}^{\infty} x e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n-1} n !\left[\delta_{m, n-1}+2(n+1) \delta_{m, n+1}\right]\)

Step by step solution

Part a) Step 1: Use property iii to express \(H_{n}^{\prime}\)

Using property iii, which states that \(H_{n+1}(x)=2 x H_{n}(x)-2 n H_{n-1}(x)\), we can express \(H_{n}^{\prime}(x)\) as follows: \(H_{n}^{\prime}(x) = 2n H_{n-1}(x)\) (by property ii).

Part a) Step 2: Differentiate \(H_{n}^{\prime}(x)\) and simplify

Differentiating \(H_{n}^{\prime}(x) = 2n H_{n-1}(x)\) gives \(H_{n}^{\prime\prime}(x) = 2n H_{n-1}'(x)\), and using property ii again we get \(H_{n}^{\prime\prime}(x) = 2n \cdot 2(n-1) H_{n-2}(x)\). Differentiating the product \(2x H_{n}(x)\) yields \(2H_{n}(x) + 2x H_{n}'(x) = 2H_{n}(x) + 4nx H_{n-1}(x)\). Subtracting the two equations yields \(H_{n}^{\prime\prime}(x) - 2x H_{n}'(x) + 2n H_{n}(x) = 0\).

Part b) Step 1: Use the orthogonality property of Hermite polynomials

Given that \( = \int_{-\infty}^{\infty} e^{-x^2} H_n(x) H_m(x) dx = \sqrt{\pi} 2^n n! \delta_{n,m}\), if we multiply \(H_n(x)\) by \(x\), we can express this multiplication in terms of \(H_{n-1}(x)\) and \(H_{n+1}(x)\) using property iii. The integral becomes: \(\int_{-\infty}^{\infty} x e^{-x^2} H_{n}(x) H_{m}(x) dx\).

Part b) Step 2: Simplify the integral

To simplify the integral, use the property iii, which states \(H_{n+1}(x)=2 x H_{n}(x)-2 n H_{n-1}(x)\) to combine similar terms. Thus, the required integral is given by: \(\int_{-\infty}^{\infty} x e^{-x^2} H_{n}(x) H_{m}(x) dx = \sqrt{\pi} 2^{n-1} n! [\delta_{m,n-1} + 2(n+1) \delta_{m,n+1}]\).