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Normalizing Hermite Polynomials
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[QUOTE="rmiller70015, post: 5756072, member: 281740"] [h2]Homework Statement [/h2] Evaluate the normalization integral in (22.15). Hint: Use (22.12) for one of the $H_n(x)$ factors, integrate by parts, and use (22.17a); then use your result repeatedly.[h2]Homework Equations[/h2] (22.15) ##\int_{-\infty}^{\infty}e^{-x^2}H_n(x)H_m(x)dx = \sqrt{\pi}2^nn!## when ##n=m## (22.12) ##H_n(x)=(-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}## (22.17a) ##H'_n(x)=2nH_{n-1}(x)## [h2]The Attempt at a Solution[/h2] I started out by writing my integral from (22.15) as: ##\int_{-\infty}^{\infty}e^{-x^2}[H_n(x)]^2dx## Expanding one of the ##H_n(x)## terms gives: ##\int_{-\infty}^{\infty}e^{-x^2}H_n(x)e^{x^2}(-1)^n\frac{d^n}{dx^n}e^{-x^2}dx## ##=(-1)^n\int_{-\infty}^{\infty}H_n(x)\frac{d^n}{dx^n}e^{-x^2}dx## Integration by parts and using equation 22.17a gives: ##(-1)^n[H_n(x)e^{-x^2}|^{\infty}_{-\infty}-\int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx## The first term goes to zero and leaves: ##(-1)^n\int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx## IBP n-1 more times gives: ##(-1)^n2^nn!\int^{\infty}_{-\infty}H_0(x)e^{-x^2}dx=(-1)^n2^nn!\int^{\infty}_{-\infty}e^{-x^2}dx = (-1)^n2^nn!\sqrt{\pi}## This is correct except I have this extra factor of ##(-1)^n## that shouldn't be there, I'm not sure how to get rid of it. [/QUOTE]
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Normalizing Hermite Polynomials
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