Can you help me prove the integral for Hermite polynomials?

In summary, the conversation discusses solving an integral involving the hermite polynomial and the distribution exp(-x^2). The speaker is trying to rewrite the integral using a different form of the hermite polynomial and has been able to solve it for certain values, but needs help proving the result formally. The expert suggests using a generating function and states the result of the integral in that form. However, they are unable to simplify the sum to the desired form.
  • #1
Gabriel Maia
72
1
Hi. I'm off to solve this integral and I'm not seeing how

[itex]\int dx Hm(x)Hm(x)e^{-2x^2}[/itex]

Where Hm(x) is the hermite polynomial of m-th order. I know the hermite polynomials are a orthogonal set under the distribution exp(-x^2) but this is not the case here.

Using Hm(x)=[itex](-1)^m e^{x^2}[/itex][itex]\frac{d^m}{dx^m}e^{-x^2}[/itex] I was able to rewrite the integral as

[itex]\int \left(\frac{d^m}{dx^m}e^{-x^2}\right)^2 dx[/itex]

I have calculated this integral for m=0,1,2,3,4,5 with mathematica and the result seems to be [itex](2m-1)!\sqrt{\frac{\pi}{2}}[/itex] but I need to prove it formally. Can you help me?


Thank you.
 
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  • #2
I haven't worked this out completely, but I'll suggest you try using the generating function
$$g(x,t) = \exp(-t^2+2tx) = \sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}$$
 
  • #3
Using the generating function, it was pretty straightforward to show that
$$\int H_m(x)H_m(x) e^{-2x^2}\,dx = \sum_{m=2p+q} \frac{1}{q!}\left(\frac{m!}{p!2^p}\right)^2,$$ which appears to reproduce the result you found. I don't see how to simplify that sum to ##(2m-1)!## however.
 
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Related to Can you help me prove the integral for Hermite polynomials?

1. What are Hermite polynomials and how are they used in integrals?

Hermite polynomials are a set of orthogonal polynomials that are commonly used in mathematics and physics to solve integrals involving Gaussian functions. They are named after the French mathematician Charles Hermite and can be expressed as a series of powers of x multiplied by a constant coefficient.

2. How do Hermite polynomials differ from other types of polynomials?

Hermite polynomials are unique in that they have the property of being orthogonal, meaning that their inner product is equal to zero. This makes them particularly useful in solving integrals involving Gaussian functions, which are known for their bell-shaped curves.

3. What is the formula for the Hermite polynomial integral?

The formula for the Hermite polynomial integral is given by ∫e-x2Hn(x)Hm(x)dx = √π2nn!δnm, where Hn(x) and Hm(x) are Hermite polynomials of orders n and m, and δnm is the Kronecker delta function.

4. How are Hermite polynomials used in quantum mechanics?

In quantum mechanics, Hermite polynomials are used to solve the Schrödinger equation, which describes the behavior of particles at the atomic and subatomic level. They are also used to determine the energy levels of particles in a harmonic oscillator potential.

5. Are there any real-world applications of Hermite polynomials?

Yes, Hermite polynomials have various real-world applications, particularly in physics and engineering. They are used in signal processing to analyze and filter data, in probability theory to model random variables, and in computer graphics to create smooth curves and surfaces.

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