Generation of Hermite polynomials with Gram-Schmidt procedure

Question

I want to use the Gram-Schmidt procedure to generate the first three Hermite polynomials. Given the set of linearly independent vectors $\{1,x,x^2,...\}$ in the Hilbert space $L^2(R,e^{-x^2}dx)$, I apply the orthogonalisation procedure as follows:

$u_{0}(x)=1$

$u_{1}(x)=x-\frac{\langle u_{0}|x \rangle}{\langle u_{0}|u_{0}\rangle} \cdot u_{0}=x-\frac{\langle 1|x\rangle}{\langle 1|1\rangle}\cdot 1 =x $

$u_{2}(x)=x^2-\frac{\langle u_{0}|x^2 \rangle}{\langle u_{0}|u_{0}\rangle} \cdot u_{0}-\frac{\langle u_{1}|x^2 \rangle}{\langle u_{1}|u_{1}\rangle} \cdot u_{1}=x^2-\frac{\langle 1|x^2\rangle}{\langle 1|1\rangle}\cdot 1-\frac{\langle x|x^2\rangle}{\langle x|x\rangle}\cdot x =x^2 -\frac{1}{2} $

The inner product I use is the one of $L^2(R,e^{-x^2}dx)$, namely:

$\langle u|v \rangle:=\int_{R}\overline{u(x)}v(x)e^{-x^2}dx$

Then I divide the functions I found by their norm in order to get orthonomal vectors:

$\tilde{H_0}(x)=\frac{1}{\sqrt{\langle u_{0}|u_{0} \rangle}}u_{0}(x)=\frac{1}{\pi^{1/4}}$

$\tilde{H_1}(x)=\frac{1}{\sqrt{\langle u_{1}|u_{1} \rangle}}u_{1}(x)=\frac{\sqrt{2}}{\pi^{1/4}}x$

$\tilde{H_2}(x)=\frac{1}{\sqrt{\langle u_{2}|u_{2} \rangle}}u_{2}(x)=\frac{\sqrt{2}}{\pi^{1/4}}(x^2-\frac{1}{2})$

I can't understand why the coefficients I get are different from the ones that can be found on Wikipedia (https://en.wikipedia.org/wiki/Hermite_polynomials) or in every list of Hermite's polynomials. Where is the mistake?

Answer 1

$u_{0}(x)=1 \quad u_{1}(x)=x \quad $ Let $\mathrm{H}_{0}(x), \mathrm{H}_{1}(x)$ be first two Hermite polynomials. If $\frac{A H^{\prime}_{n}(x)}{\sqrt{\int_{0}^{\prime} H_{n}(x) H_{n}^{\prime}(x) e^{-x^{2} }d x}}=H_{n}(x)$ then $\int_{-\infty}^{\infty} H_{n}(x) H_{n}(x) e^{-x^{2}} d x=2^{n} n ! \pi^{1 / 2}$

From here we can simply find out that $A=\sqrt{2^{n} n ! \pi^{1 / 2}}$

Let $H_{0}^{\prime}(x)=u_{0}(x)=1$ $\int_{-\infty}^{\infty} e^{-x^{2}} d x=\sqrt{\pi}$

$H_{0}(x)=\sqrt{2^{0} 0! \sqrt{\pi}} \frac{H_{0}^{\prime}(x)}{\sqrt{\sqrt{\pi}}}$ $H_{0}(x)=1$ $\begin{aligned} H_{1}^{\prime}(x) &=u_{1}(x)-H_{0}(x) \int_{-\infty}^{\infty} H_{0}(x) u_{1}(x) e^{-x^{2}} d x \\ &=x-\int_{-\infty}^{\infty} x e^{-x^{2} d x} \end{aligned}$

$H_{1}^{\prime}(x)=x$ $\int_{-\infty}^{\infty} H_{1}^{\prime}(x) \quad H_{1}^{\prime}(x) e^{-x^{2}} d x=\int_{-\infty}^{\infty} x^{2} e^{-x^{2}} d x=\frac{\sqrt{\pi}}{2}$

$\begin{aligned} H_{1}(x) &=\sqrt{2^{\prime} 1 ! \sqrt{\pi}} \frac{H_{1}^{\prime}(x)}{\sqrt{\sqrt{\pi} / 2}} \\ &=2 x \end{aligned}$

Answer 2

Actually, I think using Gram-Schmidt orthogonalization you are only expected to find polynomials that are proportional to Hermite's polynomials, since by convention you can define the Hermite polynomials to have a different coefficient than the one you find using this method.

You can find the detailed workout in this pdf doc:

http://bingweb.binghamton.edu/~suzuki/Math-Physics/LN-8_Gram_Schmidt_Orgonalization.pdf

Answer 3

The only mistake is that Hermite polynomials are NOT normalized, in fact they satisfy

$$\int H_{m}(x)H_{n}(x){\frac {e^{-x^{2}}}{\sqrt {\pi }}}dx=2^{n}n!\delta _{mn}.$$

The polynomials you found are similar to the called "Hermite functions" and are like the normalized version in $L^2(\mathbb{R},dx)$ (but without the weight function). $$\psi _{n}(x)=\left(2^{n}n!{\sqrt {\pi }}\right)^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{2}}}H_{n}(x),$$

to find Hermite Polynomials, you only need to multiply by $\left(2^{n}n!{\sqrt {\pi }}\right)^{{\frac {1}{2}}}$ ;)

Answer 4

You are generating the Legendre polynomials with this procedures, i.e., if you take $\{1,x,x^2, x^3, ...\}$ and apply Gram-Schmidt on it you'll end up with Legendre polynomials.