Orthogonality property of Hermite polynomial

We discussed about Orthogonality property of Hermite polynomial in this topic. Differential equation occurs in various branch of physics, among them Hermite differential equation occur while solving quantum harmonic oscillator in quantum physics. It helps to find energy value and various kind of wave function. The form of Hermite differential equation $$y''-2xy'+2ny=0\dotsm(1)$$

Summary

We discussed about Orthogonality property of Hermite polynomial in this topic. Differential equation occurs in various branch of physics, among them Hermite differential equation occur while solving quantum harmonic oscillator in quantum physics. It helps to find energy value and various kind of wave function. The form of Hermite differential equation $$y''-2xy'+2ny=0\dotsm(1)$$

Things to Remember

These are the important equation to be remember in this chapter

  • $$\frac{d^2y}{dx^2}-2x\cdot\frac{dy}{dx}+2ny=0$$
  • $$\sum_{n=0}^\infty \frac{H_n(x) t^n}{n!}= e^{2xt- t^2}$$
  • $$\int_{-\infty}^{+\infty} e^{-x^2}H_n(x) H_m(x) dx= \sqrt{\pi}2^n n! \delta_{mn}$$

    This is orthogonality relation of Hermite polynomial.

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Orthogonality property of Hermite polynomial

Orthogonality property of Hermite polynomial

Orthogonality property of Hermite Polynomial:

We have Hermite differential equations as:

$$\frac{d^2y}{dx^2}-2x\cdot\frac{dy}{dx}+2ny=0\dotsm(1)$$

Since \(H_n(x)\) is a solution of equation (1), then we can write

$$\frac{d^2H_n(x)}{dx^2}- 2x\cdot \frac{dH_n(x)}{dx}+2nH_n(x)=0\dotsm(2)$$

Hints:

\(\frac{d^2H_n(x)}{dx^2}-2x\cdot \frac{dH_n(x)}{d(x)}+2nH_n(x)=0\)

Put\( \frac{dH_n(x)}{dx}=\phi\)

\(or,\;\; \frac{d\phi}{dx}-2x\phi=Q\)

\(\frac{dy}{dx}+Py=Q\), I.F =\(e^{\int pdx}\)

\(I.F= e ^{\int 2xdx}=e^{-2}\cdot \frac{x^2}{2}=e^{-x^2}\)

Multiplying equation (2) on both side by \(e^{-x^2}\) we get,

$$ e^{-x^2}\frac{d^2H_n(x)}{dx^2}- e^{-x^2}\cdot 2x\frac{dH_n(x)}{dx}+2ne^{-x^2}H_n(x)=0$$

$$or,\;\; \frac{d}{dx}\biggl[e^{-x^2} \frac{dH_n(x)}{dx}\biggr] + 2ne^{-x^2} H_n(x)=0\dotsm(*)$$

For m order solution i.e. \(y= H_m(x)\) we can write

$$\frac{d}{dx}\biggl[e^{-x^2}\frac{dH_m(x)}{dx}\biggr] + 2me^{-x^2}H_m(x)=0\dotsm(3)$$

Multiplying equation (*) with \(H_m(x) \) and (3) with \(H_n(x)\) and subtract (*) we get,

$$or\;\; H_m(x)\cdot \frac{d}{dx}\biggl[\frac{e^{-x^2} dH_n(x)}{dx}\biggr]- H_n(x) \frac{d}{dx}\biggl[e^{-x^2}\frac{dH_m(x)}{dx}\biggr]+ 2(n-m) e^{-x^2} H_n(x) H_m(x)=0$$

$$or,\;\; \frac{d}{dx}\biggl[H_m(x)e^{-x^2} \frac{dH_n(x)}{dx}-H_n(x) e^{-x^2}\frac{dH_m(x)}{dx}\biggr]+2(n-m) e^{-x^2} H_n(x)H_m(x)=0$$

Because: \( H_m(x) \frac{d}{dx}\biggl[e^{-x^2}\frac{dH_n(x)}{d(x)}\biggr]+ e^{-x^2} \frac{dH_n(x)}{dx}\cdot \frac{dH_m(x)}{dx}-H_n(x)\frac{d}{dx}\biggl[e^{-x^2}\frac{dH_m(x)}{dx}\biggr]-\frac{e^{-x^2} d_m(x)}{dx}\cdot\frac{dH_n(x)}{dx}\)

Integrating above equation from \(-\infty\;to\;+\infty\) we get,

$$\int_{-\infty}^{+\infty} \frac{d}{dx}\biggl[H_m(x) e^{-x^2}\frac{dH_n(x)}{dx}-H_n(x)e^{-x^2}\frac{dH_m(x)}{dx}\biggr]dx+2(n-m) \int_{-\infty}^{+\infty}e^{-x^2} H_n(x)H_m(x) dx=0$$

$$or,\;\; H_m(x) e^{-x^2}\frac{dH_n(x)}{dx}- H_n(x) e^{-x^2} \frac{dH_m(x)}{dx}\biggl]_{-\infty}^{+\infty}+ 2(n-m) \int_{-\infty}^{+\infty}e^{-x^2} H_n(x) H_m(x)dx=0$$

$$or,\;\; 2(n-m) \int_{-\infty}^{+\infty} e^{-x^2}H_n(x)H_m(x) dx=0\dotsm(4)$$

Case I: When \(n\ne m\)

$$\int_{-\infty}^{+\infty} e^{-x^2}H_n(x) H_m(x) dx=0\dotsm(5)$$

Case II: when n=m

$$\int_{-\infty}^{+\infty} e^{-x^2} H_n(x) H_n(x) dx=?\dotsm(6)$$

We have generating function of Hermite polynomial

$$\sum_{n=0}^\infty \frac{H_n(x) t^n}{n!}= e^{2xt- t^2}$$

Squaring both side we get,

$$,\;\;\sum_{n=0}^\infty \frac{t^{2n} H_n(x) H_n(x)}{(n!)^2}= e^{4xt-2t^2}$$

Multiplying both side by \(e^{-x^2}\) we get,

$$or,\;\; \sum_{n=0}^\infty \frac{t^{2n}}{(n!)^2} e^{-x^2} H_n(x) H_n(x)= e^{-x^2}\cdot e^{4xt-2t^2}$$

On integrating both sides from \(-\infty \) to \(\infty\) we get,

$$or,\;\; \sum_{n=0}^\infty\frac{t^2n}{(n!)^2}\int_{-\infty}^{+\infty} e^{-x^2} H_n(x) H_n(x) dx=\int_{-\infty}^{+\infty} e^{-x^2+4xt-2t^2}dx$$

$$=\int_{-\infty}^{+\infty} e^{-(x-2t)^2+2t^2}dx$$

$$=e^{2t^2}\int_{-\infty}^{+\infty} e^{-(x-2t)^2}dx= e^{2t^2}\sqrt{\pi}$$

Put \(y= x-2t\)

$$dy= dx$$

$$or,\;\; \int_{-\infty}^{+\infty} e^{-x^2}H_n(x) H_n(x) dx= e^{2t^2}\sqrt{\pi}$$

$$= \sqrt{\pi} \biggl[1+\frac{2t^2}{1!}+\frac{(2t^2)^2}{2!}+\dotsm+\frac{(2t^2)^n}{n!}+\dotsm\biggr]$$

Now collecting the coefficient of \(t^2n\) on both side

$$\frac{1}{(n!)^2}\int_{-\infty}^{+\infty} e^{-x^2} H_n(x) H_n(x) dx=\sqrt{\pi} \dotsm\frac{2^n}{n!}$$

$$or,\;\; \int_{-\infty}^{+\infty} e^{-x^2} H_n(x)H_n(x)dx= \sqrt{\pi}2^n n!\dotsm(7)$$

From (6) and (7)

$$or,\;\; \int_{-\infty}^{+\infty} e^{-x^2} H_n(x) H_m(x) dx= \sqrt{\pi}2^n n! \delta_{mn}$$

Where,

\(\delta_{mn}\)=1 for m=n

\(\delta_{mn}=0\) for m\(\ne\)n

Combining equation (5) and (6) we get,

$$\int_{-\infty}^{+\infty} e^{-x^2}H_n(x) H_m(x) dx= \sqrt{\pi}2^n n! \delta_{mn}$$

This is orthogonality relation of Hermite polynomial.

Reference:

  1. Dass, H. K.Mathematical Physics. New Delhi: S. Chand, 2005.
  2. Vaughn, Michael T.Introduction to Mathematical Physics. Weinheim: Wiley-VCH, 2007
  3. Butkov, Eugene.Mathematical Physics. Reading, MA: Addison-Wesley Pub., 1968.
  4. Carroll, Robert W.Mathematical Physics. Amsterdam: North-Holland, 1988.
  5. Adhikari, Pitri Bhakta, and Dya Nidhi Chhatkuli.A Textbook of Physics. Third Revised Edition ed. Vol. III. Kathmandu: SUKUNDA PUSTAK BHAWAN, 2072.

Lesson

Differential equations

Subject

Physics

Grade

Bachelor of Science

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