## Hermite polynomials – generation

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Chapter 2, Post 16.

In the solution of the Schrödinger equation for the harmonic oscillator, we found that the wave function can be expressed as a power series:

$\displaystyle \psi(y)=e^{-y^{2}/2}\sum_{j=0}^{\infty}a_{j}y^{j}$

where ${y}$ was introduced as a shorthand variable:

$\displaystyle y\equiv\sqrt{\frac{m\omega}{\hbar}}x$

and the coefficients ${a_{j}}$ satisfy the recursion relation

$\displaystyle a_{j+2}=\frac{2j+1-\epsilon}{(j+1)(j+2)}a_{j} \ \ \ \ \ (1)$

where ${\epsilon}$ is another shorthand variable for the energy:

$\displaystyle \epsilon\equiv\frac{2E}{\hbar\omega}$

By requiring the recursion relation to terminate at various values of ${j}$ we can generate the polynomials for the various energy states, which turn out to be the Hermite polynomials.

For example, if we take the highest value of ${j}$ to be 5, then we must have

 $\displaystyle a_{7}$ $\displaystyle =$ $\displaystyle \frac{11-\epsilon}{42}a_{5}$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0$ $\displaystyle \epsilon$ $\displaystyle =$ $\displaystyle 11$ $\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{11}{2}\hbar\omega$

To get the coefficients, we can start with ${a_{0}=0}$ (since all even terms must be zero if we want ${a_{5}\ne0}$) and ${a_{1}=1}$. Then we get

$\displaystyle a_{j+2}=\frac{2j-10}{\left(j+1\right)\left(j+2\right)}a_{j}$

so ${a_{3}=-\frac{4}{3}}$, ${a_{5}=\left(-\frac{1}{5}\right)\left(-\frac{4}{3}\right)=\frac{4}{15}}$. The Hermite polynomial is

$\displaystyle H_{5}(x)=A_{5}\left[\frac{4}{15}x^{5}-\frac{4}{3}x^{3}+x\right]$

where ${A_{5}}$ is a constant that is set by convention to make the coefficients satisfy some specified rule.

If we require the coefficient of the highest power ${n}$ to be ${2^{n}}$ we can multiply this polynomial by 120 to get

$\displaystyle H_{5}(x)=32x^{5}-160x^{3}+120x$

For ${H_{6}}$, all the odd terms are zero, and we require ${a_{8}=0}$, so we get

 $\displaystyle \epsilon$ $\displaystyle =$ $\displaystyle 13$ $\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{13}{2}\hbar\omega$ $\displaystyle a_{j+2}$ $\displaystyle =$ $\displaystyle \frac{2j-12}{\left(j+1\right)\left(j+2\right)}a_{j}$

Taking ${a_{0}=1}$ and ${a_{1}=0}$, we get ${a_{2}=-6}$, ${a_{4}=4}$, ${a_{6}=-\frac{8}{15}}$. Requiring the coefficient of ${a_{6}}$ to be ${2^{6}=64}$ we get

$\displaystyle H_{6}(x)=64x^{6}-480x^{4}+720x^{2}-120$