# Differential equations II: Hermite equation and Hermite polynomials

## Course Contents

### Introduction to power series method:

Solutions of second order differential equations through power series method:

Example 1:

$\frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\omega }^{2}}y=0.$

We know the solution of the above equation already; it is

y(x) = Acosωx + Bsinωx

as can be verified by direct substitution in the differential equation.

Power series solution: Let

$y(x)=\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{x}^{n}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+\cdots +{{a}_{n}}{{x}^{n}}+\cdots }.$

We assume that the power series on the right hand side converges in a certain domain of values for x. Our solution will then be valid in that domain.

Also we assume that the derivatives are all well defined.

This is NOT a pure mathematics course for us to worry about convergence, validity of such representations etc. However, whenever we face problems, we agree to go back to the mathematician!

$\frac{dy(x)}{dx}=\sum\limits_{n=1}^{\infty }{n{{a}_{n}}{{x}^{n-1}}={{a}_{1}}+2{{a}_{2}}x+3{{a}_{3}}{{x}^{2}}+\cdots +n{{a}_{n}}{{x}^{n-1}}+\cdots }$ and

$\frac{{{d}^{2}}y}{d{{x}^{2}}}=\sum\limits_{n=2}^{\infty }{n(n-1){{a}_{n}}{{x}^{n-2}}=2.1{{a}_{2}}+3.2{{a}_{3}}x+4.3{{a}_{4}}{{x}^{2}}+\cdots +(n+2)(n+1){{a}_{n+2}}{{x}^{n}}+\cdots }$

Note that the “.” (dot, period) between numbers is actually the multiplication symbol. Substitute the above in the equation

$\frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\omega }^{2}}y=0$

and equate coefficients of every power of x to zero.

i.e., coefficient of x0 = 0, coefficient of x1 = 0 etc. Then

\begin{align} & ({{\omega }^{2}}{{a}_{0}}+2.1{{a}_{2}})+({{\omega }^{2}}{{a}_{1}}+3.2{{a}_{3}})x+({{\omega }^{2}}{{a}_{2}}+4.3{{a}_{4}}){{x}^{2}}+\cdots \\ & +[{{\omega }^{2}}{{a}_{n}}+(n+2)(n+1){{a}_{n+2}}]{{x}^{n}}+\cdots =0 \\ \end{align}

$\Rightarrow {{\omega }^{2}}{{a}_{0}}+2.1{{a}_{2}}=0$ or

${{a}_{2}}=-{{\omega }^{2}}\frac{{{a}_{0}}}{1.2}$

$({{\omega }^{2}}{{a}_{2}}+4.3{{a}_{4}})=0\Rightarrow {{a}_{4}}=-{{\omega }^{2}}\frac{{{a}_{2}}}{4.3}\,\,=-{{\omega }^{2}}(-{{\omega }^{2}}).\frac{{{a}_{0}}}{1.2.3.4}=\frac{{{\omega }^{4}}}{4!}{{a}_{0}}$

2a4 + 6.5a6) = 0

$\Rightarrow {{a}_{6}}=-\frac{{{\omega }^{6}}}{6!}{{a}_{0}}\,\,\,\,etc...$

Also ω2a1 + 3.2a3 = 0

$\Rightarrow {{a}_{3}}=-\frac{{{\omega }^{2}}}{3.2}{{a}_{1}}=-\frac{{{\omega }^{3}}}{3!}\left( \frac{{{a}_{1}}}{\omega } \right)$

ω2a3 + 5.4a5 = 0

$\Rightarrow {{a}_{5}}=-\frac{{{\omega }^{2}}{{a}_{3}}}{5.4}=+\frac{{{\omega }^{5}}}{5!}\frac{{{a}_{1}}}{\omega }\,\,\,\,etc..$

Thus we get two recurrence relations

\begin{align} & {{a}_{2n}}={{(-1)}^{n}}\frac{{{\omega }^{2n}}}{(2n)!}{{a}_{0}};\text{ } \\ & {{a}_{2n+1}}={{(-1)}^{n+1}}\frac{{{\omega }^{2n+1}}}{(2n+1)!}\left( \frac{{{a}_{1}}}{\omega } \right) \\ & n=1,2,3,\cdots \\ \end{align}.

The two constants which are left undefined are a0 and a1. This is true for all second order differential equations. They require two conditions to arrive at a particular solution.

Thus the solution is

\begin{align} & y(x)=\left( 1-\frac{{{\omega }^{2}}}{2!}{{x}^{2}}+\frac{{{\omega }^{4}}}{4!}{{x}^{4}}-\frac{{{\omega }^{6}}}{6!}{{x}^{6}}+..... \right){{a}_{0}} \\ & \text{ }+\left( \omega -\frac{{{\omega }^{3}}}{3!}{{x}^{3}}+\frac{{{\omega }^{5}}}{5!}{{x}^{5}}-\frac{{{\omega }^{7}}}{7!}{{x}^{7}}+..... \right)\left( \frac{{{a}_{1}}}{\omega } \right) \\ & \text{ }={{a}_{0}}\cos \omega x+\frac{{{a}_{1}}}{\omega }\sin \omega x \\ & \text{ }=A\cos \omega x+B\sin \omega x;\text{ }A={{a}_{0}},\text{ }B=\frac{{{a}_{1}}}{\omega } \\ \end{align}

The above example provides an illustration for a new method to solve an old problem with known solutions. The harmonic oscillator equation will be solved using this method.

The Schrödinger differential equation is

$-\frac{{{\hbar }^{2}}}{2\mu }\frac{{{d}^{2}}\psi }{d{{x}^{2}}}+\frac{1}{2}k{{x}^{2}}\psi =E\psi ;\text{ }\psi \equiv \psi (x)$

and $x=-\infty \,\,to\,\,+\infty$ to is the domain

Rewrite the above equation:

\begin{align} & \frac{{{d}^{2}}\psi }{d{{x}^{2}}}+\left( \frac{2\mu E}{{{\hbar }^{2}}}-\frac{k\mu }{{{\hbar }^{2}}}{{x}^{2}} \right)\psi =0 \\ & let\text{ }\lambda \text{ = }\frac{2\mu E}{{{\hbar }^{2}}}\text{ and }{{\alpha }^{\text{2}}}=\frac{k\mu }{{{\hbar }^{2}}} \\ & \Rightarrow \frac{{{d}^{2}}\psi }{d{{x}^{2}}}+\left( \lambda -{{\alpha }^{\text{2}}}{{x}^{2}} \right)\psi =0 \\ \end{align}

x has the dimension of length L . λ and αhave the dimension of L − 2.

\begin{align} & \Rightarrow \frac{1}{\alpha }\frac{{{d}^{2}}\psi }{d{{x}^{2}}}+\left( \frac{\lambda }{\alpha }-\alpha {{x}^{2}} \right)\psi =0 \\ & let\text{ }y\text{ = }\sqrt{\alpha }x\text{ }then\text{ }{{y}^{2}}\text{ = }\alpha {{x}^{2}} \\ & \frac{d}{dx}=\text{ }\frac{dy}{dx}\frac{d}{dy}=\text{ }\sqrt{\alpha }\frac{d}{dy}\text{ }\frac{{{d}^{2}}}{d{{x}^{2}}}\text{ = }\alpha \frac{{{d}^{2}}}{d{{y}^{2}}}\text{ } \\ & Let\text{ }\psi \left( x \right)\text{ = }\Psi \left( y \right)\text{ } \\ \end{align}

\begin{align} & \Rightarrow \frac{1}{\alpha }\frac{{{d}^{2}}\psi }{d{{x}^{2}}}+\left( \frac{\lambda }{\alpha }-\alpha {{x}^{2}} \right)\psi =0 \\ & \Rightarrow \frac{{{d}^{2}}\Psi }{d{{y}^{2}}}+\left( \frac{\lambda }{\alpha }-{{y}^{2}} \right)\Psi =0 \\ \end{align}

where ψ = ψ(x) and $\Psi =\Psi (y)\text{ = }\Psi (\sqrt{\alpha }x)$

What is the role of y?

note $y\text{ = }\sqrt{\alpha }x$ $y\text{ = }\sqrt{\alpha }x.$ What is the dimension of $\sqrt{\alpha }$ ?

It is the same as ${{\left( \frac{k\mu }{{{\hbar }^{2}}} \right)}^{\frac{1}{4}}}$

Verify that the dimension is $\frac{1}{L};\text{ (length}{{\text{)}}^{\text{-1}}}$

Since x has the dimension of length, we have introduced y so that it is dimensionless variable. Therefore in solving the differential equation we can drop all the excess baggage of constants.

Thus the differential equation that we want to solve is

$\frac{{{d}^{2}}\Psi }{d{{y}^{2}}}+\left( \frac{\lambda }{\alpha }-{{y}^{2}} \right)\Psi \left( y \right)=0$

Note that the equation is valid for all $y,\text{ (}y\text{ from + }\infty \text{ }to\text{ }-\infty ).$

### Asymptotic (large y) solution:

What does the solution look for large values of y? To do this, we can try the “asymptotic [or, limiting] equation for large y, namely,

$\frac{{{d}^{2}}\Psi }{d{{y}^{2}}}-{{y}^{2}}\Psi \left( y \right)\approx 0.$

(take y sufficiently large so that the term we dropped out, $\frac{\lambda }{\alpha }\Psi (y)$is small and can be neglected). Try the solution $\Psi \tilde{\ }{{e}^{\pm \beta {{y}^{2}}}}.$ Then

\begin{align} & \frac{d\Psi }{dy}=\pm 2\beta y{{e}^{\pm \beta {{y}^{2}}}} \\ & \frac{{{d}^{2}}\Psi }{d{{y}^{2}}}=\pm 2\beta {{e}^{\pm \beta {{y}^{2}}}}+4{{\beta }^{2}}{{y}^{2}}{{e}^{\pm \beta {{y}^{2}}}} \\ & \Rightarrow \left( \pm 2\beta +4{{\beta }^{2}}{{y}^{2}}-{{y}^{2}} \right){{e}^{\pm \beta {{y}^{2}}}}=0 \\ & \Rightarrow \left[ {{y}^{2}}\left( 4{{\beta }^{2}}-1 \right)\pm 2\beta \right]{{e}^{\pm \beta {{y}^{2}}}}=0 \\ \end{align}

For $\Psi \left( y \right)$ to vanish for large values of choose only ${{e}^{-\beta {{y}^{2}}}}$

and ${{y}^{2}}\left( 4{{\beta }^{2}}-1 \right)-2\beta =0,$ y being very large,${{y}^{2}}=\frac{2\beta }{\left( 2\beta -1 \right)\left( 2\beta +1 \right)}$

for very large $y\text{ }(y\to \pm \infty )$ $2\beta \tilde{\ }\pm 1\text{ }or\text{ }\beta \tilde{\ }\pm \frac{1}{2}$

$\text{ }\beta \tilde{\ }\frac{1}{2}\text{ }\Rightarrow \text{ }{{e}^{{{y}^{2}}/2}}\to \text{ }\infty \text{ as }y\to \infty .$ It is not acceptable. Hence the asymptotic solution must be $\Psi \left( y \right)={{e}^{-{{y}^{2}}/2}}$ for large values of y $\left( \text{ }\beta \tilde{\ }-\frac{1}{2}\text{ } \right).$

### Recurring coefficients for Hermite equation:

The complete solution be chosen as

$\Psi \left( y \right)=H\left( y \right){{e}^{-{{y}^{2}}/2}}$

where the function $H\left( y \right)$ is arbitrary and must satisfy the equation

$\frac{{{d}^{2}}\Psi }{d{{y}^{2}}}+\left( \frac{\lambda }{\alpha }-{{y}^{2}} \right)\Psi =0$

for all values of y.

Substituting for Ψ as above, we get,

$\frac{{{d}^{2}}H}{d{{y}^{2}}}-2y\frac{dH}{dy}+\left( \frac{\lambda }{\alpha }-1 \right)H=0$

This is known as the Hermite’s differential equation.

Problem: Show that Hermite’s equation is obtained when Eq. (12) is substituted in Eq.(10).

Propose a series solution as follows:

\begin{align} & H(y)=\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{y}^{n}}} \\ & \frac{dH}{dy}=\sum\limits_{n=1}^{\infty }{n{{a}_{n}}{{y}^{n-1}}} \\ & \frac{{{d}^{2}}H}{d{{y}^{2}}}=\sum\limits_{n=2}^{\infty }{n(n-1){{a}_{n}}{{y}^{n-2}}} \\ \end{align}

Substitute this in the differential equation:

\begin{align} & \frac{{{d}^{2}}H}{d{{y}^{2}}}-2y\frac{dH}{dy}+\left( \frac{\lambda }{\alpha }-1 \right)H=0\text{ }to\text{ }get \\ & \left( 2{{a}_{2}}+2.3{{a}_{3}}y+3.4{{a}_{4}}{{y}^{2}}+\cdots +\left( n+1 \right)\left( n+2 \right){{a}_{n+2}}{{y}^{n}}+\cdots \right) \\ & -2\left( {{a}_{1}}y+2{{a}_{2}}{{y}^{2}}+3{{a}_{3}}{{y}^{3}}+\cdots n{{a}_{n}}{{y}^{n}}+\cdots \right) \\ & +\left( \frac{\lambda }{\alpha }-1 \right)\left( {{a}_{0}}+{{a}_{1}}y+{{a}_{2}}{{y}^{2}}+\cdots +{{a}_{n}}{{y}^{n}}+\cdots \right)=0 \\ \end{align}

Rearrange the above in increasing powers of y to get the following,

\begin{align} & \left[ 2{{a}_{2}}+\left( \frac{\lambda }{\alpha }-1 \right){{a}_{0}} \right]+\left[ 3.2{{a}_{3}}+\left( \frac{\lambda }{\alpha }-1-2 \right){{a}_{1}} \right]y \\ & +\left[ 3.4{{a}_{4}}+\left( \frac{\lambda }{\alpha }-1-2.2 \right){{a}_{2}} \right]{{y}^{2}}+\left[ 4.5{{a}_{5}}+\left( \frac{\lambda }{\alpha }-1-2.3 \right){{a}_{3}} \right]{{y}^{3}} \\ & +\cdots +\left[ \left( n+1 \right)(n+2){{a}_{n+2}}+\left( \frac{\lambda }{\alpha }-1-2.n \right){{a}_{n}} \right]{{y}^{n}}+..... \\ & \text{ }=0 \\ \end{align}

Considering even n’s

\begin{align} & 2{{a}_{2}}=-\left( \frac{\lambda }{\alpha }-1 \right){{a}_{0}}\text{ }or \\ & {{a}_{2}}=-\frac{\left( \frac{\lambda }{\alpha }-1 \right)}{2}{{a}_{0}}; \\ & 3.4{{a}_{4}}+\left( \frac{\lambda }{\alpha }-1-2.2 \right){{a}_{2}}=0 \\ & \Rightarrow {{a}_{4}}=+\frac{\left( \frac{\lambda }{\alpha }-1-2.2 \right)\left( \frac{\lambda }{\alpha }-1 \right)}{4.3.2.1}{{a}_{0}} \\ \end{align}

Repating this for all even n’s

${{a}_{2n}}=\frac{{{\left( -1 \right)}^{n}}\left( \frac{\lambda }{\alpha }-1-2.(2n-2) \right)\left( \frac{\lambda }{\alpha }-1-2.\left( 2n-4 \right) \right)......\left( \frac{\lambda }{\alpha }-1 \right){{a}_{0}}}{\left( 2n \right)!}$

Considering odd n’s:

${{a}_{3}}=-\frac{1}{2.3}\left( \frac{\lambda }{\alpha }-1-2.1 \right){{a}_{1}}$

\begin{align} & {{a}_{5}}=-\frac{1}{4.5}\left( \frac{\lambda }{\alpha }-1-2.3 \right){{a}_{3}} \\ & =\frac{1}{4.5}\left( \frac{\lambda }{\alpha }-1-2.3 \right)\left( -\frac{1}{2.3} \right)\left( \frac{\lambda }{\alpha }-1-2.1 \right){{a}_{1}} \\ & ={{\left( -1 \right)}^{2}}\frac{1}{5!}\left( \frac{\lambda }{\alpha }-1-2.3 \right)\left( \frac{\lambda }{\alpha }-1-2.1 \right){{a}_{1}} \\ & {{a}_{2n+1}}=\frac{{{\left( -1 \right)}^{n}}}{\left( 2n+1 \right)!}\left[ \frac{\lambda }{\alpha }-1-2\left( 2n-1 \right) \right]\left[ \frac{\lambda }{\alpha }-1-2\left( 2n-3 \right) \right]......\left[ \frac{\lambda }{\alpha }-1-2.1 \right]{{a}_{1}} \\ \end{align}

The equations for ${{a}_{2n}}\,and\,\,{{a}_{2n+1}}$are known as recurrence relations.

### Hermite polynomials:

The power series $\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{y}^{n}}}$has the following asymptotic behaviour.

$\frac{{{a}_{n+2}}}{{{a}_{n}}}=\frac{-\left( \frac{\lambda }{\alpha }-1-2n \right)}{\left( n+1 \right)\left( n+2 \right)}\approx \frac{-2n}{{{n}^{2}}}=\frac{-2}{n}\to 0\text{ }as\text{ }n\to \infty$

What is the behaviour of ${{e}^{{{y}^{2}}}}$ for large values of n?

${{e}^{{{y}^{2}}}}\tilde{\ }1+{{y}^{2}}+\frac{{{y}^{4}}}{2!}+\frac{{{y}^{6}}}{3!}+........$

\begin{align} & \frac{coefficient\text{ }of\text{ }{{y}^{n+2}}}{coefficient\text{ }of\text{ }{{y}^{n}}}=\frac{\frac{1}{\left( \frac{n}{2}+1 \right)!}}{\frac{1}{\left( \frac{n}{2} \right)}!}=\frac{1}{\left( \frac{n}{2}+1 \right)} \\ & \text{ }\approx \frac{2}{n}\text{ }for\text{ }n\to \infty \\ \end{align}

Thus the absolute covergence property of the power series $\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{y}^{n}}}$ is the same as that of ${{e}^{{{y}^{2}}}}$ as n tends to infinity; Then the wave function Ψ will be such that $\Psi \left( y \right)={{e}^{-\frac{{{y}^{2}}}{2}}}H\left( y \right)$ may not be finite for all the values of y if $H\left( y \right)$ is an infinite series. However, if $H\left( y \right)$can be truncated at some finite n, then $\Psi \left( y \right)\to 0\text{ }as\text{ }y\to \infty$

That is an + 2 = 0 for some arbitrary n, but not an.

Or $\left( \frac{\lambda }{\alpha }-1-2n \right)=0$ for some n.

$n=\frac{1}{2}\left( \frac{\lambda }{\alpha }-1 \right)$

n can be any positive integer, but $\frac{1}{2}\left( \frac{\lambda }{\alpha }-1 \right)$ must be the largest n value, for a given λ and α. With the substitution

$\left( \frac{\lambda }{\alpha }-1 \right)=2n$

The power series for $H\left( y \right)$ becomes

\begin{align} & {{H}_{n}}\left( y \right)={{a}_{0}}\left[ 1-\left( \frac{2n}{2} \right){{y}^{2}}+\frac{\left( 2n \right)\left( 2n-4 \right)}{4!}{{y}^{4}}-\frac{\left( 2n \right)\left( 2n-4 \right)\left( 2n-8 \right)}{6!}{{y}^{6}}+..... \right] \\ & \text{ +}{{a}_{1}}\left[ y-\frac{\left( 2n-2 \right)}{3!}{{y}^{3}}+\frac{\left( 2n-2 \right)\left( 2n-6 \right)}{5!}{{y}^{5}}-\frac{\left( 2n-2 \right)\left( 2n-6 \right)\left( 2n-10 \right)}{7!}{{y}^{7}}+..... \right] \\ & ={{a}_{0}}\left[ 1-\frac{2n}{2}{{y}^{2}}+\frac{{{2}^{2}}n\left( n-2 \right)}{4!}{{y}^{4}}-\frac{{{2}^{3}}n\left( n-2 \right)\left( n-4 \right)}{6!}{{y}^{6}}+..... \right] \\ & +{{a}_{1}}\left[ y-\frac{2\left( n-1 \right)}{3!}{{y}^{3}}+\frac{{{2}^{2}}\left( n-1 \right)\left( n-3 \right)}{5!}{{y}^{5}}-\frac{{{2}^{3}}\left( n-1 \right)\left( n-3 \right)\left( n-5 \right)}{7!}{{y}^{7}}+..... \right] \\ \end{align}

For n even, the series multiplying a0 terminates so it is the appropriate result for Hn(y). a1 is necessarily zero.

For n odd, the series multiplying a1 terminates so it is the appropriate result for Hn(y). a0 is necessarily zero.

The traditional method of writing Hermite polynomials is such that the coefficient of highest power of y, (yn), is 2n.

Therefore for odd n,

${{a}_{1}}={{\left( -1 \right)}^{\left( n-1 \right)/2}}\frac{2\left( n! \right)}{\left( \frac{n-1}{2} \right)!}$

For even n,

${{a}_{0}}={{\left( -1 \right)}^{n/2}}\frac{n!}{\left( \frac{n}{2} \right)!}$

Therefore,

\begin{align} & {{H}_{0}}(y)=1 \\ & {{H}_{1}}(y)=2y \\ & {{H}_{2}}(y)=\frac{\left( -1 \right)2!}{1!}\left( 1-2{{y}^{2}} \right) \\ & =4{{y}^{2}}-2 \\ & {{H}_{3}}(y)=\left( -1 \right)\times \frac{2\times 3!}{1!}\left( y-\frac{2}{3}{{y}^{3}} \right)=8{{y}^{3}}-12y \\ & {{H}_{4}}(y)={{\left( -1 \right)}^{2}}\times \frac{4!}{2!}\left( 1-4{{y}^{2}}+\frac{4.4.2}{4!}{{y}^{4}} \right) \\ & \text{ =12}\left( 1-4{{y}^{2}}+\frac{4}{3}{{y}^{4}} \right) \\ & \text{ =}\left( \text{12-48}{{y}^{2}}+16{{y}^{4}} \right) \\ & {{H}_{5}}(y)={{\left( -1 \right)}^{2}}\times \frac{2\times 5!}{2!}\left( y-\frac{2.4}{3!}{{y}^{3}}+\frac{{{2}^{2}}.4.2}{5!}{{y}^{5}} \right) \\ & \text{ =120}y-160{{y}^{3}}+32{{y}^{5}} \\ \end{align}

### Harmonic oscillator wave functions. Summary:

\begin{align} & {{\psi }_{0}}(y)={{e}^{-\frac{{{y}^{2}}}{2}}} \\ & {{\psi }_{1}}(y)=2y{{e}^{-\frac{{{y}^{2}}}{2}}} \\ & {{\psi }_{2}}(y)=\left( 4{{y}^{2}}-2 \right){{e}^{-\frac{{{y}^{2}}}{2}}} \\ & {{\psi }_{3}}(y)=\left( 8{{y}^{3}}-12y \right){{e}^{-\frac{{{y}^{2}}}{2}}} \\ & {{\psi }_{4}}(y)=\left( 12-48{{y}^{2}}+16{{y}^{4}} \right){{e}^{-\frac{{{y}^{2}}}{2}}} \\ & {{\psi }_{5}}(y)=\left( 120y-160{{y}^{3}}+32{{y}^{5}} \right){{e}^{-\frac{{{y}^{2}}}{2}}} \\ \end{align}

These wave functions are NOT normalised. Hence to get the harmonic oscillator wave functions given in the lecture on harmonic oscillator, we must multiply by the normalization constant.

### Normalization constant:

The relation is

$\int\limits_{-\infty }^{+\infty }{{{e}^{-{{y}^{2}}}}dy\text{ }{{H}_{n}}\left( y \right){{H}_{k}}\left( y \right)={{2}^{n}}n!\sqrt{\pi }{{\delta }_{nk}}}$

A derivation of this relation is found in some standard quantum mechanics text books, for example, “Quantum Mechanics” by Eugen Merzbacher, John Wiley & Sons, 1999 which is an excellent text book on this subject. Refer to page 87 of the text book, in particular equations 5.36-5.38.

For the harmonic oscillator wave functions recall:

$y=\sqrt{\alpha }x={{\left( \frac{k\mu }{{{\hbar }^{2}}} \right)}^{\frac{1}{4}}}x$

The harmonic oscillator wave functions are normalized as follows:

\begin{align} & {{\psi }_{n}}\left( x \right)={{e}^{-\alpha {{x}^{2}}/2}}{{H}_{n}}\left( \sqrt{\alpha }x \right) \\ & \int\limits_{-\infty }^{+\infty }{\psi _{n}^{*}\left( x \right){{\psi }_{k}}\left( x \right)dx=}\text{ }\int\limits_{-\infty }^{+\infty }{{{e}^{-\alpha {{x}^{2}}}}{{H}_{n}}\left( \sqrt{\alpha }x \right){{H}_{k}}\left( \sqrt{\alpha }x \right)dx} \\ \end{align}

Transform the variable and do the integral.

\begin{align} & \sqrt{\alpha }x=y, \\ & \sqrt{\alpha }dx=dy \\ & \int\limits_{-\infty }^{\infty }{dx=}\text{ }\int\limits_{-\infty }^{\infty }{\frac{dy}{\sqrt{\alpha }}} \\ & \Rightarrow \int\limits_{-\infty }^{\infty }{\psi _{n}^{*}\left( x \right){{\psi }_{k}}\left( x \right)dx=\frac{1}{\sqrt{\alpha }}\int\limits_{-\infty }^{\infty }{{{e}^{-{{y}^{2}}}}{{H}_{n}}\left( y \right){{H}_{k}}\left( y \right)dy}} \\ & \text{ =}\frac{1}{\sqrt{\alpha }}{{2}^{n}}n!\sqrt{\pi }{{\delta }_{nk}}={{N}^{2}} \\ & \Rightarrow {{\psi }_{n}}\left( x \right)\Rightarrow \frac{{{\psi }_{n}}\left( x \right)}{N}\text{ }where\text{ }N={{\left( \frac{\pi }{\alpha } \right)}^{\frac{1}{4}}}{{2}^{\frac{n}{2}}}{{\left( n! \right)}^{\frac{1}{2}}} \\ \end{align}

A few examples of normalized harmonic oscillators wave functions are:

${{\psi }_{\text{0}}}(x)={{\left( \frac{\alpha }{\pi } \right)}^{\frac{1}{4}}}{{e}^{-\alpha {{x}^{2}}/2}}$

$\text{ }{{\psi }_{\text{1}}}(x)=\frac{1}{\sqrt{2}}{{\left( \frac{\alpha }{\pi } \right)}^{\frac{1}{4}}}{{e}^{-\alpha {{x}^{2}}/2}}{{H}_{1}}\left( \sqrt{\alpha }x \right)$

${{\psi }_{\text{2}}}(x)=\frac{1}{2\sqrt{2}}{{\left( \frac{\alpha }{\pi } \right)}^{\frac{1}{4}}}{{e}^{-\alpha {{x}^{2}}/2}}{{H}_{2}}\left( \sqrt{\alpha }x \right)$

### Recursion formula:

The Hermite differential equation is

$\frac{{{d}^{2}}H}{d{{y}^{2}}}-2y\frac{dH}{dy}+\left( \frac{\lambda }{\alpha }-1 \right)H=0$

In this section we want to obtain a relation satisfied among Hermite polynomials, other than the differential equation, namely,

${{H}_{n+1}}\left( y \right)-2y{{H}_{n}}\left( y \right)+2n{{H}_{n-1}}\left( y \right)=0$

Recall that

$H\left( y \right)={{H}_{n}}\left( y \right),\text{ }n=0,1,2,.....$

Therefore the differential equation above is satisfied for each polynomial so that

$\frac{{{d}^{2}}{{H}_{n}}}{d{{y}^{2}}}-2y\frac{d{{H}_{n}}}{dy}+\left( \frac{\lambda }{\alpha }-1 \right){{H}_{n}}=0\text{ }for\text{ }n=0,1,2,....etc$

The above can be verified trivially for ${{H}_{0}}(y),\,{{H}_{1}}(y)\,and\,{{H}_{2}}(y),$

To derive the recursion formula, we will first verify the following identity

$\frac{d{{H}_{n}}\left( y \right)}{dy}=2n{{H}_{n-1}}\left( y \right)$

It must be kept in mind that when n is even, (n-1) is odd and vice versa. Consider Hn(y) for even n and therefore Hn − 1(y) for odd (n-1) as

${{H}_{n}}\left( y \right)=\frac{n!}{\left( \frac{n}{2} \right)!}{{\left( -1 \right)}^{\frac{n}{2}}}\left[ 1-\frac{2n}{2}{{y}^{2}}+\frac{{{2}^{2}}n\left( n-2 \right)}{4!}{{y}^{4}}-\frac{{{2}^{3}}n\left( n-2 \right)\left( n-4 \right)}{6!}{{y}^{6}}+.... \right]$

{{H}_{n-1}}\left( y \right)={{\left( -1 \right)}^{\frac{n-1-1}{2}}}\frac{2\left( n-1 \right)!}{\left[ \frac{\left( n-2 \right)}{2} \right]!}\left[ \begin{align} & y-\frac{2\left( n-2 \right)}{3!}{{y}^{3}}+\frac{{{2}^{2}}\left( n-2 \right)\left( n-4 \right)}{5!}{{y}^{5}}- \\ & \frac{{{2}^{3}}\left( n-2 \right)\left( n-4 \right)\left( n-6 \right)}{7!}{{y}^{7}}+.... \\ \end{align} \right]

Then,

\begin{align} & \frac{d{{H}_{n}}\left( y \right)}{dy}=\frac{n!}{\left( \frac{n}{2} \right)!}{{\left( -1 \right)}^{\frac{n}{2}}}\left[ -\frac{2.2n}{2}y+\frac{{{4.2}^{2}}n\left( n-2 \right)}{4!}{{y}^{3}}-\frac{{{6.2}^{3}}n\left( n-2 \right)\left( n-4 \right)}{6!}{{y}^{5}}+\cdots \right] \\ & =\frac{n(n-1)!}{\frac{n}{2}\left( \frac{n}{2}-1 \right)!}{{\left( -1 \right)}^{\frac{n}{2}-1}}\left[ 2ny-\frac{n\left( n-2 \right){{2}^{2}}}{3!}{{y}^{3}}+\frac{n\left( n-2 \right)\left( n-4 \right){{2}^{3}}}{5!}{{y}^{5}}-\cdots \right] \\ & =2n{{H}_{n-1}}(y) \\ \end{align}

Likewise one can prove the relation

$\frac{d{{H}_{n}}\left( y \right)}{dy}=2n{{H}_{n-1}}\left( y \right)$ starting with odd n and even (n-1).

The above is important in the derivation of a relation between Hermite polynomials since

\begin{align} & \frac{{{d}^{2}}H}{d{{y}^{2}}}=\frac{d}{dy}\left[ \frac{d{{H}_{n}}}{dy} \right]=\frac{d}{dy}\left[ 2n{{H}_{n-1}}\left( y \right) \right] \\ & =2n\frac{d{{H}_{n-1}}\left( y \right)}{dy}=2n\times 2\left( n-1 \right){{H}_{n-2}}\left( y \right) \\ & \text{ }=4n\left( n-1 \right){{H}_{n-2}}\left( y \right) \\ \end{align}

Therefore the Hermite differential equation becomes

\begin{align} & 4n\left( n-1 \right){{H}_{n-2}}\left( y \right)-2y\left[ 2n{{H}_{n-1}}\left( y \right) \right]+2n{{H}_{n}}\left( y \right)=0\text{ }\because \left( \frac{\lambda }{\alpha }-1=2n \right) \\ & \Rightarrow {{H}_{n}}\left( y \right)-2y{{H}_{n-1}}\left( y \right)+2\left( n-1 \right){{H}_{n-2}}\left( y \right)=0 \\ \end{align}

This is valid for all n’s $\left( n-2\ge 0 \right).$Replacing n by n +1,

${{H}_{n+1}}\left( y \right)-2y{{H}_{n}}\left( y \right)+2n{{H}_{n-1}}\left( y \right)=0$

Thus the recursion formula for Hermite polynomials is

\begin{align} & {{H}_{n+1}}-2y{{H}_{n}}+2n{{H}_{n-1}}=0 \\ & n\ge 0,\text{ }{{H}_{0}}\left( y \right)\text{=1} \\ & {{H}_{1}}\left( y \right)=2y\text{ } \\ \end{align}

Using the above formula,

\begin{align} & {{H}_{2}}\left( y \right)=2y{{H}_{1}}-2\times 1\left( {{H}_{0}} \right) \\ & \text{ }=4{{y}^{2}}-2 \\ \end{align}

which we have obtained earlier. All other expression such as H3(y), H4(y), H5(y) etc. can be obtained by using this expression as well.

### Generating function for Hermite polynomials:

Hermite polynomials can be constructed using a generating function technique. (i.e the function generates the polynomials upon substituting the right values of n)

Let us define a new function which collects all Hermite polynomials in a series expansion as:

\begin{align} & G\left( y,s \right)=\sum\limits_{n=0}^{\infty }{\frac{{{s}^{n}}}{n!}}\text{ }{{H}_{n}}(y) \\ & then \\ & \frac{dG}{dy}=\sum\limits_{n=0}^{\infty }{\frac{{{s}^{n}}}{n!}}\text{ }\frac{d{{H}_{n}}(y)}{dy} \\ & \text{ =}\sum\limits_{n=1}^{\infty }{\frac{{{s}^{n}}2n}{n!}}\text{ }{{H}_{n-1}}\left( y \right) \\ & \text{ }=2s\sum\limits_{n=1}^{\infty }{\frac{{{s}^{n-1}}}{\left( n-1 \right)!}}\text{ }{{H}_{n-1}}\left( y \right) \\ & \text{ }=2sG\left( s,y \right) \\ \end{align}

Integrate the above equation to get

\begin{align} & G\left( y,s \right)=G\left( 0,s \right){{e}^{2sy}} \\ & from \\ & G\left( y,s \right)=\sum\limits_{n=0}^{\infty }{\frac{{{s}^{n}}}{n!}}\text{ }{{H}_{n}}(y)\text{ where }G\left( 0,s \right)=\sum\limits_{n=0}^{\infty }{\frac{{{s}^{n}}}{n!}}\text{ }{{H}_{n}}\left( 0 \right) \\ \end{align}

All the ${{H}_{n}}\left( 0 \right)$ with n being odd are zero. All the ${{H}_{n}}\left( 0 \right)$ with n being even are given by

\begin{align} & {{H}_{n}}(0)={{\left( -1 \right)}^{\frac{n}{2}}}\frac{n!}{\left( \frac{n}{2} \right)!};\text{ (even }n\text{)} \\ & \therefore G\left( 0,s \right)=\sum\limits_{n=even}^{\infty }{\frac{{{s}^{n}}{{\left( -1 \right)}^{\frac{n}{2}}}}{\left( \frac{n}{2} \right)!}} \\ & \text{ =}\sum\limits_{m=0}^{\infty }{\frac{{{s}^{2m}}{{\left( -1 \right)}^{m}}}{m!}}={{e}^{-{{s}^{2}}}}\text{ } \\ & \therefore G\left( y,s \right)=G\left( 0,s \right){{e}^{2sy}}={{e}^{-{{s}^{2}}+2sy}}={{e}^{+{{y}^{2}}-{{\left( s-y \right)}^{2}}}} \\ \end{align}

From the fact that

$G\left( y,s \right)=\sum\limits_{n=0}^{\infty }{\frac{{{s}^{n}}}{n!}{{H}_{n}}\left( y \right)}$

and expanding $G\left( y,s \right)$ as a Taylor series in s, we get

\begin{align} & G\left( y,s \right)=\sum\limits_{n=0}^{\infty }{\frac{{{s}^{n}}}{n!}}\text{ }{{\left| \frac{{{d}^{n}}G\left( y,s \right)}{d{{s}^{n}}} \right|}_{s=0}} \\ & we\text{ }note\text{ }that \\ & {{H}_{n}}\left( y \right)={{\left[ \frac{{{d}^{n}}}{d{{s}^{n}}}G\left( y,s \right) \right]}_{s=0}} \\ & \text{ }={{\left\{ \frac{{{d}^{n}}}{d{{s}^{n}}}\left[ {{e}^{{{y}^{2}}-{{\left( s-y \right)}^{2}}}} \right] \right\}}_{s=0}} \\ & {{e}^{{{y}^{2}}}}{{\left[ \frac{{{d}^{n}}}{d{{s}^{n}}}{{e}^{-{{\left( s-y \right)}^{2}}}} \right]}_{s=0}}\text{ =}{{\left( -1 \right)}^{n}}{{e}^{{{y}^{2}}}}\frac{{{d}^{n}}}{d{{y}^{n}}}\left( {{e}^{-{{y}^{2}}}} \right) \\ \end{align}

The last step is verified below for n=1 and 2 to give a feel for the equation. Try to do this generally for yourself.

\begin{align} & {{\left[ \frac{d}{ds}{{e}^{-{{\left( s-y \right)}^{2}}}} \right]}_{s=0}}={{\left. -2\left( s-y \right){{e}^{-{{\left( s-y \right)}^{2}}}} \right|}_{s=0}}=2y{{e}^{-{{y}^{2}}}}=\left( -1 \right)dy\frac{d}{dy}{{e}^{-{{y}^{2}}}} \\ & {{\left[ \frac{{{d}^{2}}}{d{{s}^{2}}}{{e}^{-{{\left( s-y \right)}^{2}}}} \right]}_{s=0}}=\frac{d}{ds}\left[ -2\left( s-y \right){{e}^{-{{\left( s-y \right)}^{2}}}} \right] \\ & \text{ }={{\left[ -2{{e}^{-{{\left( s-y \right)}^{2}}}} \right]}_{s=0}}+{{\left[ 4{{\left( s-y \right)}^{2}}{{e}^{-{{\left( s-y \right)}^{2}}}} \right]}_{s=0}} \\ & \text{ }=-2{{e}^{-{{y}^{2}}}}+4{{y}^{2}}{{e}^{-{{y}^{2}}}} \\ & \text{ }={{\left( -1 \right)}^{2}}\frac{{{d}^{2}}}{d{{y}^{2}}}{{e}^{-{{y}^{2}}}} \\ \end{align}

Likewise one can verify the general result.

Thus the final result:

${{H}_{n}}\left( y \right)={{\left( -1 \right)}^{n}}{{e}^{{{y}^{2}}}}\frac{{{d}^{n}}}{d{{y}^{n}}}{{e}^{-{{y}^{2}}}}$

The right hand side is known as the generating function as, by successive differentiation, one can generate Hermite polynomials of various orders.

### Energies of harmonic oscillator:

The boundary conditions for the wave function require that

\begin{align} & \frac{\lambda }{\alpha }=2n+1 \\ & \Rightarrow \frac{2\mu E}{{{\hbar }^{2}}}\times \frac{\hbar }{\sqrt{k\mu }}=\left( 2n+1 \right) \\ & \Rightarrow \frac{2E}{\hbar \left( \sqrt{\frac{k}{\mu }} \right)}=\left( 2n+1 \right) \\ & or\text{ }E=\left( n+\frac{1}{2} \right)\hbar \left( \sqrt{\frac{k}{\mu }} \right)=\left( n+\frac{1}{2} \right)\hbar \omega \\ & where\text{ }\omega =\sqrt{\frac{k}{\mu }} \\ \end{align}

is the expression for energy.

$w=\sqrt{\frac{k}{\mu }}$ has the dimension of angular frequency 2πυ, where

$\upsilon \text{ has the dimension }\left( \frac{1}{Time} \right)$

Thus the energy of the harmonic oscillator is quantized with n = 0, 1, 2…….. etc as possible values.