Solution to the Hermite Differential Equation

One typically finds the Hermite differential equation in the context of an infinite square well potential and the consequential solution of the Schr√∂dinger equation. However, I will consider this equation is its “raw” mathematical form viz.

\displaystyle \frac{d^{2}y}{dx^{2}}-2x\frac{dy}{dx}+\lambda y(x) =0. (1)

First we will consider the more general case, leaving \lambda undefined. The second case will consider in a future post \lambda = 2n, n\in \mathbb{Z}^{+}, where \mathbb{Z}^{+}=\bigg\{x\in\mathbb{Z}|x > 0\bigg\}.


Let us assume the solution has the form

\displaystyle y(x)=\sum_{j=0}^{\infty}a_{j}x^{j}. (2)

Now we take the necessary derivatives

\displaystyle y^{\prime}(x)=\sum_{j=1}^{\infty}ja_{j}x^{j-1}, (3)

\displaystyle y^{\prime \prime}(x)=\sum_{j=2}^{\infty} j(j-1)a_{j}x^{j-2}, (4)

where upon substitution yields the following

\displaystyle \sum_{j=2}^{\infty}j(j-1)a_{j}x^{j-2}-\sum_{j=1}^{\infty}2ja_{j}x^{j}+\sum_{j=0}^{\infty}\lambda a_{j}x^{j}=0, (5)

Introducing the dummy variable m=j-2 and using this and its variants we arrive at

\displaystyle \sum_{j=0}^{\infty}(j+2)(j+1)a_{j+2}x^{j}-\sum_{j=0}^{\infty}2ja_{j}x^{j}+\sum_{j=0}^{\infty}\lambda a_{j}x^{j}=0. (6)

Bringing this under one summation sign…

\displaystyle \sum_{j=0}^{\infty}[(j+2)(j+1)a_{j+2}-2ja_{j}+\lambda a_{j}]x^{j}=0. (7)

Since \displaystyle \sum_{j=0}^{\infty}x^{j}\neq 0, we therefore require that

\displaystyle (j+2)(j+1)a_{j+2}=(2j - \lambda)a_{j}, (8)


\displaystyle a_{j+2}=\frac{(2j-\lambda)a_{j}}{(j+2)(j+1)}. (9)

This is our recurrence relation. If we let j=0,1,2,3,... we arrive at two linearly independent solutions (one even and one odd) in terms of the fundamental coefficients a_{0} and a_{1} which may be written as

\displaystyle y_{even}(x)= a_{0}\bigg\{1+\sum_{j=0}^{j/2}\frac{(-1)^{j}(\lambda -2j)!}{(j+2)!}x^{j}\bigg\}, (10)


\displaystyle y_{odd}(x)=a_{1}\bigg\{\sum_{j=0}^{(j-1)/2}\frac{(-1)^{j}(\lambda-2j)!}{(j+2)!}x^{j}\bigg\}. (11)

Thus, our final solution is the following

\displaystyle y(x)=y_{even}(x)+y_{odd}(x), (12.1)


\displaystyle y(x)=a_{0}\bigg\{1+\sum_{j=0}^{j/2}\frac{(-1)^{j}(\lambda-2j)!}{(j+2)!}x^{j+2}\bigg\}+a_{1}\bigg\{x+\sum_{j=1}^{(j-1)/2}\frac{(-1)^{j}(\lambda-2j)!}{(j+2)!}x^{j+2}\bigg\}. (12.2)




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