Hill Determinant

A Determinant which arises in the solution of the second-order Ordinary Differential Equation

$\begin{displaymath} x^2{d^2\psi\over dx^2}+x{d\psi\over dx}+\left({{\textstyle{1... ...x^2+{\textstyle{1\over 2}}h^2-b+{h^2\over 4x^2}}\right)\psi=0. \end{displaymath}$

(1)

Writing the solution as a Power Series

$\begin{displaymath} \psi=\sum_{n=-\infty}^\infty a_n x^{s+2n} \end{displaymath}$

(2)

gives a Recurrence Relation

$\begin{displaymath} h^2a_{n+1}+[2h^2-4b+16(n+{\textstyle{1\over 2}}s)^2]a_n+h^2a_{n-1}=0. \end{displaymath}$

(3)

The value of

can be computed using the Hill determinant

$\begin{displaymath} \Delta(s)=\left\vert\matrix{ \ddots & \vdots & \vdots & \vdo... ...u}& \vdots & \vdots & \vdots & \vdots & \ddots\cr}\right\vert, \end{displaymath}$

(4)

where

$\displaystyle \sigma$	$\textstyle =$	$\displaystyle {\textstyle{1\over 2}}s$	(5)
$\displaystyle \alpha^2$	$\textstyle =$	$\displaystyle {\textstyle{1\over 4}}b-{\textstyle{1\over 8}} h^2$	(6)
$\displaystyle \beta$	$\textstyle =$	$\displaystyle {\textstyle{1\over 4}}h,$	(7)

and $\sigma$ is the variable to solve for. The determinant can be given explicitly by the amazing formula

$\begin{displaymath} \Delta(s)=\Delta(0)-{\sin^2(\pi s/2)\over\sin^2({\textstyle{1\over 2}}\pi\sqrt{b-{\textstyle{1\over 2}}h^2}\,)}, \end{displaymath}$

(8)

where

$\begin{displaymath} \Delta(0)=\left\vert\matrix{\ddots & \vdots & \vdots & \vdot... ...\vdots & \vdots & \vdots & \vdots & \ddots\cr}\right\vert,eqno \end{displaymath}$

(9)

leading to the implicit equation for ,

$\begin{displaymath} \sin^2({\textstyle{1\over 2}}\pi s)=\Delta(0)\sin^2\left({{\... ...yle{1\over 2}}\pi\sqrt{b-{\textstyle{1\over 2}}h^2}\,}\right). \end{displaymath}$

(10)

See also Hill's Differential Equation

References

Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 555-562, 1953.