Completeness

To validate the completeness of these eigenfunctions one must evaluate

$$\frac{1}{2\pi i} \oint_C G_\lambda (x;\xi)~d\lambda= \frac{1}{2\p... ...os \sqrt{\lambda} (\xi -\ell)}{\sqrt{\lambda}\sin \sqrt{\lambda}\ell } d\lambda$$ (433)

along the very large (in the limit infinite) circular contour

$$C= \{ \lambda =Re^{i\theta}:~~0<\theta <2\pi \}$$

before we deformed it into $\bigcup_{n=0}^\infty C_n$ . This evaluation is facilitated by introducing

$$\sqrt\lambda=k~~ \textrm{and hence }~\frac{d\lambda}{\sqrt\lambda}=2dk~.$$

This transforms the integration contour into a very large semicircle

$$k =\sqrt{R}e^{i\phi}$$

$$=\sqrt{R}\cos\phi+i\sqrt{R}\sin\phi,\quad 0<\phi<\pi$$ (434)

Figure 4.8: Integration contour $C$ in the $\lambda$ -plane and its semicircular image the k-plane

The integral to be evaluated is therefore

$$\frac{1}{2\pi i} \oint_C G_\lambda (x;\xi)~d\lambda= \frac{-1}{2\... ...underbrace{ \frac{\cos k x ~ \cos k (\xi -\ell)}{\sin k\ell } }_\frac{N}{D}2 dk$$ (435)

In light of Eq.(4.34) one finds that

$$\cos kx =\frac{e^{ikx}+e^{-ikx}}{2}~~\quad\quad\quad \rightarrow \quad \frac{e^{-ikx}}{2} \quad ~~\textrm{as} \quad \sqrt{R}\rightarrow \infty$$ (436)

$$\cos k(\xi-\ell) = \frac{e^{ik(\xi-\ell)}+e^{-ik(\xi-\ell)}}{2}\quad \rightarrow \quad \frac{e^{ik(\xi-\ell)}}{2} \quad \textrm{as} \quad \sqrt{R}\rightarrow \infty$$ (437)

$$\sin kx =\frac{e^{ikx}+e^{-ikx}}{2i}~~\quad\quad\quad \rightarrow \quad -\frac{e^{-ikx}}{2} \quad \textrm{as} \quad \sqrt{R}\rightarrow \infty$$ (438)
so that

$$\frac{N}{D}\quad \rightarrow\quad \frac{(e^{-ikx}/2)(e^{ik(\xi-\ell)}/2)}{-(e^{-ikx})/2}=\frac{-i}{2} e^{ik(\xi-\ell)}$$ (440)

Consequently,

$$\frac{1}{2\pi i} \oint_C G_\lambda (x;\xi)~d\lambda= \frac{-1}{2\pi i} \int_{-\sqrt R}^{\sqrt{R}} e^{ik(\xi-x)}dk$$ (441)

The integrand is analytic in the semidisk bounded by the semicircle and the the real interval $[-\sqrt{R},\sqrt{R}]$ as in the righthand picture of Figure 4.8. Thus one can use the Cauchy-Goursat theorem to deform the semicircular contour into a straight line just barely above the real $k$ -axis. This changes Eq.(4.41) into an integral along the real axis,

$$\frac{1}{2\pi i} \oint_C G_\lambda (x;\xi)~d\lambda= \frac{-1}{2\... ...^{ik(\xi-x)}dk~\rightarrow ~ -\delta(\xi-x) \textrm{ as } R\rightarrow \infty~.$$ (442)

This shows that the set of orthonormal eigenfunctions forms a complete set. Indeed, comparing this expression with Eq.(4.32) one has

$$\delta(x-\xi)=\frac{1}{\ell}+ \frac{2}{\ell}\sum_{m=1}^\infty \co... ...i x}{\ell} ~ \cos \frac{m\pi\xi}{\ell} ~~ \textrm{whenever}~~0<x,\xi <\ell~~.$$

This is the requisite completeness relation.

From a different perspective this relation is also the spectral representation of the Dirac delta function, which one may compare with that of the Green's function,

$$G_\lambda(x;\xi)= -\frac{1}{\ell(\lambda-0)} -\frac{2}{\ell}\sum_... ...s \frac{m\pi\xi}{\ell} } {\displaystyle(\lambda -\frac{m^2\pi^2}{\ell^2})} ~~.$$

Lecture 34