Contour Integration Around the Branch Cut

Lecture 37

If the poles of a Green's function coalesce into a branch cut, can one expect that the sum over the discrete eigenfunctions, Eq.(4.27), mutates into a corresponding integral? The answer is `yes', and this means that instead of representing a function as a discrete sum of eigenfunctions, one now represents a function as an integral transform. The Green's function for a semi-infinite string furnishes us with the archetypical recipe for obtaining this integral transform. It is a two step process:

1. Evaluate the contour integral of $G_\lambda(x;\xi)$ over a circle with unlimited large radius:

   $$\oint G_\lambda(x;\xi) ~d\lambda =\oint\frac{\sin \sqrt{\lambda}\,x}{\sqrt{\lambda}} e^{\displaystyle i\sqrt{\lambda} \,\xi}~d\lambda \quad 0<x<\xi$$

   
   
   When $0<\xi<x$ one interchanges $x$ and $\xi$ on the right hand side. The contour path of integration is
   $$\lambda(\theta)=Re^{i\theta}\quad 0<\theta<2\pi~.$$
   In terms of the complex variable
   $$k=\sqrt\lambda$$
   this contour integral extends over a semicircle from $k=\vert k \vert$ to $k=\vert k \vert e^{i\pi}$

   $$\oint G_\lambda ~d\lambda = \lim_{\vert k \vert \to \infty} \int_... ...vert }^{\vert k \vert e^{i\pi}} \frac{e^{ik(x+\xi)} -e^{-ik(x-\xi)} }{2i}2~dk~.$$

   
   
   The integrand is analytic in $k$ . Consequently, the semicircle can be straightened out into a line segment along the real axis. The integral becomes therefore
   

   $$\oint G_\lambda ~d\lambda = i\int_{-\infty}^\infty \left[ e^{ik(x+\xi)} -e^{-ik(x-\xi)}\right]~dk$$

   $$= 2\pi i \left[ \delta(x+\xi)- \delta(x-\xi)\right]$$

   
   For a semi-infinite string the domain variables are only positive, $0<x<\xi$ . Therefore the first Dirac delta function vanishes. We are left with

   $$\frac{1}{2\pi i}\oint G_\lambda ~d\lambda =- \delta(x-\xi)~.$$ (471)

   
   

2. The second step also starts with the closed contour integral

   $$\oint G_\lambda(x;\xi) ~d\lambda ~,$$ (472)

   
   
   but this time the circular contour gets deformed into two linear paths on either side of the positive real axis $[0,\infty)$ , the branch cut of
   $$G_\lambda(x;\xi)=\frac{\sin \sqrt{\lambda}x_<}{\sqrt{\lambda}} e^{\displaystyle i\sqrt{\lambda} \,x_>} ~.$$
   Designate the two values of $G_\lambda(x;\xi)$ on opposite sides of the branch cut by $G_+$ and $G_-$ . The integral is therefore
   

   $$\oint G_\lambda(x;\xi)~d\lambda = \int_\infty^0 G_+~d\lambda +\int^\infty_0 G_-~d\lambda$$

   $$= \int^\infty_0 [G_- - G_+]~d\lambda ~.$$ (473)

   
   To evaluate the difference $G_- - G_+$ note that

   Figure 4.15: Evaluation of the Green's function just above and just below the branch cut of $\lambda ^{1/2}$ on its first Riemann sheet.

   
   the value of $\sqrt{\lambda}$ is
   $$\sqrt{\lambda}=\left\{ \begin{array}{rl} \vert \lambda \vert ^{1/... ...t \lambda \vert ^{1/2}~\textrm{just~below~the branch~cut} \end{array} \right.$$
   Consequently, the value of the Green's function at these locations is
   

   $$G_+ = \frac{\sin \vert \lambda \vert^{1/2} x_< }{\vert\lambda\vert^{1/2} } \exp\{ i\vert\lambda\vert^{1/2} x_>\}$$

   $$G_- = \frac{\sin \vert \lambda \vert^{1/2} x_< }{\vert\lambda\vert^{1/2} } \exp\{ -i\vert\lambda\vert^{1/2} x_>\}$$

   
   The discontinuity across the branch cut is therefore
   

   $$G_+ -G_-\equiv [G_\lambda] = \frac{2i}{\vert\lambda\vert^{1/2}} \sin \vert \lambda \vert^{1/2} x_< ~\sin \vert \lambda \vert^{1/2} x_>$$

   $$= \frac{2i}{\vert\lambda\vert^{1/2}} \sin \vert \lambda \vert^{1/2} x ~\sin \vert \lambda \vert^{1/2} \xi$$ (474)

   
   Insert this result into Eq.(4.75), change the integration variable to $\vert\lambda\vert^{1/2}=k$ and obtain the result that
   

   $$\frac{1}{2\pi i} \oint G_\lambda (x;\xi)~d\lambda = \frac{-1}{\pi}\int_0^\infty \frac{\sin \vert \lambda \vert^{1/2} x ~\sin \vert \lambda \vert^{1/2} \xi} {\vert\lambda\vert^{1/2}} ~d\lambda$$

   $$= \frac{-2}{\pi}\int_0^\infty \sin kx~\sin k\xi ~dk$$ (475)

   
   

This two step procedure yields two alternative expressions, Eqs.(4.73) and (4.77) for the contour integral of the Green's function. Their equality yields the spectral representation of the Dirac delta function for a semi-infinite string,

$$\delta(x-\xi)=\frac{2}{\pi}\int_0^\infty \sin kx~\sin k\xi ~dk$$ (476)