Square Integrability

Let us determine how the location of the point $\lambda$ controls the square-integrability of the exponential solution $e^{i\lambda^{1/2}x}$ on the inteval $[0,\infty)$ .

With

$$\lambda^{1/2}\equiv \alpha+i\beta$$

the $\lambda$ -parametrized function

$$\phi_\lambda(x)=e^{i\lambda^{1/2}x}=\left\{ \begin{array}{ll... ...}x}&=e^{i\alpha x} e^{-\beta x} &\beta\le 0 \end{array}\right.$$

has entirely different integrability properties depending on whether $\lambda$ lies in the first Riemann sheet ( $\lambda^{1/2}=\sqrt{\lambda}, ~\textrm{i.e.} ~\beta >0$ ) or in the second Riemann sheet ( $\lambda^{1/2}=-\sqrt{\lambda}, \textrm{i.e.} \beta <0$ ), that is to say, in the domain of which branch of $\lambda ^{1/2}$ the point $\lambda$ happens to lie. In fact, from

$$\int_0^\infty \vert \exp\{i\lambda^{1/2} x\}\vert^2~dx= \int_0^\infty e^{-2\beta x}~dx = \frac{1}{2\beta}~~~$$ (461)

we see that $\exp\{i\lambda^{1/2}x\}$ is square-integrable $\left( ~\in ~L^2[0,\infty)~\right)$ only when $\beta>0$ , but the integral diverges whenever $\beta \le 0$ . In other words,

$\exp\{i\lambda^{1/2}x\}$ is square-integrable on $[0,\infty)$ whenever $\lambda$ lies on the 1st Riemann sheet, and not on the real $\lambda$ -axis. An analogous statement hold for the 2nd Riemann sheet and $(-\infty,0]$ . Thus the requirement of square integrability relates the Riemann sheets of $\lambda ^{1/2}$ to the two semi-infinite integration domains of $\exp(i\lambda^{1/2}x)$ :

$$\lambda \in \textrm{1st Riemann sheet} \Longleftrightarrow \exp(i\lambda^{1/2}x) \in L^2[0,\infty)$$
and

$$\lambda \in \textrm{2st Riemann sheet} \Longleftrightarrow \exp(i\lambda^{1/2}x) \in L^2(-\infty,0]$$

whenever $\lambda\ne$ real. Thus,

$$\exp(i\lambda^{1/2}x)\textrm{ is square integrable on }[0,\infty) \Rightarrow\exp(i\lambda^{1/2}x)=\exp(i\sqrt\lambda x)$$
and

$$\exp(i\lambda^{1/2}x)\textrm{ is square integrable on }(-\infty,0] \Rightarrow\exp(i\lambda^{1/2}x)=\exp(-i\sqrt\lambda x)~.$$

Lecture 36