91Ó°ÊÓ

Hilbert space:

The collection of all functions of $\mathrm{x}$constitutes a vector space, but for our purposes it is much too large. The wave function $\mathrm{Ψ}$ must be normalised to represent a hypothetical physical state:

$∫|\mathrm{Ψ}{|}^2\mathrm{dx}=1$

The set of all square integrable function, on a specified interval,

$\mathrm{f}\left(\mathrm{x}\right)\text{suchthat}{∫}_{\mathrm{a}}^{\mathrm{b}}\left|\mathrm{f}\right(\mathrm{x}){|}^2\mathrm{dx}<\mathrm{∞}$

(a)

Given the function: Hilbert space is the vector space of all square-integrable functions.

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{ν}}$

Apply the normalization we get:

Now, $⟨fâˆ\pounds fâŸ\copyright$ is finite only if $0^{2\mathrm{ν}+1}$ is finite, the value of$0^{2\mathrm{ν}+1}$ is zero, excepts if $2\mathrm{ν}+1$is less than zero, the function become infinite. Provided,$(2\mathrm{ν}+1)>0$

The function$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{v}}$ in Hilbert space, thus, the range is,$\mathrm{ν}>−\frac{1}{2}$.

b)

From part (a), for $\mathrm{v}=\frac{1}{2}>−\frac{1}{2}$, $\mathrm{f}\left(\mathrm{x}\right)$is in Hilbert space, and $\mathrm{xf}\left(\mathrm{x}\right)$is Hilbert space also, but $\mathrm{f}\text{'}\left(\mathrm{x}\right)$is not.

For $\mathrm{xf}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{v}+1}$, we have:

which is in Hilbert space for $\mathrm{ν}=\frac{1}{2}$. Now for ${\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)={\mathrm{vx}}^{\mathrm{ν}−1}$, we have:

For $\mathrm{ν}=\frac{1}{2}$the lower limit of the integration gives usdata-custom-editor="chemistry" $0°$ which is infinite, therefore $f^{\text{'}}\left(x\right)$is not in the Hilbert space.