91Ó°ÊÓ

A particle confined in a box with indefinitely rigid walls that the particle cannot penetrate is the simplest example of such a system. This potential is known as an infinite square well and is given by:

_{$\mathit{Ψ}\left(x,0\right)\mathbf{=}\{\begin{array}{ll}Ax& 0≤x≤\frac{a}{2}\\ A\left(a-x\right)& \frac{a}{2}≤x≤a\end{array}$}

Clearly, where the potential is infinite, the wave function must be zero.

(a)

Consider problem 2.7, the instance of a particle in an infinite square well with a triangular starting wave function:

$\mathrm{Ψ}(\mathrm{x},0)=\{\begin{array}{ll}\mathrm{Ax}& 0≤\mathrm{x}≤\frac{\mathrm{a}}{2}\\ \mathrm{A}(\mathrm{a}-\mathrm{x})& \frac{\mathrm{a}}{2}≤\mathrm{x}≤\mathrm{a}\end{array}$

Determine A from normalization as:

${∫}_0^{\mathrm{a}}|\mathrm{Ψ}{|}^2\mathrm{dx}=1$

The modules of the wave function are:

${\left|\mathrm{Ψ}(\mathrm{x},0)\right|}^2=\{\begin{array}{ll}{A}^2{x}^2& 0≤\mathrm{x}≤\frac{\mathrm{a}}{2}\\ {\mathrm{A}}^2{(\mathrm{a}-\mathrm{x})}^2& \frac{\mathrm{a}}{2}≤\mathrm{x}≤\mathrm{a}\end{array}$

$$\begin{gathered} {∫}_0^a{A}^2{x}^{2}dx+{∫}_{a/2}^a{A}^2{\left(a-x\right)}^{2}dx=A^2\left[\left({\overline{)\frac{x^3}{3}}}_0^{a/2}\right)+\left(\frac{x^3}{3}-a{x}^2+a^2{\overline{)x}}_{a/2}^a\right)\right] \\ 1=A^2\left[\frac{a^3}{24}+\frac{a^3}{24}\right] \\ 1=A^2\left[\frac{2a^3}{24}\right] \end{gathered}$$

A can be calculated as:

$A=\sqrt{\frac{12}{a^3}}$

The first derivative of the wave function is:

$\frac{d\mathrm{Ψ}\left(x,0\right)}{dx}=-A\left(2\mathrm{θ}\left(x-a/2\right)-1\right)$

where,

$\mathrm{θ}\left(x\right)=\left\{\begin{array}{ll}1& x<0\\ 0& x>0\end{array}\right.$

Hence, the first derivative in terms of step function is $-A\left(2\mathrm{θ}\left(x-a/2\right)-1\right)$.

(b)

In terms of the delta function, we now take the second derivative, which is the derivative of the result of part (a),

use :

$$\begin{gathered} \frac{∂}{∂x}\mathrm{θ}\left(x\right)=\mathrm{δ}\left(x\right) \\ \frac{d^2\mathrm{Ψ}\left(x,0\right)}{d{x}^2}=-2A\mathrm{δ}\left(x-\frac{a}{2}\right) \end{gathered}$$

Thus,the second derivative of the function in delta is $-2A\mathrm{δ}\left(x-\frac{a}{2}\right)$ .

(c)

Using the result of component (b), we must now determine the following integral:

as:

using:

${∫}_{-∞}^{∞}dxf\left(x\right)\mathrm{δ}\left(x-a\right)=f\left(a\right),$

we get:

But we know that:

${\mathrm{Ψ}}^{*}\left(\frac{a}{2}\right)=\frac{aA}{2}$

thus:

${\mathrm{Ψ}}^{*}\left(\frac{a}{2}\right)=\frac{aA}{2}$

Now substitute with A, and we get:

Thus, the integral form of the width of the well is _.