Question

In Problem 2.7 (d), you got the expectation value of the energy by summing the series in Equation 2.39, but 1 warned you (in footnote 15 not to try it the "old fashioned way,"$<H>\mathbf{=}\mathbf{\int }\mathit{\Psi }{\left(x,0\right)}^{\mathbf{*}}\mathit{H}\mathit{\Psi }\left(x,0\right)\mathit{d}\mathit{x}$, because the discontinuous first derivative of$\mathit{\Psi }\mathbf{(}\mathbf{x}\mathbf{.}\mathbf{0}\mathbf{)}$renders the second derivative problematic. Actually, you could have done it using integration by parts, but the Dirac delta function affords a much cleaner way to handle such anomalies.

(a) Calculate the first derivative of $\mathit{\Psi }\mathbf{(}\mathbf{x}\mathbf{.}\mathbf{0}\mathbf{)}$(in Problem 2.7), and express the answer in terms of the step function, $\mathit{\theta }\left(x-c1/2\right)$defined in Equation (Don't worry about the end points-just the interior region

(b) Exploit the result of Problem 2.24(b) to write the second derivative of $\mathit{\Psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}$in terms of the delta function.

(c) Evaluate the integral $\mathbf{\int }\mathit{\Psi }{\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}}^{\mathbf{*}}\mathit{H}\mathit{\Psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}\mathit{d}\mathit{x}$and check that you get the same answer as before.

Short answer

1. The first derivative in terms of step function is$-A\left(2\theta \left(x-a/2\right)-1\right)$.
2. The second derivative of the function in delta is $-2A\delta \left(x-\frac{a}{2}\right)$.
3. The integral form of the width of the well is

Step-by-step solution

Define the infinite square well in triangular form;

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:

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

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

Find the first derivative

(a)

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

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

Determine A from normalization as:

${\int }_0^{\mathrm{a}}|\mathrm{\Psi }{|}^2\mathrm{dx}=1$

The modules of the wave function are:

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

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

A can be calculated as:

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

The first derivative of the wave function is:

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

where,

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

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

Find the second derivative of the function in delta form.

(b)

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

use :

$$\begin{gathered} \frac{\partial }{\partial x}\theta \left(x\right)=\delta \left(x\right) \\ \frac{d^2\Psi \left(x,0\right)}{d{x}^2}=-2A\delta \left(x-\frac{a}{2}\right) \end{gathered}$$

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

Find the integral  ∫Ψ(x,0)*HΨ(x,0)dx

(c)

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

as:

using:

${\int }_{-\infty }^{\infty }dxf\left(x\right)\delta \left(x-a\right)=f\left(a\right),$

we get:

But we know that:

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

thus:

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

Now substitute with A, and we get:

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