Problem Set 4 - Siavash Davani's Website

Exercise 1 (Fourier transformation)

Assume that a quantum free particle is in the state

\Psi_a(x) = \begin{cases} \frac{1}{\sqrt{2a}}, & -a \le x \le a \\ 0, & \text{otherwise} \end{cases}.

(1)

Calculate the Fourier transform of $\Psi_a(x)$
$\Phi_a(k) = \frac{1}{\sqrt{2\pi}} \int \Psi_a(x) e^{-ikx} \, dx.$
(2)
Solution
We have
$\begin{align} \Phi_a(k) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \Psi_a(x) e^{-ikx} \, dx \\ &= \frac{1}{\sqrt{2\pi}} \int_{-a}^a \frac{1}{\sqrt{2a}} e^{-ikx} \, dx \\ &= \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{2a}} \int_{-a}^a e^{-ikx} \, dx \\ &= \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{2a}} \, \frac{e^{-ikx}}{-ik} \, \Bigg|_{x=-a}^a \\ &= \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{2a}} \, \frac{e^{-ika}-e^{ika}}{-ik}\\ &= \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{2a}} \, \frac{-2i\sin(ak)}{-ik}\\ &= \frac{1}{\sqrt{\pi a}} \, \frac{\sin(ak)}{k}\\ \end{align}$
(3)
Try to roughly plot the functions $\Psi_a(x)$ and $\Phi_a(k)$ for three cases of $a=0.5$ , $a=1$ , and $a=2$ .
Solution
Figures will be added later. I could not figure out an easy way to transfer the tikz graphs to the website!
Using your plots from part $(b)$ and without further calculations, answer the following questions. How does $\Phi_a(k)$ change in case of different $a$ ? Do the $\Phi_a(k)$ functions get narrower or broader as $a$ increases? Compare this behavior to the corresponding $\Psi_a(k)$ functions.
Solution
We can see from the figures that when $a$ increases, the $\Phi_{a}(x)$ functions get narrower and narrower. This behavior is the opposite to $\Psi_{a}(x)$ where increasing $a$ makes the function broader.

Exercise 2 (Operators and eigenfunctions)

In the lectures, we introduced the concept of operators acting on functions. An operator takes a function as an input and transforms it to another function

f(x) \quad\overset{\hat{O}}{\longrightarrow}\quad g(x) = \hat{O} f(x).

(4)

There is an important concept related to operators known as an eigenfunction. There can be functions for an operator which have the special feature that the action of the operator on them does not change the function, except for possibly multiplying it by a scalar number; for example

\hat{O}f(x) = \lambda f(x),

(5)

where λ is in general a complex number. In this case, we say $f(x)$ is an eigenfunction of the $\hat{O}$ operator with the eigenvalue of λ.

Below, you are given 5 operators $(a-e)$ and 5 functions $(1-5)$ . Find out which function is an eigenfunction of which operator and write down the corresponding eigenvalue!

\def\arraystretch{1.5} \begin{array}{cc|cc} (a) & (1-x^2)\ppdv{}{x} - x\pdv{}{x} & (1) & \sin(-2x)+\cos(2x) \\ (b) & \ppdv{}{x} & (2) & 4x^4 - 12x^2 + 3 \\ (c) & \ppppdv{}{x} & (3) & 4x^3 - 3x \\ (d) & \ppdv{}{x}-2x\pdv{}{x} & (4) & x^2 - 4x + 2\\ (e) & x\ppdv{}{x}+(1-x)\pdv{}{x} & (5) & e^{8x} + e^{-8x} \end{array}

(6)

Hint: As an example, the operator $x^2\pdv{}{x}$ acting on a function $f(x)$ is evaluated as $x^2\pdv{f(x)}{x}$ ; meaning that first, the the partial derivative is calculated and next, the result is multiplied by $x^2$ .

Solution

The way to solve this problem is by trial and error!

$(a) \longleftrightarrow (3)$ : We have

\begin{align} \pdv{}{x} \Big( 4x^3 - 3x \Big) &= 12x^2 - 3 \\ \ppdv{}{x} \Big( 4x^3 - 3x \Big) &= 24x \end{align}

(7)

Then we can see that

\begin{align} (1-x^2)\ppdv{f}{x} - x\pdv{f}{x} &= (1-x^2)(24x) - x(12x^2 - 3) \\ &= 24x - 24x^3 - 12x^3 + 3x \\ &= - 36x^3 + 27x \\ &= -9 \Big( 4x^3 - 3x \Big) \end{align}

(8)

Therefore the eigenvalue is $\lambda=-9$ .

$(b,c) \longleftrightarrow (5)$ : We have

\begin{align} \ppdv{}{x} \Big( e^{8x} + e^{-8x} \Big) &= \pdv{}{x} \Big( 8e^{8x} - 8 e^{-8x} \Big) \\ &= 64e^{8x} + 64 e^{-8x} \\ &= 64 \Big( e^{8x} + e^{-8x} \Big) \end{align}

(9)

Therefore the eigenvalue is $\lambda=64$ .

Because the function is an eigenfunction of $\ppdv{}{x}$ , it is trivially also an eigenfunction of $\ppppdv{}{x}$ but with the eigenvalue is $\lambda^2=2^{12}$ .

$(b,c) \longleftrightarrow (1)$ : We have

\begin{align} \ppdv{}{x} \Big( \sin(-2x)+\cos(2x) \Big) &= \pdv{}{x} \Bigg( -2\cos(-2x)-2\sin(2x) \Bigg) \\ &= -4\sin(-2x)-4\cos(2x) \\ &= -4 \Big( \sin(-2x)+\cos(2x) \Big) \end{align}

(10)

Therefore the eigenvalue is $\lambda=-4$ .

Because the function is an eigenfunction of $\ppdv{}{x}$ , it is trivially also an eigenfunction of $\ppppdv{}{x}$ but with the eigenvalue is $\lambda^2=16$ .

$(d) \longleftrightarrow (2)$ : We have

\begin{align} \pdv{}{x} \Big( 4x^4 - 12x^2 + 3 \Big) &= 16x^3 - 24x \\ \ppdv{}{x} \Big( 4x^4 - 12x^2 + 3 \Big) &= 48x^2 - 24 \end{align}

(11)

Then we can see that

\begin{align} \ppdv{f}{x}-2x\pdv{f}{x} &= 48x^2 - 24 - 2x (16x^3 - 24x) \\ &= 48x^2 - 24 - 32x^4 + 48x^2 \\ &= -32x^4 + 96x^2 - 24 \\ &= -8 \Big( 4x^4 - 12x^2 + 3 \Big) \end{align}

(12)

Therefore the eigenvalue is $\lambda=-8$ .

$(e) \longleftrightarrow (4)$ : We have

\begin{align} \pdv{}{x} \Big( x^2 - 4x + 2 \Big) &= 2x - 4 \\ \ppdv{}{x} \Big( x^2 - 4x + 2 \Big) &= 2 \end{align}

(13)

Then we can see that

\begin{align} x\ppdv{f}{x}+(1-x)\pdv{f}{x} &= 2x + (1-x)(2x-4) \\ &= 2x -2x^2 + 6x - 4 \\ &= -2x^2 + 8x - 4 \\ &= -2 \Big( x^2 - 4x + 2 \Big) \end{align}

(14)

Therefore the eigenvalue is $\lambda=-2$ .