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What is the Fourier transform of \(\delta(x)\) ? Using Plancherel's theorem, show that $$\delta(x)=\frac{1}{2 \pi} \int_{-\infty}^{+\infty} e^{i k x} d k.$$ Comment: This formula gives any respectable mathematician apoplexy. Although the integral is clearly infinite when \(x=0\), it doesn't converge (to zero or anything else \()\) when \(x \neq 0,\) since the integrand oscillates forever. There are ways to patch it up (for instance, you can integrate from \(-L\) to \(+L\), and interpret Equation 2.147 to mean the average value of the finite integral, as \(L \rightarrow \infty\). The source of the problem is that the delta function doesn't meet the requirement (squareintegrability) for Plancherel's theorem (see footnote \(\underline{42}\) ). In spite of this, Equation 2.147 can be extremely useful, if handled with care.

Short Answer

Expert verified
The Fourier transform of \(\delta(x)\) is 1, and representing \(\delta(x)\) using exponential integrals requires limit approaches.

Step by step solution

01

Understanding the Fourier Transform of Dirac Delta

The Dirac delta function, \( \delta(x) \), is defined such that it is zero everywhere except at \( x = 0 \), where it's infinitely high, and integrates to one over the entire real line. The Fourier transform of \( \delta(x) \) is given by:\[ \mathcal{F}\{\delta(x)\}(k) = \int_{-\infty}^{\infty} \delta(x) e^{-ikx} \, dx. \]
02

Evaluating the Fourier Transform Integral

Since \( \delta(x) \) is zero everywhere except at \( x = 0 \), the integral simplifies at \( x = 0 \):\[ \int_{-\infty}^{\infty} \delta(x) e^{-ikx} \, dx = e^{-i \cdot 0 \cdot k} = 1. \]Thus, the Fourier transform of \( \delta(x) \) is 1 for all \( k \).
03

Applying Plancherel's Theorem

Plancherel's theorem is a property of the Fourier transform relating the total energy (or norm) of a function in both spatial and frequency domains. However, the delta function is not square-integrable, so Plancherel's theorem directly can't apply.Despite this, if we attempt the application naively, we find:\[ \int_{-\infty}^{\infty} \delta(x) f(x) \, dx = \frac{1}{2\pi} \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} e^{ikx} f(x) \, dx \right) \, dk. \]This can be visualized as representing the delta function as a limit of oscillating exponential functions.
04

Understanding the Integral Representation

The claimed equation is \( \delta(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{ikx} \cdot dk \). This integral diverges everywhere, but for practical purposes (as a limiting process), can approximate the behavior of the delta function, particularly through limiting processes such as integrating over finite limits and considering limits to infinity.
05

Conclusion

The analysis shows \( \mathcal{F} \{\delta(x)\} = 1 \), and the theoretical application of integrating an exponential over infinite bounds represents \( \delta(x) \) as a formal identity under very strict interpretations or limit processes, though directly it defies conventional integrability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dirac delta function
The Dirac delta function, denoted as \( \delta(x) \), is not a function in the traditional sense. It's often described as a "function" that is zero everywhere except at \( x = 0 \) where it is infinitely high. However, its most important characteristic is that its integral over the entire real line is equal to one:\[ \int_{-\infty}^{\infty} \delta(x) \, dx = 1. \]
This property allows it to "pick out" the value of a function at a specific point during integration. Mathematically, for any well-behaved function \( f(x) \):
\[ \int_{-\infty}^{\infty} \delta(x) f(x) \, dx = f(0). \]
This makes the Dirac delta essential in signals and systems for representing an impulse at a moment in time. It is used extensively in engineering and physics to model idealized point-like sources or distributions. Speaking intuitively, the Dirac delta "transforms" a real line into a pivotal point. Despite its utility, careful handling is needed as it doesn't fit into the conventional function spaces like the square-integrable functions to which many theorems apply.
Plancherel's theorem
Plancherel's theorem is a fundamental result in Fourier analysis, named after the Swiss mathematician Michel Plancherel. It states that for any square-integrable function, the energy (total norm) in the spatial domain is equal to the energy in the frequency domain.
  • Mathematically, if \( f(x) \) is square-integrable, then:
    \[ \int_{-\infty}^{\infty} |f(x)|^2 \, dx = \int_{-\infty}^{\infty} |\mathcal{F}{f}(\xi)|^2 \, d\xi, \]
where \( \mathcal{F} \) denotes the Fourier transform.
This idea is crucial because it assures us that the energy contained within the original signal is preserved in its frequency representation. Essentially, Plancherel's theorem forms a bridge between time-based and frequency-based views of signals.
However, the Dirac delta function doesn't meet the square-integrability condition, which complicates direct application of Plancherel's theorem. Despite this, by examining limiting processes and treating certain scenarios with care, mathematical techniques can still yield powerful results akin to those derived using the theorem.
integral representation
Integral representation of the Dirac delta function helps to express it using integral equations. One of the common formulations is:
\[ \delta(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{ikx} \, dk. \]
At first glance, this integral seems problematic. If you try to calculate it directly, it appears to diverge. This is because the function \( e^{ikx} \) is oscillatory and doesn't settle to zero. However, this expression is a useful formal tool.
Instead, it represents a kind of mathematical limit. For instance, you could integrate over a finite interval \([-L, L]\) and let \( L \rightarrow \infty \). This process involves considering the 'average' behavior over such intervals, smoothing out the oscillations to approximate \( \delta(x) \).
In essence, this integral representation is a convenient way to handle and manipulate the delta function in various contexts, especially when dealing with transform techniques in signal processing or in solving differential equations. Recognizing it as a limit rather than a standard integral is the key to properly utilizing this form.

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Most popular questions from this chapter

A free particle has the initial wave function $$\Psi(x, 0)=A e^{-a x^{2}}, $$ where \(A\) and \(a\) are (real and positive) constants. (a) Normalize \(\Psi(x, 0)\) (b) Find \(\Psi(x, t) .\) Hint: Integrals of the form $$\int_{-\infty}^{+\infty} e^{-\left(a x^{2}+b x\right)} d x$$ can be handled by "completing the square": Let \(y \equiv \sqrt{a}[x+(b / 2 a)]\) and note that \(\left(a x^{2}+b x\right)=y^{2}-\left(b^{2} / 4 a\right) .\) (c) Find \(|\Psi(x, t)|^{2}\). Express your answer in terms of the quantity $$w \equiv \sqrt{a /\left[1+(2 \hbar a t / m)^{2}\right]}.$$ Sketch \(|\Psi|^{2}(\text { as a function of } x)\) at \(t=0,\) and again for some very large \(t\) Qualitatively, what happens to \(|\Psi|^{2},\) as time goes on? (d) \(\quad\) Find \(\langle x\rangle,\langle p\rangle,\left\langle x^{2}\right\rangle,\left\langle p^{2}\right\rangle, \sigma_{x},\) and \(\sigma_{p .}\) Partial answer: \(\left\langle p^{2}\right\rangle=a \hbar^{2},\) but it may take some algebra to reduce it to this simple form. (e) Does the uncertainty principle hold? At what time \(t\) does the system come closest to the uncertainty limit?

Consider a particle of mass \(m\) in the potential $$V(x)=\left\\{\begin{array}{ll}\infty & x<0, \\ -32 \hbar^{2} / m a^{2} & 0 \leq x \leq a, \\\0 & x>a.\end{array}\right.$$ (a) How many bound states are there? (b) In the highest-energy bound state, what is the probability that the particle would be found outside the well \((x>a)\) ? Answer: 0.542 , so even though it is "bound" by the well, it is more likely to be found outside than inside!

A free particle has the initial wave function $$\Psi(x, 0)=A e^{-a|x|},$$ where \(A\) and \(a\) are positive real constants. (a) Normalize \(\Psi(x, 0)\) (b) Find \(\phi(k)\) (c) Construct \(\Psi(x, t),\) in the form of an integral. (d) Discuss the limiting cases ( \(a\) very large, and \(a\) very small).

The \(1 / x^{2}\) potential. Suppose $$V(x)=\left\\{\begin{array}{ll}-\alpha / x^{2}, & x>0, \\ \infty, & x \leq 0.\end{array}\right.$$ where \(a\) is some positive constant with the appropriate dimensions. We'd like to find the bound states-solutions to the time-independent Schrödinger equation $$-\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}-\frac{\alpha}{x^{2}} \psi=E \psi$$ with negative energy \((E<0)\) (a) Let's first go for the ground state energy, \(E_{0}\). Prove, on dimensional grounds, that there is no possible formula for \(E_{0}\) -no way to construct (from the available constants \(m, \hbar,\) and \(\alpha\) ) a quantity with the units of energy. That's weird, but it gets worse \(\ldots .\) (b) For convenience, rewrite Equation 2.190 as $$\frac{d^{2} \psi}{d x^{2}}+\frac{\beta}{x^{2}} \psi=\kappa^{2} \psi, \text { where } \beta \equiv \frac{2 m \alpha}{\hbar^{2}} \text { and } \kappa \equiv \frac{\sqrt{-2 m E}}{\hbar}.$$ Show that if \(\psi(x)\) satisfies this equation with energy \(E\), then so too does \(\psi(\lambda x),\) with energy \(E^{\prime}=\lambda^{2} E,\) for any positive number \(\lambda .[\) This is a catastrophe: if there exists any solution at all, then there's a solution for every (negative) energy! Unlike the square well, the harmonic oscillator, and every other potential well we have encountered, there are no discrete allowed states - and no ground state. A system with no ground state-no lowest allowed energy-would be wildly unstable, cascading down to lower and lower levels, giving off an unlimited amount of energy as it falls. It might solve our energy problem, but we'd all be fried in the process.] Well, perhaps there simply are no solutions at all \(\ldots .\) (c) (Use a computer for the remainder of this problem.) Show that $$\psi_{\kappa}(x)=A \sqrt{x} K_{i g}(\kappa x),$$ satisfies Equation 2.191 (here \(K_{i g}\) is the modified Bessel function of order \(i g,\) and \(g \equiv \sqrt{\beta-1 / 4}\) ). Plot this function, for \(g=4\) (you might as well let \(_{K}=1\) for the graph; this just sets the scale of length). Notice that it goes to 0 as \(x \rightarrow 0\) and as \(x \rightarrow \infty\). And it's normalizable: determine \(A\). How about the old rule that the number of nodes counts the number of lower-energy states? This function has an infinite number of nodes, regardless of the energy (i.e. of \(\kappa\) ). I guess that's consistent, since for any \(E\) there are always an infinite number of states with even lower energy. (d) This potential confounds practically everything we have come to expect. The problem is that it blows up too violently as \(x \rightarrow 0\). If you move the "brick wall" over a hair, $$V(x)=\left\\{\begin{array}{ll}-\alpha / x^{2}, & x>\epsilon>0, \\\\\infty, & x \leq, \epsilon \end{array}\right.$$ it's suddenly perfectly normal. Plot the ground state wave function, for \(g=4\) and \(\epsilon=1\) (you'll first need to determine the appropriate value of \(\kappa\) ), from \(x=0\) to \(x=6 .\) Notice that we have introduced a new parameter \((\epsilon),\) with the dimensions of length, so the argument in (a) is out the window. Show that the ground state energy takes the form $$E_{0}=-\frac{\alpha}{\epsilon^{2}} f(\beta),$$ for some function \(f\) of the dimensionless quantity \(\beta,\)

Suppose $$V(x)=\left\\{\begin{array}{ll}m g x, & x>0, \\\\\infty, & x \leq 0.\end{array}\right.$$ (a) Solve the (time-independent) Schrödinger equation for this potential. Hint: First convert it to dimensionless form: \(-y^{\prime \prime}(z)+z y(z)=\epsilon y(z)\) by letting \(z \equiv a x\) and \(y(z) \equiv(1 / \sqrt{a}) \psi(x)\) (the \(\sqrt{a}\) is just so \(y(z)\) is normalized with respect to \(z\) when \(\psi(x)\) is normalized with respect to \(x\) ) What are the constants \(a\) and \(\varepsilon\) ? Actually, we might as well set \(a \rightarrow 1-\) this amounts to a convenient choice for the unit of length. Find the general solution to this equation (in Mathematica DSolve will do the job). The result is (of course) a linear combination of two (probably unfamiliar) functions. Plot each of them, for \((-15

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