/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Delta functions live under integ... [FREE SOLUTION] | 91Ó°ÊÓ

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Delta functions live under integral signs, and two expressions \(\left(D_{1}(x) \text { and } D_{2}(x)\right)\) involving delta functions are said to be equal if $$\int_{-\infty}^{+\infty} f(x) D_{1}(x) d x=\int_{-\infty}^{+\infty} f(x) D_{2}(x) d x,$$ for every (ordinary) function \(f(x).\) (a) Show that $$\delta(c x)=\frac{1}{|c|} \delta(x).$$ where \(c\) is a real constant. (Be sure to check the case where \(c\) is negative.) (b) \(\operatorname{Let} \theta(x)\) be the step function: $$\theta(x) \equiv\left\\{\begin{array}{ll}1, & x>0 \\\0, & x<0\end{array}\right.$$ (In the rare case where it actually matters, we define \(\theta(0)\) to be \(1 / 2\).) Show that \(d \theta / d x=\delta(x).\)

Short Answer

Expert verified
(a) \(\delta(cx) = \frac{1}{|c|} \delta(x)\). (b) \(d\theta/dx = \delta(x)\).

Step by step solution

01

Understand the Delta Function Property

The delta function \(\delta(x)\) is defined by the integral property: \(\int_{-\infty}^{+\infty} f(x) \delta(x) \, dx = f(0)\) for any function \(f(x)\). This property emphasizes the 'sifting' effect of delta function, selecting the value of \(f(x)\) at the origin.
02

Show Transformation of Delta Function (Part a)

Consider the transformation \(\delta(cx)\). Using the substitution \(u = cx\), resulting in \(du = c\,dx\), we have: \[\int_{-\infty}^{+\infty} f(x) \delta(cx) \, dx = \int_{-\infty}^{+\infty} f\left(\frac{u}{c}\right) \delta(u) \frac{du}{c} = \int_{-\infty}^{+\infty} f(0) \frac{du}{c} = \frac{f(0)}{|c|}.\] To ensure the equation holds for any \(f(x)\), \(\delta(cx)\) must equal \(\frac{1}{|c|} \delta(x)\).
03

Define and Differentiate the Step Function (Part b)

The step function \(\theta(x)\) is defined as 1 for \(x > 0\), 0 for \(x < 0\), and 0.5 for \(x = 0\). The focus is on differentiating \(\theta(x)\): for any interval around \(x = 0\), \(\theta(x)\) steps from 0 to 1, which implies that its derivative \(\frac{d}{dx}\theta(x) = \delta(x)\), illustrated by a sharp (infinite) spike at the origin.

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Key Concepts

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

Delta Function
In quantum mechanics, the delta function, often written as \(\delta(x)\), acts as an important theoretical concept. Its primary role is to pick out specific values from within functions. The delta function is unique as it is zero everywhere except at the origin, where it reaches an 'infinite peak'. Mathematically, this is expressed through the integral property \(\int_{-\infty}^{+\infty} f(x) \delta(x) \, dx = f(0)\). This is often described as a 'sifting' effect, where \(\delta(x)\) captures the value of \(f(x)\) precisely at \(x = 0\).
Understanding this function is crucial for many applications in physics, as it allows for the simplification of complex mathematical calculations, especially when dealing with point charges in electromagnetism or probability distributions in statistics.
Integral Calculus
Integral calculus is a branch of mathematics essential for analyzing accumulation processes, like finding areas and solving differential equations. When dealing with delta functions, integrals play a key role in defining their behavior.
For instance, when we change variables in integrals involving the delta function, such as \(\delta(cx)\), an understanding of substitution is needed. This involves recognizing how differentials also change during substitution \(u = cx; \, du = c \, dx\). It is these integral changes that enable us to demonstrate properties like \(\delta(cx) = \frac{1}{|c|} \delta(x)\).
This property shows that the delta function's 'accumulative effect' over a scaled interval is proportionate to the reciprocal of the magnitude of the constant \(c\). Thus, understanding integral calculus is crucial for mastering applications involving delta functions in physics and engineering.
Step Function
The step function, commonly denoted as \(\theta(x)\), is a basic piecewise function used in various fields to model systems with sudden changes. It is defined as \(\theta(x) = 1\) if \(x > 0\), \(\theta(x) = 0\) if \(x < 0\), and \(\theta(0) = 0.5\).
One interesting property of the step function is its relationship with the delta function. As \(\theta(x)\) transitions from 0 to 1 at \(x = 0\), its derivative exhibits a sharp change, which can be represented by \(\delta(x)\). Thus, differentiating the step function over an interval straddling zero yields the delta function, showcasing a powerful link between these two mathematical concepts.
This link provides insights into how sudden changes in values can be expressed using the spike-like characteristics of the delta function, enriching our understanding of both functions in theoretical and practical applications.
Differentiation
Differentiation is a fundamental concept in calculus, used for finding how a function changes at any given point. It deals with the concept of the derivative, which indicates the function's rate of change.
In the context of the step function \(\theta(x)\), differentiation is used to show that the sharp change from 0 to 1 at \(x = 0\) is captured by the delta function \(\delta(x)\). Differentiating \(\theta(x)\) presents an 'infinite spike' at \(x = 0\), a peculiarity that matches the definition of \(\delta(x)\).
This relationship demonstrates how differentiation can equate to a representation of highly localized behavior in functions, crucial for various areas in science where instantaneous changes are modeled, such as in signal processing and control systems. Understanding this concept helps students appreciate not just mathematical theory, but also its practical implications.

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Most popular questions from this chapter

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.

(a) Show that the wave function of a particle in the infinite square well returns to its original form after a quantum revival time \(T=4 m a^{2} / \pi \hbar\) That is: \(\Psi(x, T)=\Psi(x, 0)\) for any state (not just a stationary state). (b) What is the classical revival time, for a particle of energy \(E\) bouncing back and forth between the walls? (c) For what energy are the two revival times equal?

Determine the transmission coefficient for a rectangular barrier (same as Equation 2.148 , only with \(V(x)=+V_{0}>0\) in the region \(-aV_{0}\) (note that the wave function inside the barrier is different in the three cases). Partial answer: for \(E

Show that \(\left[A e^{i k x}+B e^{-i k x}\right]\) and \([C \cos k x+D \sin k x]\) are equivalent ways of writing the same function of \(x\), and determine the constants \(C\) and \(D\) in terms of \(A\) and \(B\), and vice versa. Comment: In quantum mechanics, when \(V=0,\) the exponentials represent traveling waves, and are most convenient in discussing the free particle, whereas sines and cosines correspond to standing waves, which arise naturally in the case of the infinite square well.

Consider the potential $$V(x)=-\frac{\hbar^{2} a^{2}}{m} \operatorname{sech}^{2}(a x),$$ where \(a\) is a positive constant, and "sech" stands for the hyperbolic secant. (a) Graph this potential. (b) Check that this potential has the ground state $$\psi_{0}(x)=A \operatorname{sech}(a x),$$ and find its energy. Normalize \(\psi_{0},\) and sketch its graph. (c) Show that the function $$\psi_{k}(x)=A\left(\frac{i k-a \tanh (a x)}{i k+a}\right) e^{i k x},$$ (where \(k \equiv \sqrt{2 m E} / \hbar,\) as usual) solves the Schrödinger equation for any (positive) energy \(E .\) since \(\tanh z \rightarrow-1\) as \(z \rightarrow-\infty,\) \(\psi_{k}(x) \approx A e^{i k x}, \quad\) for large negative \(x.\) This represents, then, a wave coming in from the left with no accompanying reflected wave (i.e. no term \(\exp (-i k x)\) ). What is the asymptotic form of \(\psi_{k}(x)\) at large positive \(x\) ? What are \(R\) and \(T\), for this potential? Comment: This is a famous example of a reflectionless potential-every incident particle, regardless its energy, passes right through. 62.

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