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Chapter 2: Problem 22

Evaluate the following integrals: (a) \(\int_{-3}^{+1}\left(x^{3}-3 x^{2}+2 x-1\right) \delta(x+2) d x.\) (b) \(\int_{0}^{\infty}[\cos (3 x)+2] \delta(x-\pi) d x.\) (c) \(\int_{-1}^{+1} \exp (|x|+3) \delta(x-2) d x.\)

Short Answer

Expert verified
(a) -25, (b) 1, (c) 0.

Step by step solution

01

Understanding Dirac Delta Function

The Dirac delta function, represented as \( \delta(x-a) \), is a special function that is zero everywhere except at \( x=a \), where it is infinitely high, and its integral over the entire real line is 1. In practical terms, it "picks out" the value of another function at \( x=a \) when used under an integral.
02

Evaluate Integral (a)

For the integral \( \int_{-3}^{+1}(x^{3}-3x^{2}+2x-1) \delta(x+2) \, dx \), we identify that \( \delta(x+2) \) affects the function at \( x = -2 \). Since \( -2 \) is within the range of integration, we substitute \( x = -2 \) into the polynomial function: \((-2)^3 - 3(-2)^2 + 2(-2) - 1 = -8 - 12 - 4 - 1 = -25 \). Therefore, the value of the integral is \(-25\).
03

Evaluate Integral (b)

For the integral \( \int_{0}^{\infty}[\cos(3x)+2] \delta(x-\pi) \, dx \), \( \delta(x-\pi) \) evaluates the function at \( x = \pi \). However, since \( x=\pi \) is not within the limits of 0 to \( \pi \) (as the upper limit is infinity in the improper sense and might not immediately suggest \( x = \pi \)), we treat \( \pi \) as a valid limit. Substituting, we have \( \cos(3\pi) + 2 = -1 + 2 = 1 \). Thus, the integral evaluates to \( 1 \).
04

Evaluate Integral (c)

For \( \int_{-1}^{+1} \exp(|x|+3) \delta(x-2) \, dx \), we see that \( \delta(x-2) \) evaluates the function at \( x=2 \). Since \( x=2 \) is not within the given interval from \(-1\) to \(+1\), the delta function does not "act" on any value within this domain, resulting in the integral being \(0\).
05

Final Understood Values

We have calculated the integrals by applying the property of the Dirac delta function. For part (a), the value is \(-25\), for part (b), the value is \(1\), and for part (c), the value is \(0\) due to the Dirac delta's location outside the integration interval.

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Key Concepts

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

Integrals Evaluation
The process of evaluating integrals involves calculating the area under the curve of a given function within specified limits. In this context, the Dirac delta function simplifies the process because it effectively reduces the process to evaluating the function at a specific point.
  • For integral (a), \( \int_{-3}^{+1}(x^{3}-3x^{2}+2x-1) \delta(x+2) \, dx \), we observe how the delta function evaluates the polynomial at \( x = -2 \). Since \( x = -2 \) is within the range of integration, the integral results in \(-25\).
  • In integral (b), \( \int_{0}^{\infty}[\cos(3x)+2] \delta(x-\pi) \, dx \), the function is evaluated at \( x = \pi \), leading to a result of 1 because \( x = \pi \) is considered part of the acceptable range when extending from 0 to infinity.
  • For integral (c), \( \int_{-1}^{+1} \exp(|x|+3) \delta(x-2) \, dx \), the integral results in 0 because the point \( x = 2 \) is outside the specified limits.
These integrals illustrate how the Dirac delta function "picks out" values at precise points, effectively simplifying the integration process by reducing it to a function evaluation.
Properties of Delta Function
Understanding the properties of the Dirac delta function is essential in solving integrals involving it. The delta function, denoted as \( \delta(x-a) \), has several unique characteristics which are useful in mathematical and physical applications.
  • It is zero everywhere but at \( x = a \).
  • The integral of \( \delta(x-a) \) over the entire real line is 1, symbolizing a 鈥榗harge鈥� concentrated at a single point.
  • In an integral \( \int f(x) \delta(x-a) \, dx \), the delta function evaluates \( f(x) \) at \( x = a \), essentially pulling out the function's value at that point.
These properties allow us to simplify otherwise complex integrals and focus solely on the point where the delta function is active. This approach is particularly advantageous in physics and engineering where impulse forces or instantaneous events are analyzed.
Integration Limits and Bounds
The limits and bounds of an integral play a crucial role when dealing with the Dirac delta function. Since this function is only non-zero at one singular point, the choice of bounds directly determines whether the integral will have a non-zero outcome.
  • In cases where the delta function's point falls outside the integration limits, such as in integral (c), the resulting integral will be zero.
  • If the point is within the limits, as demonstrated in integral (a), the delta function evaluates the function at that specific point, resulting in a finite value based on the function's expression at that point.
  • Improper limits, such as those in integral (b) extending to infinity, require careful consideration to ensure points like \( x = \pi \) are recognized within the integral evaluation framework.
By understanding how limits dictate the behavior of the Dirac delta within an integral, we can effectively determine the outcome of an evaluation, acknowledging where the delta 鈥榓cts鈥� and where it remains inert.

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Most popular questions from this chapter

The gaussian wave packet. 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 \(\left(a x^{2}+b x\right)=y^{2}-\left(b^{2} / 4 a\right) .\) Answer (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?

Reads $$\left(1-x^{2}\right) \frac{d^{2} f}{d x^{2}}-2 x \frac{d f}{d x}+\ell(\ell+1) f=0,$$ where \(\ell\) is some (non-negative) real number. (a) Assume a power series solution, $$f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}$$ and obtain a recursion relation for the constants \(a_{n}\) (b) Argue that unless the series truncates (which can only happen if \(\ell\) is an integer \(),\) the solution will diverge at \(x=1\) (c) When \(\ell\) is an integer, the series for one of the two linearly independent solutions (either \(f\) even or \(f\) odd depending on whether \(\ell\) is even or odd) will truncate, and those solutions are called Legendre polynomials \(P_{\ell}(x)\) Find \(P_{0}(x), P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) from the recursion relation. Leave your answer in terms of either \(a_{0}\) or \(a_{1} .^{69}\)

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).

In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials. (a) The Rodrigues formula says that $$H_{n}(\xi)=(-1)^{n} e^{\xi^{2}}\left(\frac{d}{d \xi}\right)^{n} e^{-\xi^{2}}.$$ Use it to derive \(H_{3}\) and \(H_{4}.\) (b) The following recursion relation gives you \(H_{n+1}\) in terms of the two preceding Hermite polynomials: $$H_{n+1}(\xi)=2 \xi H_{n}(\xi)-2 n H_{n-1}(\xi).$$ Use it, together with your answer in (a), to obtain \(H_{5}\) and \(H_{6}\). (c) If you differentiate an \(n\) th-order polynomial, you get a polynomial of order \((n-1) .\) For the Hermite polynomials, in fact, $$\frac{d H_{n}}{d \xi}=2 n H_{n-1}(\xi).$$ Check this, by differentiating \(H_{5}\) and \(H_{6}\). (d) \(H_{n}(\xi)\) is the \(n\) th \(z\) -derivative, at \(z=0\), of the generating function \(\exp \left(-z^{2}+2 z \xi\right) ;\) or, to put it another way, it is the coefficient of \(z^{n} / n !\) in the Taylor series expansion for this function: $$e^{-z^{2}+2 z \xi}=\sum_{n=0}^{\infty} \frac{z^{n}}{n !} H_{n}(\xi).$$ Use this to obtain \(H_{1}, H_{2},\) and \(H_{3}.\)

Consider the double delta-function potential $$V(x)=-\alpha[\delta(x+a)+\delta(x-a)],$$ where \(a\) and \(a\) are positive constants. (a) Sketch this potential. (b) How many bound states does it possess? Find the allowed energies, for \(\alpha=\hbar^{2} / m a\) and for \(\alpha=\hbar^{2} / 4 m a,\) and sketch the wave functions. (c) What are the bound state energies in the limiting cases (i) \(a \rightarrow 0\) and (ii) \(a \rightarrow \infty\) (holding a fixed)? Explain why your answers are reasonable, by comparison with the single delta-function well.
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