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Chapter 2: Problem 21

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).

Short Answer

Expert verified
(a) \( A = \sqrt{a} \). (b) \( \phi(k) = \frac{2\sqrt{a}}{\sqrt{2\pi}} \frac{a}{a^2 + k^2} \). (c) Time-dependent \( \Psi(x, t) \) is an integral of \( \phi(k) \). (d) Large/small \( a \) affects spread of \( \Psi(x) \) and \( \phi(k) \).

Step by step solution

01

Normalize the Initial Wave Function

To normalize the wave function, we need to find the constant \( A \) such that the total probability is 1. This means we need to solve the integral\[ \int_{-\infty}^{\infty} \left| \Psi(x, 0) \right|^2 dx = 1. \]For the given wave function, we have:\[ \int_{-\infty}^{\infty} |A|^2 e^{-2a|x|} dx = |A|^2 \int_{-\infty}^{\infty} e^{-2a|x|} dx = 1. \]Breaking this into two integrals from \(-\infty\) to 0 and from 0 to \( \infty \):\[ |A|^2 \left( \int_{-\infty}^{0} e^{2ax} dx + \int_{0}^{\infty} e^{-2ax} dx \right) = 1. \]Calculating each integral, we find:\[ \int_{-\infty}^{0} e^{2ax} dx = \left[ \frac{1}{2a} e^{2ax} \right]_{-\infty}^{0} = \frac{1}{2a}, \]\[ \int_{0}^{\infty} e^{-2ax} dx = \left[ -\frac{1}{2a} e^{-2ax} \right]_{0}^{\infty} = \frac{1}{2a}. \]Thus,\[ |A|^2 \left( \frac{1}{2a} + \frac{1}{2a} \right) = 1, \]\[ |A|^2 \frac{1}{a} = 1. \]Therefore, \( |A|^2 = a \) or \( A = \sqrt{a} \).
02

Find the Fourier Transform, \( \phi(k) \)

The Fourier transform of the wave function \( \Psi(x, 0) \) gives \( \phi(k) \). It is defined by:\[ \phi(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Psi(x, 0) e^{-ikx} dx. \]Substitute \( \Psi(x, 0) = A e^{-a|x|} \):\[ \phi(k) = \frac{A}{\sqrt{2\pi}} \left( \int_{-\infty}^{0} e^{(a + ik)x} dx + \int_{0}^{\infty} e^{-(a + ik)x} dx \right). \]Compute each integral:\[ \int_{-\infty}^{0} e^{(a + ik)x} dx = \frac{1}{a+ik}, \]\[ \int_{0}^{\infty} e^{-(a + ik)x} dx = \frac{1}{a-ik}. \]Thus,\[ \phi(k) = \frac{A}{\sqrt{2\pi}} \left( \frac{1}{a+ik} + \frac{1}{a-ik} \right). \]Substituting \( A = \sqrt{a} \) and simplifying:\[ \phi(k) = \frac{2\sqrt{a}}{\sqrt{2\pi}} \frac{a}{a^2 + k^2}. \]
03

Construct the Time-Dependent Wave Function \( \Psi(x, t) \)

The time evolution of the wave function is given by the inverse Fourier transform:\[ \Psi(x, t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \phi(k) e^{i(kx - \hbar k^2 t / 2m)} dk. \]Using \( \phi(k) \) from Step 2:\[ \Psi(x, t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \left( \frac{2\sqrt{a}}{\sqrt{2\pi}} \frac{a}{a^2 + k^2} \right) e^{i(kx - \hbar k^2 t / 2m)} dk. \]This is the time-dependent wave function expressed in integral form.
04

Discuss Limiting Cases

**Case 1: \(a\) very large.**As \(a\) becomes very large, the wave function \( \Psi(x, 0) \) becomes sharply peaked around \( x = 0 \). The Fourier transform \( \phi(k) \) will then spread out, meaning the particle has a wide range of momenta.**Case 2: \(a\) very small.**If \( a \) is very small, \( \Psi(x, 0) \) becomes spread out in space (a gentle slope), meaning the particle is in a more definitive spatial location but has a narrow range of momenta. This reflects the uncertainty principle, where precise location leads to uncertain momentum and vice versa.

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Key Concepts

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

Free Particle Wave Function
When thinking about a free particle in quantum mechanics, we describe its behavior using a wave function. A wave function can be complex, but for a free particle, it often simplifies down to expressions that are easier to work with. In this exercise, the initial wave function is given by:\[ \Psi(x, 0) = A e^{-a|x|} \]Here, \(A\) is a constant that you can think of as an initial scale factor, and \(a\) is a parameter that influences how quickly the wave function decreases as you move away from the origin. This function suggests that the particle is more likely to be found near \(x = 0\) and less likely as you move farther away. The negative exponential ensures that probabilities remain finite, which is essential for physical interpretations. Understanding this setup is crucial, as it dictates how we proceed with normalizing and transforming the function.
Normalization of Wave Function
In quantum mechanics, normalization ensures that the total probability of finding a particle is 1. This means all the possibilities add up to certainty. For our wave function, normalization involves calculating the integral:\[ \int_{-\infty}^{\infty} \left| \Psi(x, 0) \right|^2 dx = 1 \]Breaking this down, it means we sum over all the probabilities across every possible position \(x\). With our given function, the integral splits into two parts, from negative infinity to zero, and zero to infinity. Calculating this, we get:\[ |A|^2 \int_{-\infty}^{\infty} e^{-2a|x|} dx = 1 \]Solving this gives \( |A|^2 = a \), meaning \(A = \sqrt{a}\). This step is about ensuring the mathematics doesn't just work but aligns with the probabilities we expect in real life.
Fourier Transform in Quantum Mechanics
The Fourier Transform is a powerful tool in physics, used to understand functions in terms of their frequency components. In quantum mechanics, this helps us move from position space to momentum space. For our wave function, the Fourier transform, \( \phi(k) \), is:\[ \phi(k) = \frac{A}{\sqrt{2\pi}} \left( \frac{1}{a+ik} + \frac{1}{a-ik} \right) \]Simply put, it converts the position-based description of a particle to one based on momentum. With \(A = \sqrt{a}\), this function describes how the particle's momentum is distributed. The link between these spaces is key to understanding behaviors like interference and diffraction, as well as particle motion.
Time-Dependent Wave Function
Wave functions change over time, regardless of whether a particle is free or bound. Constructing a time-dependent wave function involves incorporating time evolution into the initial state:\[ \Psi(x, t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \phi(k) e^{i(kx - \hbar k^2 t / 2m)} dk \]This expression lets us understand how the particle's wave function evolves. It shows how a free particle's wave packet spreads out with time due to dispersion 鈥� a key concept when dealing with wave packets in quantum terms. This evolution aligns with our intuition that particles move and expand in a space over time.
Uncertainty Principle in Quantum Mechanics
The Uncertainty Principle, a cornerstone of quantum mechanics, states there's a trade-off between knowing a particle's position and momentum. For our wave function, different values of \(a\) illustrate this beautifully:- **Large \(a\):** Wave function sharply peaked, meaning position is very certain but momentum is quite spread out.- **Small \(a\):** The wave function is broad, indicating less certainty in position but more in momentum.This principle reflects the inherent limits of precision in measurements鈥攁 fundamental departure from classical physics. It explains why we can't predict everything with absolute certainty, shaping how we interpret particles' behavior in quantum mechanics.

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Most popular questions from this chapter

This problem is designed to guide you through a "proof" of Plancherel's theorem, by starting with the theory of ordinary Fourier series on a finite interval, and allowing that interval to expand to infinity. (a) Dirichlet's theorem says that "any" function \(f(x)\) on the interval \([-a,+a]\) can be expanded as a Fourier series: $$ f(x)=\sum_{n=0}^{\infty}\left[a_{n} \sin (n \pi x / a)+b_{n} \cos (n \pi x / a)\right] $$ Show that this can be written equivalently as $$ f(x)=\sum_{n=-\infty}^{\infty} c_{n} e^{i n \pi x / a} $$ What is \(c_{n}\), in terms of \(a_{n}\) and \(b_{n}\) ? (b) Show (by appropriate modification of Fourier's trick) that $$ c_{n}=\frac{1}{2 a} \int_{-a}^{+a} f(x) e^{-i n \pi x / a} d x $$ (c) Eliminate \(n\) and \(c_{n}\) in favor of the new variables \(k=(n \pi / a)\) and \(F(k)=\) \(\sqrt{2 / \pi} a c_{n} .\) Show that (a) and (b) now become $$ f(x)=\frac{1}{\sqrt{2 \pi}} \sum_{n=-\infty}^{\infty} F(k) e^{i k x} \Delta k ; \quad F(k)=\frac{1}{\sqrt{2 \pi}} \int_{-a}^{+a} f(x) e^{-i k x} d x $$ where \(\Delta k\) is the increment in \(k\) from one \(n\) to the next. (d) Take the limit \(a \rightarrow \infty\) to obtain Plancherel's theorem. Comment: In view of their quite different origins, it is surprising (and delightful) that the two formulas-one for \(F(k)\) in terms of \(f(x)\), the other for \(f(x)\) in terms of \(F(k)\) - have such a similar structure in the limit \(a \rightarrow \infty\).

A particle of mass \(m\) in the infinite square well (of width \(a\) ) starts out in the left half of the well, and is (at \(t=0\) ) equally likely to be found at any point in that region. (a) What is its initial wave function, \(\Psi(x, 0)\) ? (Assume it is real. Don't forget to normalize it.) (b) What is the probability that a measurement of the energy would yield the value \(\pi^{2} \hbar^{2} / 2 m a^{2}\) ?

A particle in the harmonic oscillator potential starts out in the state $$ \Psi(x, 0)=A\left[3 \psi_{0}(x)+4 \psi_{1}(x)\right] $$ (a) Find \(A\). (b) Construct \(\Psi(x, t)\) and \(|\Psi(x, t)|^{2}\). (c) Find \(\langle x\rangle\) and \(\langle p\rangle\). Don't get too excited if they oscillate at the classical frequency; what would it have been had I specified \(\psi_{2}(x)\), instead of \(\psi_{1}(x)\) ? Check that Ehrenfest's theorem (Equation \(1.38\) ) holds for this wave function. (d) If you measured the energy of this particle, what values might you get, and with what probabilities?

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 Schr枚dinger equation for any (positive) energy \(E\). Since \(\tanh z \rightarrow-1\) as \(z \rightarrow-\infty\), $$ \psi_{k}(x) \approx A e^{i k x} . \quad \text { for large negative } x . $$ 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 of its energy, passes right through. \({ }^{47}\)

*Problem 2.7 A particle in the infinite square well has the initial wave function 15 $$ \Psi(x, 0)=\left\\{\begin{array}{ll} A x, & 0 \leq x \leq a / 2 \\ A(a-x) . & a / 2 \leq x \leq a (c) What is the probability that a measurement of the energy would yield the value \(E_{1}\) ? (d) Find the expectation value of the energy. \end{array}\right. $$ (a) Sketch \(\Psi(x, 0)\), and determine the constant \(A\). (b) Find \(\Psi(x, t)\).
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