91Ó°ÊÓ

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?

Step by step solution

01

Normalize the Initial Wave Function

To normalize the wave function \( \Psi(x, 0) = A e^{-a x^2} \), we need to ensure that \( \int_{-\infty}^{+\infty} |\Psi(x, 0)|^2 \, dx = 1 \). Substituting the given wave function:\[ |\Psi(x, 0)|^2 = A^2 e^{-2ax^2} \]Using the Gaussian integral \( \int_{-\infty}^{+\infty} e^{-\alpha x^2} \, dx = \sqrt{\frac{\pi}{\alpha}} \), we find:\[ A^2 \int_{-\infty}^{+\infty} e^{-2ax^2} \, dx = A^2 \sqrt{\frac{\pi}{2a}} = 1 \]Solving for \( A \), we get:\[ A = \left(\frac{2a}{\pi}\right)^{1/4} \]

02

Find the Time-Dependent Wave Function

The wave function \( \Psi(x, t) \) can be found by applying the time evolution of a free particle, which involves the Fourier transform of \( \Psi(x, 0) \). After some complex algebra involving the Fourier transform, finding the phase factor, and completing the square, we have:\[ \Psi(x, t) = \left(\frac{a}{\pi}\right)^{1/4} \frac{1}{\sqrt{1+\frac{2i\hbar at}{m}}} \exp\left(-\frac{ax^2}{1+\frac{2i\hbar at}{m}}\right) \]

03

Compute the Probability Density

The probability density \( |\Psi(x, t)|^2 \) is given by the square of the modulus of \( \Psi(x, t) \):\[ |\Psi(x, t)|^2 = \left(\frac{a}{\pi}\right)^{1/2} \frac{1}{\sqrt{1+(\frac{2\hbar at}{m})^2}} \exp\left(-\frac{2ax^2}{1+(\frac{2\hbar at}{m})^2}\right) \]In terms of \( w = \sqrt{a/[1+(2\hbar at/m)^2]} \), this can be rewritten as:\[ |\Psi(x, t)|^2 = \left(\frac{w}{\pi}\right)^{1/2} \exp(-wx^2) \]

04

Analyze the Behavior of |\Psi|^2

The sketch of \( |\Psi(x, t)|^2 \) as a function of \( x \) at \( t = 0 \) shows a narrow Gaussian peak. As \( t \to \infty \), \( |\Psi(x, t)|^2 \) broadens, indicating the spreading of the wave packet. With increasing time, the wave packet's width grows, and its peak height decreases.

05

Calculate Expectation Values and Uncertainties

\( \langle x \rangle = 0 \) because \( \Psi(x, t) \) is an even function, so it is centered at zero. \( \langle p \rangle = 0 \) as momentum expectation is initially zero in this symmetric potential.The expectation values \( \langle x^2 \rangle \) and \( \langle p^2 \rangle = a\hbar^2 \) are obtained from integrating \( |\Psi(x, t)|^2 \) times \( x^2 \) or \( p^2 \), with \( p = -i\hbar \frac{\partial}{\partial x} \).The uncertainties are given by: \( \sigma_x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} \) and \( \sigma_p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2} \).

06

Verify the Uncertainty Principle

The uncertainty product \( \sigma_x \sigma_p = \hbar/2 \) at \( t=0 \), confirming the uncertainty principle holds initially. As time progresses, \( \sigma_x \) grows and \( \sigma_p \) remains bounded, increasing \( \sigma_x \sigma_p \). The system comes closest to the uncertainty limit at \( t=0 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

wave function normalization

Wave function normalization ensures the probability of finding a particle somewhere on a given scale is 1, meaning the particle must exist somewhere in space. For the Gaussian wave packet \(\Psi(x, 0) = A e^{-a x^2}\), normalization requires integrating \(|\Psi(x, 0)|^2\) over all space to equal 1.

We start by finding the square of the wave function: \(|\Psi(x, 0)|^2 = A^2 e^{-2ax^2}\). Then, we integrate:

• \(\int_{-\infty}^{+\infty} |\Psi(x, 0)|^2 \, dx = A^2 \int_{-\infty}^{+\infty} e^{-2ax^2} \, dx\).

The integral of a Gaussian function \(\int_{-\infty}^{+\infty} e^{-\alpha x^2} \, dx = \sqrt{\frac{\pi}{\alpha}}\) helps solve this. Substituting \(\alpha = 2a\), it becomes \(\sqrt{\frac{\pi}{2a}}\). Thus, \(A^2 \sqrt{\frac{\pi}{2a}} = 1\). Solving for \(A\), we find \(A = \left(\frac{2a}{\pi}\right)^{1/4}\).

This normalization ensures that the total probability of existence of the particle in space remains constant.

time evolution in quantum mechanics

Time evolution allows us to understand how wave functions change over time in quantum mechanics. For a free particle's wave function \(\Psi(x, t)\), we use the Schrödinger equation, which describes how quantum states evolve.

To find \(\Psi(x, t)\), it involves using a Fourier transform of the initial state \(\Psi(x, 0)\), followed by adjustments with phase factors and complex algebra to simplify results. The solution unfolds as follows:

• Start with the normalized wave function.
• Employ the Fourier transform and complete the square method to find time dependence.

After some complex algebra, we derive:

• \(\Psi(x, t) = \left(\frac{a}{\pi}\right)^{1/4} \frac{1}{\sqrt{1 + \frac{2i\hbar at}{m}}} \exp\left(-\frac{ax^2}{1 + \frac{2i\hbar at}{m}}\right)\)

This representation shows the wave packet's dispersion over time, indicating how probability densities shift and change in the quantum context.

quantum uncertainty principle

The uncertainty principle is a fundamental concept that restricts our precise knowledge of conjugate variables like position and momentum. It essentially states that increasing the precision of one quantity inherently decreases the precision of the other.

In the Gaussian wave packet context, we calculate position and momentum uncertainties, \(\sigma_x\) and \(\sigma_p\):

• \(\sigma_x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}\)
• \(\sigma_p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2}\)

For this wave packet, the calculus results show that \(\langle x \rangle = 0\) and \(\langle p \rangle = 0\) due to its symmetry.

At \(t=0\), the product \(\sigma_x \sigma_p = \hbar/2\), which meets the minimum boundary set by the uncertainty principle. As time progresses, \(\sigma_x\) increases while \(\sigma_p\) remains nearly constant, leading to an increase in the total product, emphasizing how uncertainty evolves over time.

expectation values in quantum mechanics

Expectation values give us averages of a physical quantity for a wave function, akin to mean values in statistics. They help us derive important properties of quantum systems.

For the Gaussian wave packet:

• \(\langle x \rangle = 0\) because the wave function is centered due to its symmetric nature.
• \(\langle p \rangle = 0\) as initial momentum is zero.
• \(\langle x^2 \rangle\) and \(\langle p^2 \rangle = a\hbar^2\) are calculated through integrating \(|\Psi(x, t)|^2\) with respective \(x^2\) or \(p^2\), noting that \(p = -i\hbar \frac{\partial}{\partial x}\).

Since \(\langle x^2 \rangle\) contributes to \(\sigma_x\), and \(a\hbar^2\) is the important term for \(\sigma_p\), these computations indicate how both position and momentum distribution influence the wave packet's characteristics. Understanding and verifying these expectation values aid in confidently applying quantum principles to dynamic systems.