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(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?

Short Answer

Expert verified
(a) Quantum revival at \(T = 4ma^2/\pi\hbar\). (b) Classical revival \(= 4a\sqrt{m/2E}\). (c) Energy \(E = \pi^2\hbar^2/8ma^2\).

Step by step solution

01

Step 1: Define the system

We consider a particle of mass \(m\) in an infinite square well potential. The potential \(V(x)\) is 0 for \(0 \leq x \leq a\) and infinite otherwise. The wave function inside this well satisfies the time-dependent Schrödinger equation.
02

Write the wave function

The general wave function for a particle in the infinite square well can be written in terms of its energy eigenstates: \[\Psi(x, t) = \sum_n c_n \psi_n(x) e^{-iE_nt/\hbar} \]where \(\psi_n(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi x}{a}\right)\) and \(E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}\). \(c_n\) are the coefficients of the wave function.
03

Evaluate at quantum revival time, \(T\)

At the time \(T = \frac{4ma^2}{\pi\hbar}\) (quantum revival time), substitute into the exponential factor:\[ e^{-iE_nT/\hbar} = e^{-i n^2\pi/2} \]Since \(e^{-i n^2\pi/2} = i^{-n^2}\), this simplifies the expression. Notice that for all \(n\), this equates to \(1\) or -1 depending on \(n\), so the original wave function is recovered modulo a global phase.
04

Conclusion for Part (a)

At time \(T\), the wave function becomes:\[ \Psi(x, T) = \sum_n c_n \psi_n(x) i^{-n^2} = \Psi(x, 0) \] (up to a global phase). Hence, \(\Psi(x, T) = \Psi(x, 0)\). The wave function returns to its original form.
05

Calculate classical revival time

The classical revival time for a particle bouncing between the walls of the well is the time taken for the particle to travel a distance of \(4a\) back and forth. Given the speed \(v\) of the particle, this time is:\[ T_{\text{classical}} = \frac{4a}{v} \] with \(v = \sqrt{\frac{2E}{m}}\). Thus, \[ T_{\text{classical}} = \frac{4a \sqrt{m}}{\sqrt{2E}} \]
06

Equate quantum and classical times

To find the energy when both times are equal:\[ T_{\text{quantum}} = T_{\text{classical}} \]\[ \frac{4ma^2}{\pi\hbar} = \frac{4a \sqrt{m}}{\sqrt{2E}} \]Square both sides and solve for \(E\):\[ E = \frac{\pi^2\hbar^2}{8ma^2} \]

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Key Concepts

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

Infinite Square Well
Let's begin with understanding the Infinite Square Well, a cornerstone concept in quantum mechanics. This model is a simplified system where a particle is confined within an impenetrable boundary, usually between two walls separated by a distance, denoted as "a."
The potential energy within this well is zero (\(V(x) = 0\)) when the particle's position x is between 0 and a, and infinite outside of these limits.
This scenario essentially traps the particle, restricting its movement to within the well and preventing it from escaping.In the Infinite Square Well, the solutions to the Schrödinger equation are called energy eigenstates. The wave functions (\(\psi_n(x)\)) are sinusoidal within the well and zero outside.
These wave functions represent the probabilities of finding the particle at various positions within the well.
The eigenstates are quantized, meaning that only certain discrete energy levels (\(E_n\)) are allowed for the particle, given by:\[E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}\]where \(n\) is an integer (n = 1, 2, 3,...), \(\hbar\) is the reduced Planck's constant, and \(m\) is the mass of the particle. This setup lays the foundation for exploring more complex quantum systems.
Quantum Revival
A fascinating phenomenon in quantum systems like the Infinite Square Well is Quantum Revival. Quantum Revival refers to the periodic nature when the wave function (\(\Psi(x, t)\)) of a particle returns to its initial form after a certain time period, known as the quantum revival time.For a particle in the infinite square well, the quantum revival time can be calculated as:\[T = \frac{4ma^2}{\pi\hbar}\]At this time, the wave function across all populated energy states reconstructs itself to mimic its original configuration, up to a global phase. This revival occurs because each component of the wave function evolves in a coherent manner, cycling back through phases to return to the starting configuration. Unlike classical mechanics, this revival manifests thanks to quantum mechanics' inherent wave-like properties.Understanding quantum revival enhances our comprehension of the dynamic behavior of quantum systems and proves useful in fields like quantum computing, where managing temporal wave function evolution is crucial.
Classical Revival Time
In contrast to the quantum revival, Classical Revival Time pertains to how a particle would behave if it were solely governed by classical mechanics. Picture a particle moving back and forth between two walls. In a classical scenario, devoid of quantum effects, the particle would travel at a certain speed until it hits one wall, reverses direction, and travels back.
The Classical Revival Time is the time it takes for the particle to complete this back-and-forth motion.Mathematically, this time is:\[T_{\text{classical}} = \frac{4a}{v}\]Substituting the classical velocity, \(v = \sqrt{\frac{2E}{m}}\), we get:\[T_{\text{classical}} = \frac{4a \sqrt{m}}{\sqrt{2E}}\]This equation provides insight into how energy affects the dynamics of classical motion. While classical physics often fails to capture behaviors in quantum systems, studying its overlaps with quantum revival offers a compelling edge to our understanding of both domains.
Wave Function
At the heart of quantum mechanics lies the Wave Function, symbolized often by \(\Psi(x, t)\). This function encapsulates the probabilistic nature of quantum systems. Within the infinite square well, the wave function describes the likelihood of finding a particle at a given position and time.The general form of the wave function within the infinite square well is expressed as:\[\Psi(x, t) = \sum_n c_n \psi_n(x) e^{-iE_nt/\hbar}\]where \(\psi_n(x)\) are the stationary states' matrix and \(c_n\) are the coefficients indicating each state's contribution. The exponential factors illustrate how each component of the wave function evolves over time.By inserting specific times into this equation, one can explore how the probability distribution changes. For instance, at quantum revival time (\(T\)), the \(e^{-iE_nT/\hbar}\) term alters back to a state closely matching its initial form. Of course, understanding and manipulating wave functions is central to predicting and controlling quantum behaviors, forming a core competency necessary for delving into advanced topics of modern physics.

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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.

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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