91Ó°ÊÓ

Consider a potential well defined as \(U(x)=\infty\) for \(x < 0, U(x)=0\) for \(0 < x < L,\) and \(U(x)=U_{0} > 0\) for \(x > L\) (Fig. \(\mathrm{P} 40.70 ) .\) Consider a particle with mass \(m\) and kinetic energy \(E < U_{0}\) that is trapped in the well. (a) The boundary condition at the infinite wall \((x=0)\) is \(\psi(0)=0 .\) What must the form of the function \(\psi(x)\) for \(0 < x < L\) be in order to satisfy both the Schrodinger equation and this boundary condition (b) The wave function must remain finite as \(x \rightarrow \infty .\) What must the form of the function \(\psi(x)\) for \(x>L\) be in order to satisfy both the Schrodinger equation and this boundary condition at infinity? (c) Impose the boundary conditions that \(\psi\) and \(d \psi / d x\) are continuous at \(x=L .\) Show that the energies of the allowed levels are obtained from solutions of the equation \(k \cot k L=-\kappa,\) where \(k=\sqrt{2 m E} / \hbar\) and \(\kappa=\sqrt{2 m\left(U_{0}-E\right) / \hbar}\)

Step by step solution

01

Understanding the Potential Well Problem

We are considering a particle in a potential well with three regions: Region 1 (\(x < 0\)) where \(U(x) = \infty\), Region 2 (\(0 < x < L\)) where \(U(x) = 0\), and Region 3 (\(x > L\)) where \(U(x) = U_0\). We need to find the wave functions \(\psi(x)\) for the particle in each region and the equations relating their properties.

02

Wave Function in Region 2 (\(0 < x < L\))

In Region 2, the potential energy \(U(x)\) is zero, so the time-independent Schrödinger equation is:\[-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi\]which simplifies to\[\frac{d^2 \psi}{dx^2} = -k^2 \psi\]where \(k = \frac{\sqrt{2mE}}{\hbar}\). The general solution for \(\psi(x)\) is \(A \sin(kx) + B \cos(kx)\). The boundary condition \(\psi(0) = 0\) implies \(B = 0\), hence \(\psi(x) = A \sin(kx)\).

03

Wave Function in Region 3 (\(x > L\))

For Region 3, where \(U(x) = U_0\), the Schrödinger equation is:\[-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = (E - U_0) \psi\]which simplifies to\[\frac{d^2 \psi}{dx^2} = \kappa^2 \psi\]where \(\kappa = \frac{\sqrt{2m(U_0 - E)}}{\hbar}\). The general solution is \(C e^{\kappa x} + D e^{-\kappa x}\). The requirement that \(\psi(x)\) remains finite as \(x \rightarrow \infty\) implies \(C = 0\), so \(\psi(x) = D e^{-\kappa x}\).

04

Continuity Conditions at \(x = L\)

We use continuity of the wave function and its derivative at \(x = L\). The conditions are:1. \(A \sin(kL) = D e^{-\kappa L}\)2. \(A k \cos(kL) = -D \kappa e^{-\kappa L}\)Dividing the second equation by the first gives:\[k \cot(kL) = -\kappa\]This condition must be satisfied for the allowed energy levels.

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Key Concepts

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

Schrödinger equation

The Schrödinger equation is a cornerstone of quantum mechanics, much like Newton's laws in classical mechanics. It mathematically describes how the quantum state of a physical system changes over time. In the time-independent form, which is relevant for this potential well problem, the equation is:\[-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + U(x) \psi = E \psi\]In simpler terms:- \(\psi\) is the wave function, representing the probability amplitude of finding the particle in a given position.- \(U(x)\) denotes the potential energy.- \(E\) indicates the total energy of the system.Here, the equation splits into separate forms for different regions due to varying potential energy \(U(x)\). This is why we need to solve it specifically for regions where the potential changes.

wave function

A wave function, \(\psi(x)\), contains all the information about a particle's state in quantum mechanics. It's essentially a complex-valued function that tells us the probability of finding a particle at a particular point in space when squared.In a potential well scenario:- Inside the well (Region 2), the wave function takes the form \(A \sin(kx) + B \cos(kx)\). However, at the boundary where \(x=0\), the condition \(\psi(0) = 0\) requires \(B = 0\), resulting in \(\psi(x) = A \sin(kx)\).- In the area after the well (Region 3), where \(U(x) = U_0\), the function becomes \(C e^{\kappa x} + D e^{-\kappa x}\). For physical reasons (to keep the wave function finite as \(x\) goes to infinity), \(C\) must be zero, leaving us with \(\psi(x) = D e^{-\kappa x}\).These forms are determined by solving the Schrödinger equation in each region, adapting to the conditions imposed by the boundaries and nature of the potential well.

potential well

A potential well is a concept used to describe regions where a particle is confined by potential energy barriers. In this problem, the well is defined as follows:- **Region 1**: \(U(x) = \infty\) for \(x < 0\), meaning the particle cannot exist there.- **Region 2**: \(U(x) = 0\) for \(0 < x < L\). Here, the particle is free to move with no potential energy, only kinetic.- **Region 3**: \(U(x) = U_0 > E\) for \(x > L\), representing a potential barrier higher than the particle's energy.These regions set up a scenario where the particle is trapped within \(0 < x < L\), with the potential well dictating different behavior for its wave function depending on which region it occupies. Understanding the interactions between these regions is critical for solving the Schrödinger equation and predicting the particle's behavior.

boundary conditions

Boundary conditions are essential constraints in solving differential equations that arise in quantum mechanics. For the potential well problem, they ensure our solutions are physically sensible.- At **\(x = 0\)**, the condition \(\psi(0) = 0\) ensures the wave function is zero at the infinitely high potential barrier. This is because there can be no probability of finding the particle in a region with infinite potential.- At **infinity** \(x \to \infty\), the requirement that \(\psi(x)\) remains finite helps prevent non-physical solutions like infinitely exploding wave functions. Thus, the solution must decay, leading to \(\psi(x) = D e^{-\kappa x}\).- At **\(x = L\)**, the conditions require the wave function and its derivative to be continuous. Mathematically, this is observed through:

• \(A \sin(kL) = D e^{-\kappa L}\)
• \(A k \cos(kL) = -D \kappa e^{-\kappa L}\)

These continuity conditions help relate the solutions across regions and are essential for finding the allowed energy levels.