/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 In this problem you will show th... [FREE SOLUTION] | 91Ó°ÊÓ

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In this problem you will show that the number of nodes of the stationary states of a one-dimensional potential always increases with energy. Consider two (real, normalized) solutions \(\left(\psi_{n} \text { and } \psi_{m}\) ) to the time- \right. independent Schrödinger equation (for a given potential \(V(x)\) ), with energies \(E_{n}>E_{m}\) (a) Show that $$\frac{d}{d x}\left(\frac{d \psi_{m}}{d x} \psi_{n}-\psi_{m} \frac{d \psi_{n}}{d x}\right)=\frac{2 m}{\hbar^{2}}\left(E_{n}-E_{m}\right) \psi_{m} \psi_{n}$$ (b) Let \(x_{1}\) and \(x_{2}\) be two adjacent nodes of the function \(\psi_{m}(x) .\) Show that $$\psi_{m}^{\prime}\left(x_{2}\right) \psi_{n}\left(x_{2}\right)-\psi_{m}^{\prime}\left(x_{1}\right) \psi_{n}\left(x_{1}\right)=\frac{2 m}{\hbar^{2}}\left(E_{n}-E_{m}\right) \int_{x_{1}}^{x_{2}} \psi_{m} \psi_{n} d x$$ (c) If \(\psi_{n}(x)\) has no nodes between \(x_{1}\) and \(x_{2},\) then it must have the same sign everywhere in the interval. Show that (b) then leads to a contradiction. Therefore, between every pair of nodes of \(\psi_{m}(x), \psi_{n}(x)\) must have at least one node, and in particular the number of nodes increases with energy.

Short Answer

Expert verified
The number of nodes in the stationary states increases with energy.

Step by step solution

01

Differentiate the Expression

We start with the given expression from part (a): \( \frac{d}{d x}\left(\frac{d \psi_{m}}{d x} \psi_{n}-\psi_{m} \frac{d \psi_{n}}{d x}\right) \). Our goal is to differentiate this expression. Use the product rule to reach the RHS given in the problem. For \(E_n > E_m\), exploit the time-independent Schrödinger equation properties for each wave function. The resulting RHS is \( \frac{2 m}{\hbar^{2}}\left(E_{n}-E_{m}\right) \psi_{m} \psi_{n} \), confirming our objective.
02

Evaluate at Two Adjacent Nodes

For part (b), let's consider two nodes \( x_1 \) and \( x_2 \) of \( \psi_m(x) \). Here, \( \psi_m(x_1) = 0 \) and \( \psi_m(x_2) = 0 \). Use these conditions in the differentiated result from Step 1. By integrating from \( x_1 \) to \( x_2 \), we get \[ \psi_{m}^{\prime}\left(x_{2}\right) \psi_{n}\left(x_{2}\right)-\psi_{m}^{\prime}\left(x_{1}\right) \psi_{n}\left(x_{1}\right) = \frac{2 m}{\hbar^{2}}\left(E_{n}-E_{m}\right) \int_{x_{1}}^{x_{2}} \psi_{m} \psi_{n} d x \].
03

Analyze for Nodes and Signs

In part (c), assume \( \psi_n(x) \) has no nodes between \( x_1 \) and \( x_2 \). Consequently, \( \psi_n(x) \) keeps a constant sign throughout the interval \( x_1 \) to \( x_2 \). Because the integral of \( \psi_m \psi_n \) over the interval is zero (from Step 2), and since \( \psi_n \) does not change sign, this is a contradiction. The only resolution is \( \psi_n \) must have a node within \( x_1 \) and \( x_2 \). This implies every adjacent pair of nodes in \( \psi_m \) demands at least one node in \( \psi_n \), proving an increase in the number of nodes with increasing energy.

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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 core element of quantum mechanics. It describes how quantum states evolve over time. Specifically, for one-dimensional potentials, the time-independent form is:\[-\frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} + V(x) \psi(x) = E \psi(x)\]This equation relates the wave function \(\psi(x)\), which contains information about the probability distribution of a particle's position, to the potential energy \(V(x)\) and the total energy \(E\) of the system.
  • \(\hbar\) is the reduced Planck's constant.
  • \(m\) is the mass of the particle.
  • \(V(x)\) is the potential energy as a function of position \(x\).
  • \(E\) is the energy eigenvalue, specific for each solution or state.
The solutions to the Schrödinger Equation, \(\psi(x)\), are the wave functions that tell us about "where" we might find particles at any given energy level under the role of the potential.
Wave Function Nodes
In quantum mechanics, nodes are points where the wave function \(\psi(x)\) equals zero. At these points, the probability of finding a particle is zero. Nodes are crucial because they help define the structure of wave functions and influence the behavior of quantum states.In a one-dimensional potential, nodes can be easily visualized:
  • For a wave function like a sine or cosine, nodes happen at the intersections with the horizontal axis.
  • These intersections represent the points where the probability of finding the electron is zero.
  • As energy levels increase, nodes usually do too, leading to more complex wave shapes.
Understanding nodes is fundamental, as shown in the exercise, which illustrates that wave function nodes increase with energy by proving that higher-energy states have more nodes than lower-energy ones.
Energy Levels
Energy levels in a quantum system are discrete and quantized, meaning they come in distinct values. Each solution to the Schrödinger Equation corresponds to an energy level. Higher order wave functions \((\psi_n)\) correspond to higher energies \((E_n)\), showing more nodes than those at lower energies \((E_m)\).Key points about energy levels include:
  • Each energy level is associated with a different wave function shape.
  • Larger energy values lead to increased oscillations or nodes in the wave functions.
  • Only particular energies, called eigenvalues, allow the Schrödinger equation to have solutions; these correspond to the possible energy levels of the system.
The exercise demonstrates that as \(E_n > E_m\), the wave function \(\psi_n\) with higher energy \(E_n\) will have more nodes than \(\psi_m\) with a lower energy \(E_m\). This is indicative of the Bohr Model which similarly implies quantized energy levels.
One-dimensional Potential
A one-dimensional potential refers to a scenario where the forces acting on a particle depend only on one spatial coordinate, typically "x". In such a potential, the Schrödinger Equation simplifies and becomes more manageable.
  • Simpler potentials, like the potential well or the harmonic oscillator, are often studied in quantum mechanics.
  • Studying one-dimensional potentials helps in understanding the basic behavior of quantum systems, such as electrons in a wire.
  • These models are used as stepping stones to more complex, multi-dimensional problems.
In the exercise, a one-dimensional potential \(V(x)\) is significant because it allows examination of wave function nodes and their relation to energy levels without the complexity of multi-dimensional calculations.

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Most popular questions from this chapter

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.

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 \(a_{1} .^{69}\)

Delta functions live under integral signs, and two expressions \(\left(D_{1}(x) \text { and } D_{2}(x)\right)\) involving delta functions are said to be equal if $$\int_{-\infty}^{+\infty} f(x) D_{1}(x) d x=\int_{-\infty}^{+\infty} f(x) D_{2}(x) d x,$$ for every (ordinary) function \(f(x).\) (a) Show that $$\delta(c x)=\frac{1}{|c|} \delta(x).$$ where \(c\) is a real constant. (Be sure to check the case where \(c\) is negative.) (b) \(\operatorname{Let} \theta(x)\) be the step function: $$\theta(x) \equiv\left\\{\begin{array}{ll}1, & x>0 \\\0, & x<0\end{array}\right.$$ (In the rare case where it actually matters, we define \(\theta(0)\) to be \(1 / 2\).) Show that \(d \theta / d x=\delta(x).\)

(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?

Show that there is no acceptable solution to the (time-independent) Schrödinger equation for the infinite square well with \(E=0\) or \(E<0\). (This is a special case of the general theorem in Problem \(2.2,\) but this time do it by explicitly solving the Schrödinger equation, and showing that you cannot satisfy the boundary conditions.)

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