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Chapter 37: Problem 8

State whether each of the following statements is true or false a) In a one-dimensional quantum harmonic oscillator, the energy levels are evenly spaced. b) In an infinite one-dimensional potential well, the energy levels are evenly spaced. c) The minimum total energy possible for a classical harmonic oscillator is zero. d) The correspondence principle states that because the minimum possible total energy for the classical simple harmonic oscillator is zero, the expected value for the fundamental state \((n=0)\) of the one-dimensional quantum harmonic oscillator should also be zero. e) The \(n=0\) state of the one-dimensional quantum harmonic oscillator is the state with the minimum possible uncertainty \(\Delta x \Delta p\)

Short Answer

Expert verified
Answer: True

Step by step solution

01

a) In a one-dimensional quantum harmonic oscillator, the energy levels are evenly spaced.

True. The energy levels of a one-dimensional quantum harmonic oscillator follow the formula \(E_n = (n+\frac{1}{2})\hbar\omega\), where \(n = 0, 1, 2, 3,...\) is the quantum number, \(\hbar\) is the reduced Planck's constant, and \(\omega\) is the angular frequency. As we can see, the energy levels differ by a constant value of \(\hbar\omega\) for each step in \(n\). Therefore, the energy levels are evenly spaced.
02

b) In an infinite one-dimensional potential well, the energy levels are evenly spaced.

False. The energy levels in an infinite one-dimensional potential well follow the formula \(E_n = \frac{n^2 \pi^2\hbar^2}{2mL^2}\), where \(n = 1, 2, 3, ...\) is the quantum number, \(\hbar\) is the reduced Planck's constant, \(m\) is the mass of the particle, and \(L\) is the width of the well. As \(n\) increases, the difference between consecutive energy levels increases. Therefore, the energy levels are not evenly spaced.
03

c) The minimum total energy possible for a classical harmonic oscillator is zero.

True. For a classical harmonic oscillator, the total energy consists of kinetic energy and potential energy. The minimum total energy occurs when the potential energy is at its minimum and the kinetic energy is zero (the particle is at rest). In this situation, the minimum total energy is zero.
04

d) The correspondence principle states that because the minimum possible total energy for the classical simple harmonic oscillator is zero, the expected value for the fundamental state \((n=0)\) of the one-dimensional quantum harmonic oscillator should also be zero.

False. The correspondence principle states that the behavior of a quantum system should approach the classical behavior in the limit of large quantum numbers. This doesn't mean that the minimum possible total energy for the quantum harmonic oscillator should be zero. In fact, the ground state energy of a one-dimensional quantum harmonic oscillator is given by the formula \(E_0 = \frac{1}{2}\hbar\omega\), which is non-zero.
05

e) The \(n=0\) state of the one-dimensional quantum harmonic oscillator is the state with the minimum possible uncertainty \(\Delta x \Delta p\)

True. According to the Heisenberg uncertainty principle, the product of the uncertainties in position and momentum has a minimum value \(\Delta x \Delta p \geq \frac{1}{2}\hbar\). For a one-dimensional quantum harmonic oscillator, the ground state (when \(n=0\)) achieves this minimum possible uncertainty. In this state, the wave function represents a Gaussian distribution, which optimally satisfies the uncertainty principle.

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Most popular questions from this chapter

Consider the energies allowed for bound states of a half-harmonic oscillator, namely, a potential that is $$ U(x)=\left\\{\begin{array}{l} \frac{1}{2} m \omega_{0}^{2} x^{2} \\ \infty \end{array}\right. \text { for }\left\\{\begin{array}{l} x>0 \\ x \leq 0 \end{array}\right. $$ Using simple arguments based on the characteristics of good wave functions, what are the energies allowed for bound states in this potential?

Two long, straight wires that lie along the same line have a separation at their tips of \(2.00 \mathrm{nm}\). The potential energy of an electron in the gap is about \(1.00 \mathrm{eV}\) higher than it is in the conduction band of the two wires. Conduction-band electrons have enough energy to contribute to the current flowing in the wire. What is the probability that a conduction electron in one wire will be found in the other wire after arriving at the gap?

Suppose a quantum particle is in a stationary state (energy eigenstate) with a wave function \(\psi(x, t) .\) The calculation of \(\langle x\rangle,\) the expectation value of the particle's position, is shown in the text. Calculate \(d\langle x\rangle / d t(\operatorname{not}\langle d x / d t\rangle)\).

Find the probability of finding an electron trapped in a one-dimensional infinite well of width \(2.00 \mathrm{nm}\) in the \(n=2\) state between 0.800 and \(0.900 \mathrm{nm}\) (assume that the left edge of the well is at \(x=0\) and the right edge is at \(x=2.00 \mathrm{nm}\) ).

Consider an attractive square-well potential, \(U(x)=0\) for \(x<-\alpha, U(x)=-U_{0}\) for \(-\alpha \leq x \leq \alpha\) where \(U_{0}\) is a positive constant, and \(U(x)=0\) for \(x>\alpha .\) For \(E>0,\) the solution of the Schrödinger equation in the 3 regions will be the following: For \(x<-\alpha, \psi(x)=e^{i \kappa x}+R e^{-i \kappa x}\) where \(\kappa^{2}=2 m E / \hbar^{2}\) and \(R\) is the amplitude of a reflected wave. For \(-\alpha \leq x \leq \alpha, \psi(x)=A e^{i \kappa^{\prime} x}+B e^{-i \kappa^{\prime} x}\) and \(\left(\kappa^{\prime}\right)^{2}=2 m\left(E+U_{0}\right) / \hbar^{2}\). For \(x>\alpha, \psi(x)=T e^{i \kappa x}\) where \(T\) is the amplitude of the transmitted wave. Match \(\psi(x)\) and \(d \psi(x) / d x\) at \(-\alpha\) and \(\alpha\) and find an expression for \(R\). What is the condition for which \(R=0\) (that is, there is no reflected wave)?
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