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Chapter 40: Problem 56

A harmonic oscillator consists of a 0.020-kg mass on a spring. The oscillation frequency is 1.50 Hz, and the mass has a speed of 0.480 m/s as it passes the equilibrium position. (a) What is the value of the quantum number n for its energy level? (b) What is the difference in energy between the levels \(E_n\) and \(E_{n+1}\)? Is this difference detectable?

Short Answer

Expert verified
(a) The quantum number \( n \approx 1.55 \times 10^{31} \). (b) The energy difference \( \Delta E \approx 9.92 \times 10^{-34} \, \text{J} \) is not detectable.

Step by step solution

01

Calculate Angular Frequency

The angular frequency \( \omega \) can be found from the oscillation frequency \( f \). Use the formula: \[ \omega = 2\pi f \] Substitute \( f = 1.50 \, \text{Hz} \) into the formula: \[ \omega = 2\pi \times 1.50 = 3\pi \, \text{rad/s} \]
02

Compute Maximum Speed

The maximum speed \( v_{max} \) of the mass at the equilibrium position is given as 0.480 m/s. This speed is equal to the angular frequency times the amplitude \( A \) of motion: \[ v_{max} = \omega A \] Using the previously calculated \( \omega = 3\pi \), solve for \( A \): \[ 0.480 = 3\pi A \Rightarrow A = \frac{0.480}{3\pi} \approx 0.0509 \, \text{m} \]
03

Calculate Classical Energy

The total mechanical energy \( E_{classical} \) of the harmonic oscillator is given by: \[ E_{classical} = \frac{1}{2} m v_{max}^2 \] Where \( m = 0.020 \, \text{kg} \) and \( v_{max} = 0.480 \, \text{m/s} \). Substitute the values: \[ E_{classical} = \frac{1}{2} \times 0.020 \times (0.480)^2 = 2.304 \times 10^{-3} \, \text{J} \]
04

Determine Quantum Number n

In quantum mechanics, the energy of a level \( E_n \) is given by: \[ E_n = \left(n + \frac{1}{2}\right)\hbar\omega \] Solve for \( n \) when \( E_n \approx E_{classical} = 2.304 \times 10^{-3} \, \text{J} \), using \( \hbar = 1.0545718 \times 10^{-34} \, \text{J路s} \): \[ 2.304 \times 10^{-3} \approx \left(n + \frac{1}{2}\right) \times 1.0545718 \times 10^{-34} \times 3\pi \] Solving for \( n \), we find \( n \approx \frac{2.304 \times 10^{-3}}{\pi \times 3.1637154 \times 10^{-34}} - \frac{1}{2} \approx 1.55 \times 10^{31} \).
05

Calculate Energy Difference

The energy difference between consecutive quantum levels is given by: \[ \Delta E = E_{n+1} - E_n = \hbar \omega \] Using \( \omega = 3\pi \) and \( \hbar = 1.0545718 \times 10^{-34} \, \text{J路s} \), find \( \Delta E \): \[ \Delta E = 1.0545718 \times 10^{-34} \times 3\pi \approx 9.92 \times 10^{-34} \, \text{J} \]
06

Assess Detectability

Determine if the energy difference \( \Delta E = 9.92 \times 10^{-34} \, \text{J} \) is detectable. Given current measurement technology, this difference is extremely small and not detectable, as it is far below the thermal energy quantum ~(10^{-21} \, \text{J})~ in everyday conditions.

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Key Concepts

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

Energy Levels
In quantum mechanics, the energy levels of a quantum harmonic oscillator are discrete. This means energy comes in "quanta," which are fixed amounts rather than continuous possibilities. Unlike classical systems, where energy can vary smoothly, quantum systems like the harmonic oscillator have specific allowable energy levels.
The formula for these energy levels is:
  • \( E_n = \left(n + \frac{1}{2}\right)\hbar\omega \)
Here, \( n \) is the quantum number, \( \hbar \) (h-bar) is the reduced Planck's constant, and \( \omega \) is the angular frequency.
These levels begin at \( n=0 \) for the ground state, which is the lowest energy state of the system, and increase as the quantum number increases. Understanding these discrete energy levels is crucial, as it underpins the stepwise nature of quantum systems and calls for a different approach than classical mechanics.
Angular Frequency
Angular frequency, denoted by \( \omega \), is a measure of how quickly an object oscillates in a circular motion. It is related to the physical oscillation frequency \( f \) through the formula
  • \( \omega = 2\pi f \)
For example, if the frequency of oscillation is 1.50 Hz, the angular frequency would be calculated as \( \omega = 2\pi \times 1.50 \approx 3\pi \, \text{rad/s} \).
The angular frequency is integral to understanding the dynamics of oscillatory systems, including calculating the energy levels for quantum oscillators. It serves as a bridge between linear and rotational motion, adapting the concept of frequency into radians, which are the natural units for angles in physics.
Quantum Number
In the context of quantum mechanics, the quantum number \( n \) determines the specific energy level of a system. For a quantum harmonic oscillator, \( n \) is a non-negative integer \( (0, 1, 2, \,...) \), each corresponding to a different energy state.
The energy associated with each quantum number is defined by the formula:
  • \( E_n = \left(n + \frac{1}{2}\right)\hbar\omega \)
As \( n \) increases, the energy of the system also increases, portraying that energy is quantized. Understanding the quantum number is essential to distinguish between different quantum states.
For complex systems like molecules or atoms, quantum numbers help describe electron arrangement, influencing chemical properties and reactions.
Energy Difference Detection
Energy difference detection refers to measuring the gap between energy levels in a quantum system. For a quantum harmonic oscillator, the energy difference between consecutive levels is given by:
  • \( \Delta E = E_{n+1} - E_n = \hbar \omega \)
In the provided example, the energy difference is around \( 9.92 \times 10^{-34} \, \text{J} \), an extremely small value. Current techniques and devices can measure changes in energy down to about \( 10^{-21} \, \text{J} \), making this energy difference impossible to detect with present-day technology.
This inability to detect such small differences highlights the boundary between theoretical predictions and practical applications. Advances in detection technology are necessary to explore these subtle quantum features in greater detail.

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Most popular questions from this chapter

The ground-state energy of a harmonic oscillator is 5.60 eV. If the oscillator undergoes a transition from its \(n\) = 3 to \(n\) = 2 level by emitting a photon, what is the wavelength of the photon?

Protons, neutrons, and many other particles are made of more fundamental particles called \(quarks\) and \(antiquarks\) (the antimatter equivalent of quarks). A quark and an antiquark can form a bound state with a variety of different energy levels, each of which corresponds to a different particle observed in the laboratory. As an example, the \(\psi\) particle is a low-energy bound state of a so-called charm quark and its antiquark, with a rest energy of 3097 MeV; the \(\psi\)(2S) particle is an excited state of this same quark- antiquark combination, with a rest energy of 3686 MeV. A simplified representation of the potential energy of interaction between a quark and an antiquark is \(U(x) = A\mid x \mid\) , where \(A\) is a positive constant and \(x\) represents the distance between the quark and the antiquark. You can use the WKB approximation (see Challenge Problem 40.64) to determine the bound-state energy levels for this potential-energy function. In the WKB approximation, the energy levels are the solutions to the equation $$\int ^b _a \sqrt{2m[E - U(x)]} dx = {nh \over 2} \space (n = 1, 2, 3, . . .)$$ Here \(E\) is the energy, \(U(x)\) is the potential-energy function, and \(x\) = a and \(x\) = b are the classical turning points (the points at which \(E\) is equal to the potential energy, so the Newtonian kinetic energy would be zero). (a) Determine the classical turning points for the potential \(U(x) = A \mid x \mid\) and for an energy \(E\). (b) Carry out the above integral and show that the allowed energy levels in the WKB approximation are given by $$E_n = {1 \over2m} ( {3mAh \over 4} ) ^{2/3} n^{2/3} \space (n = 1, 2, 3, . . .)$$ (\(Hint\): The integrand is even, so the integral from \(-x\) to \(x\) is equal to twice the integral from 0 to \(x\).) (c) Does the difference in energy between successive levels increase, decrease, or remain the same as \(n\) increases? How does this compare to the behavior of the energy levels for the harmonic oscillator? For the particle in a box? Can you suggest a simple rule that relates the difference in energy between successive levels to the shape of the potential-energy function?

An electron is in the ground state of a square well of width \(L = 4.00 \times 10^{-10}\) m. The depth of the well is six times the ground-state energy of an electron in an infinite well of the same width. What is the kinetic energy of this electron after it has absorbed a photon of wavelength 72 nm and moved away from the well?

A particle of mass \(m\) in a one-dimensional box has the following wave function in the region \(x\) = 0 to \(x = L\) : $$\Psi(x, t) = {1\over \sqrt2} \psi_1(x)e^{-iE_1t/\hslash} + {1\over \sqrt 2} \psi_3(x)e^{-iE_3t/\hslash}$$ Here \(\psi_1(x)\) and \(\psi_3(x)\) are the normalized stationary-state wave functions for the \(n\) = 1 and \(n\) = 3 levels, and \(E_1\) and \(E_3\) are the energies of these levels. The wave function is zero for \(x <\) 0 and for \(x > L\). (a) Find the value of the probability distribution function at \(x = L\)/2 as a function of time. (b) Find the angular frequency at which the probability distribution function oscillates.

An electron is moving as a free particle in the -\(x\)-direction with momentum that has magnitude 4.50 \(\times\) 10\(^{-24}\) kg \(\bullet\) m/s. What is the one- dimensional time-dependent wave function of the electron?
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