. Estimate the ground state energy of the quantum harmonic oscillator by Heisenberg's uncertainty principle. Start by assuming that the product of the uncertainties \(\Delta x\) and \(\Delta p\) is at its minimum. Write \(\Delta p\) in terms of \(\Delta x\) and assume that for the ground state \(x \approx \Delta x\) and \(p \approx \Delta p\) then write the ground state energy in terms of \(x .\) Finally, find the value of \(x\) that minimizes the energy and find the minimum of the energy.

Short Answer

Expert verified
The ground state energy of the quantum harmonic oscillator can be estimated using Heisenberg's uncertainty principle as \(E_{min} = \frac{1}{2}\hbar\omega\), where \(\hbar\) is the reduced Planck constant and \(\omega\) is the angular frequency of the harmonic oscillator. To find this value, we first express the uncertainties in position and momentum in terms of each other, then minimize the energy expression with respect to the position \(x\). The minimum energy is achieved at \(x = \sqrt[4]{\frac{\hbar^2}{4m^2\omega^2}}\), where \(m\) is the mass of the oscillating particle.

Step by step solution

01

Understand Heisenberg's uncertainty principle

Heisenberg's uncertainty principle states that the product of the uncertainties in position \(\Delta x\) and momentum \(\Delta p\) is greater than or equal to the reduced Planck constant divided by 2. Mathematically, this is expressed as: \(\Delta x \Delta p \geq \frac{\hbar}{2}\), where \(\hbar\) is the reduced Planck constant.
02

Express \(\Delta p\) in terms of \(\Delta x\)

Since we want to find the minimum of the product of the uncertainties, we can set it equal to the given bound: \(\Delta x \Delta p = \frac{\hbar}{2}\). Now we need to solve for one of the uncertainties in terms of the other. Let us solve for \(\Delta p\): \(\Delta p=\frac{\hbar}{2\Delta x}\).
03

Find the ground state energy expression in terms of \(x\)

We know that the ground state energy of the quantum harmonic oscillator is given as: \(E = \frac{1}{2}m\omega^2x^2 + \frac{p^2}{2m}\), where \(m\) is the mass and \(\omega\) is the angular frequency of the harmonic oscillator. Now, we're given that in the ground state, \(x \approx \Delta x\) and \(p \approx \Delta p\). Thus, we can substitute these values into the energy expression: \(E \approx \frac{1}{2}m\omega^2(\Delta x)^2 + \frac{(\Delta p)^2}{2m}\). Next, we substitute our expression for \(\Delta p\) obtained in step 2: \(E \approx \frac{1}{2}m\omega^2(\Delta x)^2 + \frac{\left(\frac{\hbar}{2\Delta x}\right)^2}{2m}\). Now, we can simplify this expression: \(E \approx \frac{1}{2}m\omega^2(\Delta x)^2 + \frac{\hbar^2}{8m(\Delta x)^2}\). Let's replace \(\Delta x\) by \(x\) to get the ground state energy in terms of \(x\): \(E(x) \approx \frac{1}{2}m\omega^2x^2 + \frac{\hbar^2}{8mx^2}\).
04

Minimize the ground state energy with respect to \(x\)

To find the minimum energy, we need to find the value of \(x\) that minimizes the energy expression \(E(x)\). We do this by differentiating \(E(x)\) with respect to \(x\) and setting it equal to zero: \(\frac{dE}{dx} = m\omega^2x - \frac{\hbar^2}{4mx^3} = 0\). Now, let's solve this equation for \(x\): \(m\omega^2x^4 = \frac{\hbar^2}{4}\). Thus, \(x^4 = \frac{\hbar^2}{4m^2\omega^2}\). Taking the fourth root, we get: \(x = \sqrt[4]{\frac{\hbar^2}{4m^2\omega^2}}\).
05

Calculate the minimum energy

Now, we substitute the value of \(x\) that minimizes the energy back into the expression for \(E(x)\): \(E_{min} = \frac{1}{2}m\omega^2\left(\sqrt[4]{\frac{\hbar^2}{4m^2\omega^2}}\right)^2 + \frac{\hbar^2}{8m\left(\sqrt[4]{\frac{\hbar^2}{4m^2\omega^2}}\right)^2}\), which simplifies to \(E_{min} = \frac{1}{2}\hbar\omega\). This is the ground state energy of the quantum harmonic oscillator estimated using Heisenberg's uncertainty principle.

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Key Concepts

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

Heisenberg's Uncertainty Principle
Understanding Heisenberg's uncertainty principle is fundamental in quantum mechanics. It posits that there is a limit to how precisely we can know both the position and momentum of a particle simultaneously. The more accurately we know one, the less accurately we can know the other. This isn't due to technological limitations but is a fundamental property of the universe.

Mathematically, Heisenberg's principle is represented as \( \Delta x \Delta p \geq \frac{\hbar}{2} \), where \( \Delta x \) is the uncertainty in position, \( \Delta p \) is the uncertainty in momentum, and \( \hbar \) is the reduced Planck constant. This relationship underscores a key concept of quantum mechanics that breaks away from classical physics' deterministic worldview, introducing inherent probabilities into the very fabric of reality.
Ground State Energy
In quantum mechanics, every system has a lowest possible energy state, known as the ground state. This is the energy level occupied by the system when it is at its most stable, with no excess energy to move to a higher state. The concept of ground state is particularly interesting in the case of the quantum harmonic oscillator. Unlike in classical mechanics, where a stationary object has zero kinetic energy, quantum systems have non-zero ground state energy due to the uncertainty principle.

For a quantum harmonic oscillator, the ground state energy can be estimated using Heisenberg's uncertainty principle, which leads to the surprisingly simple expression \( E_{min} = \frac{1}{2}\hbar\omega \). This minimum energy is a direct consequence of the wave-like nature of particles in quantum mechanics; even in the most stable state, there’s a 'zero-point energy' due to their inherent quantum fluctuations.
Quantum Mechanics
Quantum mechanics is a branch of physics that explores physical phenomena at nanoscopic scales, where the action is on the order of the Planck constant. It is here that classical concepts such as definite position and momentum break down. The theory introduces a probabilistic approach to understanding how particles behave. Key features include wave-particle duality, quantization of certain physical properties, and entanglement.

Quantum mechanics reshapes our understanding of nature, providing rules that govern the behavior of atoms and subatomic particles. It is essential for explaining why chemistry and electronics work the way they do and is foundational to modern technologies like semiconductors and lasers.
Momentum and Position Uncertainty
The Heisenberg uncertainty principle highlights the intrinsic uncertainty in the measurements of momentum and position of a particle. These aren’t just uncertainties in measurement, but rather reflect fundamental properties of particles in a quantum state. For a particle in its ground state, where it has the lowest possible energy, the uncertainties in position \( \Delta x \) and momentum \( \Delta p \) reach a compromise or balance that minimizes the system's energy.

Using the uncertainty principle to estimate the ground state energy of a quantum harmonic oscillator begins by acknowledging these uncertainties. This foundation lays the groundwork for the derivation of the ground state energy and helps to explain the non-zero energy level of particles at rest — a concept that challenges our classical understanding of the world.

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Most popular questions from this chapter

Find the expectation value \(\left\langle x^{2}\right\rangle\) of the square of the position for a quantum harmonic oscillator in the ground state. Note: \(\int_{-\infty}^{+\infty} d x x^{2} e^{-a x^{2}}=\sqrt{\pi}\left(2 a^{3 / 2}\right)^{-1}\)

A diatomic molecule behaves like a quantum harmonic oscillator with the force constant \(12.0 \mathrm{N} / \mathrm{m}\) and mass \(5.60 \times 10^{-26} \mathrm{kg}\). (a) What is the wavelength of the emitted photon when the molecule makes the transition from the third excited state to the second excited state? (b) Find the ground state energy of vibrations for this diatomic molecule.

Assume that a proton in a nucleus can be treated as if it were confined to a one-dimensional box of width 10.0 fm. (a) What are the energies of the proton when it is in the states corresponding to \(n=1, n=2,\) and \(n=3\) ? (b) What are the energies of the photons emitted when the proton makes the transitions from the first and second excited states to the ground state?

A particle with mass 0.030 kg oscillates back-andforth on a spring with frequency \(4.0 \mathrm{Hz}\). At the equilibrium position, it has a speed of \(0.60 \mathrm{m} / \mathrm{s}\). If the particle is in a state of definite energy, find its energy quantum number.

A 0.20-kg billiard ball bounces back and forth without losing its energy between the cushions of a \(1.5 \mathrm{m}\) long table. (a) If the ball is in its ground state, how many years does it need to get from one cushion to the other? You may compare this time interval to the age of the universe. (b) How much energy is required to make the ball go from its ground state to its first excited state? Compare it with the kinetic energy of the ball moving at \(2.0 \mathrm{m} / \mathrm{s}\).

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