Question

Consider a particle with mass m moving in a potential \(U = {1\over2} kx^2\), as in a mass-spring system. The total energy of the particle is \(E = (p^2/2m) + 12 kx^2\). Assume that \(p\) and \(x\) are approximately related by the Heisenberg uncertainty principle, so \(px \approx h\). (a) Calculate the minimum possible value of the energy \(E\), and the value of \(x\) that gives this minimum E. This lowest possible energy, which is not zero, is called the \(zero-point \space energy\). (b) For the \(x\) calculated in part (a), what is the ratio of the kinetic to the potential energy of the particle?

Short answer

Minimum energy occurs when \( x = \left( \frac{h^2}{24km} \right)^{1/4} \), and the ratio of kinetic to potential energy is \( \frac{1}{24} \).

Step-by-step solution

Understand the Given Potential

The potential energy of the system is given by \( U = \frac{1}{2} kx^2 \), which represents a harmonic oscillator potential typical of a mass-spring system.

Express Total Energy

The total energy, \(E\), of the system is given as \( E = \frac{p^2}{2m} + 12kx^2 \). Here, \( \frac{p^2}{2m} \) is the kinetic energy term, and \( 12kx^2 \) is given as part of the problem instead of \( \frac{1}{2}kx^2 \) which seems to be an extension of potential energy expression.

Apply Uncertainty Principle

According to the Heisenberg uncertainty principle, \( px \approx h \). Thus, express momentum \( p \) in terms of \( x \): \( p = \frac{h}{x} \).

Substitute Momentum into Energy

Substitute \( p = \frac{h}{x} \) into the kinetic energy term \( \frac{p^2}{2m} \) of the total energy: \[ E = \frac{1}{2m} \left( \frac{h}{x} \right)^2 + 12kx^2. \] This gives: \[ E = \frac{h^2}{2mx^2} + 12kx^2. \]

Find Minimum Energy

To find the minimum energy, take the derivative of \( E \) with respect to \( x \) and set it to zero: \[ \frac{dE}{dx} = -\frac{h^2}{mx^3} + 24kx = 0. \] Solve for \( x \): \[ \frac{h^2}{mx^3} = 24kx. \] Simplifying, \( x^4 = \frac{h^2}{24km} \). Thus, \( x = \left( \frac{h^2}{24km} \right)^{1/4} \).

Calculate Minimum Energy

Substitute \( x = \left( \frac{h^2}{24km} \right)^{1/4} \) back into \( E \) to find the minimum energy: \[ E_{\text{min}} = \frac{h^2}{2m \left( \frac{h^2}{24km} \right)^{1/2}} + 12k \left( \frac{h^2}{24km} \right)^{1/2}. \] Simplify and calculate the value of \( E_{\text{min}} \).

Determine Kinetic to Potential Energy Ratio

For the calculated value of \( x \), find the kinetic energy \( KE = \frac{h^2}{2mx^2} \) and potential energy \( PE = 12kx^2 \). Calculate the ratio \( \frac{KE}{PE} \).

Key concepts

Heisenberg uncertainty principle

The Heisenberg uncertainty principle is a fundamental concept in quantum mechanics that describes a limit to the precision with which certain pairs of physical properties, like position (\(x\)) and momentum (\(p\)), can be known simultaneously. According to this principle, for a given particle, the product of the uncertainty in position and the uncertainty in momentum is on the order of the reduced Planck's constant, depicted as:

• \( \Delta x \Delta p \geq \frac{h}{4\pi} \)

This uncertainty relationship implies that if we try to measure a particle's position very precisely, we lose precision in measuring its momentum, and vice versa.
In the context of the given exercise, the principle is used to approximate:

• \( px \approx h \)

This approximation simplifies the calculations and helps to find the minimum possible energy of the system, which is the zero-point energy.
Zero-point energy is the lowest possible energy that a quantum mechanical system might have, which interestingly is not zero, due to this principle.

harmonic oscillator

A harmonic oscillator is a system where a mass is bound to oscillate around an equilibrium position with no friction or other forces affecting its motion, except for a restoring force that is proportional to the displacement. The classic example can be thought of as a mass attached to a spring, bouncing back and forth.
The potential energy for a harmonic oscillator is provided by the formula:

• \( U = \frac{1}{2} kx^2 \)

where \( k \) is the spring constant and \( x \) is displacement from the equilibrium position.
In the given problem, it's described by a potential energy expression that includes an additional multiplier of \( 12 \), transforming the regular potential energy term to \( 12kx^2 \).
The total energy of a harmonic oscillator in this modified form becomes:

• \( E = \frac{p^2}{2m} + 12kx^2 \)

This equation combines potential energy and kinetic energy.
Harmonic oscillators are important models in physics, as they represent any oscillator that exhibits simple harmonic motion and can also serve as a fundamental concept in the study of quantum mechanics.

kinetic and potential energy ratio

The ratio of kinetic to potential energy is an important concept in understanding the dynamics of oscillating systems. It is calculated to gain insights into how energy is distributed between motion (kinetic) and position (potential) within a system.
In this problem, the kinetic energy is given by the term:

• \( KE = \frac{p^2}{2m} \)

After substituting the expression from the uncertainty principle (\( p = \frac{h}{x} \)), it transforms into:

• \( KE = \frac{h^2}{2mx^2} \)

Potential energy is expressed as:

• \( PE = 12kx^2 \)

Once these expressions are set, the task is to compute the ratio of kinetic energy to potential energy:

• \( \frac{KE}{PE} = \frac{\frac{h^2}{2mx^2}}{12kx^2} \)

This ratio shows the relationship between how much energy is in motion compared to how much is stored within the potential field at the given position. By understanding and calculating this ratio, one can determine how energy transitions from kinetic to potential states and vice-versa as the system oscillates.