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Chapter 35: Problem 4

The ground-state wave function for a quantum harmonic oscillator has a single central peak. Why is this at odds with classical physics?

Short Answer

Expert verified
In classical physics, a harmonic oscillator spends most of its time at the extremes of its motion where potential energy is maximum, not in the middle. Quantum mechanics, however, predicts that the oscillator (particle) has the greatest probability of being found at the centre, as shown by the ground-state wave function. The discrepancy between these two predictions is why this quantum phenomenon is at odds with classical physics.

Step by step solution

01

Understand the quantum harmonic oscillator

In quantum mechanics, the quantum harmonic oscillator is a particle subject to a force proportional to the distance from a fixed point. The ground-state wave function of this system has a Gaussian (bell-shaped) form and is highest at the center. This peak represents the most probable position of the particle.
02

Understand the classical harmonic oscillator

In classical physics, a harmonic oscillator (for example, a swinging pendulum or a mass on a spring) spends the most time at the extremes of its motion when its velocity is zero and the potential energy is maximum, not in the middle.
03

Compare the predictions of quantum mechanics and classical physics

The predictions of quantum mechanics and classical physics for this system are not the same. Quantum mechanics predicts a central peak in the wave function while classical physics predicts the particle spends most time at the ends of its motion. This discrepancy between the two models is what makes this situation seemingly at odds with classical physics.

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Most popular questions from this chapter

A particle's wave function is \(\psi=A e^{-x / a^{2}},\) where \(A\) and \(a\) are constants. (a) Where is the particle most likely to be found? (b) Where is the probability per unit length half its maximum value?

One reason we don't notice quantum effects in everyday life is that Planck's constant \(h\) is so small. Treating yourself as a particle (mass \(60 \mathrm{kg}\) ) in a room-sized one-dimensional infinite square well (width \(2.6 \mathrm{m}\) ), how big would \(h\) have to be if your minimum possible energy corresponded to a speed of \(1.0 \mathrm{m} / \mathrm{s} ?\)

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