# Difference between expectation value of ##x## and classical amplitude of oscillation for an harmonic oscillator

• Gabri110
Gabri110
Homework Statement
The oscillator is in the state ##\lvert \psi (t)\rangle = \dfrac{1}{\sqrt{2}} \left( e^{-i (n-\frac{1}{2})\omega t}\lvert n-1 \rangle + e^{-i (n+\frac{1}{2})\omega t}\lvert n \rangle \right)##.

Calculate the amplitude of oscillation of a classical oscillator of this frequency and energy ##E = \langle\psi (t)\rvert H \lvert\psi (t)\rangle## and show that it differs from your result for ##\langle\psi (t)\rvert x \lvert\psi (t)\rangle## by a factor independent of ##n##.
Relevant Equations
##\lvert \psi (t)\rangle = \dfrac{1}{\sqrt{2}} \left( e^{-i (n-\frac{1}{2})\omega t}\lvert n-1 \rangle + e^{-i (n+\frac{1}{2})\omega t}\lvert n \rangle \right)##
Using the ladder operators I can easily compute ##E = \langle H\rangle = \hbar \omega n##, so I can find the amplitude of the classical oscillator, as ##E = \frac{1}{2} m \omega^2 x_{max}^2##, thus, ##x_{max} = \sqrt{\dfrac{2 E}{m \omega^2}} = \sqrt{\dfrac{2\hbar n}{m \omega}}##.

The expectation value of ##x## can be also easily computed using the ladder operators. I find ##\langle x\rangle = \sqrt{\dfrac{2\hbar n}{m \omega}}\cos{\omega t}##. This is clearly a problem, as I find that ##\langle x\rangle## is time dependent (and the classical solution isn't!). The difference is ##x_{max} - \langle x\rangle = \sqrt{\dfrac{2\hbar n}{m \omega}} (1 - \cos{\omega t})##, which isn't independent of ##n##, as the exercise statement says.

Can someone help me find where I have made a mistake?

Last edited:
Reread the question. What does “factor” mean?

Gabri110
vela said:
Reread the question. What does “factor” mean?
Oh my... Thank you, I feel so dumb right now...

Gabri110 said:
I find ##\langle x\rangle = \sqrt{\dfrac{2\hbar n}{m \omega}}\cos{\omega t}##.
I got a slightly different result for the expression inside the square root. Of course, I might be the one making a mistake. But I calculated it two ways: using the ladder operators and using the known wavefunctions for the harmonic oscillator.

Also, since ##\langle x\rangle## oscillates harmonically, I wonder if it would be more appropriate to compare the amplitude, ##x_{max}##, of the classical oscillator with the amplitude of the ##\langle x\rangle## oscillation. I'm not sure.

Gabri110

TSny

## 1. What is the expectation value of ##x## in a quantum harmonic oscillator?

The expectation value of ##x## in a quantum harmonic oscillator is the average position of the particle over time. For a harmonic oscillator in its ground state or any stationary state, the expectation value of ##x## is zero because these states are symmetric about the origin. Mathematically, this is expressed as ##\langle x \rangle = \int \psi^*(x) x \psi(x) \, dx##, where ##\psi(x)## is the wavefunction of the oscillator.

## 2. What is the classical amplitude of oscillation for a harmonic oscillator?

The classical amplitude of oscillation for a harmonic oscillator is the maximum displacement from the equilibrium position that the oscillator reaches during its motion. In classical mechanics, this is determined by the initial conditions, such as the initial displacement and velocity of the oscillator. The amplitude is a fixed value and does not change over time in an ideal harmonic oscillator.

## 3. How does the expectation value of ##x## differ from the classical amplitude?

The expectation value of ##x## in a quantum harmonic oscillator represents the average position over time, which is zero for stationary states due to symmetry. In contrast, the classical amplitude represents the maximum displacement from equilibrium during oscillation. Thus, the expectation value of ##x## provides information about the average position, while the classical amplitude provides information about the extent of oscillation.

## 4. Can the expectation value of ##x## be non-zero in a quantum harmonic oscillator?

Yes, the expectation value of ##x## can be non-zero if the quantum harmonic oscillator is in a superposition of states or if it is in a coherent state. In such cases, the symmetry about the origin is broken, and the average position can shift away from zero. However, for stationary states (eigenstates of the Hamiltonian), the expectation value of ##x## remains zero.

## 5. How do quantum fluctuations affect the expectation value of ##x## compared to the classical amplitude?

Quantum fluctuations introduce an inherent uncertainty in the position and momentum of the particle, described by the Heisenberg uncertainty principle. These fluctuations mean that even if the expectation value of ##x## is zero, the particle is not stationary but exhibits probabilistic motion around the equilibrium position. In contrast, the classical amplitude is a deterministic value representing the maximum displacement, unaffected by such quantum uncertainties.

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