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

Click For Summary
The discussion revolves around the calculation of the expectation value of position, ##\langle x\rangle##, and its comparison to the classical amplitude of oscillation, ##x_{max}##, for a harmonic oscillator. Using ladder operators, the expectation value is found to be time-dependent, which contrasts with the classical solution. The difference between ##x_{max}## and ##\langle x\rangle## is highlighted, raising concerns about the calculations. Participants express confusion over the results and the proper interpretation of the term "factor." Ultimately, a mistake in the calculations is acknowledged, emphasizing the need for careful consideration of the ladder operator method.
Gabri110
Messages
3
Reaction score
1
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:
Physics news on Phys.org
Reread the question. What does “factor” mean?
 
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.
 
Yeah, my bad, I had forgotten to divide by 2 the ladder operators...
 
At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...