Classical vs quantum infinite potential well

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 5K views
IHateMayonnaise
Messages
87
Reaction score
0

Homework Statement


This is a problem from Merzbacher.

Assuming a particle to be in one of the stationary states of an infinitely high one-dimensional box, calculate the uncertainties in position and momentum, and show that they agree with the Heisenberg uncertainty relation. Also show that in the limit of very large quantum numbers the uncertainty in $x$ equals the root-mean-square deviation of the position of a particle moving in the enclosure classically with the same energy.


Homework Equations





The Attempt at a Solution



For most of the grunt work please see the attached pdf that I texed.

The first part of this question is quite straightforward and poses no issues, however I'm having problems when it comes to the last part. Specifically I don't know how to find the standard deviation (<x> and <x^2>). Isn't <x> just a/2 (the middle of the well)? And if this is the case how could I find this quantitatively? My guess would be to start with v=x/T, where T is the period and x=2a, but not terribly sure where to go from here. Thoughts??

Thanks!
 
Last edited:
Physics news on Phys.org
You could argue that classically the particle is equally likely to be found anywhere in the well and calculate <x> and <x2> using a uniform distribution, or you could write the particle's position as a function of time and find the time average of x and x2 over one period.
 
vela said:
You could argue that classically the particle is equally likely to be found anywhere in the well and calculate <x> and <x2> using a uniform distribution, or you could write the particle's position as a function of time and find the time average of x and x2 over one period.

yes the continuous uniform distribution seems to work.. don't know why i didn't see that. thanks so much!