What is the expected value of x^2 for a wave packet in momentum representation?

Click For Summary
The discussion revolves around calculating the expected value of x^2 for a wave packet in momentum representation, specifically defined by the function Φ(p). The user derived the wave function Ψ(x) and calculated the expected values <x> and <p>, but encountered an issue when trying to compute <x^2>, resulting in a divergent integral. Another participant confirmed that the wave function does not decay sufficiently at infinity, leading to the divergence of the expected value for x^2. The conversation highlights the challenges in dealing with wave functions that do not have rapid decay, affecting the validity of certain integrals. The topic emphasizes the importance of understanding the behavior of wave functions in quantum mechanics.
rgalvao
Messages
1
Reaction score
0

Homework Statement


My teacher made up this question, but I think there's something wrong.

Consider the wave packet in momentum representation defined by Φ(p)=N if -P/2<p<P/2 and Φ(p)=0 at any other point. Determine Ψ(x) and uncertainties Δp and Δx.

Homework Equations


Fourier trick and stuff...

The Attempt at a Solution


I found Ψ(x)=(2ħN/x√2πħ)sin(Px/2ħ), where N=±√1/P
<x>=0, <p>=0, <p^2>=(P^2)/12

But when I try to calculate <x^2>, I get a strange integral, which goes to infinity. Am I doing anything wrong?
 
Physics news on Phys.org
Hello rgalvao, welcome to PF :smile: !

I see you didn't get a reply yet, so perhaps I can put in my five cents:

Your wave function ##sin x\over x## is the Fourier transform of a rectangular function and you can indeed see that ##\int \Psi^* \,x \, \Psi dx ## yields zero, but ##\int \Psi^* \, x^2 \, \Psi dx ## diverges.

I don't see anything wrong with what you do. The wave function simply doesn't fall off fast enough with |x| to give a finite expectation value for x2.
Let us know if you or teacher finds otherwise (i.e. correct me if I am wrong ... :wink: ) !
 
I'm not even sure about <x>. Sure, you can argue with symmetry, but the integral is not well-defined.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 31 ·
2
Replies
31
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K