How do I evaluate <x> with the k-space representation?

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Homework Help Overview

The discussion revolves around evaluating the expectation value of position using the k-space representation of a wave function. Participants are exploring the relationship between k-space and position-space representations in quantum mechanics, particularly focusing on the implications of the wave function's form in k-space.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss transforming the k-space representation into the position-space representation and the implications of doing so for calculating expectation values. There is a focus on the correct form of integrals and the role of operators in different representations.

Discussion Status

Some participants have provided guidance on the necessity of computing integrals in k-space rather than transforming to position space. There is an ongoing exploration of how the position operator acts in k-space, with some agreement on its form as i∂/∂k. Multiple interpretations of the problem are being discussed, particularly regarding the use of different bases in vector space.

Contextual Notes

Participants are navigating the constraints of the problem, including the requirement to compute expectation values in k-space and the potential confusion regarding operator actions in different representations. There is mention of lecture notes that may provide additional context for understanding these concepts.

Cracker Jack
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Homework Statement


Given the following k-space representation of the wave function:
Ψ(k,t) = Ψ(k)e-iħk2t/2m

use the wave number representation to show the following:

<x>t=<x>0 + <p>0t/m

<p>t=<p>0

Homework Equations


<x>=∫Ψ*(x,t)xΨ(x,t)dx
<p>=∫Ψ*(x,t)(-iħ ∂/∂x)Ψ(x,t)dx

The Attempt at a Solution


I have tried to make the k-space representation into the normal Ψ(x,t) representation then taking Ψ*(x,t) and taking that integral where Ψ(x,t)=1/√(2pi)∫Ψ(k)e-iħk2t/2m*ei(kx-ħk2t/2m)dkHowever, I get caught up when calculating <x> at the following integral: <x>=1/2pi*∫∫(Ψ*(k)Ψ(k) x dk)dx
 
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Cracker Jack said:
Ψ(x,t)=1/√(2pi)∫Ψ(k)e-iħk2t/2m*ei(kx-ħk2t/2m)dk
You don't need to do that calculation because you are asked to compute the integral in the k space, and you have also been given the state in k space.
Cracker Jack said:
<x>=1/2pi*∫∫(Ψ*(k)Ψ(k) x dk)dx
Your equation is not correct. Even if you want to do the integral in x space by first expressing ##\psi(x,t)## in its Fourier integral, there should be three integral signs appearing there. But again you don't need to do this. Simply calculate
$$
\int_{-\infty}^{\infty} \psi^*(k,t)x\psi(x,t) \hspace{2mm}dk
$$
Your first task is to find out how the operator ##x## acts in k space.
Hint: the expression of ##x## in k space is similar to that of ##p## in position space.
 
blue_leaf77 said:
You don't need to do that calculation because you are asked to compute the integral in the k space, and you have also been given the state in k space.

Your equation is not correct. Even if you want to do the integral in x space by first expressing ##\psi(x,t)## in its Fourier integral, there should be three integral signs appearing there. But again you don't need to do this. Simply calculate
$$
\int_{-\infty}^{\infty} \psi^*(k,t)x\psi(x,t) \hspace{2mm}dk
$$
Your first task is to find out how the operator ##x## acts in k space.
Hint: the expression of ##x## in k space is similar to that of ##p## in position space.
Thank you, I think this helps. I was thinking x could only act on Ψ if it was Ψ(x,t). Looking through lecture notes, I think that x operates in k space as i∂/∂k. Is that correct?
 
Cracker Jack said:
I think that x operates in k space as i∂/∂k. Is that correct?
Yes correct.
Cracker Jack said:
I was thinking x could only act on Ψ if it was Ψ(x,t).
States are vector in vector space and the operators are the linear maps in the vector space. Since vector space can have more than one bases, the operators can also have different form depending on which bases being used. When position bases is being used ##x## becomes simply a number, in k space ##x## acts such that it has the form you wrote there.
 
blue_leaf77 said:
Yes correct.

States are vector in vector space and the operators are the linear maps in the vector space. Since vector space can have more than one bases, the operators can also have different form depending on which bases being used. When position bases is being used ##x## becomes simply a number, in k space ##x## acts such that it has the form you wrote there.
Thank you this helped a lot, and now I think I've gotten the right answer.
 

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