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

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1. Feb 15, 2016

### Cracker Jack

1. The problem statement, all variables and given/known data
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

2. Relevant equations
<x>=∫Ψ*(x,t)xΨ(x,t)dx
<p>=∫Ψ*(x,t)(-iħ ∂/∂x)Ψ(x,t)dx
3. 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)dk

However, I get caught up when calculating <x> at the following integral: <x>=1/2pi*∫∫(Ψ*(k)Ψ(k) x dk)dx

2. Feb 15, 2016

### blue_leaf77

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.

3. Feb 15, 2016

### Cracker Jack

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?

4. Feb 15, 2016

### blue_leaf77

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.

5. Feb 15, 2016

### Cracker Jack

Thank you this helped a lot, and now I think I've gotten the right answer.