# Calculating the momentum operator in a quantum state

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1. Apr 29, 2016

### DarkMatter5

1. The problem statement, all variables and given/known data
A gaussian wave packet is given by the formula:
Ψ(x)=(1/(π1/4d1/2))eikx-(x2/2d2)

Calculate the expectation value in this quantum state of the momentum squared.
2. Relevant equations
<p2>=-ħ∫Ψ*(X) (d2Ψ(x)/dx2) dx
∫e(-x2/d2) dx= d√π
∫xe(-x2/d2) dx =0
∫x2e(-x2/d2) dx = (d3√π)/2
3. The attempt at a solution
Here is my attempt at a solution.

I got ħ2k2. The correct answer is ħ2k2 + ħ2/(2d).
All help is very much appreciated.

2. Apr 29, 2016

### blue_leaf77

Working in momentum space by first Fourier transforming the given wavefunction might help minimize the possibility of error during calculation.

3. Apr 29, 2016

### DarkMatter5

Thank you. Unfortunately I have no idea how to do that. Did you see an error in my calculation?

4. Apr 29, 2016

### blue_leaf77

I didn't go through the detail but it is strange that you still have $x$ after finishing the integral, although the terms containing it eventually cancel in the final step. You were doing a definite integral over $x$, so logically this variable should not appear after the integral is evaluated.

5. Apr 29, 2016

### DarkMatter5

I was doing a definite integral? Where? I shouldn't be doing definite integrals.

6. Apr 29, 2016

### blue_leaf77

You are, in calculating the expectation value you integrate from $-\infty$ to $+\infty$.

7. Apr 30, 2016

### DarkMatter5

Oh. I thought that because the limits were infinite that it would be an indefinite integral.

8. Apr 30, 2016

### blue_leaf77

The second derivative of $\psi(x)$ is not correct.

9. May 2, 2016

### DarkMatter5

Thank you for that correction. e should be to the power of ikx-(x2/2d2.) But it still doesn't change the answer. The mark scheme said that for some reason (ik-(x/d2))2 simplifies to -1/d2. I don't know how this is possible!

10. May 2, 2016

### blue_leaf77

You seem to be not getting the correct second derivative yet. Ok, could you please show how you got the first derivative $\frac{d}{dx}\psi(x)$?

11. May 2, 2016

### PeroK

You're also missing a trick in that $\int_{-\infty}^{\infty} x e^{-\lambda x^2} dx = 0$.

You should expand your quadratic term and the term in $x$ will vanish under the integral. That's the simplification you're missing.

But, as @blue_leaf77 says, you need to get your differentiation right first.

12. May 3, 2016

### DarkMatter5

I tried the derivative again but it doesn't simplify to what you are saying for some reason:

13. May 3, 2016

### PeroK

The second derivative looks right. But, you're still taking an indefinite integral for some reason:

$<p^2> =\int_{-\infty}^{\infty} \dots dx$

Which is a number, not a function of $x$.

14. May 3, 2016

### DarkMatter5

The final answer should be ħ2k2+(ħ2/(2d2)). But I get the following:

15. May 3, 2016

### blue_leaf77

You made a mistake when calculating $v$. If you want to do this problem via integration by part, then you should have left $v$ as a function, not a number. In other words you shouldn't evaluate it yet with the given limits. I strongly suggest that you compute $\langle p^2 \rangle$ directly term by term using the relations you already have under the relevant equations part.

16. May 3, 2016

### DarkMatter5

They gave me the equation ∫e(-x2/d2) dx= d√π. That is what I used to calculate v. I'm not sure how to get the other equations in my working out. I don't get ∫xe(-x2/d2) dx =0 anywhere in my answers so I can' use it.

17. May 3, 2016

### blue_leaf77

That's resulting in a number, right? Not a function, as how $v(x)$ should be.
You have to solve the following integral
$$\langle p^2 \rangle = \frac{-\hbar^2}{d\sqrt{\pi}} \left( \frac{1}{d^4}\int_{-\infty}^\infty x^2 e^{-x^2/d^2} dx - \left(k^2+\frac{1}{d^2}\right) \int_{-\infty}^\infty e^{-x^2/d^2} dx - \frac{2ik}{d^2} \int_{-\infty}^\infty x e^{-x^2/d^2} dx \right)$$
You can completely compute those three integrals using the formula in your original post.

18. May 3, 2016

### DarkMatter5

Hmm okay. How did you derive that integral? I'm confused as to how you got that equation.

19. May 3, 2016

### blue_leaf77

You derived the integral, at the bottom of the image in post #12. I merely rearrange the terms and add $-\hbar^2$ and the necessary integration element $dx$.

20. May 3, 2016

### DarkMatter5

Sorry, I really am not understanding this. Please could you explain how my equation turns into the integral? I can't figure out how to get from what I wrote in post #12 to the integral. I don't think I was taught how to do that.