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Calculating the momentum operator in a quantum state

  1. Apr 29, 2016 #1
    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.
    physics_1.jpg

    I got ħ2k2. The correct answer is ħ2k2 + ħ2/(2d).
    All help is very much appreciated.
     
  2. jcsd
  3. Apr 29, 2016 #2

    blue_leaf77

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    Working in momentum space by first Fourier transforming the given wavefunction might help minimize the possibility of error during calculation.
     
  4. Apr 29, 2016 #3
    Thank you. Unfortunately I have no idea how to do that. Did you see an error in my calculation?
     
  5. Apr 29, 2016 #4

    blue_leaf77

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    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.
     
  6. Apr 29, 2016 #5
    I was doing a definite integral? Where? I shouldn't be doing definite integrals.
     
  7. Apr 29, 2016 #6

    blue_leaf77

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    You are, in calculating the expectation value you integrate from ##-\infty## to ##+\infty##.
     
  8. Apr 30, 2016 #7
    Oh. I thought that because the limits were infinite that it would be an indefinite integral.
     
  9. Apr 30, 2016 #8

    blue_leaf77

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    The second derivative of ##\psi(x)## is not correct.
     
  10. May 2, 2016 #9
    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!
     
  11. May 2, 2016 #10

    blue_leaf77

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    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)##?
     
  12. May 2, 2016 #11

    PeroK

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    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.
     
  13. May 3, 2016 #12
    I tried the derivative again but it doesn't simplify to what you are saying for some reason:
    1462286207396172051231.jpg
     
  14. May 3, 2016 #13

    PeroK

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    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##.
     
  15. May 3, 2016 #14
    The final answer should be ħ2k2+(ħ2/(2d2)). But I get the following:

    1462288327861_1646868398.jpg
     
  16. May 3, 2016 #15

    blue_leaf77

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    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.
     
  17. May 3, 2016 #16
    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.
     
  18. May 3, 2016 #17

    blue_leaf77

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    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.
     
  19. May 3, 2016 #18
    Hmm okay. How did you derive that integral? I'm confused as to how you got that equation.
     
  20. May 3, 2016 #19

    blue_leaf77

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    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##.
     
  21. May 3, 2016 #20
    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.
     
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