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What is <x_1-x_2> for two particles in a 1-D harmonic oscillator

  1. Jul 25, 2014 #1
    if we have two non-interacting particles of mass M in a one-dimensional harmonic oscillator potential of frequency ω, with the wavefunction defined as:

    $$\Psi\left(x_1,x_2\right) = \psi_n\left(x_1\right) \psi_m\left(x_2\right)$$

    where x_1 and x_2 are two particle co-ordinates. and ψ_n is the nth harmonic oscillator eigenfunction.

    a) will:
    $$\psi_n(x_1)= (\frac{\frac{M*\omega}{\hbar}}{\pi})^{1/4}*H_n(x_1)*e^{-\frac{M*x_1^2*\omega}{2*h}}$$

    or will it be in this format:


    b) What is <x_1-x_2>??
    Last edited: Jul 25, 2014
  2. jcsd
  3. Jul 25, 2014 #2
    B) Would it be the same as <x_1> - <x_2> as they are non-interacting?
  4. Jul 25, 2014 #3
    A) what is x?
  5. Jul 25, 2014 #4
    Since im new to quantum physics I want to even know what <x_1> represents. sorry that had to be x_1 or x_2 in the second formulas, these are the positions of the two particles.
  6. Jul 25, 2014 #5

    Vanadium 50

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    I think you need to think harder about what it is you are trying to calculate. <x1 - x2> should be zero. <|x1 - x2|> won't be zero, nor will sqrt(<(x1-x2)^2>.
  7. Jul 25, 2014 #6
    Can you please tell me what something inside < > represents? Like the thing im trying to solve needs me to calculate <(x_1-x_2)^2> ?
  8. Jul 26, 2014 #7
    It means the expectation value. (What it might be found to be on average.) So <x1-x2> is the expectation value of the vector distance between the two particles and <(x1-x2)^2> is the expectation value of the square of the distance between the two particles. The first one will be zero as the first particle will expected to be the left of the second particle as often as it is the right of it etc.
  9. Jul 26, 2014 #8
    Thanks, So incase I wanted to make sure that <x_1-x_2> is zero then will I have to do this integral? :

    $$<(x_1-x_2)> = \int_{-\inf}^{\inf}{(\psi_n^{*}(x_1)*x_1*\psi_n(x_1))dx_1} - \int_{-\inf}^{\inf}{(\psi_m^{*}(x_2)*x_2*\psi(x_2))dx_2}$$
  10. Jul 26, 2014 #9
    And If I want to calculate $$<(x_1-x_2)^2>$$ then can i write it as $$<x_1^2>-<2*x_1><x_2>+<x_2^2>$$ ?
  11. Jul 26, 2014 #10
    Well they are non-interacting particles, and when I solve these I do get 0.
  12. Jul 26, 2014 #11
    Yes, I'm pretty sure it will work that way if x1 and x2 are independent. I haven't studied this though, so I am surmising......
  13. Jul 26, 2014 #12
    Here is the work I did, two pages full of calculations, am I correct?

    Attached Files:

  14. Jul 26, 2014 #13
    Mniazi, can you send PDF, I'm on an iPad here?
  15. Jul 26, 2014 #14
    sure, here you go, :)

    Attached Files:

  16. Jul 26, 2014 #15
    Thanks! The last equation on the first page is not right. You need to sandwich x^2 between the ψ s in the integral, rather than x and squaring the whole thing.
  17. Jul 26, 2014 #16
    so the <x_1^2> and <x_2^2> integrals are wrong. Oh! ok, do you mean to say that the wavefunctions should stay as they are, and only x_1 should be changed to x_1^2 respectively? Thanks! I shall correct this imediately.
  18. Jul 26, 2014 #17
    Yes, and then you don't need to square the whole thing.
  19. Jul 26, 2014 #18
    When I do the integral from -infinity to infinity, it doesnt converge? Edit: I think there is a problem with my formula, wait, let me fix it .
  20. Jul 26, 2014 #19
    yeah, in the doc I forgot to put a square over the x in the wavefunction exponent.
  21. Jul 26, 2014 #20
    FIXED!! here it is updated:

    Attached Files:

  22. Jul 26, 2014 #21
    here is the more corrected one, I mistakenly redirected M as the wavefunction modulus with variable, which could cause confusion with the Mass

    Attached Files:

  23. Jul 27, 2014 #22
    Good morning Mniazi. What is H(y)? Aren't these the hermite polynomials? I think they should be a function of x.
  24. Jul 27, 2014 #23
    Yes! These are Hermite Polynomials, but since theey are a function of $$\sqrt{\frac{M*\omega}{\pi}*x}$$, so I wrote it with respect to y, which the document states equals to $$\sqrt{\frac{M*\omega}{\pi}*x}$$. :)
  25. Jul 27, 2014 #24
    OK, I missed that. If that's the case why are they appearing in your final expressions? Shouldn't they be evaluated as part of the integral?
  26. Jul 27, 2014 #25
    I even checked the integration on Mathematica, It is coming the same. should they not appear in the final equation?
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