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Box normalization

  1. Jun 17, 2013 #1
    In one dimension the normalized momentum eigenstate for a particle with periodic boundary conditions of length L is: [tex]\psi_k(x)=\frac{1}{\sqrt{L}}e^{ikx} [/tex].

    Is the completeness relation obvious:

    [tex]\Sigma \psi_k(x)\psi_{k}(0)=\frac{1}{L}\Sigma e^{ikx}e^{-ik0}=\frac{1}{L}\Sigma e^{ikx}=\delta(x) [/tex]

    where the sum is over discrete eigenstates k?

    How would you go about proving that sum?
     
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  3. Jun 17, 2013 #2

    Simon Bridge

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  4. Jun 18, 2013 #3

    Bill_K

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    Hm, well the first thing to do is recognize that what you've written is false. The LHS is a periodic function with period L, while the RHS is not.
     
  5. Jun 18, 2013 #4
    The schrodinger equation is a Sturm–Liouville equation and thus its solutions must form a complete basis.
    So what is wrong with his equation Bill?
     
  6. Jun 18, 2013 #5

    Bill_K

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    The equation he's trying to prove is purely a mathematical identity involving nothing but exponentials and delta functions. Whether it is true or not cannot depend on how we were led to it, e.g. via Sturm-Liouville theory or Schrodinger quantum mechanics.

    When you say the solutions must form a complete basis, the question is, basis in what space. And the answer is only the space of periodic functions with period L. The completeness relation you write down must adhere to this.
     
  7. Jun 18, 2013 #6
    I can see that if you have a periodic function, and integrate it with respect to the delta function representation sum that I have, you'll get the value of the function at x=0 (you just write the periodic function as a sum of the allowed plane waves and use orthonormality).

    I just don't find it obvious that the infinite sum is zero everywhere outside of x=0. I guess you are adding a lot of phases and they have the potential to cancel. O well.
     
  8. Jun 18, 2013 #7

    Bill_K

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    It isn't. It is nonzero at all points x = 0, ±L, ±2L, etc. What you should have written on the RHS is an infinite sum of delta functions, ∑ δ(x + nL). I know you are thinking that x is bounded in the range 0 to L, but a delta function δ(x) does not know that!
     
  9. Jun 18, 2013 #8
    You are right. I should have been more careful.
     
  10. Jun 18, 2013 #9

    Avodyne

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    To "prove" the original identity: given a function f(x) on an interval -L/2 < x <L/2, we can compute the coefficients of a Fourier series that represents this function on this interval, and is periodic with period L outside of the interval. If we do this computation for δ(x), we get the OP's completeness relation (with appropriate factors of 2π).

    I put "prove" in quotes because we have to worry about convergence of the series and other mathematical details.
     
  11. Jun 18, 2013 #10

    Bill_K

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    Nonsense, Avodyne. The function that is "δ(x) on the interval -L/2 < x <L/2 and periodic with period L outside" is not what anyone else would recognize as an ordinary delta function.
     
  12. Jun 19, 2013 #11

    Bill_K

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    Here's a brief note on the properties of this sum of exponentials, pointing out as I said that it has an infinite number of equally spaced peaks.
     
  13. Jun 19, 2013 #12
    If I remember it correctly,then it can be proved by going to L→∞,In that way the sum may be represented by fourier integral using Ʃ→∫Ldk/2∏ and hence
    1/L∫Ldk/2∏ e(ikx) which is δ(x).But this integral conversion can only be justified on convergence and periodicity condition.
     
  14. Jun 19, 2013 #13

    Bill_K

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    But of course L=infinity is only a special case, and does not help to address the more general question when L is finite.
     
  15. Jun 19, 2013 #14
    yes,you are right.I will have to see some reference for the finite case.
     
  16. Jun 19, 2013 #15

    Avodyne

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    I meant the original identity restricted to -L/2 < x <L/2.
     
  17. Jun 20, 2013 #16
    Here is how we will prove it.We take an arbitrary well behaved function which can be expanded in terms of the eigenfunctions one got.We will take the normalized eigenfunction so it will be (1/√L)eikx,which we will denote by ψn(x) for different k.So
    f(x)=Ʃncnψn(x),we multiply by ψm*(x) on both sides and integrate and use the orthonormality condition on right side and we get,
    cn=∫ψn*(x')f(x')dx',putting into f(x) we have
    f(x)=Ʃn∫ψn*(x')ψn(x)f(x')dx',if we are allowed to change the order of summation and integration we have,
    f(x)=∫f(x')dx'Ʃnψn*(x')ψn(x),and so it is evident that
    Ʃnψn*(x')ψn(x)=δ(x-x'),putting the ψn(x) as we have written and putting x'=0,we get
    (1/L)Ʃneikx=δ(x)
     
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