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Stat. States, Orthonorm. expansion coefficients etc qn?

  1. Feb 20, 2016 #1
    I'm reading through some lecture notes for QM in a subsection about stationary states where the definition of orthonormality involving a kronecker delta [tex] \int_{-\infty}^{\infty}\psi_n(x) \ \psi_m^*(x)dx=\delta_{m,n}[/tex]and the formula for some wavefunction that is a superposition of energy eigenstates with expanded coefficients (I think I said that all right? Maybe?) [tex]\psi(x)= \sum_{n=1}^{\infty}b_n\psi_n(x) [/tex] are related to solve for b_n, which turns out to be: [tex]b_n=\sum_{n=1}^{\infty}\psi^*_n(x)\psi(x)[/tex] which is substituted back into the second formula to give: [tex] \psi(x) = \sum_{n=1}^{\infty} \bigg( \int_{-\infty}^{\infty} \psi^*_n(x') \psi(x') \bigg) dx' \ \psi_n(x)= \int_{-\infty}^{\infty} \bigg( \sum_{n=1}^{\infty} \psi^*_n(x') \psi_n(x) \bigg) dx' \ \psi(x') [/tex]Where the order of integration and summation is switched and x' is another variable, not a derivative-It is mentioned that the above formula is of the form: [tex]f(x) = \int_{-\infty}^{\infty}K(x',x)f(x')dx[/tex] and is supposed to hold for any function x.

    What I don't understand is that it says that,

    "It is intuitively clear that K(x',x) must vanish for x' =/= x, for otherwise we could cook up a contradiction by choosing a peculiar function f(x)";

    What exactly is even having the point of two different variables x' and x if it vanishes for all instances where they aren't equal? Isn't this basically saying, since you're multiplying the integrand by zero in all instances where x' =/= x, that the only nonzero result you have is when x' = x? What is the purpose of having the two separate variables? Is it to "preserve" one of them so one of the functions-that based on only f(x) and not at all on f(x')-is un-integrated and is only evaluated later, or something? Or is it more related to the orthonormality bit of this, so it only shows up in one exact circumstance?
     
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  3. Feb 21, 2016 #2

    blue_leaf77

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    A derivation of that kind is usually done to show that
    $$
    \sum_{n=1}^{\infty}\psi^*_n(x')\psi_n(x) = \delta(x-x')
    $$
     
  4. Feb 21, 2016 #3
    It does end up showing that, but I have a bunch of questions revolving around "why" that I'll have to post next morning.
     
  5. Feb 21, 2016 #4

    PeroK

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    You may not be understanding the concept of a dummy variable. First, let's take one of the things you wrote:

    [tex]b_n=\sum_{n=1}^{\infty}\psi^*_n(x)\psi(x)[/tex]

    You can see immediately that this is not right, as ##n## is a free (unmatched) index on the left and a dummy index on the right. The right-hand side here is independent of ##n##, so would give the same result for all ##b_n##. In fact, it should be:

    [tex]b_n=\int_{-\infty}^{\infty}\psi^*_n(x)\psi(x) dx[/tex]

    To get this, you need to be careful with your summation indices, as follows:

    [tex]\psi(x)= \sum_{n=1}^{\infty}b_n\psi_n(x) [/tex] (note that here ##n## is a dummy index, so we can change it to:

    [tex]\psi(x)= \sum_{m=1}^{\infty}b_m \psi_m(x) [/tex]

    Now, for a given ##n## we can multiply both sides by ##\psi^*_n(x)## and integrate:

    [tex]\int_{-\infty}^{\infty}\psi^*_n(x)\psi(x) dx = \int_{-\infty}^{\infty}\psi^*_n(x) \sum_{m=1}^{\infty}b_m \psi_m(x)dx = \sum_{m=1}^{\infty} \delta_{nm} b_m = b_n[/tex]

    The same approach is needed for integrals, where you have a number defined as an integral, like this:

    [tex]b_n = \int_{-\infty}^{\infty}\psi^*_n(x)\psi(x) dx = \int_{-\infty}^{\infty}\psi^*_n(x')\psi(x') dx'[/tex]

    Because in this case ##x## (or ##x'## or whatever you choose) is a dummy variable. If you are going to use this expression for ##b_n## in an equation that involves ##x## as a free variable, then you must change the dummy variable from ##x## to something else, in the same way that we had to change ##n## to something else before.
     
    Last edited: Feb 21, 2016
  6. Feb 21, 2016 #5
    Sorry, the series notation on b_n was just a really egregious typo. I'm going to comb over this post and fix any errors that were just unintenionally writing the wrong symbols, and leave any "intentional" mistakes that are indicative of my current understanding alone.

    Well...apparently I can't, because I'm unable to edit my OP or the following comment, and I'm not sure why. Any idea why that could've happened? I have permissions to edit this comment, but not those two.
     
  7. Feb 21, 2016 #6

    PeroK

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    I wouldn't worry about that. Is the final step still unclear?
     
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