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In QM: How to derive <x|f> from f_n(x)?

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data

    If you have a function f_n(x), how do you get the equivalent representation <x|f>?


    2. Relevant equations

    I have a system with a given Hamiltonian (not in matrix-form), from which I derived the specter of energy eigenvalues E_n, and the corresponding energy eigenfunctions f_n(x). However, I am asked to derive the eigenstates in the form <x|f> also, how do I do that?


    3. The attempt at a solution
     
  2. jcsd
  3. Oct 13, 2008 #2

    gabbagabbahey

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    f_n(x) is equivalent to [itex]\left< x | f_n \right>[/itex]
     
  4. Oct 13, 2008 #3

    HallsofIvy

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    By the way, it is "spectrum", not "specter". Although, with Halloween coming, it may be appropriate!

    As gabbagabbahey said, f_n(x) IS [itex]\left< x | f_n \right>[/itex] . <x, f> is the sum of [itex]\left< x | f_n \right>[/itex] over all n.
     
  5. Oct 13, 2008 #4
    Thanks for the answers! So what is then meant by

    “When you have found the spectrum of energy-eigenvalues, find the corresponding energy-eigenstates, both the abstract number basis and the concrete position-representation <x|f>.”?

    Is “the abstract number basis” the same as f_n(x) (the one I have found)? And is “the concrete position representation <x|f>” then the sum of f_n(x) over all n?

    And thanks for the correction of my misspelling, English is not my mother tounge…
     
  6. Oct 13, 2008 #5

    gabbagabbahey

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    Suppose one of the energy eigenstates in some abstract basis was

    [tex]|f_1 \rangle =\frac{1}{\sqrt{2}} |x \rangle -\frac{i}{\sqrt{2}} |y \rangle[/tex]

    Then in the concrete basis it would be [itex]f_1(x)=\langle x|f_1 \rangle=\frac{1}{\sqrt{2}}[/itex]
     
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