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Quantum Mechanics: k basis

  1. Mar 4, 2013 #1
    1. The problem statement, all variables and given/known data

    write the following in K basis:

    A=∫|x><x|dx where the integral limits are from -a to a


    2. Relevant equations



    3. The attempt at a solution

    I tried solving it by inserting the identity
    I=∫|k><k|dk where the integral limits are from -∞ to +∞

    but then I do not know how to proceed from there. What to do about the two integrals with varying limits!
     
  2. jcsd
  3. Mar 4, 2013 #2

    CompuChip

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    Why is it a problem that the integrals have different limits?

    More relevant question to help your forward: what is <k|x> ?
     
  4. Mar 4, 2013 #3
    <k|x>= exp(-ikx)/(2*pi)^0.5
     
  5. Mar 4, 2013 #4
    I am getting a very weird answer.
     
  6. Mar 4, 2013 #5
    I introduced the identity twice and on simplifying, I get 1/2pi ∫∫dk dx ??
     
  7. Mar 4, 2013 #6

    CompuChip

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    If you introduce the identity twice, you should use two different integration variables. So I expect a triple integration, e.g. over x, k and k'.
     
  8. Mar 4, 2013 #7
    Yes, I know that. I simplified things and I got that answer.
     
  9. Mar 4, 2013 #8
    Could you please solve the solve question and suggest the steps?
     
  10. Mar 4, 2013 #9

    CompuChip

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    OK, so I was thinking

    [tex]\int |x\rangle \langle x | \, dx =
    \iiint |k\rangle \langle k | x\rangle \langle x | |k'\rangle \langle k' | \, dx \, dk \, dk'
    \propto \iiint e^{-i(k - k')x}|k\rangle \langle k' | \, dx \, dk \, dk'[/tex]

    Is that where you got to as well?

    And then you go on to use
    [tex]\int e^{i(k - k')x} \, dx \propto \delta(k - k')[/tex]
    but I don't see how the |k> <k'| disappeared from your suggested answer... after all, what you should get is similar in form to |x> <x|.
     
    Last edited: Mar 4, 2013
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