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Propagator for a particle in free space

  1. Jul 13, 2007 #1
    Sorry for not following the template but as I'm not answering a problem it didn't seem apropriate. Hopefully this is the right place to put this (it seems somewhere between introductory and advanced).

    Just when I thought I was getting my head round this stuff I'm completly stuck on how the two lines in the section of the book I'm reading follow from the first line. Any explaination would be appreciated please.

    For context it's derived the propagator for a particle in free space (H=P^2/2m) (all in 1 dimension) so:
    |E> = a|p=(2mE)^1/2> + b|p=-(2mE)^1/2>
    for arbitary a,b.
    Giving a propagator:
    U(t)=INT( |p><p|exp(-iEt/hbar) from minus infinity to plus infinity where E is the energy eigenvalue=p^2/2m since it's degenerate.

    The book is then evaluating the propagator (U) explicitly in the X basis.
    I've done the rest in math type to hopefully make it readable:
    http://i196.photobucket.com/albums/aa266/plmokn_02/prop.jpg
    Thanks very much
     
  2. jcsd
  3. Jul 13, 2007 #2

    George Jones

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    What does <x|p><p'|x> equal?
     
  4. Jul 13, 2007 #3
    I'm not sure, sorry. I know the kets labled |p> are the energy eigenvectors, but I'm not sure where that leads.
    The second line doesn't look dissimilar to a delta function but there are some extra factors of hbar that have appeared.
    Thanks
     
  5. Jul 13, 2007 #4

    George Jones

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    Have you ever seen [itex]\psi \left( x \right) = \left< x | \psi \right>?[/itex]

    If you have, then replace [itex]\psi[/itex] by the momentum eigenstate [itex]p.[/itex]
     
  6. Jul 13, 2007 #5
    letting psi take the momentum eigenstate:
    [itex] p( x) = \left< x | p \right>[/itex]
    so
    <x|p><p|x'>=p(x)p(x')
    but then I'm not sure how to get from this to an integrated exponential?
    Thank you.
     
    Last edited: Jul 13, 2007
  7. Jul 13, 2007 #6

    George Jones

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    Good. Now, what is the explicit form of wavefunction p(x) that is a momentum eigenfunction?
     
  8. Jul 13, 2007 #7
    (all the h should be hbar)
    [itex]-i h d\psi/dx = p\psi[/itex]
    giving [itex]\psi=Ae^{ipx/h}[/itex]
    and then when we choose A to normalise it the factor we need comes out. Then writing the second inner product as <p|x'>=complex conjugate(<x'|p>) we get the result in the second line.
    Then the last bit is a straightforward integral once the arguement of the exponential is written in completed square form.
    Thank you
     
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