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Dirac notation

  1. Nov 3, 2007 #1
    Hey guys,
    I am having difficulty interpreting M(x,x') into dirac notation. How do i write M(x,x') in dirac notation? The actual problem is to write the following in dirac notation:

    int { int { m(x)* M(x,x') g(x') } dx} dx'

    I would appreciate your help.
     
  2. jcsd
  3. Nov 3, 2007 #2
    1. The problem statement, all variables and given/known data
    Hey guys,
    I am having difficulty interpreting M(x,x') into dirac notation. How do i write M(x,x') in dirac notation? The actual problem is to write the following in dirac notation:
    I would appreciate your help.


    2. Relevant equations
    int { int { m(x)* M(x,x') g(x') } dx} dx'





    3. The attempt at a solution
    i tried to use M(x,x') = m(x) m(x'). would that be appropriate?
     
  4. Nov 3, 2007 #3
    could any one help me on this ?
     
  5. Nov 3, 2007 #4

    mjsd

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    what is what in
    ?
     
  6. Nov 3, 2007 #5
    int means integration
     
  7. Nov 3, 2007 #6

    mjsd

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    and what about m , M, g? are the states, operators .. what is what?
     
  8. Nov 4, 2007 #7

    CompuChip

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    Consider the inner product [itex]\langle m | M | g \rangle[/itex] and write it out in the position basis (hint, insert the completeness relation twice, once at each vertical bar).
     
  9. Nov 4, 2007 #8

    CompuChip

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    For your attempt at solution, I don't know what M and m are supposed to be, but if they are what I think they are (M is some sort of propagator and m is a state) then I don't think you can write this.
     
    Last edited: Nov 4, 2007
  10. Nov 4, 2007 #9
    I think m(x) = <x|M> and m(x)* = <M|x>. But i don't know what to write for M(x,x') in dirac notation. Here x and x' are two different bases. Also, g(x') = <x'|g>.
     
  11. Nov 4, 2007 #10

    CompuChip

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    Have you tried that already?
     
  12. Nov 4, 2007 #11
    Ok, now i tried it. Following what you said, I found above integral equals to <m|M|g>. But i don't understand what you mean by M(x,x') is propagator? In dirac notation does M(x,x') equal to <x|M|x'> . It works out fine if i make that assumption.
     
  13. Nov 4, 2007 #12

    malawi_glenn

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    What does your original integral equal? And what do you mean by "I think m(x) = <x|M> .. " ? You dont know?
     
  14. Nov 4, 2007 #13
    No. I was not sure about that statement. But now it's clear. m(x) = <x|m> , not <x|M>. Thanks for your help guys. I get the answer <m|M|g>. But i still don't understand why M(x,x') = <x|M|x'>?
     
  15. Nov 4, 2007 #14

    CompuChip

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    OK, let me put it this way: do you know what [itex]\int \mathrm{d}x |x\rangle\langle x|[/itex] is?
     
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