Dirac notation

  • Thread starter raj2004
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  • #1
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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.
 

Answers and Replies

  • #2
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Homework Statement


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.


Homework Equations


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





The Attempt at a Solution


i tried to use M(x,x') = m(x) m(x'). would that be appropriate?
 
  • #3
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could any one help me on this ?
 
  • #4
mjsd
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what is what in
int { int { m(x)* M(x,x') g(x') } dx} dx'
?
 
  • #5
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int means integration
 
  • #6
mjsd
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and what about m , M, g? are the states, operators .. what is what?
 
  • #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).
 
  • #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:
  • #9
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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>.
 
  • #10
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).

Have you tried that already?
 
  • #11
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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.
 
  • #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?
 
  • #13
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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'>?
 
  • #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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