Quesion on dirac notation

  • Thread starter Marthius
  • Start date
  • #1
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I have recently finished reading a section on this notation, and while i though i understood it, i now find myself lost

The question is to show that
<m|x|n>
Is zero unless m = n + or - 1


As I understand it so far <m| and |n> correspond to the eigenstates of an arbitrary system and x is just supposed to be the x operator

The only thing i could think to do with his was plug into
[tex]\int[/tex]m*xn dx but that did not help me

I also susspect i need to use that
<m|n>=[tex]\delta[/tex]mn

If anyone could give me a nudge in the correct direction it would be much appreciated.
 

Answers and Replies

  • #2
367
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This question is meant to be about the energy eigenstates of the harmonic oscillator, not an arbitrary system.
 
  • #3
37
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I'm sry, but I still do not really understand where to go with this, I suppose I can use |n> = the nth eigenstate of the harmonic oscillator, but isn't the x operator just x?
 
  • #4
367
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Haven't you seen the "ladder operators" a^+ and a^-?
 
  • #5
37
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Haven't you seen the "ladder operators" a^+ and a^-?

I have, and if x were a ladder operator this would be trivial, but I thought it was supposed to be the x (position) operator....
 
  • #6
367
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You can write the x operator in terms of the ladder operators. Your question is trivial, too.
 
  • #7
37
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So then would i write
[tex]x=.5*\sqrt{2h/mw}(A+A^{+})[/tex]?
and distribute out getting
(<m|A|n>+<m|A^+|n>) * some constant
 
  • #8
367
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So then would i write
[tex]x=.5*\sqrt{2h/mw}(A+A^{+})[/tex]?
and distribute out getting
(<m|A|n>+<m|A^+|n>) * some constant

Indeed! And all you need is to show for what m,n combinations <m|A|n>+<m|A^+|n> is nonzero, which as you said is trivial.
 
  • #9
37
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It makes sense now, thank you for your help
 

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