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Eigenvalue of position operator and delta function.

  1. Aug 31, 2012 #1
    I'd like to show that if there exists some operator [itex] \overset {\wedge}{x} [/itex] which satisfies [itex] \overset {-}{x} = <\psi|\overset {\wedge}{x}|\psi> [/itex], [itex] \overset {\wedge}{x}|x> = x|x> [/itex] be correct.

    [itex]\overset {-}{x} = \int <\psi|x> (\int<x|\overset {\wedge}{x}|x'><x'|\psi> dx')dx = \int <\psi|x> (x<x|\psi>)dx = \int <\psi|x> (x\int<x|x'><x'|\psi> dx')dx[/itex]

    So I concluded that [itex]<x|\overset {\wedge}{x}|x'> = x<x|x'>[/itex] and if we factor out x' I can get [itex]<x|\overset {\wedge}{x} = x<x|[/itex] but not [itex] \overset {\wedge}{x}|x> = x|x>. [/itex]

    However, fortunately I could find a book saying,

    "[itex]<x|\overset {\wedge}{x}|x'> = x\delta(x-x') [/itex] where [itex]\delta(x-x') = <x'|x>[/itex]. Then you can work out the result that [itex] \overset {\wedge}{x}|x> = x|x>. [/itex]"

    I was confused by the definition of delta function in this book and confirmed [itex]\delta(x-x') = <x|x'> [/itex] in some documents on the internet. But I cannot convince myself that the definition [itex]\delta(x-x') = <x'|x>[/itex] is an error of the book, because the definition is used in the book many times. Furthermore the book don't have any ideas of hermitian operators.

    Here, I come to have two questions.

    1. [itex]\delta(x-x') = <x|x'> or <x'|x> ? [/itex] Which is correct answer?
    2. How to prove [itex] \overset {\wedge}{x}|x> = x|x> [/itex] from [itex]<x|\overset {\wedge}{x}|x'> = x\delta(x-x') [/itex] without the properties of hermitian operator?
     
    Last edited: Aug 31, 2012
  2. jcsd
  3. Aug 31, 2012 #2

    George Jones

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    What is the relation between [itex]\delta(x-x')[/itex] and [itex]\delta(x'-x)[/itex]?
     
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