Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What is the relation between [itex]\delta(x-x')[/itex] and [itex]\delta(x'-x)[/itex]?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook