# Eigenvalue of position operator and delta function.

1. Aug 31, 2012

### maser

I'd like to show that if there exists some operator $\overset {\wedge}{x}$ which satisfies $\overset {-}{x} = <\psi|\overset {\wedge}{x}|\psi>$, $\overset {\wedge}{x}|x> = x|x>$ be correct.

$\overset {-}{x} = \int <\psi|x> (\int<x|\overset {\wedge}{x}|x'><x'|\psi> dx')dx = \int <\psi|x> (x<x|\psi>)dx = \int <\psi|x> (x\int<x|x'><x'|\psi> dx')dx$

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

However, fortunately I could find a book saying,

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

I was confused by the definition of delta function in this book and confirmed $\delta(x-x') = <x|x'>$ in some documents on the internet. But I cannot convince myself that the definition $\delta(x-x') = <x'|x>$ 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. $\delta(x-x') = <x|x'> or <x'|x> ?$ Which is correct answer?
2. How to prove $\overset {\wedge}{x}|x> = x|x>$ from $<x|\overset {\wedge}{x}|x'> = x\delta(x-x')$ without the properties of hermitian operator?

Last edited: Aug 31, 2012
2. Aug 31, 2012

### George Jones

Staff Emeritus
What is the relation between $\delta(x-x')$ and $\delta(x'-x)$?