# Understanding operators for Green's function derivation

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1. Feb 16, 2016

Dear All,

I am trying to understand what operators actually mean when deriving the definition of green's function. Is this integral representation of an operator in the $x-basis$ correct ?

$D = <x|\int dx|D|x>$

I am asking this because the identity operator for non-denumerable or infinite dimensional basis sets is given as

$I = \int |x><x| dx$

But i cannot seem to use this knowledge directly. For eg. if i start by saying $D.f(x)=g(x)$, I can first write this in the bra-ket notation as:

$D |f> = |g>$

where $D$ is a differential operator. I can express the same, (given $D$ is invertible), as:

$|f> = D^{-1}|g>$

Now if i project all of them on the $x-basis$, I can simply write the same equation as:

$<x|f> = <x|D^{-1}|g>$

Now, I am allowed to introduce an Identity operator in the middle as:

$<x|f> = <x|D^{-1}|I|g> = <x|D^{-1}| \int dx'|x'><x'|g> = \int dx'<x|D^{-1}|x'><x'|g>$

Clearly $<x|f> = f(x)$, $<x'|g> = g(x')$ and I introduce $<x|D^{-1}|x'> = G(x,x')$. then I get:

$f(x) = \int dx' G(x,x') g(x')$

Now, the books say that to solve for $G(x,x')$, one can actually note that:

$D|G> = D<x|D^{-1}|x'> = \delta (x-x')$

.... which is the step i couldn't understand!!!! I know that $<x'|x> = <x|x'>= \delta (x-x')$, but how to reproduce these from the expression of the operator ? When does an integral kick in and when not ?

When i use my definition of the operator in terms of an integral, which i defined right at the top then I can get:

$D|G> = <x| \int D|x>dx <x|D^{-1}|x'> = <x|D \int dx |x><x| D^{-1}|x'> = <x|x'>$

Which will lead me to the delta function. But this was just cooked up by looking back. Whats the definition of an operator when expressed in terms on an integral? if this definition is incorrect (which it 100% is), how do I get to the last step that $D|G>= \delta (x-x')$