I’d like to add some details about the answer of Cosmos Zachos, dated 25th Feb. 2023. These details, might make his answer be useful to more people.
Details: 1
On the part that says
you may think of (1) as $(\hat A -\lambda I)~ (\hat A -\lambda
I)^{-1}= I~$
When learning about equations involving “bare” delta functions, such as (1), I picked up two ideas
- Such an equation, did not make complete sense
- It was “unfinished”, without putting the delta function inside some integral
If you turn (1) into complete sense by integrating both sides with a function $f(\mathbf{y} )$, you obtain
\begin{equation*}
\int d\mathbf{y} ~(\hat{A} -\lambda) G_\lambda (\mathbf{x},\mathbf{y}) f(\mathbf{y} )= \int d\mathbf{y} \delta^{(3)}(\mathbf{x}-\mathbf{y}) f(\mathbf{y})
\end{equation*}
Interchanging the order of $\int$ and $(\hat{A} -\lambda)$
\begin{equation*}
(\hat{A} -\lambda) \int~ d\mathbf{y} G_\lambda (\mathbf{x},\mathbf{y}) f(\mathbf{y} )= \int d\mathbf{y} \delta^{(3)}(\mathbf{x}-\mathbf{y}) f(\mathbf{y}) =f(\mathbf{x})
\end{equation*}
Defining, $G_x$ by
\begin{equation*}
G_x~f(\mathbf{x}) = \int~ d\mathbf{y} G_\lambda (\mathbf{x},\mathbf{y}) f(\mathbf{y} )
\end{equation*}
we can say, for any of our $f(\mathbf{x})$, the integrated form of (1) is the same as
\begin{equation*}
(\hat{A} -\lambda)G_x~f(\mathbf{x}) =~f(\mathbf{x})
\end{equation*}
This means that
\begin{equation*}
(\hat{A} -\lambda)G_x =I
\end{equation*}
and
\begin{equation*}
G_x=(\hat{A} -\lambda)^{-1}
\end{equation*}
Hence our integrated form of (1), " the complete sense form of (1) " , is equivalent to the operator equality
\begin{equation*}
(\hat{A} -\lambda) (\hat{A} -\lambda)^{-1}=I
\end{equation*}
Details:2
The answer starts with
Yes, it is the same formula
This opening phrasing, takes quite some justification. It is saying that
\begin{equation*}
(~I - \lambda A~)~(~I - \lambda A~)^{-1}=I~~~\tag{2}
\end{equation*}
is the same thing as
\begin{equation*} (\hat{A} -\lambda)G_\lambda
(\mathbf{x},\mathbf{y})=\delta^{(3)}(\mathbf{x}-\mathbf{y}) \tag{1}
\end{equation*}
Now, in ‘Details:1’, it is shown that the complete sense version of (1), can be thought of as the same as
\begin{equation*} (\hat A -\lambda I)~ (\hat A -\lambda I)^{-1}= I~~\tag{3}
\end{equation*}
Hence, we show that (2) is the same as (1), if we show that the LHS’s of (2) and (3), are the same expression.
NB: I use the rule that for operators , including multiplication by a scalar, $(BC)^{-1}=C^{-1}B^{-1}$.
For the LHS of (2) we have
\begin{equation*}
(~I - \lambda A~)~(~I - \lambda A~)^{-1}= (-1)(~ \lambda A-I~) (~-1 (~ \lambda A-I~)~)^{-1}
\end{equation*}
\begin{equation*}
= (~-1~)(~ \lambda A-I~) (~ \lambda A-I~)^{-1}~(~-1 ~)
\end{equation*}
\begin{equation*}
=\lambda~\left(~A-\frac{ I }{\lambda} \right) \left( \lambda~(~A-\frac{ I }{\lambda})\right)^{-1}
\end{equation*}
\begin{equation*}
=\lambda~\left(~A-\frac{ I }{\lambda} \right) \left( ~A-\frac{ I }{\lambda} \right)^{-1}
\frac{1}{\lambda}
\end{equation*}
Cancel the $\lambda$’s and put $\mu = 1/\lambda$
\begin{equation*}
=(~A-\mu I ) (~A-\mu I ) ^{-1}
\end{equation*}
Introduce a new $\lambda$ by $\lambda= \mu$, this finishes off the transformation of the LHS of (2) into the LHS of (3).
That is, we can think of (2), as the same as the complete sense form of (1).
This justifies the start of the answer.
