Basic Question - Green's Functions in Quantum Mechanics

Question

I am trying to learn about Green's functions as part of my graduate studies and have a rather basic question about them:

In my maths textbooks and a lot of places online, the basic Greens function G for a linear differential operator L is stated as

$$ L G = \delta (x-x') $$

which is all well and good. I am now reading Economou's text on GF in Quantum Physics where he goes to define Green's functions as solutions of inhomogenous DE of the type:

$$ [z - L(r)]G(r,r';z) = \delta (r-r') $$

Where $z = \lambda + is$ and L is a time independent, linear, hermitian differential operator that has eigenfunctions $\phi_n (r)$

$$ L(r) \phi_n (r) = \lambda_n \phi_n (r)$$

Where these $\lambda_n$ are the eigenvalues of L. Where does this z come from in the second equation and what is the link between this and the first one?

Edit: see my post below for a new couple of questions.

Answer 1

$z$ is the frequency form the Fourier transform of the time-axis, it appears when you solve the time-dependent Schrodinger equation:

$$\left (i \frac{\partial}{\partial t} - L(x) \right ) G(x,t; x't')= \delta (x-x') \delta(t-t')$$

For time-independent $L$, $G$ is a function of difference $t-t'$ only, so you write:

$$G(x;x'; z) = \int_{-\infty}^{+\infty} e^{i z (t-t')} G(x,t; x't') d t$$.

For the retarded Green function, $G^{R}(x,t; x,t')=0$ if $t<t'$ and the integral converges if $\rm{Im} \, z >0$. For the advanced Green function $G^{A}(x,t; x,t')=0$ if $t>t'$ and the integral converges for $\rm{Im} \, z =s <0$. Thus the resolvent $G(x;x';z)$ conveniently encodes both:

$$G^{R/A}(x,t; x',t') = \lim_{s \to \pm 0} \frac{1}{2\pi} \int e^{-i z(t-t')} G(x;x';z) d z$$ with $+$ sign for the retarded, and $-$ sign for the advanced Green function. For finite systems $G(z)$ is analytic on the whole plane except the discrete set of singularities on the real axis. For an infinite system there is a cut on the real axis corresponding to the continuous part of the spectrum.

Substituting the inverse transform into the equation gives: $$\left ( z - L(x) \right ) G(x;x';z) = \delta (x-x')$$

as in the Economou text.

Answer 2

$z\in\mathbb{C}$ is a complex parameter, or if you wish, a spectral parameter. When $z\in\mathbb{C}\backslash {\rm Spec}(L) $ is not in the spectrum of the operator $L$, then the operator $L-zI$ is invertible, and we can form the resolvent, $$(L-zI)^{-1}. $$ So Economou introduces a $1$-parameter family $(G(z))_{z\in\mathbb{C}}$ of Greens functions. When $z=0$, the second equation in the question(v1) reduces to the first equation (if we ignore the different sign convention and different notation).