# Basic Question - Green's Functions in Quantum Mechanics

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.

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Beside Econoumou text, I recoomend Supriyo Datta's books, especially the older "Electronic Transport in Mesoscopic Systems" –  Slaviks Apr 3 '12 at 18:36

$z$ 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.

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Okay this seems to follow the lecture slides and the book a bit more closely. In your first equation do we not include the time derivative part as part of the L operator because we want to keep the spatial and time parts seperately? Basically am I to understand that the only reason z is around is to account for the time dependency?For example in Poissons equation we only have to solve $LG = \delta (x-x')$, there is no z, because there is no time involved in the differential equation?? And this in turn leads to no eigenvalues? –  Josh Apr 3 '12 at 19:07
You must keep the spatial and time parts separately because your Green's function depends on $\mathbf{r}$ and $t$. As you have a conservative system it only depends on the time difference t-t' and it is easy to solve the equation in the frequency domain $z$. Poisson's equation is for electrostatics, so obviously there is ni time dependence. –  DaniH Apr 3 '12 at 20:05
@Josh "the only reason z is around is to account for the time dependency" - yes. In Poisson case you use GF to get the potential of an arbitrary charge distribution, in the Schrodinger case, you get full $x$ and $t$ dependence of the wave function for arbitary initial condition. –  Slaviks Apr 4 '12 at 6:38
The concept is the same for both cases, but $t$ adds an exta dimension with a special symmetry (time-translation invariance). –  Slaviks Apr 4 '12 at 6:40
Thank you ever so much, I think I am finally getting this. So for a free particle we get the resolvent $1/(z-L)$ which is also our Greens Function, when using Dirac notation $\sum_n |\phi_n><\phi_n| = 1$ we pop this bit in the numerator and get $G = \frac{\sum_n |\phi_n><\phi_n|}{z - \lambda}$. For a free particle the eigenvalues are $k^2 /2m$ (I guess we are working in units where $\hbar=1$ which in turn we can say $z = E$) meaning we have as a result $G = \frac{1}{E - \frac{k^2}{2m}}$ –  Josh Apr 4 '12 at 8:48
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$z\in\mathbb{C}$ 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).

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Thanks for the quick reply. So if I understand this correctly, the spectrum of L is simply its set of eigenvalues ($\lambda_n$ in this case). When $z =/= \lambda_n$ then it is possible to invert $L-z$ which is defined as the resolvent? This resolvent is analytic wherever z is not in the spectrum of L? When/why would we ever set z=0 and what exactly does that mean, only one eigenvalue? –  Josh Apr 3 '12 at 16:03
@Josh "This resolvent is analytic wherever z is not in the spectrum of L?" - yes, it is. "When would we ever set z=0" - never if we are solving the Shrodinger equation and not, e.g., Poisson. –  Slaviks Apr 3 '12 at 18:33