# Analytical problems with Green's function

I have a question about the right definition of the Green's function in physics. Why do we introduce (or not) an infinitesimal, positive number $\eta$ to the following definition:

$$\left[ i\hbar\frac{\partial}{\partial t} - \hat{H}(\mathbf{r}) \pm i\eta\right]G(\mathbf{r},t;\mathbf{r'},t') = \delta(\mathbf{r} - \mathbf{r'})\delta(t-t')$$

• Such infinitesimal additions are almost always a prescription how to close some contour integral in the complex plane, since the naive real integral would hit some poles (cf. advanced vs. retarded vs. Feynman propagator). – ACuriousMind Jul 8 '14 at 0:42
• The subject to read/learn about to understand @ACuriousMind's comment is called "complex analysis" (more specifically the residual theorem or the method of residuals). Many physics departments include it in a course called something like "mathematical methods in physics". – dmckee Jul 8 '14 at 0:58
• See also Propagators in Wikipedia – Trimok Jul 8 '14 at 8:51
• The choice of the $i\eta$ prescription depends on the physical problem you are considering. For instance, in Quantum field theory, one is using the Feynmann propagator. – Trimok Jul 8 '14 at 8:54

A Green's function is nothing but the (generally distributional) integral kernel of the inverse of a given operator. The point is that the operator $$A:= \left[ i\hbar\frac{\partial}{\partial t} - \hat{H}(\mathbf{r})\right]$$ does not admit a unique inverse. Conversely, $$A_{\pm\eta}:=\left[ i\hbar\frac{\partial}{\partial t} - \hat{H}(\mathbf{r}) \pm i\eta\right]$$ admits a unique inverse (for every choice of the sign) and it is given by an operator whose integral kernel is $G_{\pm \eta}$. It turns out that $\lim_{\eta \to 0^+}G_{\pm \eta} f$ exist and these limits select a pair of inverse operators (among the class of inverse operators) of $A$, whose physical meaning is relevant (advanced and retarded solutions).
In practice, the computation of the limits above can be performed in the complex plane using the residuum theory, after having written down $G_{\pm \eta}$ in terms of a Fourier expansion. Within this picture, the appearance of several inverses of $A$ is described in terms of the various ways to surround the singularity in the complex plane.
This type of problem can be conveniently treated using the complex Laplace transform. For a function $f(t)$ with $t\geqslant 0$ it is defined as \begin{equation*} \hat{f}(z)=\int_{0}^{\infty }dt\exp [izt]f(t),\;Imz>0. \end{equation*} Setting $z=\omega +i\eta$ ($\eta >0$ but arbitrary otherwise) we have, with $\theta (t)$ the Heaviside step function ($\theta (t)=1$ for $t\geqslant 0$ and $0$ otherwise), \begin{equation*} \hat{f}(\omega +i\eta )=\int_{-\infty }^{+\infty }dt\exp [i\omega t]\theta (t)\exp [-\eta t]f(t). \end{equation*} Thus $\hat{f}(\omega +i\eta )$ is the Fourier transform of $\theta (t)\exp [-\eta t]f(t)$ and hence \begin{eqnarray*} \theta (t)\exp [-\eta t]f(t) &=&\frac{1}{2\pi }\int_{-\infty }^{+\infty }d\omega \exp [-i\omega t]\hat{f}(\omega +i\eta ) \\ f(t) &=&\frac{1}{2\pi }\int_{-\infty }^{+\infty }d\omega \exp [-i(\omega +i\eta )t]\hat{f}(\omega +i\eta ) \\ &=&\frac{1}{2\pi }\int_{\Gamma }dz\exp [-izt]\hat{f}(z),\;t\geqslant 0, \end{eqnarray*} where $\Gamma$ is the line $(-\infty +i\eta ,+\infty +i\eta )$.
Now the problem at hand. Setting $\hslash =1$ we deal with the Schr\"{o} dinger equation \begin{equation*} \partial _{t}\psi (t)=-iH\psi (t). \end{equation*} Since, by partial integration (note $Imz>0)$, \begin{equation*} \int_{0}^{\infty }dt\exp [izt]\partial _{t}\psi (t)=-\psi (0)-iz\hat{\psi} (z), \end{equation*} we obtain, after some rearranging, \begin{eqnarray*} i[z-H]\hat{\psi}(z) &=&\psi (0), \\ \hat{\psi}(z) &=&-i[z-H]^{-1}\psi (0), \\ \psi (t) &=&\frac{1}{2\pi i}\int_{\Gamma }dz\exp [-izt][z-H]^{-1}\psi (0)=\exp [-iHt]\psi (0),\;t\geqslant 0 \end{eqnarray*} The object $[z-H]^{-1}$ is known as the resolvent of $H$ and plays a key role in mathematical investigations. The Green's function is the corresponding kernel in coordinate representation \begin{equation*} \hat{G}(\mathbf{x}_{1},\mathbf{x}_{2},z)=\langle \mathbf{x}_{1}|[z-H]^{-1}|% \mathbf{x}_{2}\rangle \end{equation*} and \begin{eqnarray*} G(\mathbf{x}_{1},\mathbf{x}_{2},t_{1},t_{2}) &=&\frac{1}{2\pi i}\int_{\Gamma }dz\exp [-iz(t_{1}-t_{2})]\hat{G}(\mathbf{x}_{1},\mathbf{x}_{2},z), \\ \psi (\mathbf{x}_{1},t_{1}) &=&\int d\mathbf{x}_{2}dt_{2}G(\mathbf{x}_{1},% \mathbf{x}_{2},t_{1},t_{2})\psi (\mathbf{x}_{2},t_{2}),\;t_{1}\geqslant t_{2}. \end{eqnarray*} Formally $G(\mathbf{x}_{1},\mathbf{x}_{2},t_{1},t_{2})$ satisfies the equation in the question but now we have a precise description about the $z$ - integral.
I know complex analysis and the residuum theorem. This kind of thinks are introduce to evaluate some integrals of the form $\int\limits_{-\infty}^{+\infty}$ with some function which have a pole in $0$ and then You add an infinitesimal to evaluate an integral using residuum theorem. My question is why can't You just go through zero (where the pole is) make a small circle and then the Cauchy principal value limit ? There are some special problems with analytical properties of those Green's functions if I don't intrduce $\eta$?