Advanced Green's function for the classical forced harmonic oscillator

Question

In Ashok Das' Field theory: A path integral approach (second edition, World Scientific, 2006), page 47, for the forced harmonic oscillator with Newton's equation of motion $$ \ddot{x} + m\omega^2x = F(t), $$ the retarded and advanced Green's functions are defined as $$ G^{R,A}(k) = \lim_{\epsilon\rightarrow 0+}\frac{1}{\sqrt{2\pi}}\frac{1}{(k\pm i\epsilon)^2 - \omega^2}. $$ I know that quantum mechanically, both Green's functions are needed, as for instance in here. But classically, is there a physical motivation to define $G^A$? If I'm not mistaken, once one performs the Fourier transforms one finds $G^R(t-t') = G^A(t-t')$, so they are pretty much the same.

Answer 1

$G^R$ is used to evolve initial conditions forward in time. $G^A$ is used to evolve final conditions back in time. Classically, we're normally more interested in evolving initial conditions forward in time to describe causal evolution of a system. However, you could ask "what did this radiation field look like at an earlier time," then you would use the advanced propagator.

The Fourier transforms are very much not the same. The have the same $k$ dependence, but the location of the poles due to the $i\epsilon$ is different (lower half plane for retarded propagator, upper half plane for advanced propagator). The singularities of the propagator are closely tied to causality and there's a lot of information in where they occur. The Kramers-Kronig relations, for example, use causality to infer analytic properties of the Fourier transform of linear response functions.

Answer 2

The physical motivation comes from the following: the force $F(t)$ is turned on only for an interval starting from some time $t_0$.

That's the reason you want the retarded green's function, which gives you the correct solution satisfying boundary conditions: $x(t)=x_0(t)+\int GF~dx'$ where $x_0(t)$ is the solution to the problem for $F(t)=0$. Your reduced green's function is only non-zero for $t>t_0$.

As you correctly stated both have the same Fourier transform, but their values will differ depending on which contour you integrate over. There are recipes to differentiate between the two, and the recipe is encoded in the $\lim_{\epsilon\to 0}(k\pm i\epsilon)$.

Answer 3

Green's function of a classical oscillator is defined as: $$ \frac{d^2}{dt^2}G(t,t')+\omega_0^2G(t,t')=\delta(t-t'). $$ Differential equation itself is not sufficient to unambiguously determine the solution - it needs to be complemented by initial or boundary conditions. These depend on the problem we are dealing with - when studying time evolution these are usually boundary conditions in time:

• Green's function is retarded, if we demand $G(t,t')=0$ for $t<t'$
• It is advanced, when we demand that $G(t,t')=0$ for $t>t'$

A consistent way of solving linear equations with initial conditions is actually by using Laplace transform instead of Fourier transform. But out of Fourier transform is often used one-fits-all method out of sloppiness/laziness, which usually results in no ambiguities. To my knowledge, Engineers in the field of signal processing actually prefer Laplace, which is why it features prominently in classical reference works Abramovitz&Stegun, Gradshtein&Ryzhik, etc.

But classically, is there a physical motivation to define $G^A$? If I'm not mistaken, once one performs the Fourier transforms one finds $G^R(t−t′)=G^A(t−t′)$, so they are pretty much the same.

Whenever dealing with stationary problems (homogeneous in time) the relation between retarded and advanced Green's functions is reduced to trivial complex or hermitian conjugation and possible sign change - this is equally true for quantum and classical cases (just as the example Green's function given in the Q.) E.g., in Keldysh formalism one generally needs three Green's functions for describing the processes - often taken to be Retarded, Advanced, and Keldysh/Lesser - but for stationary problems two functions - Retarded and Keldysh/Lesser - are sufficient.

Related:
Damped oscillator retarded Green function
What is propagated by the time-ordered Green's function in quantum field theory?

Appendix: Freshman's derivation of retarded and advanced Green's functions
Due to the time translational invariance we can write the equation for the Green's function as: $$\ddot{G}(t)+\omega_0^2G(t)=\delta(t).$$ For $t\neq 0$ the oscillation is a homogeneous oscillator equation (no driving term) and we can write its solution as: $$ G(t)=\begin{cases} Ae^{i\omega_0t}+Be^{-i\omega_0t}, t>0,\\ Ce^{i\omega_0t}+De^{-i\omega_0t}, t<0. \end{cases} $$ The solutions have to be joined via the boundary conditions at $t=0$. The first is $G(0^+)=G(0^-)$. The condition for derivatives can be obtained by integrating the oscillator equation over an infinitesimal region around $t=0$: $$ \lim_{\epsilon\rightarrow 0^+}\left[ \int_{-\epsilon}^{+\epsilon}\ddot{G}(t)+ \omega_0^2\int_{-\epsilon}^{+\epsilon}G(t)\right]= \lim_{\epsilon\rightarrow 0^+}\int_{-\epsilon}^{+\epsilon}\delta(t),\\ \lim_{\epsilon\rightarrow 0^+}\left[\dot{G}(+\epsilon)-\dot{G}(-\epsilon) + \omega_0^2\int_{-\epsilon}^{+\epsilon}G(t)\right]= \lim_{\epsilon\rightarrow 0^+}1,\\ \dot{G}(+\epsilon)-\dot{G}(-\epsilon)=1. $$ (This is not unlike the manipulations used when solving for a Schrödinger equation with a delta potential.) Thus, we have the following boundary conditions: $$ G(0^+)=G(0^-),\\ \dot{G}(+\epsilon)-\dot{G}(-\epsilon)=1. $$

Retarded Green's function
If we demand that $G(t)=0$ for $t<0$, we have $C=D=0$, so that the boundary conditions give us $$ A+B=0,\\ i\omega_0(A-B)=1, $$ and the Green's function becomes $$ G^R(t)=\Theta(t)\frac{\sin\omega_0t}{\omega_0}.$$ Fourier transform requires here introducing a small imaginary part to assure convergence (i.e., it becomes de facto Laplace transform): $$ \tilde{G}^R(\omega)= \int_{-\infty}^{+\infty}dte^{i\omega t}G^R(t)\rightarrow \int_{0}^{+\infty}dte^{i\omega t-\eta t}G^R(t)= \frac{1}{\omega_0^2 -(\omega + i\eta)^2}\approx \frac{1}{\omega_0^2-2i\eta\omega-\omega^2} $$ (This is actually identical with the retarded Green's function of a damped harmonic oscillator with damping rate $\eta$: $\ddot{G}(t)+2\eta\dot{G}(t)+\omega_0^2G(t)=\delta(t)$.)

Advanced Green's function The advanced Green's function can be derived in the same way, except that now we have to assure convergence for $t\rightarrow -\infty$, by taking $\eta$ with a different sign: $$ G^A(t)=-\Theta(-t)\frac{\sin\omega_0t}{\omega_0} \tilde{G}^A(\omega)= \int_{-\infty}^{+\infty}dte^{i\omega t}G^A(t)\rightarrow \int_{-\infty}^{0}dte^{i\omega t+\eta t}G^R(t)= \frac{1}{\omega_0^2 -(\omega - i\eta)^2}\approx \frac{1}{\omega_0^2+2i\eta\omega-\omega^2} $$ (Note that this corresponds to a harmonic oscillator with negative damping: $\ddot{G}(t)-2\eta\dot{G}(t)+\omega_0^2G(t)=\delta(t)$.)
As expected, the Green's functions differ only by the sign of the infinitesimal imaginary part (which is also often the case for Green's functions in QFT.)

Other Green's functions
We could recall other relations between various Green's functions in QFT, e.g., that $$G^R(t)=\left[G^>(t)-G^<(t)\right]\Theta(t)$$ and identify $$ G^>(t)=\frac{e^{i\omega_0t}}{2i\omega_0}, G^<(t)=\frac{e^{-i\omega_0t}}{2i\omega_0}, $$ whereas the usual time-ordered/causal QFT Green's function then corresponds to $$ G^T(t)=G^>(t)\Theta(t)-G^<(t)\Theta(-t)=\frac{e^{i\omega_0t}\Theta(t)-e^{-i\omega_0t}\Theta(-t)}{2i\omega_0} $$