quantum field theory - Complex integration by shifting the contour

In section 12.11 of Jackson's Classical Electrodynamics, he evaluates an integral involved in the Green function solution to the 4-potential wave equation. Here it is:

$$\int_{-\infty}^\infty dk_0 \frac{e^{-ik_0z_0}}{k_0^2-\kappa^2}$$

where $k$ and $z_0$ are real constants.

Jackson considers two open contours: one above and one below the real axis. I understand that in order to use Jordan's lemma, when $z_0 < 0$ we have to close the contour in the upper half of the complex plane whereas if $z_0 > 0$ we have to close the contour in the lower half plane.

What I don't understand is why it's OK to consider contours above and below the real axis when the original integral is along the real axis. As I understand it, the necessity to deal with poles like this also arises a lot in QFT, so perhaps it is well understood from that point of view.

Answer

Suppose we want to analyse the problem of a forced harmonic oscillator. Denote as $\phi(t)$ the time dependent position of the oscillator. The oscillator experiences two forces, the spring force $-k\phi(t)$ and an external force $F_{\text{ext}}(t)$. Newton's law says

$$\begin{align} F(t) &= m a(t) \\ -k \phi(t) + F_{\text{ext}}(t) &= m \ddot{\phi}(t) \\ F_{\text{ext}}(t)/m &= \ddot{\phi}(t) + (k/m) \phi(t) \\ j(t) &= \ddot{\phi}(t) + \omega_0^2 \phi(t) \tag 1 \end{align}$$

where $\omega_0$ is the free oscillation frequency and $j(t)\equiv F_{\text{ext}}/m$. We use the following Fourier transform convention: $$\begin{align} f(t) &= \int_\omega \tilde{f}(\omega) e^{i\omega t} \frac{\mathrm d\omega}{2\pi} \\ \tilde{f}(\omega) &= \int_t f(t) e^{-i\omega t}~\mathrm dt . \end{align}$$

With this convention on Eq. $(1)$, and defining $$\omega_{\pm} \equiv \pm \omega_0,$$ we find $$\tilde{\phi}(\omega) = \frac{\tilde{j}(\omega)}{\omega_0^2-\omega^2} = \frac{-\tilde{j}(\omega)}{(\omega-\omega_+)(\omega-\omega_-)}. \tag 2$$ From Eq. $(2)$ we see that the Green's function is $$\tilde{G}(\omega) = \frac{-1}{(\omega-\omega_+)(\omega-\omega_-)}$$ which has poles on the real axis. If we want to compute $\phi(t)$ we do a Fourier transform

$$\phi(t) = \int_\omega \frac{-\tilde{j}(\omega)e^{i\omega t}}{(\omega-\omega_+)(\omega-\omega_-)} \frac{\mathrm d\omega}{2\pi} = \int_\omega e^{i\omega t}\tilde{j}(\omega) \tilde{G}(\omega)\frac{\mathrm d\omega}{2 \pi}. \tag{*}$$

This integral is tricky because of the poles on the axis. The solution everyone knows is to push the poles off the axis by adding an imaginary part to $\omega_{\pm}$, or by moving the contour above or below the real axis, but what does this actually mean physically? How do we choose which direction to push the poles or move the contour?

In a real system, we always have some damping. In our oscillator model, this could come in the form of a velocity dependent friction $F_{\text{friction}} = -\mu \dot{\phi}(t)$. Defining $2\beta = \mu/m$, the equation of motion becomes

$$\ddot{\phi}(t) + 2\beta \dot{\phi}(t) + \omega_0^2\phi(t) = j(t) . \tag 3$$

Fourier transforming everything again leads to Eq. $(2)$ but now with $$\begin{equation} \omega_{\pm} = \pm \omega_0' + i\beta \end{equation}$$ where $$\begin{equation} \omega_0' = \omega_0\sqrt{1-(\beta/\omega_0)^2}. \end{equation}$$

Therefore, we see that adding damping moves the poles a bit toward the origin along the real axis, but also gives them a positive imaginary component. In the limit of small damping (i.e. $\beta \ll \omega_0$), we find $\omega_0' \approx \omega_0$. In other words, the frequency shift of the poles due to the damping is small. So let's ignore that and focus on the added imaginary part.

Ok, suppose we want to do the integral $(*)$ in the case that $j(t)$ is a delta function at $t=0$. In that case, $\tilde{j}=1$ (I'm ignoring units) and we have $$\phi(t) = \int_\omega \frac{e^{i\omega t}}{(\omega-\omega_+)(\omega-\omega_-)} \frac{\mathrm d\omega}{2\pi}$$ As you noted, for $t<0$ you have to close the contour in the lower plane in order to use Jordan's lemma. There aren't any poles in the lower half plane, so we get $\phi(t<0)=0$. This makes complete sense: the driving force is a delta function at $t=0$ and there shouldn't be any response of the system before the driving happens. This means that our introduction of friction imposed a causal boundary condition to the system! For $t>0$, you close in the upper half plane where there are poles, and so you get some response out of the integral.

In many cases, you don't naturally have damping in the system. For example, the Green's function from the question, $$\int^{\infty}_{-\infty}~\mathrm dk_0 \frac{e^{−ik_0 z_0}}{k_0^2 − \kappa^2}$$ doesn't have any damping and thus the poles sit on the real axis. So what you do is just bump the contour a bit up or down, or equivalently add $\pm i \beta$ to the poles (most people write $i \epsilon$ instead of $i \beta$), then do the integral, and finally take $\beta \rightarrow 0$. In doing this, you're solving the problem in the presence of damping (or anti-damping), and then taking the damping to zero in the end to recover the no-damping case.

Choosing to push the contour up or down, or equivalently choosing the sign of $\pm i \beta$, corresponds to imposing either friction or anti-friction, causal or anti-causal boundary conditions. If you pick the "causal" boundary condition, you find that the response of the system to a delta function in time and space is an outgoing spherical wave which starts at the delta function source. This gives you the so-called "retarded Green's function". If you pick the other condition, you find that the solution for a point source is actually an incoming spherical wave which converges right onto the point of the source. This gives you the so-called "Advanced Green's function".

The thing is, you can solve a problem using either Green's function. You're "allowed" to push the contour up or down (or add $+i\beta$ or $-i\beta$ to the poles) because you invented that as a trick to do the integral; it's not representing a real factor in your physical system. Of course, in problems where there is damping, the choice is made for you. When you have damping, you can't have fields at infinity; they'd be damped away by the time they interact with your sources.

I hope this was helpful, and I really hope if someone finds mistakes they'll jump in and fix 'em.