The idea of analytical continuation method to solve the Klein-Gordon equation, how and why?

Question

For simplicity, let's consider a two dimensional version of Klein-Gorden equation: $$ (\partial_t^2-\partial_x^2-\partial_y^2+m^2) G(\vec{x},t) = -\delta(\vec{x})\delta(t) $$

From the previous posts:

• How to obtain the explicit form of Green's function of the Klein-Gordon equation?

• How do I Derive the Green's Function for $-\nabla^2 + m^2$ in $d$ dimensions?

both the answers suggest the analytical continuation and especially follow the answer given by @Sean Lake, we can solve this equation simply by analytical continue to a familiar equation then convert back.

attempt 1

Here is the outline of the procedure:

1. let $t\to iz', x\to x', y\to y'$, we have $lhs=-(\nabla'^2-m^2)G$, $rhs=-\delta(iz')\delta(x')\delta(y')=i\delta(\vec{r}')$. The equation now reads: $$ (\nabla'^2-m^2)G=-i\delta(\vec{r}') $$

2. The above equation is the screened Poission equation, the solution can be easily get as: $$ G=\frac{i\,e^{-mr'}}{4\pi r'} $$

3. Convert back using $z'\to -it, x'\to x, y'\to y$, we have: $$ G=\frac{i\, e^{-m\sqrt{x^2+y^2-t^2}}}{4\pi\sqrt{x^2+y^2-t^2}} $$ for $x^2+y^2>t^2$.

4. when $t^2<x^2+y^2$, we can analytically continue it to: $$ G=\frac{e^{-im\sqrt{t^2-x^2-y^2}}}{4\pi\sqrt{t^2-x^2-y^2}} $$ where we have used the $\sqrt{-1}=i$.

Two question regarding the above procedure:

question 1: since we can also change $t\to -iz'$, the lhs of the equation unchanged, while the rhs of the equation has an additional minus sign, because: $rhs=-\delta(-iz')\delta(x')\delta(y')=-i\delta(\vec{r}')$, therefore the final answer differ by an overall minus sign!

question 2: in step 4, we have used that $\sqrt{-1}=i$, but what if I use $\sqrt{-1}=-i$, it seems that it will lead to: $$ G=-\frac{e^{im\sqrt{t^2-x^2-y^2}}}{4\pi\sqrt{t^2-x^2-y^2}} $$ when $t^2>x^2+y^2$.

attempt 2

the outline:

1. let $t\to z', x\to ix', y\to iy'$, we have $lhs=(\nabla'^2+m^2)G$, $rhs=-\delta(z')\delta(ix')\delta(iy')=\delta(\vec{r}')$. The equation now reads: $$ (\nabla'^2+m^2)G=\delta(\vec{r}') $$

2. the above equation is the Helmholz equation, the solution is: $$ G=-\frac{e^{imr'}}{4\pi r'} $$

3. Convert back, we have $$ G=-\frac{e^{im\sqrt{t^2-x^2-y^2}}}{4\pi\sqrt{t^2-x^2-y^2}} $$ when $x^2+y^2<t^2$.

4. when $x^2+y^2>t^2$, we analytical continue the results, we get: $$ G=i\frac{e^{im\sqrt{x^2+y^2-t^2}}}{4\pi\sqrt{x^2+y^2-t^2}} $$ we have used $\sqrt{-1}=i$.

5. If I use $\sqrt{-1}=-i$, then when $x^2+y^2>t^2$, we have: $$ G=-i\frac{e^{-im\sqrt{x^2+y^2-t^2}}}{4\pi\sqrt{x^2+y^2-t^2}} $$

attempt 2 has the same problem as attemp 1. Also, does the two attemps consistent?

In summary, I am confused about the idea of analytical continuation here, which way to do and why to do so is my question. In my point of view, the above substitution can't be all correct, there must be some point I missed by the mindless substitution of variables.

In fact, I remember the solution to Helmholtz equation has two solutions, which are: $G=-\frac{e^{\pm imr'}}{4\pi r'}$, similar to Screened Poisson equation I think. This would lead to more complications (more results).

Answer 1

I have the feeling that your problem comes from the way you pick a solution to the Poisson equation. In your attempt 1, you claim that the solution is $$G=\frac{i\,e^{-mr'}}{4\pi r'}$$ but you could as well have claimed that it is $$G=\frac{i\,e^{mr'}}{4\pi r'}$$ because $m^2 = (-m)^2$ (the Poisson equation ignores if you pick $m$ or $-m$). The choice of the physical solution is determined by the Sommerfeld condition, that is, the behavior of the solution at infinity. Once this choice is done in a consistent way all along your calculus, you should get rid of any contradiction. Note that I wouldn't employ the word "analytical continuation" when changing the sign of real $r$ in $\sqrt{r}$, the function square root being not analytical at $0$.

Answer 2

You have an error in attempt 1. Let's be explicit about the steps: $$\begin{array}{lrl} & \left[\frac{\partial^2}{\partial t^2} - \nabla^2 + m^2\right] G(r, t) & = \delta(t) \delta(\mathbf{r}) \\ t\rightarrow iz' \Rightarrow & \left[-\frac{\partial^2}{\partial z'^2} - \nabla^2 + m^2\right] G(r, t) & = \delta(iz') \delta(\mathbf{r}) \\ & \left[-\nabla'^2 + m^2\right] G & = \frac{\delta(\mathbf{r}')}{i} \\ & \left[\nabla'^2 - m^2\right] G & = i\delta(\mathbf{r}') \end{array}$$

What this does is it puts the oscillating part back where it belongs - inside of the forward and backward light cones, leaving the exponentially damped parts in the space-like separated region. Note that, $$\frac{\operatorname{e}^{-mr}}{4\pi r} = \frac{1}{(2\pi)^{3/2}}\sqrt{\frac{m}{r}} K_{1/2}(mr),$$ in agreement with the formula presented in my post.

For attempt 2, where you analytically continue the real space variables, yes, the Helmholtz equation has both inbound and outbound wave solutions. In this case you resolve the problem by requiring the Green's function to go to zero as $r\rightarrow \infty$ after you rotate back.

In both cases you should find that you have exponentials with real arguments when $x^2 + y^2 > t^2$, and imaginary arguments otherwise. If you get something else it's because you made an error in the algebra somewhere.