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:

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).

  • $\begingroup$ Have you considered the Wick rotation? The analytic continuation in attempt 1 looks very similar to the Wick rotation to me. $\endgroup$ Sep 24, 2016 at 5:00
  • $\begingroup$ @flippiefanus It is, but my confusion is about the details about the analytical continuation. The listed questions and the concerns. For example, are the second attempt correct? why not equal to the first one. Why $\sqrt{-1}=i$ and $\sqrt{-1}=-1$ give different result, which one to choose, etc... $\endgroup$ Sep 24, 2016 at 6:43
  • 1
    $\begingroup$ Part of the problem could be the Dirac delta with an imaginary argument. What happens if one expresses Dirac delta in terms of its Fourier transform and then changes the variable to an imaginary number? The result does not seem to give a Dirac delta anymore. $\endgroup$ Sep 24, 2016 at 7:58
  • $\begingroup$ You can have a Dirac delta function that works with a rotation in complex coordinates, but it is decidedly non-analytic: $$\delta(x) = \lim_{\sigma\rightarrow 0} \frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{x^*x}{2\sigma^2}\right). $$ It, also, would not spit out a factor of $i$. $\endgroup$ Sep 26, 2016 at 3:23

2 Answers 2


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 $0$.

  • $\begingroup$ @buzhidao A comment on inhomogenous linear differential equations: the space of solutions is an affine space of dimension the order of the differential equation, here two. This is in reply to one of your comments, but I am not allowed to reply to it directly, having not enough reputation. $\endgroup$
    – user130529
    Sep 30, 2016 at 16:47

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.

  • $\begingroup$ My attempt 1 is OK, note that I have a minus sign at RHS. $\endgroup$ Sep 26, 2016 at 1:31
  • $\begingroup$ It seems not make sense of two solutions, which just opposition in signs, to the same inhomogenous differential equation. $\endgroup$ Sep 26, 2016 at 1:34
  • $\begingroup$ I think we could be making a mistake in not applying this identity correctly: $$\delta(ax) = \frac{\delta(x)}{|a|}.$$ Though, then I have no way to explain the factor of $i$ in the Feynman propagator. Regardless of whether that absolute value covers when $a$ is complex, we do know that $\delta(-x) = \delta(x)$, so there should be no sign difference. I was wrong there. I'll have to get back to you, when I have time. $\endgroup$ Sep 26, 2016 at 2:52
  • $\begingroup$ For what you want to do, I recommend looking at page 27 and following of "Quantum Field Theory: From Operators to Path Integrals" by Huang. You should be able to examine the Fourier transform based proof in the free preview. $\endgroup$ Sep 26, 2016 at 3:59

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