What's the Green's function of d'Alembert operator in 4D Minkowski spacetime? I mean the solution to the equation $$ \Box G \equiv \frac{\partial^2 G}{\partial t^2}-\nabla^2 G = \delta^{4}(x)~? $$

I've had some thoughts about this question.

If it was in Euclidean space, the 4D Green's function is (the source point set to be the origin) $$ G(\boldsymbol x,0)=-\frac{1}{4\pi^2}\frac{1}{r^2}, $$ where $r$ is the distance from field point $\boldsymbol x$ to the source point $0$: $$ r^2 = \boldsymbol x^2 = \sum_{i=1}^{4}(x^i)^2. $$ It's easy to show that $$\nabla^2 \left(-\frac{1}{4\pi^2}\frac{1}{r^2}\right) = 0$$ at any point other than the origin point. And we can integrate it over a spherical volume centered at the origin point: $$ \int \nabla^2\left(-\frac{1}{4\pi^2}\frac{1}{r^2}\right) \mathrm d^4x = \int \nabla \left(-\frac{1}{4\pi^2}\frac{1}{r^2}\right) \cdot \mathrm d\boldsymbol S = 1. $$

A natural analogy in Minkowski spacetime would be $$ G(x,0) = -\frac{1}{4\pi^2}\frac{1}{x^2} = -\frac{1}{4\pi^2}\frac{1}{t^2-\boldsymbol x^2}, $$ where $\boldsymbol x$ here represents spatial part of $x$. Actually, this expression appears in Problem 6.1 of Matthew Schwartz's QFT textbook.

I have verified that $$\Box \left(-\frac{1}{4\pi^2}\frac{1}{x^2}\right)=0$$ as long as $x$ is not the origin point. But I can't verify the $\delta$-funcion part. I can't do the integral as in Euclidean space because the function always diverge on the light cone $t^2-\boldsymbol x^2=0$. No matter how the integral ball is, there will always be many points in the ball making the funciton ill.

Is the function above the correct Green's function? How to demonstrate that the correct Green's function can give a $\delta$-function?

  • $\begingroup$ It seems that the Feynman propagator is the Green's function of d'Alembert operator and it's easy to verify it in the momentum space. But as far as I know, there's some converging problem with Feynman propagator? $\endgroup$ Jun 4, 2022 at 14:49
  • $\begingroup$ Maybe the Feynman propagator is enough, since finally we'll always multiply the Green's function by some other function and integrate them? Then I don't need to worry integrating the Green's function alone in a ball. $\endgroup$ Jun 4, 2022 at 14:54

1 Answer 1


In order to make the Green's function unique, you need to specify a boundary condition.

For the boundary condition $\lim_{t\to -\infty}G(\mathbf{x},t)=0$ (which is probably the most often used one) the solution is $$G(\mathbf{x},t)=\frac{1}{4\pi r}\Theta(t)\delta(t-r)$$ where $\Theta$ is Heaviside's step function. See also d'Alembert operator - Green's function.

For other boundary conditions you will get other Green's functions.


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