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The retarded propagator for a massless scalar field is $$ G_R(t,\mathbf{x} ;t',\mathbf{x}' ) = \frac{ \Theta(t-t') \delta\big( - (t-t')^2 + |\mathbf{x} - \mathbf{x}'|^2 \big)}{2\pi} \tag{1} $$ which is supposed to satisfy the equation $$ ( - \partial_t^2 + \nabla_{\mathbf{x}}^2 )G_R(t,\mathbf{x} ;t',\mathbf{x}' ) = - \delta^{(4)}(x-x')\ . \tag{2} $$

How does one actual going about proving that the solution (1) actually satisfies (2)? I have tried inserting (1) into (2), and I get a lot of $\Theta$ and $\delta$ functions, as well as some derivatives $\delta'$ and $\delta''$, which I can not understand how this is supposed to result in the RHS of (2)?

EDIT: Being explicit, I find that (where $x^2 := -t^2 + |\mathbf{x}|^2$) $$ \begin{align} ( - \partial_t^2 + \nabla_{\mathbf{x}}^2 )G_R(t,\mathbf{x} ;0,\mathbf{0} ) & = ( - \partial_t^2 + \nabla_{\mathbf{x}}^2 ) \big\{ \frac{\Theta(t) \delta(x^2 )}{2\pi} \big\} \\ & = \frac{ \delta'(t) \delta(x^2) + 4 t \delta(t) \delta'(x^2) + 4 \Theta(t) \big[ 2 \delta'(x^2) + x^2 \delta''(x^2) \big] }{ 2\pi } \end{align} $$ This final expression does not seem to reduce to the claimed RHS.

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    $\begingroup$ Try working in Fourier space. Btw, Klein-Gordon equation for a massless particle is also known as wave equation. $\endgroup$ Sep 12, 2020 at 8:27
  • $\begingroup$ @Vadim I know that in fourier space the propagator is just $1/ p^2$ with an $i \epsilon$ in different places for different types of propagators, and there it is easier to show that $\Box e^{i p x} / p^2$ gives the desired answer. But it still interests me how the position-space representation explicitly works out, it does not seem obvious $\endgroup$ Sep 13, 2020 at 3:53
  • $\begingroup$ Related : Why is the propagator the Green's function for Schrodinger equation?. $\endgroup$
    – Frobenius
    Jan 30, 2021 at 23:32
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    $\begingroup$ It should reduce to the correct form (if you made no mistake). One simply needs some more distribution "algebra" here. Observe that $t \delta(t) = 0$ or $\Theta(t) \delta'(x^2) = -\delta(t) \delta(x^2) = 0$, etc. Use the relation $f(t) \delta'(t) = -f'(t) \delta(t)$ several times. $\endgroup$
    – Nikodem
    Jan 31, 2021 at 0:12

1 Answer 1

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First, we need the expression for the propagator. Let's start with the expression for the Greens function \begin{align} \square G = \delta^4(x)\,. \end{align} We now perform a Fourier transform, so the right hand side of the equation becomes \begin{align} \int{\rm d}^4 x e^{{\rm i}(\omega t - \vec k\cdot\vec x)}\delta^4(x) = 1\,. \end{align} The left hand side of the equation is \begin{align} \square G(x) = (\partial_t^2 - \nabla^2)G(x)\to (-\omega^2 + (\vec k)^2)G(k)\,. \end{align} Thus, \begin{align} G(k) = \frac{-1}{\omega^2 - (\vec k)^2}\,. \end{align} The retarded propagator is obtained by the following prescription, $\omega\to \omega + {\rm i}\epsilon$. Thus, we can perform the inverse Fourier transform to obtain the position space representation of $G$: \begin{align} G_\text{ret}(r) = \int\frac{{\rm d}^4 k}{(2\pi)^4} \frac{e^{-{\rm i}(\omega t - \vec k\cdot\vec x)}}{(\omega + {\rm i}\epsilon)^2 - (\vec k)^2}\,. \end{align} Importantly, the poles of this function are in the lower half-plane. Therefore, the integral must close in the lower half plane in order to be non-zero (see the residue theorem for an explanation). When $t > 0$, the coefficient of $\omega$ is negative, so we close the integral in the lower half plane. Similarly, when $t < 0$, we close the integral in the upper half plane. The latter case does not contribute, which we express with a Heaviside function \begin{align} G_\text{ret} &= \theta(t)2\pi{\rm i}\int\frac{{\rm d}^3 k}{(2\pi)^4}\left( \frac{e^{-{\rm i}(-k t - \vec k\cdot\vec x)}}{2k} + \frac{e^{-{\rm i}(k t - \vec k\cdot\vec x)}}{2k}\right)\,,\\ &=\theta(t)\int\frac{{\rm d}^3 k}{(2\pi)^3}e^{{\rm i} k x}\frac{\sin kt}{k} \end{align} We now take the angular integral \begin{align} G_\text{ret} &= \theta(t)\int \frac{4\pi k^2{\rm d} k}{(2\pi)^3}j_0(k r)\frac{\sin kt}{k}\,,\\ &=\frac{\theta(t)}{r}\int \frac{4\pi {\rm d} k}{(2\pi)^3}\sin(k r)\sin(k t)\,. \end{align} We now observe that this is just an expression of the $\delta$-function \begin{align} G_\text{ret} &=\frac{\theta(t)}{4\pi r}\delta(t - r)\,. \end{align}

Now that we have the correct expression for the Greens function, let's apply the D'Alembertian to it and see that this is indeed the Greens function. \begin{align} \square\frac{\theta(t)}{4\pi r}\delta(t - r) &= \left(\partial_t^2 - \nabla^2\right)\frac{\theta(t)}{4\pi r}\delta(t - r)\,,\\ &=\frac{1}{4\pi r}\partial_t^2\theta(t)\delta(t - r) - \theta(t)\nabla^2\frac{1}{4\pi r}\delta(t - r)\,,\\ &=\frac{1}{4\pi r}\partial_t(\delta(t - r)\delta(t) + \theta(t)\delta'(t - r)) - \theta(t)\left(\delta(t - r)\nabla^2\frac{1}{4\pi r} + \frac{1}{4\pi r}\nabla^2\delta(t - r) + 2\partial_r\delta(t - r)\partial_r\frac{1}{4\pi r}\right)\,,\\ &=\frac{1}{4\pi r}(2\delta'(t - r)\delta(t)+\delta(t - r)\delta'(t) +\theta(t)\delta''(t - r)) - \theta(t)\left(-\delta(t - r)\delta(r) + \frac{1}{4\pi r}\delta''(t - r)- \frac{2}{4\pi r^2}\delta'(t - r) + 2\delta'(t - r)\frac{1}{4\pi r^2}\right)\,,\\ &=\frac{1}{4\pi r}(2\delta'(t - r)\delta(t)+\delta(t - r)\delta'(t) +\theta(t)\delta''(t - r)) - \theta(t)\left(-\delta(t - r)\delta(r) + \frac{1}{4\pi r}\delta''(t - r)\right)\,,\\ &=\frac{1}{4\pi r}(2\delta'(t - r)\delta(t)+\delta(t - r)\delta'(t)) - \theta(t)\left(-\delta(t - r)\delta(r)\right)\,,\\ &=\theta(t)\delta(t - r)\delta(r)\,. \end{align} Note, the identity that $\nabla^2 1/r = -4\pi\delta(r)$ is well explained elsewhere (I'll provide a link later).

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  • $\begingroup$ Thanks for the answer. A small nitpick --- the way I wrote the propagator is equivalent to the way you did, although yours definitely seems to be the better version to prove the identity (to see why note $\delta(-t^2 + r^2) = [ \delta(t-r) + \delta(t+r) ] / ( 2r )$, with the second $\delta$ vanishing once you take the step-function into account). $\endgroup$ Sep 15, 2020 at 19:26
  • $\begingroup$ I have two remaining questions: (1) how do you see that $2 \delta'(t-r) \delta(t) + \delta(t-r)\delta'(t)=0$ in the second-last line and (2) how do you see that $\Theta(t) \delta(t-|\mathbf{x}|) \delta^{(3)}(\mathbf{x}) = \Theta(t) \delta(t) \delta^{(3)}(\mathbf{x}) $ in the last line? $\endgroup$ Sep 15, 2020 at 19:29
  • $\begingroup$ Ah good point about the identity. Regarding the first question in your second comment: note that the heaviside function requires $t > 0$. This ties in to your second question: I was wrong - the answer is fixed now! $\endgroup$
    – Guy
    Sep 18, 2020 at 0:11
  • $\begingroup$ I gave you the bounty but will leave the question unanswered, since the result was not proven although is close $\endgroup$ Sep 19, 2020 at 0:37

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