What is the solution to this equation in $d$ dimensions: $$(-\nabla_d^2 + m^2)G(\mathbf{x}, \mathbf{x}') = A \delta(\mathbf{x} - \mathbf{x}'),$$ with the boundary condition that $\lim_{|\mathbf{x} - \mathbf{x}'|\rightarrow \infty} G(\mathbf{x}, \mathbf{x}') = 0$?


1 Answer 1


The first step is to recognize that equation is invariant under $d$-dimensional rotations around $\mathbf{x} - \mathbf{x}' = \mathbf{0}$ and simultaneous identical translations of $\mathbf{x}$ and $\mathbf{x}'$, so we can make the following step: $$\begin{align}(-\nabla_d^2 + m^2)G(\mathbf{x}, \mathbf{x}') &= A \delta(\mathbf{x} - \mathbf{x}') \\ & \mathrm{let: }\ r \equiv |\mathbf{x} - \mathbf{x}'| \Rightarrow \\ \frac{A}{\Omega_d r^{d-1}} \delta(r) &= - \frac{1}{r^{d-1}} \frac{\partial}{\partial r} \left[r^{d-1} \frac{\partial G(r)}{\partial r}\right] + m^2 G(r),\end{align}$$ where the factor $\Omega_d r^{d-1}$ comes from the volume element $\operatorname{d} V = \Omega_d r^{d-1} \operatorname{d}r$ and the derivatives on the right hand side (rhs) are the radial term of $\nabla_d^2$ in $d$-dimensional spherical coordinates (wiki link).

The next step is to integrate both sides of the equation over a spherical volume centered at the origin with radius $r$, then take the limit as $r \rightarrow 0$. This yields the normalization condition for $G$: $$\lim_{r\rightarrow 0} \left[r^{d-1} \frac{\partial G}{\partial r}\right] = -\frac{A}{\Omega_d},$$ and handles the part of the equation where the delta function is non-zero.

The region where the delta function is zero, the homogeneous region, becomes: $$0 = \frac{\partial^2 G}{\partial r^2} + \frac{d-1}{r} \frac{\partial G}{\partial r} - m^2 G.$$ The equation in the homogeneous region can be brought into a more familiar form by the function substitution $G(r) = f(r) r^{-(d/2 - 1)}$ giving: $$0 = r^2 \frac{\partial^2 f}{\partial r^2} + r \frac{\partial f}{\partial r} - \left(\frac{d}{2} - 1\right)^2 f - m^2 r^2 f.$$ The familiar form to this equation is the modified Bessel's equation. The most general solution to this equation is: $$f(r) = C K_{d/2-1}(mr) + D I_{d/2-1}(mr),$$ with $I_{d/2-1}$ and $K_{d/2-1}$ modified Bessel functions of the first and second kind, respectively, and $C$ and $D$ constants fixed by the boundary conditions.

The boundary condition at $r\rightarrow \infty$ requires $D=0$, giving the following form for $G$: $$G(r) = \frac{C}{r^{d/2-1}} K_{d/2-1}(mr).$$ Plugging our solution for $G$ into the left hand side (lhs) of the $r \rightarrow 0$ boundary condition derived above gives: $$\lim_{r\rightarrow0} \left[ r^{d-1} \frac{\partial G}{\partial r}\right] = -\Gamma\left(\frac{d}{2}\right) 2^{d/2-1} m^{1-d/2} C,$$ after application of the small argument limit form of $K_\nu$. This implies that: $$\begin{align}C &= \frac{A m^{d/2-1}}{2^{d/2-1}\Gamma(d/2) \Omega_d} \\ & = \frac{A m^{d/2-1}}{2^{d/2} \pi^{d/2}},\end{align}$$ where the explicit form of $\Omega_d = S_{d-1}$ has been inserted.

Finally, replacing $C$ gives: $$G(r) = \frac{A}{(2\pi)^{d/2}} \left(\frac{m}{r}\right)^{d/2-1} K_{d/2-1}(mr).$$

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    $\begingroup$ Excellent answer, resourceful and enjoyable to follow. One question about the analytical continuation: we can only analytical continued to Feynman propagator, right? If yes, why Feynman propagator is justified to do so while other propagator are not? If no, how to get the other propagators? $\endgroup$ Commented Sep 15, 2016 at 8:05
  • $\begingroup$ Another annoying question is what cause the missing term $\delta(r_{d-1}^2-(ct)^2)$, how to fix this part to make your derivation perfect... $\endgroup$ Commented Sep 15, 2016 at 8:07
  • $\begingroup$ Excellent questions. I'd love to hear any answers you have, because I'm still working on this one in my spare time. I can tell you why this one gets the Feynman propagator, though. In the Fourier transform technique the Feynman propagator comes from moving one pole up and one pole down from the integration axis. What we've done here is the same as starting with with a vertical axis and rotating it down, so the relation of the axis to the poles is the same. The key will be redoing the boundary conditions in Minkowski space, I think. $\endgroup$ Commented Sep 15, 2016 at 8:09
  • $\begingroup$ The possible answer to the first question I can come up is: Feynman propagator is allowed to analytical continued because by doing this we might have explicitly deformed the contour from real axis to the imaginary ones. While others are not because the placement of the two poles prevent the contour from rotating(they will pass the poles). See the other answer to my original post. $\endgroup$ Commented Sep 15, 2016 at 8:16
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    $\begingroup$ @user143410 If the differential equation and the boundary conditions all share a symmetry ($d$-dimensional rotations, in this case), then the solution will also share that symmetry. I don't know a proof off of the top of my head, so you might consider consulting math.stackexchange.com for details. $\endgroup$ Commented Oct 16, 2017 at 17:21

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