# [Physics] How to Derive the Green’s Function for $-\nabla^2 + m^2$ in $d$ dimensions

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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 $\lim_{|\mathbf{x} – \mathbf{x}'|\rightarrow \infty} G(\mathbf{x}, \mathbf{x}') = 0$?

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