homework and exercises - How do I Derive the Green's Function for $-nabla^2 + m^2$ in $d$ dimensions?

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

Answer

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: $$\begin{align}C &= \frac{A m}{2^{d/2-1}\Gamma(d/2) \Omega_d} \\ & = \frac{A m^{d/2-1}}{2^{d/2} \pi^{d/2}},\end{align}$$ where the explicit form of $\Omega_d = S_{d-1}$ has been inserted.

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