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

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

correlation-functionsfield-theorygreens-functionshomework-and-exercisespropagator

#### Related Solutions

I'm aware this is an old question and may be considered somewhat obsolete, but let me answer it nevertheless - if only for the sake of completeness.

The position space representation of the Klein-Gordon Green function (propagator) clearly looks intimidating. The trick is to do the calculation in momentum space, where the propagator is just a rational function. Before actually doing this let me point out that in the massless case, $m = 0$, and for a static source, $\rho = \rho (\mathbf{x})$, one is actually solving the Poisson equation. If the source is radially symmetric, $\rho = \rho(r)$, as suggested in the question, the solution is the Coulomb potential, $\phi = \phi(r) \sim 1/r$. Allowing for non-vanishing mass, one obtains a Yukawa potential, $\phi \sim \exp(-mr)/r$.

This can be shown explicitly in terms of Fourier integrals. First, transform the field,

$$ \phi(k) = \int d^4 x \, e^{i k\cdot x} \, \phi (x) \; , $$

and similarly the the source, $\rho \to \rho(k)$. The momentum space solution of the KG equation is then

$$ \phi(k) = \frac{\rho(k)}{k^2 - m^2 + i \epsilon} \; , $$

with $k^2 = k_0^2 - \mathbf{k}^2$ and the causal $i\epsilon$-prescription. (Change appropriately for retarded/advanced solutions.) Then transform back to position space,

$$ \phi (x) = \int \frac{d^4 k}{(2\pi)^4} \, e^{-i k\cdot x} \, \frac{\rho(k)}{k^2 - m^2 + i \epsilon} \; . \quad (**)$$

As an example, take a Gaussian source, $\rho(r) = \rho_0 \exp(-\alpha r^2)$, with 'normalisation' $ \rho_0 = (\alpha/\pi)^{3/2}$. Its Fourier transfrom is $\rho(k) = 2\pi i \, \delta(k^0) \exp (-k^2/4\alpha)$. The $k^0$-integral in $(**)$ is hence trivial, and the $d^3 k$ integration can be done with the usual residue technique writing $\mathbf{k}^2 + m^2 \equiv (\kappa+im)(\kappa-im)$. The result is

$$ \phi(r) = e^{m^2/4\alpha} \, \frac{e^{-mr}}{4\pi r} \; .$$

In the point source limit ($\alpha \to \infty$) we reobtain the standard Yukawa potential.

For time dependent sources there will be energy transfer (no $\delta(k^0)$), and the integral $(**)$ will normally be harder. Typically, one can do the $k^0$-integration via residues and the remaining one(s) using stationary phase as e.g. in Ch. 2.1 of Peskin and Schroeder.

"My problem here is the following: to find G we need boundary conditions of the problem. I can't understand, though, what boundary conditions we should impose here.

So to solve scattering problems using Green's functions like that, what are the boundary conditions we need to impose to compute the Green's function?"

Why do you want to set a boundary condition? Clearly here your problem has no *boundary* since you can detect some particles very far away from the place where the interaction with the potential takes place. However, surely you will need an *intitial condition* provided by $\psi^0_\textbf{k}(\textbf{r})$.

I feel (tell me if I'm wrong) that what you mean by *"boundary condition"* refers actually to what kind of solution you are interested in. As you pointed out, the equation
$$
(\nabla^2+k^2)G(\textbf{r},\textbf{r}')=-4\pi\delta(\textbf{r}-\textbf{r}')
$$
has two solution called *retarded* $G^+$ and *advanced* $G^-$ Green functions. This gives rise to two different kind of solutions to your differential equation :

The causal solution : $$ \psi_\textbf{k}(\textbf{r})=\psi^{\text{(in)}}_\textbf{k}(\textbf{r})-\frac{1}{4\pi}\int\mathrm{d}\textbf{r}'\,G^+(\textbf{r}-\textbf{r}')\,U(\textbf{r}')\,\psi_\textbf{k}(\textbf{r}') $$ where $\psi^{\text{(in)}}_\textbf{k}\equiv \psi^{\text{0}}_\textbf{k}$ is interpreted as the

*incoming field*,*i.e.*the asymptotic field you obtain when taking the limit $t\rightarrow -\infty$ and which evolves freely with time. This is usually the solution people are interested in because it describes the scattering event as a result of the interaction with the potential.The anti-causal solution : $$ \psi_\textbf{k}(\textbf{r})=\psi^{\text{(out)}}_\textbf{k}(\textbf{r})-\frac{1}{4\pi}\int\mathrm{d}\textbf{r}'\,G^-(\textbf{r}-\textbf{r}')\,U(\textbf{r}')\,\psi_\textbf{k}(\textbf{r}') $$ where $\psi^{\text{(out)}}_\textbf{k}$ is interpreted as the

*outcoming field*,*i.e.*the asymptotic field you obtain when taking the limit $t\rightarrow +\infty$ and which had evolved freely with time from the past.

In both of these cases, the limits $t\rightarrow\pm\infty$ allow you to get rid of the integral terms and give rise to different interpretations of the solutions.

- You may also be only interested in the radiated field which is defined as : $$ \psi^{\text{(rad)}}_\textbf{k}(\textbf{r})=\psi^{\text{(out)}}_\textbf{k}(\textbf{r})-\psi^{\text{(in)}}_\textbf{k}(\textbf{r})=\frac{1}{4\pi}\int\mathrm{d}\textbf{r}'\,\mathcal{G}(\textbf{r}-\textbf{r}')\,U(\textbf{r}')\,\psi_\textbf{k}(\textbf{r}') $$ with $$ \mathcal{G}(\textbf{r})=G^+(\textbf{r})-G^-(\textbf{r}) $$

"I've seem this method in some notes and as things are written it seems the only imposed condition is that $G(\textbf{r},\textbf{r}')=G(\textbf{r}-\textbf{r}')$."

It seems to me that this is not directly related to your initial problem. One can show independently that the property
$$
G(\textbf{r},\textbf{r}')=G(\textbf{r}-\textbf{r}')
$$
is just a consequence of the fact that the dispersion relation $\langle\textbf{k}| H|\textbf{k}\rangle =E(\textbf{k})$ only depends on the *magnitude* of the impulsion, *i.e.* $E(\textbf{k})\equiv E(k)$, which is a consequence of the fact that your system is invariant under translation.

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

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