A Result for the Laplacian Green Function

Sun 27 January 2019
Physics
#green-function, #laplacian

In $d$ -dimensional euclidean space, the laplacian operator is

$\begin{equation*} \partial^{2} \equiv \delta^{ab} \partial_{a} \partial_{b} \end{equation*}$

The laplacian operator is the trace of the $\partial_{a} \partial_{b}$ operator. The Green function for the laplacian operator is such that

$\begin{equation*} \partial^{2} G = - A \delta(x) \end{equation*}$

Here $A$ is a constant and $\delta(x)$ is a $d$ -dimensional Dirac delta function.

In this post I will evaluate the following matrix:

$\begin{equation*} D_{ab} \equiv \partial_{a} \partial_{b} G \end{equation*}$

Note that $D_{ab}$ is a symmetric matrix. The trace of $D_{ab}$ must give the Dirac delta:

$\begin{equation*} \delta^{ab} D_{ab} = -A \delta(x) \end{equation*}$

So maybe $D_{ab}$ can be split into a traceless and non-traceless parts:

$\begin{equation*} D_{ab} = -\frac{A}{d}\delta_{ab}\delta(x) + T_{ab} \end{equation*}$

Here $T_{ab}$ is symmetric and traceless. Maybe it can be written as

$\begin{equation*} T_{ab} = F(x) \left(\delta_{ab} - d\frac{x_{a}x_{b}}{x^{2}}\right) \end{equation*}$

Now, dimensional analysis suggest

$\begin{equation*} F(x) = B\left(\frac{1}{x^{2}}\right)^{d/2} \end{equation*}$

Here, $B$ is a constant. Thus, apparently, you would have

$\begin{equation*} D_{ab} = B\left(\frac{1}{x^{2}}\right)^{d/2} \left(\delta_{ab} - d\frac{x_{a}x_{b}}{x^{2}}\right) - \frac{A}{d}\delta_{ab}\delta(x) \end{equation*}$

The last step is to find the values of $A$ and $B$ .