# M.E. Irizarry-Gelpí

Physics impostor. Mathematics interloper. Husband. Father.

# A Result for the Laplacian Green Function

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 $$B$$.