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 \(B\).