Finding the Green function of an operator in QFT I'm working on some quantum field theory and have to operate on a field with the following operator:
$$
(x^\mu \partial_\mu + 1)^{-1}
$$
I've been trying to find an explicit form of this operator, but it has proved a challenge. Calling it $\phi$, I know it has to obey:
$$
\int d^4x\> (x^\mu \partial_\mu + 1)\phi = 1
$$
but all my attempts have gotten me nowhere. Can anyone point me in the right direction?
 A: I would not presume pointing you in the right direction, but you might consider the evident Lorentz-invariant eigenfunctions of your operator, namely  the powers $r^n$,
for $r\equiv \sqrt{x^\mu x_\mu}$, 
$$
\frac{1}{1+x^\mu\partial_\mu}  ~ r^n = \frac{1}{1+n}     r^n ~.
$$
It is also easy to quantify the directional eigenfunctions $(a\cdot x)^n$ as well, since the kernel $x^\mu\partial_\mu$ is just a power counter,
$$
x^\mu\partial_\mu  r^n= x^\mu (2 x_\mu)  \frac{n}{2} (r^2)^{n/2-1} =n ~r^n.
$$
A: Building on Cosmas' answer. You can extend his definition to a basis of functions on $\mathbb{R}^d$. This is the reason behind my comment before. Any function can be written as a radial function times a spherical harmonic
$$
\varphi(x) = f(|x|) \,Y_{l_1,\ldots l_{d-1}}(\theta_1,\ldots,\theta_{d-1})\,.
$$
The spherical harmonics can be equivalently represented as polynomials of degree $l_{d-1}$ (where $l_{d-1}\geq l_{d-2}\geq\cdots \geq |l_1|$)
$$
Y_{l_1,\ldots l_{d-1}}(\theta_1,\ldots,\theta_{d-1}) = \frac{1}{|x|^{l_{d-1}}}C(l_1,\ldots,l_{d-1})^{\mu_1\ldots \mu_{l_{d-1}}} x^{\mu_1}\cdots x^{\mu_{l_{d-1}}}\,,
$$
where the coefficients are symmetric traceless tensors. Clearly in this representation
$$
x^\mu \partial_\mu Y_{l_1,\ldots l_{d-1}}(\theta_1,\ldots,\theta_{d-1}) = 0
$$
because the function is homogeneous of degree zero. So now your operator acts only on $f$ and we have reduced the problem to one dimension.
For any sufficiently regular function ($L^2(\mathbb{R})$ should be enough) one can define the Mellin (inverse) transform as
$$
f(x) = (\mathcal{M}^{-1}g)(x) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \mathrm{d}s\,x^{-s}\,g(x)\,.
$$
Check the link for the conditions on $c$ and the direct Mellin transform used to find $g(x)$. Now for any sufficiently well behaved function $F$ one can define
$$
F\big(-x_\mu\partial^\mu\big)f(|x|) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i \infty}\mathrm{d}s \,F(s)\,|x|^{-s}g(|x|)\,.
$$
Your question regards the special case
$$
F(s) = \frac{1}{1-s}\,.
$$
To summarize, you can decompose any field into a radial and an angular part, do the inverse Mellin transform on the radial part, apply the operator in Mellin space and transform back.
