# 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?

• Explicit form how? Would you be happy to know how it works on a convenient basis or you need like an integral expression? – MannyC Apr 21 '19 at 1:11
• @MannyC anything to evaluate it! (So really the former) – Craig Apr 21 '19 at 2:02
• A way is writing down the inverse operator as a series $\sum_{k=0}^\infty (-1)^k (x^\mu \partial_\mu)^k$. It works on real analytic functions a least... – Valter Moretti Apr 21 '19 at 10:30

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, $$x^\mu\partial_\mu r^n= x^\mu (2 x_\mu) \frac{n}{2} (r^2)^{n/2-1} =n ~r^n.$$

• Hi Cosmas, how is this fact derived/obvious to you? Am I missing something basic? – Craig Apr 21 '19 at 2:03
• OK, expanding the point. – Cosmas Zachos Apr 21 '19 at 2:46
• Ahh okay I was over thinking the act of inverting the operator. Thanks a ton! – Craig Apr 21 '19 at 2:55

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.

• I do not think it works as it stands: you are using the Euclidean metric, whereas the post refers to the Minkowskian one so that what you indicate as the norm of $x$ has actually a sign. – Valter Moretti Apr 21 '19 at 10:09
• This does not matter for the second part of my answer obviously because $f$ is still a one dimensional function. It becomes a bit messier to define the spherical harmonics but it can be done. They will have a continuous label and one of the angles (being the rapidity) will be unbounded. – MannyC Apr 21 '19 at 11:34
• I'm a little wary of the signs, can you elaborate on why theres a minus sign in the argument of the capital F, second last equation. Why does F(-O) correspond to O in the integral? – Craig Apr 23 '19 at 7:14
• It's just because the Mellin transform is defined with $x^{-s}$. So $-\partial_\mu x^\mu$ becomes multiplication by $s$. – MannyC Apr 23 '19 at 14:27