In Equation (8) of this paper by Groote et. al., we are given the following Euclidean identity: $$ \int \frac{d^{4}\mathbf{p}_{\mathrm{E}}}{(2\pi)^{4}} \frac{e^{ i \mathbf{p}_{\mathrm{E}} \cdot \mathbf{x}_{\mathrm{E}} }}{ \left( |\mathbf{p}_{\mathrm{E}}|^2 + m^2 \right)^{\mu + 1}} \ = \ \frac{1}{4 \pi^2} \frac{1}{2^{\mu} \Gamma(\mu + 1)} \left( \frac{m}{|\mathbf{x}_{\mathrm{E}}|} \right)^{1 - \mu} K_{1 - \mu}\big( m |\mathbf{x}_{\mathrm{E}}| \big). \tag{8} $$

The above identity holds in Euclidean space where $\mathbf{p}_{\mathrm{E}} = (\mathbf{p}, p_4)$ and so on. I should say that $K_{\nu}(x)$ is the modified Bessel function of the second kind (McDonald function) of order $\nu$.

I am pretty sure that I can just replace the above vectors with their Minkowski counterparts meaning put $\mathbf{p}_{\mathrm{E}} \mapsto p = (p_0, \mathbf{p})$ where $p_0 = i p_4$, giving the Minkowski identity: $$ \int \frac{d^{4}p}{(2\pi)^{4}} \frac{e^{ i p \cdot x }}{ \left( p^2 + m^2 \right)^{\mu + 1}} \ = \ \frac{1}{4 \pi^2} \frac{1}{2^{\mu} \Gamma(\mu + 1)} \left( \frac{m}{\sqrt{x^2} } \right)^{1 - \mu} K_{1 - \mu}\big( m \sqrt{x^2} \big). $$

But I am not sure exactly how to prove this. How can one do this?

My thoughts were to start with the Euclidean identity and note the rotational invariance and pick the vector $\mathbf{x}_{\mathrm{E}} = ( |\mathbf{x}_{\mathrm{E}}|, 0, 0, 0 )$. Maybe then this allows for a Wick-rotation in the $p_4$ variable, but I am unsure exactly how to proceed since this Euclidean identity has no '$i\epsilon$' regulator, and so the $p_4$ poles lie right on the axis where we're integrating.


1 Answer 1


Let $$I_\mu = \int\,\frac{d^4p}{(2\pi)^4}\frac{e^{ipx}}{(p^2+m^2)^{\mu+1}}$$ Differentiating, we obtain the following identity:

$$I_{\mu+1} =-\frac{1}{2m(\mu+1)}\frac{\partial I_\mu}{\partial m}$$

$I_0$ is the Green's function for the Klein-Gordon equation in a spacetime with signature $(-+++)$. This has two poles; performing the integration according to the Feynman prescription and assuming $x$ is timelike, $$I_0 = \lim_{\epsilon\to 0} \int\,\frac{d^4p}{(2\pi)^4}\frac{e^{ipx}}{p^2+m^2+i\epsilon}=-\frac{m}{8\pi\sqrt{-x^2}}H_1^{(2)}(m\sqrt{-x^2})$$ where $H_1^{(2)}$ is a Hankel function. In the case of spacelike $x$, you would get a modified Bessel function of the second kind.

Both the Bessel functions of the first and second kind satisfy the following identity. Since $H^{(2)}$ is a linear combination of these two, it satisfies the identity as well: $$\frac{d}{dz}[z^\nu H^{(2)}_\nu(z)]=z^\nu H^{(2)}_{\nu-1}(z)$$ Taking a look at the Euclidean identity, one can guess the following expression: $$I_\mu = \frac{1}{8\pi} \frac{1}{2^\mu \mu!} \left(\frac{-m}{\sqrt{-x^2}}\right)^{1-\mu} H^{(2)}_{1-\mu}(m\sqrt{-x^2})$$ which can be proved via induction.

In contrast, for spacelike $x$ one obtains $$I_0 = \lim_{\epsilon\to 0} \int\,\frac{d^4p}{(2\pi)^4}\frac{e^{ipx}}{p^2+m^2+i\epsilon}=\frac{im}{4\pi^2\sqrt{x^2}}K_1(m\sqrt{x^2})$$ which leads to the solution $$I_\mu = \frac{1}{4\pi^2} \frac{1}{2^\mu \mu!} \left(\frac{-m}{\sqrt{-x^2}}\right)^{1-\mu} K_{1-\mu}(m\sqrt{-x^2})$$ again proved by induction.

  • $\begingroup$ Thanks for your thoughts. There is a subtle thing, exemplified for $\mu = 0$ in that $\lim\limits_{\epsilon \to 0^{+}}\frac{1}{p^2+m^2+i \epsilon} = \mathscr{P}\frac{1}{p^2 +m^2} - i \pi \delta(p^2+m^2)$. I believe that $-\frac{m}{8\pi \sqrt{-x^2}}H_{1}^{(2)}(m\sqrt{-x^2})$ is the part of the integral coming from $\mathscr{P}$, and we should have a light-cone delta $+\frac{1}{4\pi }\delta( -x^2 )$ added to your formula. It is the role of the delta that I am confused about - I guess that for general $\mu$, we have a factor $+\frac{(-1)^\mu}{4\pi \Gamma(\mu-1)} \delta^{(\mu)}(-x^2)$ added... $\endgroup$ Commented Apr 23, 2018 at 16:56
  • $\begingroup$ @Greg.Paul The delta is 0 unless $x^2 =0$. I assumed $x$ is timelike, so $x^2<0$ $\endgroup$
    – John Donne
    Commented Apr 23, 2018 at 17:30

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