I am reading Davies' and Birrell's book on QFT in Curved Spacetimes. In Chapter 2 and specifically in the subchapter 2.7, the authors derive an expression for the Feynman propagator $$G_F(x,x')=\frac{-i\pi}{(4\pi i)^{n/2}}\Big(\frac{2m^2}{-\sigma+i\epsilon}\Big)^{(n-2)/4} H^{(2)}_{\frac{1}{2}n-1}\Big\{[2m^2(\sigma-i\epsilon)]^{1/2}\Big\}\tag{2.77}$$ by performing the contour integration and by choosing one of the poles to lie on the upper half of the $\text{Im}(k^0)$ - $\text{Re}(k^0)$ plane and the other one to lie on the lower half of the abovementioned plane (just like any similar derivation in QFT textbooks). This is the first time I see the Feynman propagator in this form and I do not know how to derive it, so any tip would be appreciated...

My attempt revolves around performing the $k^0$ integral first, from the expression $$G_F(x,x')=\int\frac{d^nk}{(2\pi)^n}\frac{e^{-ik\cdot x}}{k^2-m^2}$$ (where $k^2=k\cdot k$ with the $\cdot$ denotes a four-vector multiplication) and then trying to bring the resulting expression to the form of the integral representation of the Hankel function, namely $$H^{(2)}_{\nu}(x)=\frac{1}{\pi i}\int_0^{\infty e^{i\pi}}dt \frac{e^{x/2(t-1/t)}}{t^{\nu+1}}$$

So, I wanted to ask whether or not

  1. I am headed to the right direction and if yes, what would my $t$ variable be? The three momentum vector magnitude perhaps? (I know this sounds like a "do my derivation for me" question, but I wouldn't want that. Some tips would be nice... Specifically, there is a gap between the integral representation expression and the result I get from performing the $k^0$ contour integral and it would be nice if somehow someone would bridged that gap)

  2. is there another approach for deriving the propagator without the integral representation of the Hankel function... For instance, I thought of writting the Kleing-Gordon equation for the propagator in spherical coordinates and then solve it. But then I would need boundary conditions and hence another dead end...

Any help/comments on my attempts would be appreciated.

  • 1
    $\begingroup$ Nice question. Nice answer. I found them both helpful for something I was doing. My comment here is somewhat sideways and completely irrelevant from a physics perspective .... but after I read the phrase "it would be nice if somehow someone would breach that gap" I spent 2 minutes wondering why (a) it sounded odd, and (b) whether it should have been a "dam" or a "wall" that should have been breached instead of a "gap". Eventually I realised that I was solving the wrong problem: the gap should have been "bridged" not "breached". Now I can sleep ... $\endgroup$ Commented Feb 18 at 15:05

1 Answer 1


There is a useful class of dummy integration variable trick, which applies also later when calculating Feynman diagrams. You can make the following observation: $$ \frac{1}{x} = \int_0^{+\infty}dt e^{-xt} $$ Thanks to the definition of the Feynman propagator, which makes you avoid the poles in a particular way, using contour integration, you can modify the contour to integrate $k^0$ over the imaginary axis without crossing any poles. You get: $$ G_F = \int\frac{d^nk}{(2\pi)^n}\frac{e^{-ikx}}{k^2-m^2+i\epsilon} \\ = -i\int\frac{d^nk_E}{(2\pi)^n}\frac{e^{ik_Ex_E}}{k_E^2+m^2} $$ with $k^0 = ik_E^0$ and $x^0 = ix_E^0$ and the spatial components unchanged, so that the new dot product in the exponential is the usual euclidean dot product. Now you can apply the trick: $$ G_E=-i\int_0^{+\infty}dt\int\frac{d^nk_E}{(2\pi)^n}\exp(ik_Ex_E-t(k_E^2+m^2)) \\ =-i\int_0^{+\infty}dt\frac{1}{(2\pi)^n}\sqrt{\frac{\pi}{t}}^ne^{-m^2t-x_E^2/4t} $$ where I completed the square and performed the gaussian integral (and swapped the integrals). By rescaling, you'll obtain the integral representation of a Bessel function. If you're keeping the same $x$, ie $x_E$ is imaginary, you'll get Hankel's function, but you'll need to be careful in the definition of the above integrals and modify the contour. The easiest way it to Wick rotate, and assume $x_E$ is real. You will get rigorously the modified Bessel functions of the second kind. Extending $G_E$ thus calculated on the imaginary axis of $x^0$ will give you Hankel's function.

If you try to directly solve the KG equation, you need to implement the pole avoidance of the Feynman propagator in the boundary conditions, which is not easy to do. One way would be to admit Wick rotation an the relation: $$ G(x) = -iG_E(x_E) $$ You can then solve for $G_E$ for real $x_E$ by solving the screened Poisson equation: $$ (-\Delta_E+m^2)G_E = \delta $$ By rotation invariance, you look for only radial solutions: $G_E(x_E) = g(r)$. You just need the formula for the Laplacian: $$ \Delta_EG_E = g''+\frac{n-1}rg' $$ and you recover the equation: $$ g''+\frac{n-1}rg'-m^2g = 0 $$ with the boundary condition: $$ \frac{2\pi^{n/2}}{\Gamma(n/2)}r^{n-1}f'(r)\xrightarrow{r\to0}1 \\ \frac{2\pi^{n/2}}{\Gamma(n/2)}r^{n-1}f'(r)\xrightarrow{r\to\infty}0\\ $$ To get the modified Bessel's equation, multiply by the correct power to absorb the extra factor $n-1$ and make the argument dimensionless with $m$: $$ f(x) = r^{(1-n)/2}g(mx) \\ x^2f''+xf'+\left(\left(\frac n2-1\right)^2-x^2\right)f = 0 $$ So $f$ is a linear combination of $I_{n/2-1},K_{n/2-1}$. From the boundary condition at infinity, there is no contribution from $I_{n/2-1}$, and the boundary condition at $0$ gives you the prefactor of $K_{n/2-1}$. The formula is given for example in tables of Green's functions. To get $G$, you use the same trick of analytic continuation of $G_E$ for imaginary times, which will give you Hankel's function.

Hope this helps.

  • $\begingroup$ thank you so much! I would most certainly not think of something like that! I will try it! $\endgroup$
    – schris38
    Commented Jul 15, 2022 at 19:41

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