12
$\begingroup$

Preliminaries: In their QFT text, Peskin and Schroeder give the KG propagator (eq. 2.50)

$$ D(x-y)\equiv\langle0|\phi(x)\phi(y)|0\rangle = \int\frac{d^3p}{(2\pi)^3}\frac{1}{2\omega_\vec{p}}e^{-ip\cdot(x-y)}, $$

where $\omega_\vec{p}\equiv\sqrt{|\vec{p}|^2+m^2}$. For lightlike separations, we can choose a frame where $x-y$ is purely in the time-direction and the propagator can be put into the form (2.51)

$$ D(x-y)=\frac{1}{4\pi^2}\int^\infty_m d\omega\sqrt{\omega^2-m^2}e^{-i\omega (y^0-x^0)} \tag{1}\label{timelike_prop}, $$

where I use the $\text{diag }\eta=(-,+,+,+)$ convention.

Now, one has the following integral representation of the modified Bessel function (http://dlmf.nist.gov/10.32.8)

\begin{align} K_1(z) &= z\int^\infty_1 dt \sqrt{t^2-1} e^{-zt} \\ &= \frac{z}{m^2} \int^\infty_m dt \sqrt{t^2-m^2} e^{-zt/m}, \tag{2}\label{int_rep} \end{align}

where we go to the second line by rescaling the integration variable $t \to t/m$. Comparing \eqref{timelike_prop} with \eqref{int_rep} suggests

$$ D(x-y)=\frac{m}{(2\pi)^2|y-x|}K_1(m|y-x|), $$

where we have written the time separation in terms of the Lorentz invariant $i (y^0-x^0)=|y-x|$. (Note: there is an issue in what I've written here in that the integral representation \eqref{int_rep} is only valid for $|arg z|<\pi/2$ and $|y-x|$ is on the imaginary axis ($|arg z|=\pi$), but I think one could infinitesimally displace $z$ off of the imaginary axis to get a convergent integral. Check me on that.)

Anyway, for spacelike separations, we can choose a frame where $y-x=\vec{y}-\vec{x}\equiv\vec{r}$. Performing the polar integrations yield

$$ D(x-y)=\frac{-i}{2(2\pi)^2 r}\int^\infty_{-\infty}dp\frac{p e^{ipr}}{\sqrt{p^2+m^2}}. $$

Finally, PS claim that taking a contour integral in the upper half plane (making sure to avoid the branch cut at +im) will give

$$ D(x-y)= \frac{1}{(2\pi)^2r}\int^\infty_m d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}, \tag{3}\label{spacelike_prop} $$ where $\rho\equiv-ip$.

Question: I know from plugging into Mathematica that the spacelike propagator \eqref{spacelike_prop} can also be expressed as a modified Bessel function $K_1$. Moreover, the integration bounds of \eqref{spacelike_prop} and \eqref{int_rep} are even the same. However, I don't see how to transform the spacelike propagator integral \eqref{spacelike_prop} into the form of \eqref{int_rep}. Any ideas?

(I'd prefer, if at all possible, to use the integral representation that I've quoted \eqref{int_rep} and used for the timelike case rather than some other representation of the modified Bessel function.)

$\endgroup$
2
  • $\begingroup$ Here are some references for similar calculations, although not exactly this one: "Quantum Electrodynamics" (2009), Walter Greiner, Joachim Reinhardt, Exercise 2.5 p.68-76 and "Introduction to the Theory of Quantized Fields" (1976), N.N. Bogoliubov, D.V. Shirkov, §16 p.147 and "Perturbative Quantum Electrodynamic and Axiomatic Field Theory" (TMP, 2000), O. Steinmann, § 5.4 p.61 ff, in particular (5.111) p.63 $\endgroup$
    – Noix07
    Commented Feb 16, 2022 at 15:46
  • 1
    $\begingroup$ I saw that you used the - + + + convention but you used the other one instead in the power of e $\endgroup$
    – Zin Zhao
    Commented Jan 20 at 1:27

1 Answer 1

13
$\begingroup$

This can be seen by partial integration

$$\frac{\partial}{\partial \rho}\sqrt{\rho^2-m^2}=\frac{\rho}{\sqrt{\rho^2-m^2}}$$

OP edit: More explicitly, we use this to write $(3)$ as

\begin{align} D(x-y) &= \frac{1}{(2\pi)^2r}\int^\infty_m d\rho \frac{\partial}{\partial \rho}\sqrt{\rho^2-m^2} e^{-\rho r} \\ &= \frac{1}{(2\pi)^2r}\left[\sqrt{\rho^2-m^2} e^{-\rho r}\right]^\infty_m-\frac{1}{(2\pi)^2r}\int^\infty_m d\rho \sqrt{\rho^2-m^2} \frac{\partial}{\partial \rho}e^{-\rho r}\\ &= \frac{1}{(2\pi)^2}\int^\infty_m d\rho \sqrt{\rho^2-m^2} e^{-\rho r}\\ &= \frac{m}{(2\pi)^2r}K_1(mr) \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.