Klein-Gordon propagator integral in the light-like case

Question

In Kerson Huang's Quantum Field Theory From Operators to Path Integrals (Amazon link), pages 28 and 29, he calculates the propagator in the following case: time-like, space-like and light-like. First he integrates the time-component of $k$, and arrive this expression: $$ \Delta_F(x)=\frac{i}{4\pi^2}\int_0^\infty dk\,\frac{k^2}{\omega_k}\frac{\sin kr}{kr}e^{i\omega_k|t|} $$

Then he gets the Bessel function in the time-like and the space-like case:

By Lorentz invariance $\Delta_F(x)$ can only depend on $$s\equiv x^2=t^2-\mathbf r^2\tag{2.83}$$ For $s>0$, we can put $\mathbf r=0$ to obtain the representation $$ \Delta_F(x)=\frac{i}{4\pi^2}\int_0^\infty dk\frac{k^2}{\omega_k}e^{i\omega_k\sqrt{s}}=\frac{m}{8\pi\sqrt{s}}H_1^{(1)}(m\sqrt{s})\tag{2.84} $$ For $s<=0$, we put $t=0$ to obtain $$ \Delta_F(x)=\frac{i}{4\pi^2}\int_0^\infty dk\frac{k^2}{\omega_k}\frac{\sin k\sqrt{-s}}{k\sqrt{-s}}=-\frac{im}{4\pi^2\sqrt{s}}K_1(m\sqrt{-s})\tag{2.85} $$

At last, he gets the result in light-like case: a delta function:

where $H_1^{(1)}$ and $K_1$ are Bessel functions. In the time-like gregion $s>0$ the function describes an outgoing wave for large $s$. This corresponds to the $i\eta$ prescription in (2.80). The $-i\eta$ prescription would have yielded an incoming wave. In the space-like region $s<0$ it damps exponentially for large $|s|$. On the light cone $s=0$ there is a delta function singularity not covered by the above formulas: $$\lim_{x^2\to0}\Delta_F(x)=-\frac{1}{4\pi}\delta(x^2)$$

Can anyone explain how he obtained the delta function? I don't understand the limit $s=0$ because the Bessel function is divergent there.

Answer 1

I'll only consider the case $s \leq 0$. Consider your original integral; $$ \Delta_{F}(s) \ = \ \frac{i}{4 \pi^2 r} \int_0^\infty dk \frac{k\ e^{i |t|\sqrt{k^2 + m^2}}}{\sqrt{k^2 + m^2 }} \sin(rk) $$

If we're careful, we notice that this integral is quite naughty; it doesn't converge for $\mathrm{Im}(|t|)=0$. To see why this is the case, change the integration variable to $E = \sqrt{k^2 + m^2}$: $$ \Delta_{F}(s) \ = \ \frac{i}{4 \pi^2 \sqrt{-s}} \int_m^\infty dE \sin\left(r\sqrt{E^2 - m^2}\right) e^{ i |t| E } $$

We just have an integrand that wildly oscillates as we take $E$ further and further to the right.

Now that we agree that the integral diverges; consider the following integral which I've stolen out of Gradshteyn and Rhyzik: $$ \int_0^\infty dz\ \frac{z e^{ - a \sqrt{ z^2 + c^2 } } }{ \sqrt{ z^2 + c^2 } } \sin( b z ) \ = \ \frac{ b c }{ \sqrt{ a^2 + b^2 } } K_{1}\left( c \sqrt{ a^2 + b^2 } \right) $$

This is integral 3.914.9 in G+R for those interested. It converges only for $\mathrm{Re}(a) > 0$ and $\mathrm{Re}(c) > 0$. Meaning if we set $a = i|t|$, $b = r$ and $c = m$, we can't use the above formula.

The way we remedy the situation, is we pick a tiny $0 < \epsilon \ll 1$ and fix $a = - \epsilon + i|t|$ (as well as $b= r$ and $c=m$). We then have an $\epsilon$-regulated $\Delta_{F}^{\epsilon}(x)$, where we understand that $\lim\limits_{\epsilon \to 0^+} \Delta_{F}^{\epsilon}(x) = \Delta_{F}(x)$. We get: $$ \Delta_{F}^{\epsilon}(x) \ = \ \frac{i}{4 \pi^2 \sqrt{-s}} \int_m^\infty dE \sin\left(r\sqrt{E^2 - m^2}\right) e^{ - ( \epsilon - i |t| ) E } $$

Using the above G+R integral we arrive at: $$ \Delta_{F}^{\epsilon}(x) \ = \ \frac{i}{4\pi^2} \frac{ m }{ \sqrt{ ( i t + \epsilon )^2 + r^2 } } K_{1}\left( m \sqrt{ (it + \epsilon)^2 + r^2 } \right) \ = \ \frac{i}{4\pi^2} \frac{ m }{ \sqrt{ - t^2 + r^2 + i \epsilon } } K_{1}\left( m \sqrt{ - t^2 + r^2 + i \epsilon } \right) $$

In the second equality, we've completed the squares, and noted that $\epsilon$ times anything will look like $\epsilon$. We see that we have the same function you gave, except for the $i\epsilon$'s.

Note that $K_1(z) \approx \frac{1}{z}$ near $z=0$. This means that near the light-cone $t^2 - r^2 \to 0$, we can write the above as; $$ \Delta_{F}^{\epsilon}(x) \ \approx \ \frac{i}{4 \pi^2} \frac{1}{-t^2 + r^2 + i \epsilon} $$

Finally, we have the principal-value prescription (see page 113 of Weinberg's 'Quantum Theory of Fields: Volume 1'): $$ \lim_{\epsilon \to 0^+} \frac{1}{z \pm i \epsilon} = \mathcal{P}\frac{1}{z} \mp i \pi \delta(z) $$

where $\mathcal{P}$ means principal value. Hence, near the light-cone; $$ \Delta_{F}(x) \ \approx \ \lim_{\epsilon \to 0^+} \frac{i}{4 \pi^2} \frac{1}{-t^2 + r^2 + i \epsilon} \ = \ \frac{i}{4\pi^2} \frac{1}{-t^2 + r^2} + \frac{1}{4\pi} \delta(-t^2 + s^2) $$

So we have extracted a delta function in the limit $\epsilon \to 0^+$. This is valid not only at the light-cone, but everywhere else, so we can now write: $$ \Delta_{F}(x) \ = \ \frac{i}{4\pi^2} \frac{ m }{ \sqrt{ - t^2 + r^2 } } K_{1}\left( m \sqrt{ - t^2 + r^2 } \right) + \frac{1}{4\pi} \delta(-t^2 + r^2) $$