I'm trying to obtain the Feynman Green Function (i.e. I'm using the Feynman Causal prescription to compute the green function) for the D'Alembertian in 1+1D, I'm finding

$$G^{(2)}_F (t; \vec x) = \frac{1}{4\pi} \frac{\Theta(t^2-\vec x^2)}{\sqrt{t^2-\vec x^2}} - \frac{i}{4\pi^2} \int_{-\infty}^\infty dz' \text{P.V.}\left(\frac{1}{t^2-\vec x^2 - z'^2}\right).$$

This result is right ou wrong? I don't have any reference with the answer. Supposing the result is right, there is some way to simplify it more?

The way I find the result.

I've used the usual procedure (as in Eleftherios Economou and in Morse Feshbach) to find the Green function in 1+1D as a potential generated by an infinite line of charge in 2+1D.

$$G^{(2)}_F (t; \vec x) = \int dt' d^3 r' G^{(3)}_C (t-t'; \vec r-\vec r') J(t',\vec r').$$

The Feynman green function in 2+1D is,

$$G^{(3)}_F (t-t'; \vec r-\vec r') = \frac{1}{4\pi} \left[\delta((t-t')^2-||\vec r - \vec r'||) - \frac{i}{\pi} \text{P.V.}\left(\frac{1}{(t-t')^2-||\vec r-\vec r'||}\right)\right]$$

as can be checked on Bogoliubov-Shirkov (Appendix II, pag 605, A2b.6)

And the source is,

$$J(t',\vec r') = \delta(x') \delta(y') \delta(t')$$

So that,

$$G^{(2)}_F (t; \vec x) = \int_{-\infty}^\infty dz'\frac{1}{4\pi} \left[\delta(t^2-\vec x^2 - z'^2) - \frac{i}{\pi} \text{P.V.}\left(\frac{1}{t^2-\vec x^2 - z'^2}\right)\right]$$

Using a basic Dirac Delta property,

$$\delta(x^2-a^2) = \frac{1}{2|a|} \left(\delta(x+a) + \delta(x-a) \right)$$ we get for the first integral,

$$\frac{1}{8\pi} \int_{-\infty}^\infty dz'\frac{1}{||\sqrt{t^2-\vec x^2}||} \left(\delta(z-\sqrt{t^2-\vec x^2}) + \delta(z+\sqrt{t^2-\vec x^2})\right)$$

For $T^2>\vec x^2$ (time-like interval) the points $\pm \sqrt{t^2-\vec x^2}$ are real and belong to interval $(-\infty,\infty)$. So we have (for the first integral),

$$\frac{1}{4\pi} \frac{\Theta(t^2-\vec x^2)}{\sqrt{t^2-\vec x^2}}$$

And, finally,

$$G^{(2)}_F (t; \vec x) = \frac{1}{4\pi} \frac{\Theta(t^2-\vec x^2)}{\sqrt{t^2-\vec x^2}} + \int_{-\infty}^\infty dz' \text{P.V.}\left(\frac{1}{t^2-\vec x^2 - z'^2}\right)$$

Just putting here what happens with my solution if we are put of the light cone singularity ($t=\pm \vec x$). I think that we can forget about the principal value at this case.

If I take the integral and solve it, I get $$\int_{-\infty}^\infty dz' \left(\frac{1}{t^2-\vec x^2 - z'^2}\right) = \frac{i\pi}{\sqrt{t^2-\vec x^2}}.$$

So I get,

$$G^{(2)}_F (t; \vec x) = \frac{1}{4\pi} \frac{1}{\sqrt{t^2-\vec x^2}} \left(\Theta(t^2-\vec x^2) + 1\right)$$

  • $\begingroup$ By the way, the PV is meant to be a symbol accompanying the integral, it is meaningless next to a function except as notation implying the integral. $\endgroup$
    – auxsvr
    Apr 21, 2014 at 21:08
  • $\begingroup$ Apparently you used $\lim_{\epsilon \to 0}\frac{1}{x\pm i \epsilon} = \text{PV} \frac{1}{x} \mp i\pi \delta(x)$, right? $\endgroup$
    – auxsvr
    Apr 21, 2014 at 21:59
  • $\begingroup$ When we attempt to find the Green's function, we observe that the integral diverges (the PV part), therefore we must take the retarded or advanced solutions instead, which are on the left of the Sokhotski-Plemelj formula. If you start from the definition of the Green's function, you'll see that you did things backwards. Also, constants in the definition of Green's function are irrelevant. $\endgroup$
    – auxsvr
    Apr 21, 2014 at 22:46

1 Answer 1


First of all, the retarded (causal) Green's function for the dalembertian is ($\vec{R} \equiv \vec{x} - \vec{x}'$, $T \equiv t - t'$, $R\equiv |\vec{R}|$), $$G^{(3)}_R(R,T) = \frac{\Theta(T)\delta(T-R/c)}{4\pi R}.$$ This leads to the retarded potential in electromagnetism. It depends only on the difference of position vectors because of the symmetry of the boundary condition in unbounded space, and on the time difference, because the defining equations and boundary condition are invariant if we substitute $t \to t-t'$.

Using the method of embedding or descent with appropriate boundary conditions, Green's function in dimension 2 is given by ($r \equiv \sqrt{ (x-x')^2 + (y-y')^2}$), $$G^{(2)}_R(r,T) = \frac{\Theta(T)}{4\pi} \int_{-\infty}^\infty \frac{\delta ( T - R/c)}{R} d z = \frac{\Theta(T)}{4\pi} \int_{-\infty}^\infty \frac{\delta\left( T - \sqrt{ r^2 + (z-z')^2}/c \right)}{\sqrt{ r^2 + (z-z')^2}} d (z-z'),$$ which equals $$G^{(2)}_R(r,T) = \frac{\Theta(T - r/c)}{2\pi \sqrt{T^2- r^2/c^2}},$$ for $T>0$, otherwise $0$.

The principal value solution can be found as the linear combination of the retarded and the advanced functions, according to the expansion $$-\text{PV} \int_{-\infty}^\infty \frac{\phi_p(x) \phi^*_p(x')}{\lambda_p} d p = G^{\pm} \mp \Lambda$$ in terms of eigenfunctions, where $\Lambda$ is the integral comprising the eigenfunctions with zero eigenvalue. You could also see this with the Sokhotski-Plemelj theorem.

  • $\begingroup$ Ow... sorry, looks like I was making a mistake. I had the idea that the causal prescription was the Feynman prescription not the retarded one! I'll look to the Sokhotski-Plemelj theorem. $\endgroup$
    – Erich
    Apr 21, 2014 at 19:32
  • $\begingroup$ Other thing, what I've find to the retarded prescription has a $\Theta(t)$ (which agrees with Hassani - Math Phys - Problem 22.26) instead of the $\Theta(t-r/c)$ you've found. $\endgroup$
    – Erich
    Apr 21, 2014 at 19:48
  • $\begingroup$ It should be $\Theta(T-r/c)$, otherwise the square root becomes imaginary, which is not allowed by the definition of the roots of the argument of the delta function. $\endgroup$
    – auxsvr
    Apr 21, 2014 at 20:35
  • $\begingroup$ I didn't mean that. What I'm trying to say is that the $G^{(3)}$ I wrote was for the Feynman prescription (which i was calling 'causal') not the retarded one. I've already corrected my question to remove the mention to 'causal' - motivated by the Bogoliubov book. The Feynman prescription means that the integral to be solved to find the green function is $\int \frac{d^4 p}{(2\pi)^4} \frac{e^{-i p_\mu x^\mu}}{-p_\mu p^\mu - i \varepsilon}$ $\endgroup$
    – Erich
    Apr 21, 2014 at 21:07
  • $\begingroup$ Well, $\int \frac{d^4 p}{(2\pi)^4} \frac{e^{ip_\mu x^\mu}}{-p_\mu p^\mu -i \epsilon}$ is the retarded function and, since the (non-converging) Green's function must be real (the dalembertian is hermitian), your integral does yield the retarded Green. $\endgroup$
    – auxsvr
    Apr 21, 2014 at 21:24

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