This question is same as the question here which is not answered yet.

By Poincare invariance, the propagator of any field theory $G(x,y)$ must be a function of $|x-y|$ and by dimensional arguments $$G(x,y)=\langle 0|\phi(x)\phi(y)|0\rangle\sim\frac{1}{|x-y|^2}\tag{1}$$ which diverges as $x\to y$. Is this divergence a pathology of the theory? Is it an indication of taking some unphysical limit? If yes, why is the limit $x\to y$ unphysical? Does it necessarily mean continuum picture of spacetime in quantum field theory has to be given up?


2 Answers 2


That is not the real-space propagator. For a massless field in $d=4$ Minkowski space-time that obeys a standard wave equation in real space the Green's functions are given by: $$G(x,y) = \frac{1}{2\pi} \delta(\tau^2) \Theta(x^0-y^0) + A + \frac{B}{2\pi} \delta(\tau^2) \operatorname{sign}(x^0-y^0),$$ where $\tau^2 = \left[x^0 - y^0\right]^2 - \left[\vec{x} - \vec{y}\right]^2$ is the standard proper time between events at $x$ and $y$.

The first term is the standard retarded Green's function from E&M, and the other two terms satisfy the homogeneous wave equation ($[\partial_0^2 - \nabla^2] f(x) = 0$) and are there to accommodate different boundary conditions.

The first term can be constructed as the zero mass limit of the retarded propagator for a massive field. The second term is an obvious solution to the homogeneous equation. The third term is the combination of retarded and advanced propagators in such a way that it produces a solution to the homogeneous equation. This is the most general function that obeys: $$\left[\partial_0^2 - \nabla^2\right] G(x,y) = \pm \delta(x-y),$$ and obeys all of the same continuous symmetries as the d'Alembert operator (space-time translations, rotations, space parity, and boosts - time reflection is often broken by boundary conditions). Note that the $\pm$ on the right hand side is there because I'm too lazy to put in the work to ensure the proper sign.

All of this is perfectly sensible and well behaved when $G(x,y)$ is treated only as the kernel of a linear operator. Where QFT runs into trouble is that $G(x,y)$ is treated as an ordinary function. In other words, it's like having a matrix $M$ where all of the powers of $M$, $M^n$, are perfectly well defined and behave well when they are applied to the relevant vectors. QFT relies on calculations analogous to $M_2 = \left[M_{ij}^2\right]$, i.e. squaring of the individual matrix elements. What does that even mean?

Regardless of how we interpret treating $G$ as a function, it is the need to do so that requires the introduction of regularization and renormalization to get results that make sense.

  • $\begingroup$ 1. Did you mean Eqn. (1) in my question is not a real-space propagator? Why do you say that? 2. What are A and B in Eqn. (A)? Does Eqn. (A) represent the generic form of the propagator for any field fermionic or bosonic, massless or massive? @Sean E. Lake $\endgroup$
    – SRS
    Jun 25, 2017 at 7:12
  • $\begingroup$ Yes, Eqn. 1 is simply wrong. The propagator corresponds to the Green's function (with some small tweaking), and the correct Green's function for massless wave equations are well known (see Jackson's "Classical Electrodynamics" Section 12.11, also Duffy's "Green's Functions with Applications" eqn 4.0.3 makes clear that in this case $G(0) = 0$ is allowed). 2. Arbitrary constants set to satisfy boundary conditions. Usual choices are: $A=0,\ B=0$ (retarded), $A=0,\ B=-1$ (advanced), and $A=0,\ B=-1/2$ (symmetric/Feynman). $\endgroup$ Jun 25, 2017 at 17:27
  • $\begingroup$ No, this propagator/Green's function only covers fields that obey the wave equation (massless) for their classical equations of motion. For the massive field propagators, see the Wikipedia article already referenced. Much of the above can be reconstructed by taking the $m\rightarrow 0$ limit of those. $\endgroup$ Jun 25, 2017 at 17:31

This is indeed a pathology for the computation of some quantities, such as any quantities derived from the stress energy tensor

$$T_{\mu\nu} = \lim_{x\to y} D_{\mu\nu} G(x,y)$$

For some differential operator $D_{\mu\nu}$. This can be easily renormalized, though, as it can easily be shown that in the coincidence limit $x = y$, the divergence term will be a constant (or depend on the curvature of the manifold, if in curved space), and this is then just a renormalization of the cosmological constant, or the gravitational constant as well in curved space.


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