The two-point function of local quantum fields on a curved spacetime exhibits a singularity of a very particular form, known as Hadamard form, for null separated points $(x,y)$ (including the coincident case where $x=y$). For example, if $\phi(x)$ is a scalar field in $4$-dimensions, then the Hadamard form of the anti-commutator correlation function (see for example http://link.springer.com/article/10.1007%2FBF01196934)

$$ G(x,y) = \langle \phi(x) \phi(y) + \phi(y) \phi(x) \rangle \, ,$$

is given by

$$ G(x,y) = \frac{u(x,y)}{\sigma} + v(x,y) \ln \sigma + w(x,y) \, , $$

where $\sigma$ is half the square of the geodesic distance between $x,y$ (the curved space analog of $(x-y)^2/2$), and $u,v,w$ are smooth functions of $x,y$. The first two terms diverge at zero and null separation, where $\sigma = 0$.

My only physical understanding of the requirement that a correlator exhibit this Hadamard form is that the short-distance singularities indicate that the state where the expectation value is being taken in is "close" to the flat space vacuum in the UV. There is no unique vacuum in curved space, but any candidate vacuum should probably look like the flat space vacuum on short enough distance scales.

My question is: are there other, simple and physical understandings of the meaning of this singularity?

  • $\begingroup$ The leading $1/\sigma$ singularity should be a 4d version of the Coulombic potential around a point charge. The log presumably comes because obtaining the Green's function involves solving a second order differential equation in in the variable $\sigma$ with characteristic exponents -1 and 0. The series expansion near $\sigma = 0$ will then naturally involve logs because the difference between the exponents is an integer. $\endgroup$ May 21 '16 at 4:00
  • $\begingroup$ This seems plausable, but I don't see why this should be true? The field in this case is a scalar, not a gauge field capable of producing a Coulombic potential, it may be charged or uncharged, and it may be massive or massless. $\endgroup$ May 21 '16 at 16:16
  • $\begingroup$ In both cases, the relevant partial differential equation is Poisson's equation. $\endgroup$ May 31 '16 at 8:49

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