I am trying to understand the paper "Off-diagonal coefficients of the DeWitt-Schwinger and Hadamard representations of the Feynman propagator" by Decanini and Folacci.

Consider a ($m$)assive scalar field $\phi$ living in a spacetime with metric $g$ (coupled to the Ricci scalar $R$ with coupling $\xi$). I specify to $D=4$ dimensions in what follows here. The paper concerns itself with the short-distance behaviour of the Feynman propagator $G^{\mathrm{F}}$ obeying $$ ( \Box_x - m^2 - \xi R ) G^{\mathrm{F}}(x,x') \ = \ - \frac{\delta^{(4)}(x-x')}{\sqrt{-g(x)}} \ , $$ which I think here is the time-ordered correlation function where $G^{\mathrm{F}}(x,x') \sim \langle \Omega | \mathscr{T}\big( \phi(x) \phi(x') \big) | \Omega \rangle$ for some vacuum state $| \Omega \rangle$.

The authors say that the Hadamard expansion for $G^{\mathrm{F}}(x,x')$ is $$ G^{\mathrm{F}}(x,x') = \frac{i}{8\pi^2} \bigg[ \frac{\sqrt{ \Delta(x,x') }}{\sigma(x,x') + i \epsilon} + V(x,x') \log\big( \sigma(x,x') + i \epsilon \big)+ W(x,x') \bigg] $$ where $\sigma(x,x')$ is the Synge world function (aka. half the square of the geodesic distance between the points $x$ and $x'$), and $\Delta(x,x')$ is the Van Vleck-Morette determinant. The parameter $\epsilon \to 0^{+}$ is a regulator with mass dimension $-2$, and is there so that the singularity structure of $G_{\mathrm{F}}$ is consistent with the definition of the Feynman propagator.

The functions $V(x,x')$ and $W(x,x')$ are symmetric biscalars, which are regular for $x \to x'$, and possess expansions of the form $$ V(x,x') = \sum_{n=0}^{\infty} V_{n}(x,x') \ \sigma(x,x')^{n} \\ W(x,x') = \sum_{n=0}^{\infty} W_{n}(x,x') \ \sigma(x,x')^{n} $$ The so-called Hadamard coefficients $V_{n}$ and $W_{n}$ are also symmetric and regular as $x \to x'$. The functions $V_{n}$ are determined via the recursion relation $$ (n+1)(2n+4) V_{n+1} + 2 (n+1) V_{n+1;\mu}\sigma^{;\mu} - 2 (n+1) V_{n+1} \Delta^{-1/2}\Delta^{1/2}_{\ ;\mu} \sigma^{;\mu} + ( \Box_{x} - m^2 - \xi R ) V_{n}=0 $$ for all $n \in \mathbb{N}$, subject to the boundary condition at $n=0$ given by $$ 2 V_0 + 2 V_{0\;\mu} \sigma^{;\mu} - 2 V_0 \Delta^{-1/2} \Delta^{1/2}_{ ;\mu} \sigma^{;\mu} + ( \Box_{x} - m^2 - \xi R )\Delta^{1/2} = 0\ . $$ Similarly the coefficients $W_{n}$ satisfy the recursion relations $$ (n+1)(2n+4) W_{n+1} + 2 (n+1) W_{n+1;\mu}\sigma^{;\mu} - 2 (n+1) W_{n+1} \Delta^{-1/2}\Delta^{1/2}_{\ ;\mu} \sigma^{;\mu} + (4n+6) V_{n+1} + 2 V_{n+1;\mu} \sigma^{;\mu} - 2 V_{n+1} \Delta^{-1/2}\Delta^{1/2}_{\ ;\mu} \sigma^{;\mu} + ( \Box_{x} - m^2 - \xi R ) W_{n}=0 $$ for $n \in \mathbb{N}$, however the biscalar $W_0(x,x')$ (boundary condition) is unrestrained by the recursion relation.

Usually, I have seen $W_0(x,x')=0$ being set, and it's somewhat implied that it is a choice to do so.

Question 1: Is the above function $G^{\mathrm{F}}(x,x')$ truely the time-ordered correlator $\sim \langle \Omega | \mathscr{T}\big( \phi(x) \phi(x') \big) | \Omega \rangle$? Sometimes in the literature I see a Hadamard expansion referred to for retarded correlation functions (see for example equation (2.4) of Ottewill and Wardell's Quasilocal contribution to the scalar self-force: Nongeodesic motion), and the use of the same phrase ``Hadamard Expansion'' for these two different functions confuses me.

Question 2: Does the choice of $W_0(x,x')$ correspond to a choice of vacuum state $|\Omega \rangle$? I am very confused about this --- for example, if I chose $W_0=0$ in Schwarszchild space, would this correspond to the Boulware vacuum, or the Hartle-Hawking vacuum, or the Unruh vacuum? It seems it would have to pick out one of those three vacuums (surely not all of them?). Is there a statement regarding this? Does some other choice of $W_0(x,x') \neq 0$ correspond to different vacuum states then? (From page 162 of "Semiclassical and Stochastic Gravity" by Hu and Verdaguer, it seems that there are restrictions on what $W_0(x,x')$ can be ie. you need to pick it such that all $W_{n}(x,x')$ are still symmetric in $x$ and $x'$)


In both cases the answer is NO.

Regarding the first issue: You are dealing with a parametrix and not with a true state. Usually the series does not even converge.

Regarding the second issue: in general it is impossible to obtain a final sum $W(x,y)=W(y,x)$ by summing the series with a suitable choice of $w_0$ as a true two-point function would deserve. This is the reason for the appearance of the trace anomaly when renormalizing the stress energy tensor with the point-splitting procedure.

Decanini and Folacci cite a paper of mine in their paper

V. Moretti: Comments on the stress-energy tensor operator in curved spacetime Commun. Math. Phys. 232, 189 (2003)

about the stress energy tensor operator, there is a discussion on all these issues therein.

Again regarding your second question. The ``almost modern'' prescription is to remove completely the part of the asympthotic series containing all the terms $W_k$ obtaining a parametrix $H$ since it is useless to fix a state. The remaining part $H$ is therefore universal and a Gaussian state is said to be of Hadamard type is its two-point function $G$ (or its Feynman propagator depending on which version of the expansion you consider) subtracted to the parametrix $$G(x,y) - H(x,y) = w(x,y)$$ is a smooth function $w$. This function, if the state is Hadamard, completely determines it.

Excluding very particular cases it is impossible to obtain $G$ as a true complete convergent expansion suitably fixing $W_0$.

The case of the vacua in Schwarzschild-Kruskal manifold is very delicate and the approach in terms of a parametrix is even more difficult. It can be tackled (to prove that the state is Hadamard) with the `'modern'' approach of mircolocal analysis of Hoermander introduced by Radzikowski (wavefront sets technology). Here is a complete discussion and rigorous construction on the Unruh state as a Hadamard state,

C. Dappiaggi, V. Moretti and N. Pinamonti: Rigorous construction and Hadamard property of the Unruh state in Schwarzschild spacetime. Adv. Theor. Math. Phys. 15, vol 2, 355-448 (2011) (93 pages!)

It could be also useful the chapter of a book I wrote with I. Khavkine,

I.Khavkine and V. Moretti: Algebraic QFT in Curved Spacetime and quasifree Hadamard states: an introduction (freely downloadable at arXiv:1412.5945)

And also this paper with T. Hack written into a more physically minded fashion

T.-P. Hack and V. Moretti: On the stress-energy tensor of QFT in curved spacetime - Comparison of different regualrization schemes and symmetry of the Hadamard/Seeley-DeWitt coefficients J. Phys. A: Math.Theor. 45 374019 (2012)

| cite | improve this answer | |
  • $\begingroup$ Thank you for your interesting response. Regarding your paper about the Unruh state: it sounds like your paper implies the existence of a function $w(x,y)$ (which is $=G(x,y) - H(x,y)$)? Is it possible to understand anything about this function, perhaps as a series in $\sigma(x,y)$? $\endgroup$ – QuantumEyedea Jun 2 at 3:10
  • $\begingroup$ Also, one last question. Your paper proves that the Unruh vacuum is Hadamard. Is it known yet whether Hartle-Hawking, or Boulware are Hadamard? If rigorous proofs do not exist, are they at least expected to be? $\endgroup$ – QuantumEyedea Jun 2 at 3:11
  • 1
    $\begingroup$ Yes, the paper of mine with Dappiaggi and Pinamonti implies what you wrote, but the details are quite delicate. The rigorous construction of Hartle-Hawking state and its Hadamard property has been established by Ko Sanders in 2015 On the Construction of Hartle–Hawking–Israel States Across a Static Bifurcate Killing Horizon Letters in Mathematical Physics volume 105, pages575–640(2015) $\endgroup$ – Valter Moretti Jun 2 at 13:17
  • 1
    $\begingroup$ The Boulware state id not Hadamard since a non-Hadamard singularity arises on the Killing horizon. $\endgroup$ – Valter Moretti Jun 2 at 13:18
  • 1
    $\begingroup$ Unfortunately I stopped with these research subjects some years ago and I do not remember where it is established. However it is a well known result...You should look in the literature quoted in the papers I pointed out. However the problem is the same as for the Rindler vacuum in Minkowski spacetime: it shows a non-Hadamard singularity on the Killing horizon and this can be proved per direct inspection using the 2-point function... $\endgroup$ – Valter Moretti Jun 5 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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