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In Birrell's and Davies' book on "Quantum Fields in Curved Space", and in particularly in Chapter 2.7, the authors claim that from the expression $$\mathcal{G}(x,x')= \int\frac{d^nk}{(2\pi)^n} \frac{e^{i\vec{k}\cdot(\vec{x}-\vec{x}')-ik^0(t-t')}} {\big[(k^0)^2-|\vec{k}|^2-m^2\big]}\tag{2.76}$$ one can obtain all of the Green functions discussed later on.

For the case of the Feynman propagator, the retarded and the advanced propagators, it is fairly understandable that from the expression above, one can shift the locations of the poles accordingly, perform the $\int dk^0$ integral using contour integration methods and specifically the Cauchy residue theorem, and obtain the relevant expressions. For the case of Pauli-Jordan or Schwinger function $$iG(x,x')=\langle0|[\phi(x),\phi(x')]|0\rangle,\tag{2.65}$$ the case of Hadamard's elementary function $$G^{(1)}(x,x')=\langle0|\{\phi(x),\phi(x')\}|0\rangle\tag{2.66}$$ and the cases of the Wightman functions $$G^+(x,x')=\langle0|\phi(x)\phi(x')|0\rangle \tag{2.68}$$ or $$G^-(x,x')=\langle0|\phi(x')\phi(x)|0\rangle,\tag{2.68}$$ it is not obvious how the integral over $k^0$ is related to the integrals over the contours depicted in the Figure (Fig. 3 of the book).

Or to phrase my question a littile bit better, how am I supposed to obtain from the expression that yields all the possible Green functions, an integral over the contours depicted in Fig. 3, such that I can perform the aforesaid integrals and obtain the relevant expressions for the Pauli-Jordan or Schwinger function, the Hadamard's elementary function and the Wightman functions?

P.S.#1: Not related to Physical Interpretation for Schwinger and Hadamard functions

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1 Answer 1

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  1. That the $\pm$ Wightman functions (2.68) correspond to closed contours encircling the $\pm$ poles, respectively, is e.g. derived in eq. (2.50) of Peskin & Schroeder.

  2. The Pauli-Jordan/Schwinger and Hadamard functions are just some linear combinations of the Wightman functions, cf. Fig. 3.

  3. The above integration contours are closed.

  4. In contrast, the Feynman, advanced & retarded propagators have open integration coutours from $-\infty$ to $+\infty$ near the real axis, as OP essentially already mentions.

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  • $\begingroup$ Hi and thanks for the answer. I am still having a difficulty to obtain the relevant expression for the Hadamard elementary function, using the information that the contour must be the one in FIg. 3. I can, however, verify that $D^{(1)}(x-y)=D^{+}(x-y)+D^{-}(x-y)$ from the contour integrals and I can also derive the expression $-i\int\frac{d^4k_E}{(2\pi)^4}\frac{e^{ik_E\cdot(x_E-y_E)}}{k^2_E}=-\frac{1}{4\pi^2(x-y)^2}$. Hence, I arrive at a contradiction: I have to perform the contour integrals to distinguish the Hadamard's elementary function from the remaining Green functions, but** $\endgroup$
    – schris38
    Sep 5, 2022 at 10:52
  • $\begingroup$ **in calculating $D(x-y)=-\frac{1}{4\pi^2(x-y)^2}$, I simply used the Wick rotation (i.e. substituting $k_E^0=ik^0,\ k^i_E=k^i$) without using any information about the contour (whether or not it corresponds to $D^+,\ D^- $ etc). What am I missing here? There is no way every Green function corresponds to the same value, namely $-\frac{1}{4\pi^2(x-y)^2}$, right? $\endgroup$
    – schris38
    Sep 5, 2022 at 10:54
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    $\begingroup$ For the massless case in 3+1D, the Hadamard function $D^{(1)}$ is given by eq. (2.79), while the Pauli-Jordan/Schwinger function $D$ is proportional to ${\rm sgn}(x^0-y^0)\delta((x-y)^2)$. $\endgroup$
    – Qmechanic
    Sep 5, 2022 at 11:15
  • $\begingroup$ My problem is that when I perform the contour integral, starting from Eq. (2.76) from Birrell and Davies, I obtain the result $D^{(1)}=D^{(+)}+D^{(-)}$. I am looking for a way to be able to start from Eq. (2.76) and result in Eq. (2.79) by appropriately choosing the relevant contour in Fig. 3. However, I find that if I attempt to perform the $k^0$ integration, my answer then, upon integration over the spatial components of the momentum, will be a function of $|\vec{x}-\vec{y}|$, rather than $|(x-y)^{\mu}|$ ... $\endgroup$
    – schris38
    Sep 5, 2022 at 11:35
  • $\begingroup$ And since I can derive the result [Eq. (2.79)] by Wick rotating the expression $\int\frac{d^4k}{(2\pi)^4}\frac{e^{-ik\cdot(x-y)}}{k^2}$, I am wondering whether or not the latter result by Wick rotation is connected to the Hadamard's elementary function or whether or not I am somewhere mistaken... And since Wick rotating does not (not in my understanding at least) need any information about the contour, I am feeling kind of lost :) $\endgroup$
    – schris38
    Sep 5, 2022 at 11:47

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