My question is a follow up question of the following one: Schwinger and Hadamard functions derivation in Birrell's and Davies' book

So, in Birrell's and Davies' book the authors derive an expression for the Hadamard's function $$D^{(1)}(x,y)=-\frac{1}{2\pi^2(x-y)^2}.\tag{2.79}$$ The authors also mention that in deriving this result, they have considered a complex integral with respect to the $k^0$ component of the momentum with respect to a closed contour that starts from the beginning of the complex plane $(\text{Im}k^0,\text{Re}k^0)$, circles around the pole to $k^0=|\vec{k}|$, returns to the begining and then circles around the pole $k^0=-|\vec{k}|$ and closes back at the beginning [Fig. 3 from the book].

My way of obtaining the same results includes starting from Eq. $$\mathcal{G}(x,y)= \int\frac{d^nk}{(2\pi)^n} \frac{e^{i\vec{k}\cdot(\vec{x}-\vec{y})-ik^0(x^0-y^0)}} {\big[(k^0)^2-|\vec{k}|^2-m^2\big]}\tag{2.76}$$ substituting $n=4$, $m=0$ and performing a Wick rotation, i.e. $k^0\rightarrow k^0_E=ik^0$ $$\mathcal{G}(x,y)= -i\int\frac{d^4k_E}{(2\pi)^4} \frac{e^{i\vec{k}_E\cdot(\vec{x}_E-\vec{y}_E)}} {\big[(k^0_E)^2+|\vec{k}_E|^2\big]}$$ and then perform the integration by expressing the volume element in spherical coordinates $$\mathcal{G}(x,y)=-i\frac{1}{(2\pi)^4} \int_0^{2\pi}d\psi\int_0^{\pi}d\phi\sin\phi\int_0^{\pi}d\theta\sin^2\theta\int_0^{\infty} dk_Ek_E^3 \frac{e^{ik_E|\vec{x}_E-\vec{y}_E|\cos\theta}} {k_E^2}$$ (where $\theta$ is the angle between the four vector $(\vec{x}_E-\vec{y}_E)$ and $\vec{k}_E$ and $k_E$ is simply the magnitude of $\vec{k}_E$. Performing the integrations with respect to $\psi$ and $\phi$ yields $$\mathcal{G}(x,y)=-\frac{i}{4\pi^3} \int_0^{\pi}d\theta\sin^2\theta\int_0^{\infty} dk_Ek_E e^{ik_E|\vec{x}_E-\vec{y}_E|\cos\theta}$$ Now, I can use the following identity about the first Bessel function: $J_1(x)=\frac{x}{\pi}\int_0^{\pi}e^{\pm ix\cos\theta}\sin^2\theta d\theta$ and hence $$\mathcal{G}(x,y)=-\frac{i}{4\pi^2|\vec{x}_E-\vec{y}_E|} \int_0^{\infty} dk_E J_1(k_E|\vec{x}_E-\vec{y}_E|)$$ and by using $\int dxJ_1(x)=1$ we have $$\mathcal{G}(x,y)=-\frac{i}{4\pi^2(\vec{x}_E-\vec{y}_E)^2}.$$ As a last step we can substitute back the Lorentzian versions for the vectors $$\mathcal{G}(x,y)=\frac{i}{4\pi^2(x-y)^2}$$

So, here lies my problem: my answer is not exactly the same as $D^{(1)}(x,y)$ of Birrell's and Davies' book. It is actually $i$ times $D^{(1)}(x,y)$. Despite that, I can not fail to notice that, up to the sign and the i factors, the result is the same. Also, I further understand that I have not made the assumption that the contour of my choosing is the one corresponding to $D^{(1)}(x,y)$, but yet I obtain a result that has a similar form.

  1. Is my derivation correct? If not, where have I gone wrong?

  2. If it is indeed correct, there has to be some unknown feature of Wick rotation that makes the Green function correspond to the Hadamard's function (with the contour chosen accordingly such that in Fig.3 ). Is this true? If so, how? And are there other ways of Wick rotating that correspond to choosing other contours and hence other Green functions, such as the Pauli-Jordan (or the Schwinger) one?

Any help is welcome, thanks.

  • 2
    $\begingroup$ Birrell & Davies explicitly say that $iG^+$ and $iG^-$ are such that $G^{(1)}=G^++G^-$. To me, it seems you have just found $iG^{(1)}$. $\endgroup$ Sep 6, 2022 at 15:33
  • $\begingroup$ Okay I can see now that probably $G^{(+)}=G^{(-)}=-\frac{1}{4\pi^2(x-y)^2}$ (I am probably missing a minus sign somewhere in my above derivation) and hence their addition yields the correct result regarding the $1/2$ factors. Does the fact that $G^{(+)}=G^{(-)}=-\frac{1}{4\pi^2(x-y)^2}$ then imply that the Schwinger function vanishes? Also, I still can not wrap my head around the fact that performing a Wick rotation is equivalent to evaluating $D^{(+)}$ or $D^{(-)}$. Why is that? I am still missing how we encode the information of "which contour we are considering" in the result's derivation.. $\endgroup$
    – schris38
    Sep 7, 2022 at 6:44

1 Answer 1


I have B&D's book at hand so I will try to give an answer to this question.

From what I've seen, you have correctly done a Wick rotation, but you forgot to include a $-1$ factor from the denominator, which gives the right sign at the end. Indeed: \begin{align} \mathcal{G}(x,y)=&\int \frac{d^nk}{(2\pi)^n}\frac{e^{ik\cdot(x-y)}}{(k^0)^2-|\vec{k}|^2} \\=&-i\int \frac{d^nk_E}{(2\pi)^n}\frac{e^{ik\cdot(x-y)}}{(k^0)^2-|\vec{k}|^2} \\=&i\int \frac{d^nk_E}{(2\pi)^n}\frac{e^{ik_E\cdot(x_E-y_E)}}{(k^0_E)^2+|\vec{k}_E|^2} \end{align}

The Wightman functions have symmetric contours, as shown in figure $3$ of B&D. So, it seems the Schwinger function vanishes identically. At least to me, I've encountered Schwinger's function only in another context and so I am no expert for this part of the question.

For the Wick rotation, it basically follows from the residue theorem. Indeed you can perform the integration on $k_E^0$ without using it, and using your method (which is also the method Srednicki employs in his book/lectures), or a 'brute force method', by choosing $x^0_E,y^0_E=0$, integrating on $k_E^0$ using the well-known formula for $\int \frac{d^nk}{(2\pi)^n} \frac{k^{2r}}{[k^2+A]^\alpha}$. Now, you can close the contour like the black one ($\gamma'$) in the following picture (the position of the poles is unimportant for the following argument):

enter image description here

The difference here is that $\omega$ and $-\omega$ are, in euclidean, right on the imaginary axis for your case. We can do this extended contour because, by the estimation lemma, the integral around the semi-circles vanishes. So evaluating the integral for $\mathfrak{Re}(p^0)$ or $\gamma'$ gives the same result. But by the residue theorem, this last integral doesn't depend on the shape of the contour. You can reduce it to a small circle around $-\omega$ (again, in your case $-\omega$ is on the imaginary axis). Wick-rotating back to Lorentzian signature, you will end up with exactly the same contour as for $iG^+$ in figure 3 of B&D.

All this to say, that indeed, performing a Wick rotation is equivalent to evaluating $iG^+$ or $iG^-$, depending on which contour you choose (but it doesn't matter since the two give the same answer here.) Strictly speaking, doing a Wick rotation gives one of the Wightman propagators but does not pick automatically the one we would want. We have to choose which contour to use in the integration.


Choosing the contour of integration can be done by looking at the estimation lemma for $\mathcal{G}(x,y)$. For convenience I rename $\gamma'\leadsto \gamma^+$ and $\gamma \leadsto \gamma^-$. We can write:

\begin{align} \mathcal{G}(x,y)=&i\int_{\mathbb{R}^n} \frac{d^nk_E}{(2\pi)^n}\frac{e^{ik_E\cdot(x_E-y_E)}}{(k^0_E)^2+|\vec{k}_E|^2} \\ =& i\int_{\gamma^{\pm}\times\mathbb{R}^{n-1}} \frac{d^nk_E}{(2\pi)^n}\frac{e^{ik_E\cdot(x_E-y_E)}}{(k^0_E)^2+|\vec{k}_E|^2}-i\int_{C^{\pm}\times\mathbb{R}^{n-1}} \frac{d^nk_E}{(2\pi)^n}\frac{e^{ik_E\cdot(x_E-y_E)}}{(k^0_E)^2+|\vec{k}_E|^2} \end{align} Where $C^\pm$ are the positive and negative semi-circles such that $C^\pm \cup \mathbb{R}=\gamma^\pm$. We introduce $C_R^\pm$ which are integration contours like $C^\pm$ but with radius $R$ (so $C^\pm=C_\infty^\pm$), and $\delta^\pm \in (0; \pm \pi)$. Now, by the estimation lemma we have: \begin{align} &\left|\int_{C^\pm}\frac{dk_E^0}{2\pi}\int_{\mathbb{R}^3} \frac{d^{n-1}k_E}{(2\pi)^{n-1}}\frac{e^{ik_E\cdot(x_E-y_E)}}{(k^0_E)^2+|\vec{k}_E|^2}\right| \\ \leq& \lim_{a\rightarrow+\infty} \frac{\pi a}{2\pi} \sup_{C^\pm_{R=a}} \left| \int_{\mathbb{R}^{n-1}} \frac{d^{n-1}k_E}{(2\pi)^{n-1}}\frac{e^{ik_E\cdot(x_E-y_E)}}{a^2+|\vec{k}_E|^2} \right| \\ \leq& \lim_{a\rightarrow+\infty} \frac{\pi a}{2\pi} \int_{\mathbb{R}^{n-1}} \frac{d^{n-1}k_E}{(2\pi)^{n-1}}\frac{e^{-a(x^0_E-y_E^0)\sin \delta^\pm}}{a^2+|\vec{k}_E|^2} \\ \stackrel{\text{sgn}(x_E^0-y_E^0)=\pm}{\leq}& \lim_{a\rightarrow+\infty} \frac{\pi a}{2\pi} \int_{\mathbb{R}^{n-1}} \frac{d^{n-1}k_E}{(2\pi)^{n-1}}\frac{e^{-a(x^0_E-y_E^0)\sin \delta^\pm}}{a^2+|\vec{k}_E|^2}=0 \end{align} So we can integrate along $\gamma^+$ ($\gamma'$ in the picture) only when $x^0_E-y^0_E >0$, and along $\gamma^-$ ($\gamma$ in the picture) when $x^0_E-y^0_E <0$.

So, from this example, we can see that if we want to derive an expression for $G^+$ with $x^0_E-y^0_E >0$, we have to pick $\gamma^+$. But if instead, we want $G^-$ together with the condition $x^0_E-y^0_E >0$, we have to choose the opposite, $\gamma^-$ (this follows from the definition of $G^-$). And of course the other way around when $x^0_E-y^0_E <0$.

  • $\begingroup$ Hi and thank you for your answer. Despite the fact that it is enlightenning, I can not understand how the Wick rotation knows about the contour selection (and hence the choice between $D^{+}$ or $D^{-}$). Also, I couldn't help but notice that the poles in the image you attach are shifted. This is not the case in B&D book, if I am not mistaken (at least for the case of $D^{+}$ or $D^{-}$). And last, why do you say that I am missing a minus sign (remember that I am using a mostly minus convention for the metric)? $\endgroup$
    – schris38
    Sep 7, 2022 at 15:12
  • 1
    $\begingroup$ @schris38 I've edited my answer, I hope I clarified some things about choosing the integration contour, the position of the poles, and what appears to me to be a sign error... $\endgroup$ Sep 7, 2022 at 17:10
  • $\begingroup$ Sorry for taking so long to reply, but now I see where the sign difference we have originates from: your Wick rotation is $k_E^0=-ik^0$, whereas mine was $k_E^0=+ik^0$. Does that mean that our final result is prone to the kind of Wick rotation we make? Or is there a proper (naturally occuring) way of Wick rotating according to whether or not the arc contrours (denoted above by $C^{\pm}$) are such that the integrals over those contours vanish? $\endgroup$
    – schris38
    Sep 16, 2022 at 7:44
  • $\begingroup$ And also, I would like to ask one more thing: when we make the Wick rotation (or any other complex transformation of the complex variable to be integrated over a contour), does the contour change? It should, right? If I substitute for example $k^0_E=ik^0$ over $-\infty<k^0<0$, then the contour corresponding to the $k^0_E$ variable is the line$-i \infty<k^0_E<0$, correct? $\endgroup$
    – schris38
    Sep 16, 2022 at 7:49

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