On Peskin & Schroeder's QFT, page 30, the scalar field propagator as the retarded Green function is defined as

$$(\partial^2+m^2)D_R(x-y)=-i\delta^4(x-y) \tag{2.56}$$ The Fourier transformation is given as $$D_R(x-y)=\int \frac{d^4p}{(2\pi)^4}e^{-ip\cdot(x-y)}\tilde{D}_R(p) \tag{2.57}$$ with $$\tilde{D}_R(p)=\frac{i}{p^2-m^2} $$

My question is, does the inverse propagator Fourier transform is consistent in the closed form? $$D_R(x-y)^{-1}=\int \frac{d^4p}{(2\pi)^4}e^{-ip\cdot(x-y)}\tilde{D}_R(p)^{-1} \tag{*} $$

My answer to this question is not. However, I am asking this question since in later chapter 11, eq.(11.97) about the effective potential $$ \tilde{D}^{-1}\left(p^2\right)=\int d^4 x e^{i p \cdot(x-y)} \frac{\delta^2 \Gamma}{\delta \phi \delta \phi}(x, y)=0 \tag{11.97} $$ the book seems uses the relation of Eq.~$*$, since $$ \left(\frac{\delta^2 \Gamma}{\delta \phi_{\mathrm{cl}}(x) \delta \phi_{\mathrm{cl}}(y)}\right)=i D^{-1}(x, y) \tag{11.90} $$ where this relations arises from (11.89) and (11.87) $$ \begin{aligned} \delta(x-y)&=\int d^4 z \frac{\delta^2 E}{\delta J(y) \delta J(z)} \frac{\delta^2 \Gamma}{\delta \phi_{\mathrm{cl}}(z) \delta \phi_{\mathrm{cl}}(x)} \\ &=\int d^4 z D(y,z) D(z,x)^{-1}. \end{aligned} \tag{11.87} $$ and $$ \left(\frac{\delta^2 E}{\delta J(x) \delta J(y)}\right)=-i\langle\phi(x) \phi(y)\rangle_{\mathrm{conn}} \equiv-i D(x, y) \tag{11.89} $$

My thoughts:

(1) The relation of $F(p)=\frac{1}{2\pi}\int dx\ e^{-ipx} f(x)$ and $F^{-1}(p)=\frac{1}{2\pi}\int dx\ e^{-ipx} f^{-1}(x)$ should not coincide with each other for a general function of $f$;

(2) However, in our case of Green function, I am not sure. Can we infer the $D_R(x-y)^{-1}$ from (2.56)?


Qmechanic's answer really make sense. But I still don't clear for

(3) Why $D^{-1}(x-y)=i(\partial^2+m^2)\delta^{(4)}(x-y) $? When I put this expression and Eq.(2.56) into Eq.(11.87), I don't obtain the correct form of $\delta(x-y)$.

(4) Why $D(x-y)$ is not invertible? Would this the same reason with the trivial photon propagator? (Like P & S Eq.(9.52)) \begin{equation} \begin{aligned} \left(\partial^2 g_{\mu \nu}-\partial_\mu \partial_\nu\right) D_F^{\nu \rho}(x-y) & =i \delta_\mu{ }^\rho \delta^{(4)}(x-y) \\ \text { or } \quad\left(-k^2 g_{\mu \nu}+k_\mu k_\nu\right) \widetilde{D}_F^{\nu \rho}(k) & =i \delta_\mu{ }^\rho, \end{aligned} \tag{9.52} \end{equation} where the Feynman propagator $D^{\nu \rho}_F$ has no solution, since the $4\times 4$ matrix $(-k^2g_{\mu \nu}+k_{\mu} k_{\nu})$ is singular.


1 Answer 1


Comments to the post (v8):

  1. There is often a slight misuse of notation in the literature: We start by defining the integral kernel$^1$ $$-iD^{-1}(x\!-\!y)~:=~(\partial^2+m^2)\delta^4(x\!-\!y)\tag{A}$$ for the Klein-Gordon operator, which despite the notation a priori is not invertible, since it has zero-modes (namely, the solutions to the Klein-Gordon equation). The various Greens functions $D(x\!-\!y)$ (retarded, Feynman, etc) are inverse integral kernels to eq. (A) with different boundary conditions$^2$.

  2. Note that $D$ in chapter 2 is the free propagator while $D$ in chapter 11 is the full/connected propagator. These 2 propagators should not be conflated.

  3. For a proof of the inverse relationship in chapter 11 between the 2-point functions for the generator $W_c[J]$ of connected diagrams and the 1PI effective/proper action, see e.g. eq. (8) of my Phys.SE answer here.


  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; p. 30-31.


$^1$ The integral kernel (A) also appear in the kinetic term of the action, cf. e.g. this Phys.SE post.

$^2$ Be aware that the retarded Greens function in eq. (2.58) does not explicitly contain the pertinent $i\epsilon$-prescription: The denominator $p^2-m^2$ should be replaced with $(p^0+i\epsilon)^2-E^2_{\rm p}$ to match the figure on p. 30.

  • $\begingroup$ Hi Qmechanic, your answer and links therein really make sense. But before I vote your answer, I still have two unclear points. I summary them in the edit post. If you can elaborate more on them, I would be really appreciate. $\endgroup$
    – Daren
    Commented Aug 24, 2023 at 7:27
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Aug 24, 2023 at 7:44
  • $\begingroup$ Thanks! I understand these 2 propagators should not be conflated. 1. And does this means that the propagator in chapter 11 cannot be represented by your equation (A)? 2. If so, how to obtain Eq.(11.97)? 3. Or how to obtain Eq.(11.92) from Eq.(11.91) in Peskin's book? $\endgroup$
    – Daren
    Commented Aug 24, 2023 at 10:49
  • 1
    $\begingroup$ 1. Yes. 2. Eq. (8) in the link. 3. For eq. (11.92) see e.g. physics.stackexchange.com/q/440789/2451 $\endgroup$
    – Qmechanic
    Commented Aug 24, 2023 at 10:59

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