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If we start from Klein Gordon Lagrangian density and work through canonical quantization, we could arrive at field operators for scalar fields. Now, if we solve for the free propagator we arrive at (in momentum basis): $$\Delta(p) = i/(p^2-m^2+i\epsilon).$$ But we also know that if we take the fourier transform of Yukawa potential of the Yukawa potential: $$\int d^3r \begin{equation} e^{-i\textbf{p*r}}e^{-im*\textbf{r}} \end{equation} = 4\pi/(\textbf{p}^2 + m^2),$$ which according to Lancaster and Blundell (pg 161) "is the Green's function for the time-independent Klein-Gordon equation."

Could someone please explain where in KG equation we assumed the presence of a potential that looks similar to Yukawa potential? I thought KG equation was one for relativistic free scalar particles. How is it that an equation for a free particle lead to a propagator that assumes the presence of a potential?

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It seems your confusion comes from the term "potential". Let me first introduce the Proca equation: \begin{equation} \partial^\mu F_{\mu \nu}+m^2 A_\nu = 0 \end{equation} Where $F$ is the usual Faraday tensor (the gauge curvature tensor). Expanding this equation leads to : \begin{equation} \square A_\nu-\partial_\nu \partial^\mu A_\mu + m^2 A_\nu=0 \end{equation} Then, imposing the Lorenz gauge condition $\partial^\mu A_\mu =0$ leaves us with: \begin{equation} (\square+m^2)A_\mu=0 \end{equation} Well, what have we done here? We considered a spin 1 field, so a field that can be responsible for a force, and we recovered the Klein-Gordon equation for all its components. Since we deal with a spin 1 particle, one can assimilate it to the 4-potential of some force. The time-independent version of this equation is: \begin{equation} (\Delta-m^2)A_\mu=0 \end{equation} When dealing with matter fields, one has a non-vanishing term at the RHS, and to solve the new equation arising from this second term, one use Green functions : \begin{equation} (\Delta-m^2)G(x-y)=-i\delta(x-y) \end{equation} This equation can be solved in the position space as shown in Wikipedia, and leads to the usual Yukawa potential: \begin{equation} G(r)=\frac{1}{4\pi r}e^{-mr} \end{equation} So to conclude, I would say that it is important to keep in mind that the Klein-Gordon equation is not restricted to the spin-0 particles and that the word "potential" in QFT refers to the spin-1 particles.

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The main point is that the mass term ${\cal V}=\frac{1}{2}m^2\phi^2$ in the KG Lagrangian density ${\cal L}$ leads to exponential suppression of the KG propagator/Green's function (a Bessel function) for asymptotic spatial separations.

See also e.g. my related Phys.SE answer here.

References:

  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; eq. (2.52).
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A pedestrian point: the wave equation in time-independent case is the Laplace equation, $$ \nabla^2 u-\partial_t u=0 \Rightarrow \nabla^2 u=0, $$ which has as a solution the potential of a point charge $$\frac{1}{r}.$$

The wave equation with a mass term (Klein-Gordon equation) similarly gives the Yukawa potential $$ \nabla^2 u-\partial_t u - m^2 u=0 \Rightarrow \nabla^2 u - m^2 u=0\Rightarrow \frac{1}{r}e^{-mr}. $$

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The Klein-Gordon equation is not necessarily for free particles. You can add electromagnetic field and sources. So, as far as I remember, one of the solutions of the time independent Klein-Gordon equation with a delta-function source is the Yukawa's solution.

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  • $\begingroup$ Yes but the field operators came from canonical quantization of KG equation without any additional terms, which should be for the case of a free particle. And the propagator using such operators lead to fourier transform of Yukawa potential which seems odd $\endgroup$
    – VVC
    Dec 7, 2021 at 3:41
  • $\begingroup$ @VVC : I am afraid I don't understand what you are talking about. The Klein-Gordon equation is a relativistic analog of the Schrödinger equation. So is the Schrödinger equation for nonrelativistic free particles only? So one starts with quantization of free field equations in perturbative quantum field theory, so what? $\endgroup$
    – akhmeteli
    Dec 7, 2021 at 4:19

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