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I'm trying to understand why the equal time propagator for a scalar field can sometimes be written in terms of the modified Bessel function $K_1(mr)$ and sometimes in terms of a Yukawa-like potential $\sim \frac{e^{-mr }}{4\pi r} $.

On the one hand, it is regularly argued that the integral that shows up in the Klein-Gordon equal time propagator can be written in terms of the modified Bessel function $K_1$: $$ D(\vec x,\vec x') \equiv \langle \vec x' | \vec x\rangle = \int \frac{ \mathrm{d }k^3 }{(2\pi)^3 2\omega_{k} } {\mathrm{e }}^{- i\vec k \cdot (\vec x ' -\vec x)} =\frac{m}{4\pi^2 r} K_1(mr),$$ where $r \equiv |\vec x - \vec x'|$.

On the other hand, the Green's function of the Klein-Gordon equation for a static field configuration $\phi$ reads $$G(\vec x,\vec x') = \int \frac{dk^3}{(2\pi)^3} \; \frac{e^{-i \vec k \cdot (\vec x - \vec x')}}{-k^2 + m^2}= \frac{e^{-mr }}{4\pi r} $$ where again $r \equiv | \vec x - \vec x'| $. (The integral is solved explicitly in Zee's book on page 29.)

Shouldn't the equal time propagator be equal to the Green's function for a static field configuration? If yes, why do we get such as simple solution (a Yukawa potential) in the second case and a much more complicated solution (a modified Bessel function) in the former case?


A useful hint is probably that the propagator for spacelike separations decays approximately like $e^{-mr }$: $$ D(\vec x,\vec x') \equiv \langle \vec x' | \vec x\rangle = \int \frac{ \mathrm{d }k^3 }{(2\pi)^3 2\omega_{k} } {\mathrm{e }}^{- i\vec k \cdot (\vec x ' -\vec x)} \sim e^{-mr }$$ (This is demonstrated at page 18 here, for example, or on page 27 in Peskin & Schröder's book)

jak
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1 Answers1

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Shouldn't the equal time propagator be equal to the Green's function for a static field configuration?

Nope. The equal time propagator lives in a $4$-d Minkowski space-time, and the Yukawa potential lives in a $3$-d Euclidean space. They are, in fact, related. You can see the relationship in the propagator to the wave equation versus the Coulomb potential.

Short version: $$G_3(\mathbf{x};\mathbf{x}') = \int_{-\infty}^\infty \mathrm{d}t\ G_4(x;x'),$$ where $G_4$ is one of (perhaps any of? maybe only the causal ones?) the position space propagators for the Klein-Gordon equation, and the Green's function for the D'Alembertion for the wave equation. $G_3$ is the Yukawa potential for the KG equation, and the Coulomb potential for the Poisson's equation.

Sean E. Lake
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  • Thanks a lot! This seems to imply there is indeed a close relationship between the modified Bessel function and the form of the Yukawa potential. I will try to figure it out in explicit terms and then get back to you. – jak Oct 28 '19 at 14:57
  • @jak Closer than you think. The Yukawa potential is a modified Bessel function of the second kind. In the Wikipedia Green's function article linked above, plug $n=3$ and $k=m$ into the $\Delta - k^2$ row of the table and you get the Yukawa potential (poke around in the "Asymptotic forms" section of the Wikipedia Bessel function article to find $K_{1/2}$). – Sean E. Lake Oct 28 '19 at 15:35
  • Ah perfect! A problem I encountered when I thought about your answer is that the time dependence drops out from the greens function for equal times. Thus I don’t see how we can integrate over t to find the yukawa potential. – jak Oct 28 '19 at 15:43
  • @jak Yeah, you can't integrate over time after evaluating, you need to do the integral with the full propagator. – Sean E. Lake Oct 28 '19 at 15:46
  • "the Yukawa potential lives in a 3-d Euclidean space." Can you please elaborate on that? Does it not experience time? Is that the same for nuclear force? – Árpád Szendrei Oct 29 '19 at 02:38
  • @ÁrpádSzendrei That's right. The same is true of the Coulomb potential, naturally. You can only use them to describe potentials when the accelerations are slow enough that the potential's source is moving with approximately constant velocity. This makes them, in some frame, approximately constant in time. You can treat that as a time-like Klein-Gordon source with infinite extent (the integral in the answer), or solve the static KG equation. Either way reduces the dimension of the space by 1. – Sean E. Lake Oct 29 '19 at 02:47