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For $\kappa \ge 0$ in the upper half-space, let $r^{\prime \ast }$ be the reflection of the interior source $r^{\prime }$ across its boundary plane. The displayed decaying Green function solves $(\Delta -{\kappa }^2)G_D=\delta (r-r^{\prime })$ with zero Dirichlet boundary condition. The direct term supplies the source; the image source lies outside the domain and its opposite sign makes the two terms cancel on the plane. The minus sign relative to the positive static Yukawa Green function comes from the operator convention. The case $\kappa =0$ is the Coulomb limit. An infinity condition is needed for uniqueness on an unbounded domain.

A free scalar Klein-Gordon equation with rest mass $m$ isHere $c$ is the speed of light and $\hslash$ the reduced Planck constant. In the static limit, comparison with the static Yukawa Green function identifiesThis $\lambda$ is the reduced Compton wavelength, rather than $h/(mc)$; the two differ by $2\pi$. It is an exponential decay length of this exchange field, not the geometric size of a nucleon or an exact radius of a bound state. This article uses $\lambda$ for a length; the 1935 paper used that letter for its inverse.

For a deliberately rounded worked example, stipulate $\hslash c=197.3\mathrm{MeV}\mathrm{fm}$ and $m{c}^2=140\mathrm{MeV}$. ThenA femtometre is $10^{-15}\mathrm{m}$. The chosen rest energy is a convenient pion-scale input; neither it nor the resulting Yukawa potential is asserted to be an exact measured nuclear interaction. Holding $\hslash c$ fixed, chosen rest energies $70$, $140$ and $280\mathrm{MeV}$ give lengths $2.819$, $1.409$ and $0.7046\mathrm{fm}$. Doubling the exchange rest mass halves the decay length.

Range alone does not guarantee binding: the potential energy must overcome the kinetic energy cost of localizing a wavefunction. The existing exponential variational bound for the Yukawa potential makes this precise. For the same toy $V$, with reduced mass $M$, choosing a normalized exponential wavefunction proportional to $e^{-\alpha r}$ with $\alpha =\kappa /2$ yieldsFor additional stipulated inputs $M{c}^2=470\mathrm{MeV}$ and $A=100\mathrm{MeV}\mathrm{fm}$, this gives $⟨H⟩\approx -3.66\mathrm{MeV}$. A negative value proves a bound state in this scalar model by the quantum variational principle; it is an upper bound on its ground-state energy, not a prediction for the deuteron. A nonnegative trial value would not prove the absence of binding.

Write $C=A\sqrt{\pi /(2k)}$, so the radial field is $\Phi =C{e}^{-kr}/r$. Integrate its equation over a ball of radius $ϵ$ centred at the source. The divergence theorem givesThe second term is $4\pi k^{2}C{\int }_0^{ϵ}r{e}^{-kr}dr=O({ϵ}^2)$ and vanishes in the limit. MeanwhileThusThe minus sign follows from the convention $(\Delta -k^2)\Phi ={\delta }^{(3)}$, opposite to the positive static Yukawa Green function for $-\Delta +k^2$. For arbitrary real $k$, replace it in the exponential by $|k|$ to select decay at infinity. The $k=0$ limit is $-1/(4\pi r)$.