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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.

For a normalized trial state $|\psi ⟩={\sum }_n{c}_n|n⟩$ in the form domain of a bounded-below Hamiltonian, the quantum variational principle gives $⟨H⟩={\sum }_n|c_n{|}^2{E}_n\ge E_0$. Thus each trial expectation is an upper bound for the ground-state energy. The same spectral argument covers a continuum: for a decaying potential whose essential spectrum begins at $0$, a negative trial expectation guarantees a negative-energy bound state.

For the stated exponential trial function, normalization follows from $4{\alpha }^3{\int }_0^{\infty }{r}^2{e}^{-2\alpha r}dr=1$. The kinetic expectation, using $\int |\nabla \psi {|}^2$, is ${\hslash }^2{\alpha }^2/(2m)$. The potential expectation is $-4A{\alpha }^3{\int }_0^{\infty }r{e}^{-(\mu +2\alpha )r}dr$. ThereforeThe exponential variational bound for the Yukawa potential at $\alpha =\mu /2$ gives $E=(\mu /8)({\hslash }^2\mu /m-A)<0$ if $\mu <Am/{\hslash }^2$. Thus a bound state necessarily exists in the stated range.

For $\mu \ge Am/{\hslash }^2$, this single-exponential family cannot prove binding: the maximum of $4A\alpha /(\mu +2\alpha {)}^2$ is $A/(2\mu )$, so every $E(\alpha )\ge 0$. This does not prove absence of bound states.

The equality case can in fact be strengthened. At $\mu =Am/{\hslash }^2$, use normalized exponentials ${\psi }_{\mu /2},{\psi }_{\mu }$ and write $Q={\hslash }^2{\mu }^2/m$. Their matrix elements areThe Rayleigh numerator for ${\psi }_{\mu /2}+ϵ{\psi }_{\mu }$ is negative for sufficiently small $ϵ>0$, and its norm is nonzero. Hence binding also holds at equality, and by continuity in a small interval beyond it. For arbitrary larger $\mu$, these trials alone do not settle existence; a nonnegative variational upper bound cannot establish a nonnegative ground-state energy.