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Use $\hslash =c=1$ and the mostly-plus Minkowski metric. Then the source's Klein-Gordon equation is $({\partial }_t^2-{\nabla }^2+m^2)\varphi =0$. For a plane wave $\varphi =C{e}^{-i\omega t+i\mathbf{k}\cdot \mathbf{x}}$, substitution givesThus a real propagating wave vector requires $|\omega |\ge m$, or $|E|\ge m{c}^2$ on restoring units. The positive-energy branch has a rest-energy threshold $m{c}^2$. At $|\omega |=m$, $\mathbf{k}=0$; frequencies below the mass threshold give evanescent spatial dependence rather than real-wave-vector propagation. Both signs of $\omega$ occur in the classical equation; a real scalar field combines complex-conjugate modes. Its group velocity is $\mathbf{k}/\omega$ and is subluminal, though the positive-branch phase speed exceeds one when $m>0$.

For a static spherically symmetric scalar field, the equation outside the source isSetting $\psi =r\varphi$ gives ${\psi }^{\prime \prime }-m^2\psi =0$, soDecay at infinity selects $B=0$, the Yukawa potential. It is singular at the origin and describes a point source distributionally; a source-free solution regular at the origin and decaying at infinity must be zero. For $m=0$, $\varphi =A/r+B$, and decay sets the constant $B$ to zero.

For a free massless Weyl spinor, yes: each component obeys the massless Klein-Gordon equation. In $c=\hslash =1$ units, take either chiral Weyl equation, for exampleApply its opposite-sign partner. The Pauli matrix multiplication identity makes the antisymmetric part vanish against the symmetric pair of spatial derivatives, soThus $\boxed{{\partial }_{\mu }{\partial }^{\mu }u=0}$ componentwise. This does not turn the field into a scalar: its components still mix under a Spinor representation of the Lorentz group. The second-order equation is a necessary consequence of the first-order equation, but does not by itself impose the chiral helicity relation.

There is no unconditional massless-wave conclusion from merely being a two-component spinor, or from the null bilinear of part d. For example, $u(t,\mathbf{x})=e^{-i\omega t}(1,0{)}^T$ has a null bilinear at every point, yet $({\partial }_t^2-{\nabla }^2)u=-{\omega }^{2}u\ne 0$ for $\omega \ne 0$. Massive free two-component chiral fields coupled through a mass term instead obey the massive Klein-Gordon equation; interacting fields can have additional terms. The affirmative result therefore uses the free, massless dynamics.

Separate the Klein-Gordon equation as $\varphi (t,x)=e^{\pm i\omega t}X(x)$. The spatial equation is $X^{\prime \prime }+k^{2}X=0$, with ${\omega }^2=k^2+m^2$. The first Dirichlet boundary condition removes the cosine solution, and the second requires $\sin (kL)=0$. ThusThere is no nonzero $n=0$ Dirichlet mode. The real field is built from both complex-conjugate time dependences, with mode functionsThe sine functions satisfy ${\int }_0^L\sin (k_{n}x)\sin (k_{m}x)dx=(L/2){\delta }_{nm}$. Consequently the factor $1/\sqrt{{\omega }_{n}L}$ gives unit Klein-Gordon inner product for each positive-frequency mode and the canonical equal-time field commutator after quantization. A general solution is a superposition of these modes, rather than a claim that a single mode exhausts all solutions.

The Euler-Lagrange equation for the real scalar field givesThus the conjugate momentum and Hamiltonian density areThe spatial Fourier transform reduces the Klein-Gordon equation to ${\ddot{\varphi }}_{\mathbf{p}}+{\omega }_{\mathbf{p}}^2{\varphi }_{\mathbf{p}}=0$, with positive frequency ${\omega }_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$. Hence a general real solution is the mode expansion of a free fieldThe two coefficients are complex conjugates because the field is real. Each pair of initial data $\varphi (0,\mathbf{x}),\pi (0,\mathbf{x})$ fixes these coefficients. The normalization anticipates canonical quantization and the relativistic normalization of one-particle states.

For a free real scalar field in four spacetime dimensions, $\mathcal{L}=(\partial \varphi {)}^2/2-m^2{\varphi }^2/2$ and $\pi =\dot{\varphi }$. Canonical quantization promotes these canonical fields to Hermitian operator-valued distributions with equal-time canonical commutation relationsExpand the solution of the Klein-Gordon equation into positive- and negative-frequency modes:Hermiticity pairs the two coefficients as adjoints. The field commutators are equivalent toFor example, inserting the expansion in $[\varphi ,\dot{\varphi }]$ yields two equal Fourier integrals, each with coefficient $i/2$, and hence $i{\delta }^{(3)}$. The vacuum obeys $a(\mathbf{k})|0⟩=0$, and repeated creation operators generate the symmetric bosonic Fock space. Normal ordering of the quadratic Hamiltonian removes its vacuum constant and gives $H=\int d{\Pi }_k{\omega }_k{a}^{†}a$; the same computation for the spatial momentum gives the stipulated four-momentum $P^{\mu }=\int d{\Pi }_k{k}^{\mu }{a}^{†}a$.

The oscillator algebra now impliesConsequentlyIndeed, ${\partial }^{\mu }{e}^{\mp ikx}=\mp i{k}^{\mu }{e}^{\mp ikx}$. For $\mu =0$, the result is $[H,\varphi ]=-i\dot{\varphi }$, which is the Heisenberg equation of motion with the correct sign. As usual, the field identities are interpreted after smearing the operator-valued distributions, or on a suitable common finite-particle domain.

With inverse-time convention $e^{i\omega t}$, the half-line Klein-Gordon equation ${\sigma }_{tt}-{\sigma }_{rr}+{\mu }^2\sigma =0$ has spatial normal modes $A{e}^{ikr}+B{e}^{-ikr}$, where $k=\sqrt{{\omega }^2-{\mu }^2}>0$ above threshold and $k=i\sqrt{{\mu }^2-{\omega }^2}$ below it. For the branch even in $\omega$, reality above threshold requires $A(\omega )=B(-\omega {)}^{\ast }$. Below threshold boundedness eliminates $B$, and reality requires $A(\omega )=A(-\omega {)}^{\ast }$. At threshold the spatial differential equation instead has the basis $1,r$; boundedness retains only the constant mode. A spectral integral with ordinary integrable functions as amplitudes ignores the isolated threshold points, while distribution amplitudes can represent the surviving monochromatic constant-spatial modes.

A free real scalar field obeys the Klein-Gordon equation and creates neutral spin-zero particles.

For the massless Klein-Gordon equation on a Kerr black hole, the covariant wave operator separates into radial and angular ordinary differential equations. With $K=(r^2+a^2)\omega -am$ and angular separation constant $\Lambda$, the radial equation is $(\Delta R^{\prime }{)}^{\prime }+[K^2/\Delta -a^2{\omega }^2+2am\omega -\Lambda ]R=0$. The angular equation is $(\sin \theta {\Theta }^{\prime }{)}^{\prime }/\sin \theta +[a^2{\omega }^2{\cos }^2\theta -m^2/{\sin }^2\theta +\Lambda ]\Theta =0$. No extra radial factor is included in the displayed ansatz.

For interaction density $g{\varphi }_1{\varphi }_2$, the combinations ${\varphi }_{\pm }={\varphi }_1\pm {\varphi }_2$ diagonalize the equations into independent Klein-Gordon equations with the displayed mass squares. When $0<g<m^2$, all Fourier frequencies are real and nonzero. Initial data in one original field generally excite both normal combinations.

In 1935, Hideki Yukawa proposed a new massive exchange quantum to explain nucleon binding. His paper began with the inadequacy of an electron–neutrino exchange description of the binding strength. It then required a potential energy falling faster with distance than a Coulomb potential, introduced an exponentially decaying field, and related its inverse range to the exchange particle's rest mass. This was a theoretical proposal, not a detection with an apparatus of his own.^[ref]

The crucial idea is that a finite rest mass supplies a length scale even in empty space. In a scalar model, the static limit of a Klein-Gordon equation produces the modified Helmholtz equation. Its decaying Green's function gives the Yukawa potential. The derivation below isolates this range mechanism; one-pion exchange also carries spin and isospin structure.

The nuclear emulsion track experiment used developed photographic emulsions exposed to cosmic rays and inspected microscopically. Stopping pions gave daughter muon tracks with closely grouped ranges. Comparing penetrating muons with pion-induced nuclear breakup distinguished their interactions; this was not a direct static force measurement. The muon is a lepton, not Yukawa's strongly interacting meson.^[ref]