--- title: The Klein-Gordon Equation module: Relativistic Wave Equations moduleNumber: 5 lessonNumber: 1 order: 501 summary: > Quantizing the relativistic energy relation $E^2 = p^2 + m^2$ produces the Klein-Gordon equation for a scalar field. Its plane-wave solutions come in positive- and negative-energy branches, and the conserved density it supplies is not positive-definite — the two difficulties that first drove physicists to seek a first-order equation. The static Klein-Gordon equation with a point source gives the Yukawa potential, and the free equation gives the scalar propagator that later modules attach to exchanged lines. topics: [Relativistic Wave Equations] draft: false sources: - book: Griffiths ref: "Ch. 7 — Quantum Electrodynamics, §7.1 (Klein-Gordon preface to the Dirac equation)" - book: Thomson ref: "§4.1–4.2 (the Klein-Gordon equation and its interpretation)" - book: Halzen & Martin ref: "Ch. 3 (wave equations and the Klein-Gordon propagator)" --- Quantum mechanics and special relativity each work superbly in their own domain, and particle physics needs both at once: the particles are fast, often ultrarelativistic, and they are quantum. The Schrödinger equation is built on the nonrelativistic energy relation $E = p^2/2m$ and cannot be the final word. The natural first attempt is to keep the quantum recipe — promote energy and momentum to operators acting on a wavefunction — but feed it the _relativistic_ energy–momentum relation. That attempt is the Klein-Gordon equation. It is the correct equation for a spinless particle, but reading it as a single-particle wave equation exposes two difficulties, negative energies and a non-positive probability density, that shaped everything that followed. This lesson derives the equation, confronts both difficulties honestly, and extracts two results that the rest of the course uses directly: the Yukawa potential and the scalar propagator. Throughout we use natural units $\hbar = c = 1$ and the metric signature $\eta^{\mu\nu} = \mathrm{diag}(+,-,-,-)$, so that $p^\mu = (E,\vec p)$ and $p_\mu p^\mu = E^2 - \vec p^{\,2} = m^2$ for a free particle of mass $m$. ## From the energy relation to a wave equation The nonrelativistic Schrödinger equation follows from the classical relation $E = \vec p^{\,2}/2m$ by the canonical substitution $$ E \;\to\; i\,\frac{\partial}{\partial t}, \qquad \vec p \;\to\; -i\,\vec\nabla, $$ each operator acting on a wavefunction $\psi(\vec x, t)$. The same substitution applied to the relativistic relation $E^2 = \vec p^{\,2} + m^2$ gives $$ -\frac{\partial^2 \psi}{\partial t^2} = \left(-\nabla^2 + m^2\right)\psi , $$ which rearranges into the **Klein-Gordon equation** $$ \left(\frac{\partial^2}{\partial t^2} - \nabla^2 + m^2\right)\psi = 0 . $$ In manifestly covariant form, writing $\partial_\mu = (\partial_t, \vec\nabla)$ and the d'Alembertian $\Box = \partial_\mu \partial^\mu = \partial_t^2 - \nabla^2$, $$ \left(\Box + m^2\right)\psi = 0 . $$ The equation is Lorentz invariant by construction: $\Box$ is a scalar operator and $m^2$ a scalar, so if $\psi$ is a scalar field the whole equation holds in every inertial frame. This is already an improvement over the Schrödinger equation, which treats time and space asymmetrically (first order in $t$, second order in $\vec x$) and cannot be covariant. The price is that the Klein-Gordon equation is _second_ order in time, and that turns out to matter. Historically the equation is older than its name suggests: Schrödinger wrote it down first, in 1925, before settling on his nonrelativistic equation, and discarded it because it gave the wrong fine structure for hydrogen (it omits spin) and because of the difficulties below. Klein and Gordon published it in 1926.[^kg-history] ## Plane-wave solutions and the two energy branches Because the equation has constant coefficients, plane waves solve it. Try $$ \psi(\vec x, t) = N\, e^{-i(E t - \vec p\cdot \vec x)} = N\, e^{-i\,p_\mu x^\mu}. $$ Substituting, each $\partial_t$ brings down $-iE$ and each $\nabla$ brings down $i\vec p$, so the Klein-Gordon operator returns $$ \left(-E^2 + \vec p^{\,2} + m^2\right)\psi = 0 \quad\Longrightarrow\quad E^2 = \vec p^{\,2} + m^2 . $$ The equation is satisfied precisely when $E$ and $\vec p$ lie on the mass shell — exactly the relativistic dispersion relation we started from, as it must be. But $E^2 = \vec p^{\,2} + m^2$ is quadratic in $E$, so for each momentum $\vec p$ there are **two** solutions, $$ E = \pm E_p , \qquad E_p \equiv +\sqrt{\vec p^{\,2} + m^2} \;\ge\; m . $$ The positive branch is the physical energy of a particle. The negative branch is the difficulty: it describes states of arbitrarily large negative energy, with no lower bound. A single particle coupled to any environment could cascade down the negative branch forever, radiating energy without limit. There is no way to discard the negative-energy solutions by hand, because a complete set of solutions is needed to expand a general field, and the positive-energy solutions alone are not complete. $$ % caption: The Klein-Gordon dispersion relation E^2 = p^2 + m^2 has two branches. % The upper branch E = +sqrt(p^2+m^2) is the particle energy, bounded below by the % rest energy m. The lower branch E = -sqrt(p^2+m^2) is unbounded below — the % negative-energy problem. A gap of width 2m separates them at p = 0. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % axes \draw[->, black!70] (-3.4,0) -- (3.4,0) node[right, font=\scriptsize] {momentum}; \draw[->, black!70] (0,-3.0) -- (0,3.0) node[above, font=\scriptsize] {energy}; % positive branch y = sqrt(x^2 + 1) (m=1) \draw[very thick] plot[domain=-3:3, samples=80] (\x, {sqrt(\x*\x + 1)}); % negative branch \draw[acc, very thick, dashed] plot[domain=-3:3, samples=80] (\x, {-sqrt(\x*\x + 1)}); % rest-energy marks \draw[black] (-0.12,1) -- (0.12,1); \draw[black] (-0.12,-1) -- (0.12,-1); \node[black, right, font=\scriptsize] at (0.15,1) {rest energy}; \node[black, right, font=\scriptsize] at (0.15,-1.35) {minus rest energy}; % gap bracket \draw[black, <->] (2.6,-1) -- (2.6,1); \node[black, right, font=\scriptsize] at (2.55,0) {gap}; \node[above, font=\scriptsize] at (-2.3, 2.5) {particle branch}; \node[acc, below, font=\scriptsize] at (-2.3,-2.5) {negative branch}; \end{tikzpicture} $$ ## The probability-density difficulty For the Schrödinger equation the quantity $\rho = |\psi|^2$ is a positive probability density satisfying a continuity equation, and its integral is conserved. The relativistic equation supplies a conserved current too, but the density is not positive. Multiply the Klein-Gordon equation by $\psi^\ast$, multiply its complex conjugate by $\psi$, and subtract. The mass terms cancel and the result is a continuity equation $\partial_\mu j^\mu = 0$ with $$ j^\mu = i\left(\psi^\ast \partial^\mu \psi - \psi\, \partial^\mu \psi^\ast\right), $$ whose time component, after restoring the conventional normalization, is the density $$ \rho = j^0 = i\left(\psi^\ast \frac{\partial \psi}{\partial t} - \psi\, \frac{\partial \psi^\ast}{\partial t}\right). $$ This $\rho$ involves a _first_ time derivative — forced on us because the equation is second order, so $\psi$ and $\dot\psi$ can be specified independently at an initial time. For a plane wave $\psi = N e^{-i(Et - \vec p\cdot\vec x)}$ it evaluates to $$ \rho = 2 |N|^2 E, $$ which is proportional to $E$ and therefore **negative for the negative-energy branch**. A quantity that can be negative cannot be a probability density. This is the second difficulty, and it is tied to the first: the same negative-energy solutions that are unbounded below also carry negative density. $$ % caption: The Klein-Gordon density rho = 2|N|^2 E is positive on the particle % branch and negative on the negative-energy branch. As a single-particle % probability density this is a contradiction; the modern reading takes rho as a % charge density, whose sign labels particle versus antiparticle. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black!70] (0,-2.2) -- (0,2.4) node[above, font=\scriptsize] {density}; \draw[->, black!70] (-3.4,0) -- (3.4,0) node[right, font=\scriptsize] {energy branch}; % positive bar \draw[very thick] (0.8,0) rectangle (2.2,1.7); \node[above, font=\scriptsize] at (1.5,1.7) {positive}; \node[below, font=\scriptsize] at (1.5,-0.05) {particle}; % negative bar \draw[acc, very thick, fill=acc!12] (-2.2,0) rectangle (-0.8,-1.7); \node[acc, below, font=\scriptsize] at (-1.5,-1.7) {negative}; \node[above, font=\scriptsize] at (-1.5,0.05) {negative branch}; \end{tikzpicture} $$ The way out, understood only after quantum field theory, is to stop reading $\psi$ as a single-particle wavefunction and stop reading $\rho$ as a probability. In the field interpretation $\psi$ is a field operator, the negative-energy solutions are reinterpreted as positive-energy antiparticles (the subject of the third lesson of this module), and $\rho$ becomes a **charge** density whose sign distinguishes a particle from its antiparticle. A negative charge density is not a contradiction. For now we keep the equation and read its solutions with this reinterpretation in mind; the negative-energy branch is not a defect to be discarded but the seed of antimatter. ## Why a first-order equation was sought Both difficulties can be traced to the same root: the equation is second order in time. The negative-energy branch appears because $E^2 = \dots$ has two roots; the indefinite density appears because a second-order equation needs $\dot\psi$ as independent initial data, forcing a density that is first order in $\partial_t$ and hence linear in $E$. Dirac's response, taken up in the next lesson, was to demand an equation _first_ order in time — and, for Lorentz invariance, first order in space as well: $$ i\,\frac{\partial \psi}{\partial t} = \left(\vec\alpha \cdot \vec p + \beta m\right)\psi . $$ A first-order-in-time equation admits a positive-definite density $\rho = \psi^\dagger \psi$, curing the probability problem. The cost is that the coefficients $\vec\alpha, \beta$ cannot be numbers — squaring the equation must reproduce $E^2 = \vec p^{\,2} + m^2$, and that forces $\vec\alpha$ and $\beta$ to be anticommuting matrices, which drags in multi-component wavefunctions and, with them, spin. The negative-energy solutions do _not_ disappear in the Dirac equation; they return in a form (spin-$\tfrac12$ antiparticles) that the field interpretation handles cleanly. The Klein-Gordon equation is not wrong — it is the right equation for a spin-$0$ particle such as the pion or the Higgs — but it is not the wavefunction equation for the electron, and its difficulties are what made that clear. ## The Yukawa potential as a static solution The Klein-Gordon equation earns its place in this course through a static solution. Consider a heavy point source at the origin emitting a scalar field, so that $\psi$ is time-independent and the equation reduces to $$ \left(-\nabla^2 + m^2\right)\psi(\vec r) = g\,\delta^3(\vec r), $$ with $g$ the coupling of the source. This is the static Klein-Gordon equation with a point source. For a spherically symmetric field, $\nabla^2 = \frac{1}{r} \frac{\d^2}{\d r^2}(r\,\cdot)$, and away from the origin the equation reads $$ \frac{1}{r}\frac{\d^2}{\d r^2}\bigl(r\,\psi\bigr) = m^2\,\psi , $$ whose decaying solution is $$ \psi(r) = -\frac{g}{4\pi}\,\frac{e^{-m r}}{r}. $$ This is the **Yukawa potential**. Two features are decisive. First, the massless limit $m \to 0$ returns the Coulomb potential $\psi \propto 1/r$: a massless mediator gives a long-range, inverse-square force, exactly electromagnetism. Second, a massive mediator gives an exponentially screened force with range $$ R = \frac{1}{m} = \frac{\hbar}{m c}, $$ the mediator's Compton wavelength. This is precisely the relation Yukawa used to predict the pion: a nuclear force of range $\sim 1.4$ fm requires a mediator of mass $\sim 140$ MeV. The Klein-Gordon equation is the field equation whose static Green's function _is_ the Yukawa potential, so this course's very first force-range argument is a Klein-Gordon calculation. $$ % caption: The Yukawa potential exp(-mr)/r (accent, solid) against the Coulomb % potential 1/r (dashed). Both diverge at the origin, but the massive mediator % screens the field beyond a range of order 1/m, while the massless Coulomb tail % falls off only as a power. The screening length shrinks as the mediator mass % grows. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black!70] (0,0) -- (5.6,0) node[right, font=\scriptsize] {separation}; \draw[->, black!70] (0,0) -- (0,4.0) node[above, font=\scriptsize] {potential}; % Coulomb 1/r, scaled; start at r=0.35 to avoid blow-up \draw[black, very thick, dashed] plot[domain=0.35:5.2, samples=90] (\x, {1.1/\x}); % Yukawa exp(-m r)/r with m ~ 1.1 \draw[acc, very thick] plot[domain=0.35:5.2, samples=90] (\x, {1.1*exp(-1.05*\x)/\x}); \node[black, right, font=\scriptsize] at (2.6,0.55) {Coulomb tail}; \node[acc, right, font=\scriptsize] at (1.35,1.4) {Yukawa}; % range marker \draw[black, dashed] (0.95,0) -- (0.95,1.2); \node[black, below, font=\scriptsize] at (0.95,0) {range}; \end{tikzpicture} $$ ## The scalar propagator The free Klein-Gordon equation also supplies the object that Feynman diagrams attach to an internal scalar line: the **propagator**. It is the Green's function of the Klein-Gordon operator, the field produced by a unit point disturbance in spacetime, $$ \left(\Box + m^2\right)\Delta(x) = -\,\delta^4(x). $$ Fourier transforming to momentum space, $\Box \to -p^2$ (with $p^2 = p_\mu p^\mu = E^2 - \vec p^{\,2}$), turns the differential equation into an algebraic one, and $$ \tilde\Delta(p) = \frac{1}{p^2 - m^2}. $$ This is the momentum-space scalar propagator. It carries two lessons that recur throughout the Feynman calculus. First, the internal line is **off shell**: a propagating virtual particle need not satisfy $p^2 = m^2$, and the propagator is large precisely when it is _near_ shell, $p^2 \approx m^2$, which is why exchanged particles "prefer" to be nearly real. Second, the pole at $p^2 = m^2$ must be handled with the Feynman $+i\epsilon$ prescription, $1/(p^2 - m^2 + i\epsilon)$, which encodes the causal ordering of the two energy branches — positive-energy solutions propagating forward in time, negative-energy solutions backward — and so ties the propagator directly to the antiparticle interpretation. The exchange of a scalar between two currents, with a factor $1/(p^2 - m^2)$ on the internal line, reproduces the Yukawa potential in the static limit, closing the circle between the two results of this lesson. $$ % caption: A scalar exchanged between two vertices. The internal line carries the % propagator 1/(p^2 - m^2) and momentum q = p_A - p_C; it is virtual (off shell). % In the static limit this single-scalar exchange reproduces the Yukawa potential. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % lower vertex \fill (0,0) circle (2pt); % upper vertex \fill (0,2.4) circle (2pt); % external legs lower \draw[very thick] (-1.8,-1.2) -- (0,0); \draw[very thick] (0,0) -- (1.8,-1.2); \node[left, font=\scriptsize] at (-1.8,-1.2) {in}; \node[right, font=\scriptsize] at (1.8,-1.2) {out}; % external legs upper \draw[very thick] (-1.8,3.6) -- (0,2.4); \draw[very thick] (0,2.4) -- (1.8,3.6); \node[left, font=\scriptsize] at (-1.8,3.6) {in}; \node[right, font=\scriptsize] at (1.8,3.6) {out}; % internal scalar line (dashed = scalar) \draw[acc, very thick, dashed] (0,0) -- (0,2.4); \node[acc, right, font=\scriptsize] at (0.15,1.2) {propagator}; \end{tikzpicture} $$ ## Summary Quantizing $E^2 = \vec p^{\,2} + m^2$ gives the Lorentz-invariant Klein-Gordon equation $(\Box + m^2)\psi = 0$ for a scalar field. Its plane waves come in two branches $E = \pm E_p$, and the conserved density $\rho = 2|N|^2 E$ changes sign between them — the negative-energy and negative-probability difficulties that sank the naive single-particle reading and motivated Dirac's first-order equation. Neither difficulty is fatal once $\psi$ is a quantum field and $\rho$ a charge density: the negative-energy branch becomes antimatter. For spinless particles the equation stands on its own, and it delivers two workhorses of the rest of the course — the Yukawa potential $e^{-mr}/r$ as its static point-source solution, and the scalar propagator $1/(p^2 - m^2)$ as its momentum-space Green's function. [^kg-history]: Griffiths, _Introduction to Elementary Particles_, 2nd ed., §7.1, presents the Klein-Gordon equation as the natural but flawed first attempt that motivates the Dirac equation; Thomson, _Modern Particle Physics_, §4.1–4.2, derives the plane-wave solutions and the indefinite density; Halzen & Martin, _Quarks and Leptons_, Ch. 3, gives the propagator and the Yukawa Green's function. Masses and the natural-unit conversion $\hbar c = 197.3$ MeV·fm follow the Particle Data Group, _Review of Particle Physics_, [pdg.lbl.gov](https://pdg.lbl.gov).