Klein-Gordon Field from Its Lagrangian

Derivation

Klein-Gordon Field from Its Lagrangian

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Statement

For a single real scalar field \(\varphi(x)\) with Lorentz-invariant Lagrangian density \(\mathcal{L}=\tfrac{1}{2}\,\partial_\mu\varphi\,\partial^\mu\varphi-\tfrac{1}{2}m^2\varphi^2\), the field Euler–Lagrange equation yields the Klein–Gordon equation \((\Box+m^2)\varphi=0\), and substituting a plane wave \(\varphi\propto e^{-ik_\mu x^\mu}\) gives the relativistic dispersion relation \(\omega^2=\mathbf{k}^2+m^2\) (natural units \(\hbar=c=1\); equivalently \(E^2=\mathbf{p}^2c^2+m^2c^4\)).

Why it matters

The Klein–Gordon equation is the simplest relativistic field equation and the template for every free field in physics: the photon, the electron (through the Dirac square), and the Higgs boson all satisfy it component-by-component. It is the direct field-theoretic statement of Einstein's mass–energy–momentum relation \(E^2=p^2c^2+m^2c^4\), promoted from a single particle to a field filling spacetime.

Deriving it from a Lagrangian rather than postulating it fixes the theory's content: the mass term \(m^2\varphi^2\) is the only local, Lorentz-scalar, quadratic potential available, so the free relativistic scalar is essentially unique. Everything that follows — quantisation into particles of mass \(m\), the propagator, the range of the Yukawa force \(\sim 1/m\) — is dictated by these two terms.

Assumptions

The field is a real Lorentz scalar.If \(\varphi\) carried an index (vector, spinor) the free Lagrangian and its equation of motion would differ (Proca, Dirac); scalarity is what makes \(\partial_\mu\varphi\,\partial^\mu\varphi\) the unique kinetic term. Flat Minkowski spacetime, metric \(\eta_{\mu\nu}=\mathrm{diag}(+,-,-,-)\).On curved spacetime \(\partial_\mu\to\nabla_\mu\) and \(\Box\to\Box_g=\tfrac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}\,g^{\mu\nu}\partial_\nu)\), plus a possible curvature coupling \(\xi R\varphi^2\); the plane-wave dispersion no longer holds globally. The Lagrangian is exactly quadratic (free field, no interactions).Adding \(\lambda\varphi^4\) or a source \(J\varphi\) makes the equation nonlinear/inhomogeneous, \((\Box+m^2)\varphi=-\lambda\varphi^3+J\), and destroys the exact plane-wave dispersion. The action is stationary under variations \(\delta\varphi\) that vanish on the spatial boundary at infinity.If \(\delta\varphi\) does not vanish on \(\partial V\), the surface term from integration by parts survives and the Euler–Lagrange equation is not recovered; well-posedness requires fields decaying (or periodic) at infinity. \(m^2>0\).If \(m^2<0\) the "mass" term is a wrong-sign potential: the \(\varphi=0\) vacuum is unstable, low-\(k\) modes grow exponentially (tachyonic instability), and the field rolls to a new minimum — spontaneous symmetry breaking, not free propagation.

Derivation

\[\mathcal{L}=\tfrac{1}{2}\,\eta^{\mu\nu}\,\partial_\mu\varphi\,\partial_\nu\varphi-\tfrac{1}{2}m^2\varphi^2,\qquad S=\int d^4x\,\mathcal{L}.\]

Write the kinetic term with the metric explicit so the two derivative factors are visibly symmetric under \(\mu\leftrightarrow\nu\); the dynamics come from demanding \(\delta S=0\). A

\[\frac{\partial\mathcal{L}}{\partial(\partial_\alpha\varphi)}=\tfrac{1}{2}\,\eta^{\mu\nu}\big(\delta^\alpha_\mu\,\partial_\nu\varphi+\partial_\mu\varphi\,\delta^\alpha_\nu\big)=\tfrac{1}{2}\big(\partial^\alpha\varphi+\partial^\alpha\varphi\big)=\partial^\alpha\varphi.\]

Differentiate the kinetic term treating \(\partial_\alpha\varphi\) as an independent variable. Both factors depend on it, giving two identical contributions; the symmetry of \(\eta^{\mu\nu}\) is exactly what cancels the \(\tfrac12\). B

\[\frac{\partial\mathcal{L}}{\partial\varphi}=-m^2\varphi.\]

Only the potential term \(-\tfrac12 m^2\varphi^2\) depends on \(\varphi\) itself (not its derivatives); ordinary differentiation of a quadratic. A

\[\partial_\alpha\!\left(\frac{\partial\mathcal{L}}{\partial(\partial_\alpha\varphi)}\right)-\frac{\partial\mathcal{L}}{\partial\varphi}=0\;\;\Longrightarrow\;\;\partial_\alpha\partial^\alpha\varphi+m^2\varphi=0.\]

Insert Steps 2–3 into the field Euler–Lagrange equation (the assumed prior result), obtained by requiring \(\delta S=0\) with \(\delta\varphi\) vanishing on the boundary. B

\[(\Box+m^2)\varphi=0,\qquad \Box\equiv\partial_\alpha\partial^\alpha=\frac{\partial^2}{\partial t^2}-\nabla^2.\]

Define the d'Alembertian \(\Box=\partial_\alpha\partial^\alpha\); with signature \((+,-,-,-)\) it expands to \(\partial_t^2-\nabla^2\). This is the Klein–Gordon equation in operator form. A

\[\varphi(x)=A\,e^{-ik_\mu x^\mu}=A\,e^{-i(\omega t-\mathbf{k}\cdot\mathbf{x})},\qquad \partial_\alpha\varphi=-ik_\alpha\,\varphi.\]

Try a plane-wave mode. Because the equation is linear with constant coefficients, each Fourier mode evolves independently, so a single \(k_\mu=(\omega,\mathbf{k})\) is an exact ansatz. B

\[\Box\varphi=(-ik_\alpha)(-ik^\alpha)\varphi=-k_\alpha k^\alpha\,\varphi=-(\omega^2-\mathbf{k}^2)\varphi.\]

Apply \(\partial_\alpha\to-ik_\alpha\) twice and contract with the metric: \(k_\alpha k^\alpha=\omega^2-\mathbf{k}^2\). Pure algebra on the exponential. A

\[\big[-(\omega^2-\mathbf{k}^2)+m^2\big]\varphi=0\;\;\Longrightarrow\;\;\boxed{\;\omega^2=\mathbf{k}^2+m^2\;}\]

For a nontrivial mode \(\varphi\neq0\), the bracket must vanish. Solving for \(\omega^2\) gives the dispersion relation; only these \((\omega,\mathbf{k})\) pairs propagate. A

Result

\[(\Box+m^2)\varphi=0\qquad\Longleftrightarrow\qquad \omega^2=\mathbf{k}^2+m^2\quad\big(E^2=\mathbf{p}^2c^2+m^2c^4\big)\]

Reading. The scalar Lagrangian forces every Fourier mode of the field to oscillate on a fixed "mass shell": its frequency and wavevector are locked by \(\omega^2=\mathbf{k}^2+m^2\). The kinetic term \(\tfrac12(\partial\varphi)^2\) supplies the \(\Box\) (the \(\omega^2-\mathbf{k}^2\)); the mass term supplies the constant \(m^2\) that gives even a mode at rest (\(\mathbf{k}=0\)) a nonzero frequency \(\omega=m\). In quantum language, restoring \(\hbar,c\), this is exactly \(E=\hbar\omega\), \(\mathbf{p}=\hbar\mathbf{k}\) plugged into Einstein's relation: the field's normal modes are relativistic particles of rest mass \(m\).

Units check. In natural units \([\varphi]=\text{mass}\), \([\partial_\mu]=\text{mass}\), so \([\partial_\mu\varphi\,\partial^\mu\varphi]=[m^2\varphi^2]=\text{mass}^4=[\mathcal{L}]\), consistent with \([S]=[\mathcal{L}]\cdot[d^4x]=\text{mass}^4\cdot\text{mass}^{-4}=1\). In the dispersion, restoring constants gives \(\hbar^2\omega^2=\hbar^2\mathbf{k}^2c^2+m^2c^4\): each term has units \((\text{J})^2\), since \([\hbar\omega]=\text{J}\), \([\hbar k c]=\text{J s}\cdot\text{m}^{-1}\cdot\text{m s}^{-1}=\text{J}\), \([mc^2]=\text{J}\). Dimensionally homogeneous.

Limiting cases

Massless limit \(m\to0\): \((\Box)\varphi=0\), the ordinary relativistic wave equation; dispersion \(\omega=|\mathbf{k}|\) (light-like), phase and group speed both \(=c\).
Rest / long-wavelength \(\mathbf{k}\to0\): \(\ddot\varphi=-m^2\varphi\), a spatially uniform field oscillating at \(\omega=m\) — the Compton frequency \(mc^2/\hbar\), the minimum energy quantum.
Non-relativistic \(|\mathbf{k}|\ll m\): \(\omega=m\sqrt{1+\mathbf{k}^2/m^2}\approx m+\tfrac{\mathbf{k}^2}{2m}\); subtracting the rest term recovers the Schrödinger dispersion \(E_{\text{kin}}=p^2/2m\).
Ultra-relativistic \(|\mathbf{k}|\gg m\): \(\omega\approx|\mathbf{k}|+\tfrac{m^2}{2|\mathbf{k}|}\to|\mathbf{k}|\); the mass is a small correction and the field behaves nearly like a massless one.

Breaks when

Interactions or self-coupling are present. With \(\mathcal{L}_{\text{int}}=-\tfrac{\lambda}{4!}\varphi^4\) the equation becomes \((\Box+m^2)\varphi=-\tfrac{\lambda}{3!}\varphi^3\): nonlinear, modes mix, and \(\omega^2=\mathbf{k}^2+m^2\) is only the free (asymptotic) dispersion, corrected by a self-energy \(m^2\to m^2+\Pi(k)\).
The medium or spacetime is not the vacuum. In a dispersive medium, a plasma, or curved/expanding spacetime, \(\Box\) acquires position-dependent coefficients; the dispersion becomes \(\omega^2=\mathbf{k}^2+m^2_{\text{eff}}(\mathbf{x},t)\) and can even develop a gap or cutoff (e.g. plasma frequency).
Wrong-sign mass \(m^2<0\). Then \(\omega^2=\mathbf{k}^2-|m^2|<0\) for \(|\mathbf{k}|<|m|\): \(\omega\) is imaginary, the mode grows exponentially rather than oscillating, and the plane-wave interpretation fails (tachyonic instability / symmetry breaking).
Boundaries or finite volume. With walls or periodic boxes, \(\mathbf{k}\) is quantised and boundary/surface terms matter; the continuous mass-shell is replaced by a discrete spectrum.

Failure modes

Sign error in \(\Box\). Writing \(\Box=\nabla^2-\partial_t^2\) (opposite convention) or mixing signatures gives \((-\Box+m^2)\varphi=0\) and a spurious sign in the dispersion. Fix the metric signature once and stay with it.
Losing the factor of 2 in Step 2. Forgetting that both derivative factors in \(\partial_\mu\varphi\,\partial^\mu\varphi\) depend on \(\partial_\alpha\varphi\) gives \(\tfrac12\partial^\alpha\varphi\) and a wrong mass term. The symmetry of the two factors restores the 2.
Differentiating \(\mathcal{L}\) with respect to \(\varphi\) and \(\partial_\mu\varphi\) as if they were dependent. In the Lagrangian formalism they are treated as independent field arguments; conflating them scrambles the Euler–Lagrange step.
Reading \(m\) as a wavenumber without \(\hbar,c\). In SI the mass term is \((mc/\hbar)^2\varphi^2\); the quantity playing the role of "\(m\)" in \(\omega^2=k^2+m^2\) is the inverse Compton length \(mc/\hbar\), not the kilogram value.
Claiming superluminal phase velocity violates relativity. \(v_p=\omega/k>c\) for massive fields is real but carries no signal; information travels at \(v_g=c^2k/\omega<c\). Confusing the two is a classic error.
Dropping the negative-frequency solutions. \(\omega^2=\mathbf{k}^2+m^2\) has two roots \(\omega=\pm E_{\mathbf{k}}\); both are needed for a real field and for the correct field expansion (particles and antiparticles).

Discussion

The derivation shows that "mass" in field theory is not an intrinsic mechanical inertia but a term in the potential energy of the field: \(\tfrac12 m^2\varphi^2\) is a harmonic restoring force per unit volume for the field amplitude. Each spatial Fourier mode is an independent oscillator whose stiffness is \(\mathbf{k}^2+m^2\), so it rings at \(\omega=\sqrt{\mathbf{k}^2+m^2}\). Quantising these oscillators (raising/lowering operators) turns each quantum of a mode into a relativistic particle with energy \(E_{\mathbf{k}}=\hbar\omega\) and momentum \(\hbar\mathbf{k}\); the dispersion relation is the particle's energy–momentum relation. This is the bridge from classical field to particle content.

The mass sets a length scale, the reduced Compton wavelength \(\lambda_C=\hbar/(mc)=1/m\). It governs the range of the static force mediated by the field: the time-independent Klein–Gordon equation \((\nabla^2-m^2)\varphi=-\rho\) has the Yukawa solution \(\varphi\sim e^{-mr}/r\), so a massive scalar produces a short-range force screened beyond \(\sim\lambda_C\), while \(m\to0\) restores the long-range \(1/r\) Coulomb-like potential. Yukawa used exactly this to predict the pion mass from the range of the nuclear force.

The Lagrangian route also exposes the symmetries by construction. \(\mathcal{L}\) is a Lorentz scalar, so the equation of motion and the dispersion are automatically Lorentz-covariant: \(k_\mu k^\mu=m^2\) is a frame-independent statement (the mass shell). For a real field \(\mathcal{L}\) has no continuous internal symmetry, but a complex scalar \(\mathcal{L}=\partial_\mu\varphi^*\partial^\mu\varphi-m^2|\varphi|^2\) is invariant under \(\varphi\to e^{i\alpha}\varphi\), and Noether's theorem then delivers a conserved current — the charge that distinguishes particle from antiparticle.

Historically the Klein–Gordon equation was Schrödinger's first (relativistic) attempt at a wave equation; he discarded it because, read as a single-particle probability equation, it gives a non-positive-definite density \(\rho=\tfrac{i}{2m}(\varphi^*\dot\varphi-\dot\varphi^*\varphi)\) and admits negative-energy solutions. The resolution is not to read \(\varphi\) as a wavefunction at all: promoted to a quantum field, the negative-frequency modes are creation operators for antiparticles, the "probability density" becomes a conserved charge density (which may legitimately be negative), and unitarity is restored in Fock space. The equation is perfectly consistent as a field equation; it was only inconsistent as a relativistic single-particle quantum mechanics.

Common misconceptions. (i) The Klein–Gordon equation is not "the relativistic Schrödinger equation" for one particle — that reading fails; it is a field equation. (ii) \(\omega^2=\mathbf{k}^2+m^2\) does not mean the scalar travels faster than light despite \(v_p>c\); signals ride the group/front velocity \(\le c\). (iii) The mass term is a potential, not a friction/damping term — it makes modes oscillate faster, never decay.

Worked examples

Compton frequency and range of a neutral pion, \(m_{\pi^0}c^2=135\ \text{MeV}\).

A pion is (to first approximation) a scalar field of mass \(m\). Find the rest-frame oscillation frequency \(\omega_0=mc^2/\hbar\) (the \(\mathbf{k}=0\) mode) and the range \(\lambda_C=\hbar/(mc)\) of the Yukawa force it mediates. B

\[\omega_0=\frac{mc^2}{\hbar}=\frac{135\ \text{MeV}}{6.582\times10^{-22}\ \text{MeV s}}=2.05\times10^{23}\ \text{s}^{-1}.\]

Set \(\mathbf{k}=0\) in \(\omega^2=\mathbf{k}^2+(mc^2/\hbar)^2\); use \(\hbar=6.582\times10^{-22}\ \text{MeV s}\). Symbols first, then numbers. A

\[\lambda_C=\frac{\hbar}{mc}=\frac{\hbar c}{mc^2}=\frac{197.3\ \text{MeV fm}}{135\ \text{MeV}}=1.46\ \text{fm}.\]

Use the convenient constant \(\hbar c=197.3\ \text{MeV fm}\); the MeV cancel leaving femtometres — the observed scale of the nuclear force. A

\[\omega_0\approx2.05\times10^{23}\ \text{s}^{-1},\qquad \lambda_C\approx1.46\ \text{fm}.\]

Reading. Even at rest the pion field cannot be static: it rings at \(\sim10^{23}\) Hz, the frequency of its rest energy. Its mass caps the nuclear force at \(\sim1.5\) fm, matching the known short range of the strong interaction between nucleons.

Units check. \([\hbar c/(mc^2)]=\text{MeV fm}/\text{MeV}=\text{fm}\); \([mc^2/\hbar]=\text{MeV}/(\text{MeV s})=\text{s}^{-1}\). Correct.

Phase and group velocity of a massive scalar mode with \(m c^2=0.500\ \text{MeV}\) carrying momentum \(pc=0.500\ \text{MeV}\) (\(pc=\hbar k c\)).

Compute \(\omega\) (via \(E=\hbar\omega\)), then \(v_p=\omega/k=E/p\) and \(v_g=d\omega/dk=c^2 p/E\), and verify \(v_p v_g=c^2\). B

\[E=\sqrt{(pc)^2+(mc^2)^2}=\sqrt{0.500^2+0.500^2}\ \text{MeV}=0.707\ \text{MeV}.\]

Directly from the dispersion \(\omega^2=\mathbf{k}^2+m^2\) multiplied by \(\hbar^2\): \(E^2=(pc)^2+(mc^2)^2\). Symbols then numbers. A

\[v_p=\frac{\omega}{k}=\frac{E}{p}=\frac{Ec}{pc}=\frac{0.707}{0.500}\,c=1.41\,c,\qquad v_g=\frac{c^2 p}{E}=\frac{pc}{E}\,c=\frac{0.500}{0.707}\,c=0.707\,c.\]

\(v_p=\omega/k\) and \(v_g=d\omega/dk\); differentiating \(\omega=\sqrt{k^2+m^2}\) gives \(v_g=k/\omega=c^2p/E\). B

\[v_p\,v_g=\frac{E}{p}\cdot\frac{c^2 p}{E}=c^2\quad\checkmark\qquad(1.41\times0.707=1.00).\]

The product is identically \(c^2\) for any \(m,k\), a hallmark of the relativistic mass shell. A

\[v_p=1.41\,c,\qquad v_g=0.707\,c,\qquad v_p v_g=c^2.\]

Reading. The phase velocity exceeds \(c\) — permissible, since a pure infinite plane wave carries no information. The group velocity, which transports energy and signals, is \(0.707\,c<c\), respecting causality. Their product is exactly \(c^2\).

Units check. \([E/p]=\text{MeV}/(\text{MeV}/c)=c\); \([c^2 p/E]=c^2\cdot(\text{MeV}/c)/\text{MeV}=c\). Both velocities in units of \(c\); their product in \(c^2\). Correct.

Problems

(A) Direct verification. Show by substitution that \(\varphi=A\cos(\omega t-\mathbf{k}\cdot\mathbf{x})\) solves \((\Box+m^2)\varphi=0\) provided \(\omega^2=\mathbf{k}^2+m^2\). State what constraint (if any) applies to the amplitude \(A\).
Solution
Compute \(\partial_t^2\varphi=-\omega^2\varphi\) and \(\nabla^2\varphi=-\mathbf{k}^2\varphi\), so \(\Box\varphi=\partial_t^2\varphi-\nabla^2\varphi=(-\omega^2+\mathbf{k}^2)\varphi\). Then \((\Box+m^2)\varphi=(-\omega^2+\mathbf{k}^2+m^2)\varphi\). This vanishes for all \((t,\mathbf{x})\) iff \(\omega^2=\mathbf{k}^2+m^2\). The equation is linear and homogeneous, so \(A\) is arbitrary (unconstrained) — no amplitude condition.
(A) Massless limit. Take \(m=0\) and find the phase speed of \(\varphi=A\,e^{-i(\omega t-kx)}\) in SI units. Interpret.
Solution
With \(m=0\), the SI dispersion \(\omega^2=c^2k^2+(mc^2/\hbar)^2\) reduces to \(\omega=ck\). Phase speed \(v_p=\omega/k=c\), independent of \(k\) (no dispersion). The massless real scalar propagates exactly like light: a single universal speed, wave equation \(\Box\varphi=0\).
(B) Compton wavelength. The Higgs boson has \(m_H c^2=125\ \text{GeV}\). Compute its reduced Compton wavelength \(\lambda_C=\hbar/(m_Hc)\) in metres, and its rest-frame frequency \(\omega_0\).
Solution
\(\lambda_C=\hbar c/(m_Hc^2)=197.3\ \text{MeV fm}/(1.25\times10^{5}\ \text{MeV})=1.58\times10^{-3}\ \text{fm}=1.58\times10^{-18}\ \text{m}\). Frequency \(\omega_0=m_Hc^2/\hbar=1.25\times10^{5}\ \text{MeV}/(6.582\times10^{-22}\ \text{MeV s})=1.90\times10^{26}\ \text{s}^{-1}\). The huge mass gives an extremely short range and correspondingly high frequency.
(B) Non-relativistic reduction. For \(|\mathbf{k}|\ll m\), expand \(\omega=\sqrt{\mathbf{k}^2+m^2}\) to first order beyond the rest term and identify the effective non-relativistic kinetic energy. Restore \(\hbar,c\).
Solution
\(\omega=m\sqrt{1+\mathbf{k}^2/m^2}\approx m+\dfrac{\mathbf{k}^2}{2m}\). Multiply by \(\hbar\) and restore \(c\): \(E=\hbar\omega\approx mc^2+\dfrac{\hbar^2\mathbf{k}^2}{2m}=mc^2+\dfrac{\mathbf{p}^2}{2m}\), with \(\mathbf{p}=\hbar\mathbf{k}\). Subtracting the constant rest energy \(mc^2\) leaves \(E_{\text{kin}}=p^2/2m\), the Schrödinger dispersion — the Klein–Gordon field's low-energy limit reproduces non-relativistic quantum mechanics.
(C) Source and Yukawa potential. A static point source gives the time-independent equation \((\nabla^2-m^2)\varphi=-g\,\delta^3(\mathbf{x})\). Show the spherically symmetric solution is \(\varphi(r)=\dfrac{g}{4\pi r}e^{-mr}\), and state the range.
Solution
For \(r>0\), spherical symmetry gives \(\nabla^2\varphi=\dfrac{1}{r}\dfrac{d^2}{dr^2}(r\varphi)\). Let \(u=r\varphi\); then \(\dfrac{d^2u}{dr^2}-m^2u=0\), with decaying solution \(u=Ce^{-mr}\), so \(\varphi=Ce^{-mr}/r\). Fix \(C\) by the source: near \(r\to0\), \(e^{-mr}\to1\) and \(\varphi\to C/r\), whose Laplacian is \(-4\pi C\,\delta^3(\mathbf{x})\); matching to \(-g\,\delta^3\) gives \(C=g/4\pi\). Thus \(\varphi(r)=\dfrac{g}{4\pi r}e^{-mr}\). The exponential screens the potential beyond the range \(\lambda_C=1/m=\hbar/(mc)\); as \(m\to0\) it becomes the long-range Coulomb form \(g/(4\pi r)\).