03 Jun 2026
Klein–Gordon Equation
Relativistic scalar wave equation, conserved current, and the limits of its single-particle interpretation.
The Schrödinger equation treats time and space differently and uses the nonrelativistic relation $E=p^2/(2m)$. A relativistic wave equation should instead reproduce
\[E^2=p^2c^2+m^2c^4.\]For a scalar wavefunction $\phi(\mathbf r,t)$, make the operator substitutions
\[E\longrightarrow i\hbar\frac{\partial}{\partial t}, \qquad \mathbf p\longrightarrow-i\hbar\nabla.\]Acting on $\phi$, the energy–momentum relation first gives
\[\left(i\hbar\frac{\partial}{\partial t}\right)^2\phi = \left[ c^2(-i\hbar\nabla)^2+m^2c^4 \right]\phi.\]Since $i^2=-1$, this is
\[-\hbar^2\frac{\partial^2\phi}{\partial t^2} =-\hbar^2c^2\nabla^2\phi+m^2c^4\phi.\]Move every term to the left and divide by $-\hbar^2c^2$:
\[\boxed{ \left( \frac{1}{c^2}\frac{\partial^2}{\partial t^2} -\nabla^2+\frac{m^2c^2}{\hbar^2} \right)\phi=0 }.\]With metric $g_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$ and
\[\Box=\partial_\mu\partial^\mu =\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2,\]this is the manifestly covariant equation
\[\boxed{\left(\Box+\frac{m^2c^2}{\hbar^2}\right)\phi=0}.\]Plane waves and the two frequency branches
For
\[\phi=Ae^{i(\mathbf k\cdot\mathbf r-\omega t)},\]the required derivatives are
\[\frac{\partial^2\phi}{\partial t^2}=-\omega^2\phi, \qquad \nabla^2\phi=-k^2\phi.\]Substitution into the Klein–Gordon equation gives
\[\left( -\frac{\omega^2}{c^2} +k^2+\frac{m^2c^2}{\hbar^2} \right)\phi=0.\]For a nonzero plane wave, the coefficient must vanish. Therefore
\[\omega^2=c^2k^2+\frac{m^2c^4}{\hbar^2}.\]Therefore $E=\hbar\omega$ has both signs,
\[E=\pm\sqrt{p^2c^2+m^2c^4}.\]The negative-frequency branch is not an algebraic accident: the equation is second order in time, so both signs are part of its complete solution space.
Conserved Klein–Gordon current
Write the equation and its complex conjugate as
\[\frac{1}{c^2}\partial_t^2\phi-\nabla^2\phi +\frac{m^2c^2}{\hbar^2}\phi=0,\] \[\frac{1}{c^2}\partial_t^2\phi^*-\nabla^2\phi^* +\frac{m^2c^2}{\hbar^2}\phi^*=0.\]Multiply the first equation by $\phi^*$, the second by $\phi$, and subtract the second result from the first. The mass terms are identical and cancel:
\[\frac{1}{c^2} \left(\phi^*\partial_t^2\phi-\phi\partial_t^2\phi^*\right) - \left(\phi^*\nabla^2\phi-\phi\nabla^2\phi^*\right) =0.\]The time term is a total derivative because
\[\begin{aligned} \partial_t \left(\phi^*\partial_t\phi-\phi\partial_t\phi^*\right) &= (\partial_t\phi^*)(\partial_t\phi)+\phi^*\partial_t^2\phi\\ &\quad -(\partial_t\phi)(\partial_t\phi^*)-\phi\partial_t^2\phi^*\\ &=\phi^*\partial_t^2\phi-\phi\partial_t^2\phi^*. \end{aligned}\]Similarly, the spatial product rule gives
\[\nabla\cdot \left(\phi^*\nabla\phi-\phi\nabla\phi^*\right) =\phi^*\nabla^2\phi-\phi\nabla^2\phi^*.\]The subtraction equation therefore becomes
\[\frac{1}{c^2}\partial_t \left(\phi^*\partial_t\phi-\phi\partial_t\phi^*\right) - \nabla\cdot \left(\phi^*\nabla\phi-\phi\nabla\phi^*\right) =0.\]Multiplication by $i\hbar/(2m)$ gives the continuity equation
\[\frac{\partial \rho_{\mathrm{KG}}}{\partial t} +\nabla\cdot\mathbf j_{\mathrm{KG}}=0,\]where
\[\rho_{\mathrm{KG}} =\frac{i\hbar}{2mc^2} \left( \phi^*\frac{\partial\phi}{\partial t} -\phi\frac{\partial\phi^*}{\partial t} \right),\]and
\[\mathbf j_{\mathrm{KG}} =-\frac{i\hbar}{2m} \left(\phi^*\nabla\phi-\phi\nabla\phi^*\right).\]For the plane wave used above,
\[\partial_t\phi=-i\omega\phi, \qquad \partial_t\phi^*=i\omega\phi^*.\]Substitution gives
\[\begin{aligned} \rho_{\mathrm{KG}} &=\frac{i\hbar}{2mc^2} \left(-i\omega|A|^2-i\omega|A|^2\right)\\ &=\frac{\hbar\omega}{mc^2}|A|^2 =\frac{E}{mc^2}|A|^2. \end{aligned}\]It is positive on the positive-frequency branch and negative on the negative-frequency branch. Thus it cannot be a probability density for a single particle, because probability must be non-negative everywhere. In relativistic field theory the same current is instead interpreted as a charge current, whose sign may legitimately distinguish particles from antiparticles.
Nonrelativistic positive-frequency limit
Remove the rapid positive-frequency rest-energy phase by writing
\[\phi(\mathbf r,t) =e^{-imc^2t/\hbar}\psi(\mathbf r,t).\]The time derivatives are
\[\frac{\partial\phi}{\partial t} =e^{-imc^2t/\hbar} \left( \frac{\partial\psi}{\partial t} -\frac{imc^2}{\hbar}\psi \right),\]and
\[\frac{\partial^2\phi}{\partial t^2} =e^{-imc^2t/\hbar} \left( \frac{\partial^2\psi}{\partial t^2} -\frac{2imc^2}{\hbar}\frac{\partial\psi}{\partial t} -\frac{m^2c^4}{\hbar^2}\psi \right).\]Substitution in the Klein–Gordon equation cancels the two rest-mass terms and leaves
\[\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2} -\frac{2im}{\hbar}\frac{\partial\psi}{\partial t} -\nabla^2\psi=0.\]Solving for the first time derivative gives
\[i\hbar\frac{\partial\psi}{\partial t} =-\frac{\hbar^2}{2m}\nabla^2\psi +\frac{\hbar^2}{2mc^2} \frac{\partial^2\psi}{\partial t^2}.\]If the envelope energy scale is $\varepsilon\ll mc^2$, then $\partial_t\psi\sim\varepsilon\psi/\hbar$ and the last term is smaller than the first-time-derivative term by order $\varepsilon/(mc^2)$. Neglecting it yields
\[\boxed{ i\hbar\frac{\partial\psi}{\partial t} =-\frac{\hbar^2}{2m}\nabla^2\psi, }\]the free Schrödinger equation.
Merits and limitations
The equation is Lorentz covariant, gives the correct relativistic dispersion relation, and describes free spin-zero particles and scalar fields. Its nonrelativistic positive-frequency limit reduces to the Schrödinger equation after the rapid rest-energy phase is removed.
| Its difficulty is specifically the single-particle interpretation. It requires both $\phi$ and $\partial_t\phi$ as initial data, admits positive- and negative-frequency solutions, and has no positive-definite conserved density analogous to $ | \psi | ^2$. Describing spin zero is not itself a defect—it is the equation’s domain—but it makes the equation unsuitable for an electron. These points motivate a relativistic equation that is first order in time and acts on a multicomponent wavefunction. |
Discussion