The Klein–Gordon Equation

The Schrödinger equation is built on the non-relativistic energy $E = p^2/2m$ . It treats time and space asymmetrically (first derivative in $t$ , second in $x$ ) and breaks Lorentz invariance. To describe fast particles we need a wave equation consistent with special relativity. The first attempt — the Klein–Gordon equation — succeeds for spinless particles and exposes, through its failures, the deep features of relativistic quantum theory.

From the relativistic energy to a wave equation

Special relativity gives the energy–momentum relation (with $c$ the speed of light)

E^2 = p^2 c^2 + m^2 c^4 .

Promote energy and momentum to operators, $E \to i\hbar\,\partial_t$ and $\mathbf{p}\to -i\hbar\nabla$ , and apply both sides to a field $\phi(\mathbf{r},t)$ :

-\hbar^2\,\partial_t^2\,\phi = \big(-\hbar^2 c^2\nabla^2 + m^2 c^4\big)\phi .

Rearranged, this is the Klein–Gordon equation:

\boxed{\;\Big(\frac{1}{c^2}\,\partial_t^2 - \nabla^2 + \frac{m^2 c^2}{\hbar^2}\Big)\phi = 0\;}

Manifestly covariant form

Using the d'Alembertian $\Box \equiv \partial^\mu\partial_\mu = \frac{1}{c^2}\partial_t^2 - \nabla^2$ and the Compton wavenumber $mc/\hbar$ , it compresses to the Lorentz-invariant statement

\big(\Box + \tfrac{m^2 c^2}{\hbar^2}\big)\phi = 0 .

Treating $t$ and $\mathbf{x}$ on equal footing — both appearing as second derivatives — is exactly what restores Lorentz invariance. Plane-wave solutions $\phi \propto e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$ obey the relativistic dispersion relation $\hbar^2\omega^2 = \hbar^2 c^2 k^2 + m^2 c^4$ .

Two problems Schrödinger never had

The Klein–Gordon equation immediately raises two difficulties that shaped all of relativistic quantum theory.

Negative energies. Solving $E^2 = p^2c^2 + m^2c^4$ gives both signs,

E = \pm\sqrt{p^2 c^2 + m^2 c^4}.

Unlike the non-relativistic case, the negative-energy branch cannot simply be discarded — it is needed for completeness, and a spectrum unbounded below seems to allow unlimited energy release.

Negative probabilities. The conserved density that comes with the equation,

\rho = \frac{i\hbar}{2mc^2}\big(\phi^*\,\partial_t\phi - \phi\,\partial_t\phi^*\big),

involves a first time derivative, so it is not positive-definite: with the freedom to choose $\phi$ and $\partial_t\phi$ independently at one instant, $\rho$ can come out negative. It cannot be a probability density in the Schrödinger sense.

What the Klein–Gordon equation is good for

Despite the interpretational subtleties, the Klein–Gordon equation correctly describes spin-0 particles — the pions and the Higgs boson are physical Klein–Gordon fields. It also reproduces relativistic kinematics and, in the non-relativistic limit $E \approx mc^2 + \varepsilon$ with $\varepsilon \ll mc^2$ , reduces to the ordinary Schrödinger equation: writing $\phi = e^{-imc^2 t/\hbar}\psi$ and dropping $\partial_t^2\psi$ against $mc^2\,\partial_t\psi$ recovers $i\hbar\,\partial_t\psi = -\tfrac{\hbar^2}{2m}\nabla^2\psi$ exactly. It is the correct theory — for particles without spin. Dirac's question was how to do the same for the electron, which has spin $\tfrac12$ , and the answer is the subject of the next lesson.