Math Infinitum

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Intuition

The quickest relativistic equation turns $E^{2}=p^{2}c^{2}+m^{2}c^{4}$ into operators: $-\hbar^{2}\partial^{2}\psi/\partial t^{2}=(-\hbar^{2}c^{2}\nabla^{2}+m^{2}c^{4})\psi$ , the Klein–Gordon equation. Schrödinger found it first, tried it on hydrogen, got the fine structure wrong and put it aside; the trouble was the electron’s spin, which it has no room for. Its plane waves come with both signs of energy, $E=\pm\sqrt{p^{2}c^{2}+m^{2}c^{4}}$ , separated by a gap of $2mc^{2}$ . Worse, because the equation is second order in time, the density that it conserves is not $|\psi|^{2}$ but a combination of $\psi$ and its time derivative, and it can be negative. So $\psi$ cannot be the wavefunction of one particle. The equation came back as the equation of a field for particles of spin zero, like the pion, whose conserved density is a charge: positive for particles, negative for antiparticles.

A balance sheet can show a negative number where a count of objects cannot. The Klein–Gordon density behaves like a balance of charge, particles minus antiparticles, not like a count of one particle’s presence.

The energies of Klein–Gordon plane waves against momentum, in units where $m=c=1$ : two branches, $E=\pm\sqrt{p^{2}c^{2}+m^{2}c^{4}}$ , separated at by the gap . The lower branch is what a single-particle reading cannot explain.

The Klein–Gordon equation

Replacing $E\to i\hbar\partial_{t}$ and $\mathbf{p}\to-i\hbar\nabla$ in $E^{}$ gives

\Bigl(\frac{1}{c^{2}}\frac{\partial^{2}}{\partial t^{2}}-\nabla^{2}+\frac{m^{2}c^{2}}{\hbar^{2}}\Bigr)\psi=0

Properties

It is Lorentz invariant: $\bigl(\partial_{\mu}\partial^{\mu}+m^{2}c^{2}/\hbar^{2}\bigr)\psi=0$ .

The conserved Klein–Gordon density

\partial_{t}\rho+\nabla\cdot\mathbf{j}=0,\qquad \rho=\frac{i\hbar}{2mc^{2}}\bigl(\psi^{*}\dot{\psi}-\psi\dot{\psi}^{*}\bigr)

Multiply the equation by $\psi^{*}$ and its conjugate by $\psi$ , and subtract: the mass terms cancel, and what is left is a time derivative plus a divergence. For a plane wave the density is proportional to $E$ , so it is negative on the lower branch: it cannot be a probability.

Proof steps

\psi^{*}\Bigl(\frac{\ddot{\psi}}{c^{2}}-\nabla^{2}\psi\Bigr)-\psi\Bigl(\frac{\ddot{\psi}^{*}}{c^{2}}-\nabla^{2}\psi^{*}\Bigr)=0

The mass terms $m^{2}c^{2}|\psi|^{2}/\hbar^{2}$ cancel.

\frac{1}{c^{2}}\partial_{t}\bigl(\psi^{*}\dot{\psi}-\psi\dot{\psi}^{*}\bigr)-\nabla\cdot\bigl(\psi^{*}\nabla\psi-\psi\nabla\psi^{*}\bigr)=0

Each bracket is a product-rule derivative.

\begin{aligned}\rho&=\frac{i\hbar}{2mc^{2}}(\psi^{*}\dot{\psi}-\psi\dot{\psi}^{*})\\ \mathbf{j}&=\frac{\hbar}{2mi}(\psi^{*}\nabla\psi-\psi\nabla\psi^{*})\end{aligned}

Multiply by $i\hbar/2m$ ; $\mathbf{j}$ is the current of the Schrödinger equation.

\psi=N\,e^{i(\mathbf{p}\cdot\mathbf{r}-Et)/\hbar}\ \Longrightarrow\ \rho=\frac{E}{mc^{2}}\,|N|^{2}

$\dot{\psi}=-iE\psi/\hbar$ : the density has the sign of $E$ .

Applications

Practice

Two Branches

Klein–Gordon plane waves have $E=\pm\sqrt{p^{2}c^{2}+m^{2}c^{4}}$ : for every momentum a positive and a negative energy.

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

Try it

A Klein–Gordon plane wave has $pc=3$ MeV and $mc^{2}=4$ MeV. What is its negative-energy value, in MeV?