Klein–Gordon Theory in External Potentials

A prescribed potential turns the Klein–Gordon equation into a linear external-field problem. Its derivatives act on both the amplitude and the potentials, and its stationary form is a quadratic energy-eigenvalue problem. Electrostatic potentials and Lorentz-scalar mass shifts can produce the same leading nonrelativistic potential while defining different relativistic theories.

Required background. Minimal Coupling derives the covariant prescription; the Conserved Current provides its density and flux. Helpful background. The Klein–Gordon Inner Product explains stationary-mode normalization and orthogonality.

The external-field differential operator

For a charged complex scalar with mass $m>0$, write $V=q\Phi$ and $\boldsymbol\pi=-i\hbar\nabla-q\mathbf A$. The equation is

$$\left[(i\hbar\partial_t-V)^2 -c^2\boldsymbol\pi^2-m^2c^4\right]\phi=0.$$

This is an operator composition, not an algebraic polynomial in commuting numbers. Explicitly,

$$\begin{aligned} (i\hbar\partial_t-V)^2\phi ={}&-\hbar^2\partial_t^2\phi-2i\hbar V\partial_t\phi\\ &-i\hbar(\partial_t V)\phi+V^2\phi, \end{aligned}$$

and

$$\begin{aligned} \boldsymbol\pi^2\phi ={}&-\hbar^2\nabla^2\phi +iq\hbar(\nabla\cdot\mathbf A)\phi\\ &+2iq\hbar\mathbf A\cdot\nabla\phi+q^2\mathbf A^2\phi. \end{aligned}$$

The derivative-of-potential terms are required even though one can choose gauges in which an individual term vanishes. Deleting them in general destroys gauge covariance. For regular prescribed coefficients the equation remains second order in time; specify an initial amplitude and its covariant time derivative, or equivalent ordinary Cauchy data in a fixed gauge. Charge conservation does not reduce these to one arbitrary function.

Stationary backgrounds and kinetic energy

In a gauge where the potentials are time independent, set $\phi=e^{-iEt/\hbar}u(\mathbf x)$. For real stationary energy $E$,

$$\left[c^2(-i\hbar\nabla-q\mathbf A)^2+m^2c^4\right]u =(E-V(\mathbf x))^2u.$$

The operator depends quadratically on the spectral parameter $E$. Treating its modes as eigenfunctions of an ordinary energy-independent Schrödinger operator can give incorrect orthogonality and normalization. Use the KG pairing, whose stationary density is $(E-V)|u|^2/(mc^2)$.

In a constant-potential region,

$$(E-V)^2=c^2|\mathbf p-q\mathbf A|^2+m^2c^4.$$

This separates canonical labels from mechanical quantities. When $|E-V|<mc^2$, a one-dimensional longitudinal momentum can become imaginary (with transverse kinetic energy raising the relevant gap). Evanescence is a property of the stationary mode equation; it is not itself a particle-production rate or a violation of conservation.

Even for $\mathbf A=0$, replacing the equation by $Eu=[V+\sqrt{c^2\mathbf P^2+m^2c^4}]u$ is generally incorrect. The square-root operator fails to commute with a varying $V$. The commutator obstruction is derived in The Square-Root Hamiltonian. For constant $V$ the obstruction vanishes and the two algebraic branches can be separated exactly.

Electrostatic coupling versus a scalar mass shift

An electrostatic potential is the time component of a four-vector. A different model introduces a real Lorentz scalar $S(x)$ with energy units through $mc^2\mapsto mc^2+S(x)$. Its equation is

$$\left[(i\hbar\partial_t-V)^2 -c^2\boldsymbol\pi^2-(mc^2+S)^2\right]\phi=0.$$

The real scalar term cancels in the current-conservation proof, but it does not transform like an electromagnetic potential. For constants and $mc^2+S>0$, the two branches are

$$E_\pm=V\pm\sqrt{c^2\boldsymbol\pi^2+(mc^2+S)^2}.$$

$V$ translates the midpoint of the branches; $S$ changes their gap. When $|S|,|\boldsymbol\pi|^2/(2m)\ll mc^2$, the positive branch begins as $E_+=mc^2+V+S+\boldsymbol\pi^2/(2m)+\cdots$. The constant $V$ shifts the energy exactly and need not be small. That shared leading limit does not make the relativistic interactions equivalent. A potential must therefore be identified by its Lorentz character, not just by its familiar nonrelativistic name.

A pure-gauge time dependence as a diagnostic

Let $\mathbf A=0$ and let $\Phi(t)$ be spatially uniform. Both electric and magnetic fields vanish. If $\phi_0$ is a free solution, then

$$\phi(t,\mathbf x)= \exp\left[-\frac{iq}{\hbar}\int^t\Phi(s)\,ds\right] \phi_0(t,\mathbf x)$$

solves the coupled equation, because $(i\hbar\partial_t-q\Phi)\phi$ equals the same phase times $i\hbar\partial_t\phi_0$. The density and current equal the free ones. This check fails if the $-i\hbar\partial_tV$ term is omitted from the expanded equation. A large or rapidly varying pure-gauge potential cannot create particles or change a gauge-invariant observable.

What the external-field approximation leaves out

The background is supplied rather than dynamically solved with the matter. Depending on the application, this omits source recoil, radiation, backreaction, and quantum fluctuations. A quantized scalar field in the same classical background can describe pair creation without quantizing the electromagnetic field. A first-quantized stationary calculation does not supply that vacuum calculation. Use field strengths, spatial extent, duration, energy transfers, and channel probabilities to assess a fixed particle-sector approximation; an absolute potential offset is not a valid criterion. When Relativistic Quantum Mechanics Is Useful organizes these model choices.

Exercises

1. Expand $(i\hbar\partial_t-V)^2f$ for an arbitrary test function. Which term is lost if $V$ is treated as a commuting number?

Solution

The result is $-\hbar^2\partial_t^2f-2i\hbar V\partial_tf -i\hbar\dot V f+V^2f$. The missing term would be $-i\hbar\dot V f$. It arises when the outer derivative acts on the inner multiplication by $V$.

2. With constant $V,S$, no vector potential, and zero momentum, compare the branch midpoint and gap. Assume $mc^2+S>0$.

Solution

$E_\pm=V\pm(mc^2+S)$, so their midpoint is $V$ and their separation is $2(mc^2+S)$. An electrostatic shift and a mass shift are distinct even though both enter the leading positive-branch energy additively.

3. In a stationary problem, add a constant $V_0$ to $V$ and $V_0$ to $E$. Determine the change in the spatial equation and current density.

Solution

Neither changes: both depend on $E-V$. The time-dependent overall gauge phase accounts for the shifted stationary label. Threshold statements based only on $|V|$ would fail this invariance test.

References

• F. Bastianelli, Relativistic Quantum Mechanics, University of Bologna lecture notes, 2023–2024, §3 — charged scalar equations and external potentials; its metric convention must be translated.
• H. Feshbach and F. Villars, “Elementary Relativistic Wave Mechanics of Spin 0 and Spin 1/2 Particles,” Reviews of Modern Physics 30, 24–45, 1958, doi:10.1103/RevModPhys.30.24 — external-field scalar dynamics.
• W. Greiner, Relativistic Quantum Mechanics: Wave Equations, 3rd ed., Springer, 2000, doi:10.1007/978-3-662-04275-5 — scalar and vector interactions.