The Klein–Gordon Conserved Current
The complex Klein–Gordon amplitude has a conserved current even in a prescribed electromagnetic background. Ordinary derivatives must be replaced by covariant derivatives in the current as well as in the equation. The resulting density depends on kinetic energy in a stationary electrostatic problem. Its conservation and gauge invariance do not make it a positive probability density.
Required background. The Klein–Gordon Equation derives the free current; Minimal Coupling fixes the signed charge and gauge package. Helpful background. Relativistic Currents supplies the geometry of integrated charge and boundary flux.
A gauge-invariant bilinear current
Section titled “A gauge-invariant bilinear current”Let , let be real, and let solve
Using the free current’s normalization, define
Both and acquire the same phase under a gauge transformation, so is invariant. Expansion gives the useful check
The two terms on the right are not separately gauge invariant. In particular, using the free expression in a nonzero vector potential gives an incorrect flux.
To prove conservation, apply the product rule with opposite covariant derivatives on the conjugate factor. The connection terms cancel in the neutral bilinear, giving
The last equality uses the wave equation and its conjugate; the real mass terms cancel. No time-translation symmetry of the background was needed. A time-dependent background can exchange energy with the scalar while this charge remains conserved. An absorptive complex potential would instead require a source or sink term and lies outside this proof.
Density and mechanical flux
Section titled “Density and mechanical flux”With and the SI potentials,
For in a static background with real ,
The canonical energy label alone does not determine this sign. Adding a constant to through a gauge transformation shifts to , leaving unchanged. Large absolute potential values in one gauge are therefore not by themselves evidence of a physical instability.
In a region of constant potentials a plane wave has kinetic momentum and kinetic energy . It obeys and has
For this single mode the ratio is the branch group velocity. It is not a general velocity field for arbitrary superpositions: the density can vanish or change sign at interference nodes. The explicit counterexample belongs to Probability-Density Problems.
A stationary flux check
Section titled “A stationary flux check”For one-dimensional motion with and real static potential energy , the stationary equation has the form
Its real coefficient implies
In a constant-potential propagating region with real , put . The interference terms have zero imaginary contribution, so
This conserved Wronskian is the reliable starting point for reflection and transmission ratios. Squared amplitude ratios alone miss wave-number factors. If a kinetic-energy branch changes sign, classify propagation using and interpret charge flux carefully before calling a coefficient a one-particle probability.
What charge conservation establishes
Section titled “What charge conservation establishes”Under suitable boundary decay, is time independent; the same charge can be evaluated on another spacelike surface as explained by the geometric current owner. The normalization here was chosen to match the nonrelativistic amplitude convention. In field theory, the physical electric current follows from the normalized action; multiplying this one-particle current by is appropriate only after its amplitude normalization has been fixed. For a neutral real field the free antisymmetric current vanishes, while the energy current need not vanish. Conserved charge and energy are distinct quantities.
Exercises
Section titled “Exercises”- Verify the density is unchanged by using and the gauge phase.
Solution
, so a stationary mode has . Then , and its modulus is unchanged.
- A stationary region has , with real . Find its current. Can an evanescent superposition carry flux?
Solution
, hence . Each real exponential alone has zero flux, but their relative complex phase can produce a nonzero constant flux across a finite forbidden region.
- For a single propagating mode on either branch, show that for . Explain why this does not prove positivity.
Solution
The mass shell gives , so . However is negative when , and superpositions need not retain the single-mode current relation. The velocity inequality and density positivity are different tests.
References
Section titled “References”- J. D. Bjorken and S. D. Drell, Relativistic Quantum Mechanics, McGraw–Hill, 1964 — charged scalar wave mechanics and flux.
- 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 — scalar charge and external fields.
- W. Greiner, Relativistic Quantum Mechanics: Wave Equations, 3rd ed., Springer, 2000, doi:10.1007/978-3-662-04275-5 — the Klein–Gordon current and stationary scattering.