Probability Density Problems

A conserved density is not automatically a probability density. The Klein–Gordon current gives a sharp example: even a superposition containing only positive frequencies can have negative local density while its integrated norm stays positive. The difficulty concerns a particular local Born interpretation, not the consistency of the scalar wave equation or the existence of a positive-energy scalar-particle Hilbert space.

Required background. The Klein–Gordon Equation supplies the scalar current; Relativistic Currents separates conservation and positivity.

Helpful background. The Covariant Dirac Equation provides the contrasting positive spinor density.

Four different questions about a density

For an ordinary position measurement, probabilities assigned to disjoint regions must be nonnegative and add to one over all space. A proposed local density must therefore answer more than the conservation question.

Question	Mathematical test	What it does not establish
Is the integral conserved?	A continuity equation and controlled boundary flux	Pointwise positivity
Is it locally nonnegative?	Nonnegative density for every allowed state	Its transformation law
Is it a relativistic current?	A four-vector current and hypersurface integral	A particle-number interpretation
Does it describe one particle?	A justified fixed-particle sector and measurement rule	Validity when pairs are produced

The current geometry explains why a local probability current should be future causal. Here the focus is a worked scalar counterexample and its interpretation.

Positive frequencies can give negative local KG density

For $m>0$ the density associated with the standard KG current is

$$\rho_{\mathrm{KG}}= \frac{i\hbar}{2mc^2} \left(\phi^*\partial_t\phi-(\partial_t\phi^*)\phi\right).$$

A single positive-frequency plane wave has positive density. For two such modes with real positive amplitudes $A,B$ and a relative phase $\theta$, the density is instead

$$\rho_{\mathrm{KG}}= \frac{1}{mc^2}\left[ E_1A^2+E_2B^2+(E_1+E_2)AB\cos\theta\right].$$

At destructive phase $\cos\theta=-1$, the numerator factors:

$$E_1A^2+E_2B^2-(E_1+E_2)AB =(A-B)(E_1A-E_2B).$$

For $E_2>E_1$, this is negative whenever $E_1/E_2<B/A<1$. The time derivative weights the two modes by different energies; the interference term cannot be reorganized into a nonnegative absolute square.

A normalized box example

Use $c=\hbar=m=1$ in this example. Choose a periodic box of volume $V$ whose $x$ period is $2\pi/\sqrt{15}$, and a field independent of the transverse coordinates:

$$\phi(t,x)=\frac{1}{\sqrt{2V}} \left(e^{-it}-\frac12e^{-4it+i\sqrt{15}x}\right).$$

The energies $1$ and $4$ obey the positive mass shell for momenta $0$ and $\sqrt{15}$. Direct substitution gives

$$\rho_{\mathrm{KG}}(t,x) =\frac1{2V}\left[2-\frac52\cos(3t-\sqrt{15}x)\right].$$

At $t=x=0$ it equals $-1/(4V)$ and is negative in a neighborhood. Yet the cosine averages to zero over the box, so $\int_V\rho_{\mathrm{KG}}d^3x=1$ at every time. This is a normalized, finite-volume positive-frequency counterexample. Plane waves in infinite volume would need wave-packet or distributional normalization; the box avoids that distraction.

Why a positive integrated inner product survives

Orthogonality removes interference between distinct momentum modes when integrating over a complete spatial slice. On the free positive-frequency subspace, the KG norm is consequently a sum or integral of positive energy-weighted mode coefficients. On the full solution space the negative frequency sector contributes with the opposite sign.

Global positivity on the restricted subspace does not imply positivity of its pointwise integrand. Nor can one set $\rho\mapsto|\rho|$ without changing the theory: the absolute value generally destroys the conservation law and changes normalization in a state-dependent way.

The alternative $|\phi|^2$ is nonnegative, but its integral is not conserved for arbitrary two-branch KG data. For example a spatially constant mode in a finite periodic box, $\phi(t)=\cos(mc^2t/\hbar)$ has oscillating $|\phi|^2$ and zero KG charge. Within the free positive-frequency subspace, an equal-time $L^2$ norm can be conserved under the square-root Hamiltonian. This equal-time $L^2$ norm is not the invariant KG norm, and $|\phi|^2$ is not the time component of the KG current. A positive-norm position representation needs separately specified transformation and localization rules. This qualification prevents an overly strong claim that no positive norm can exist.

Dirac positivity and the particle-number boundary

For the first-quantized Dirac equation, $\rho_D=\psi^\dagger\psi$ is nonnegative and its current is conserved. It includes contributions from either free energy sector with the same norm sign. This solves the local-density problem for its one-particle description, but does not make that description sufficient for pair creation.

Electric charge and particle number are also different. In field theory a particle–antiparticle pair can have total charge zero while contributing two excitations. Conserving signed charge does not conserve the number of particles. The scalar charge-current interpretation and the Dirac probability-current interpretation must therefore be stated at their respective levels of description.

Exercises

1. For $E_1=2mc^2$ and $E_2=5mc^2$, determine the interval of $B/A$ that makes the destructive-phase KG density negative.

Solution

The condition is $2/5<B/A<1$. At the endpoints the density is zero. Equal energies do not admit this interval; their interference can be written as an energy times an absolute square.

2. In the normalized box example, find the minimum and maximum local density. Does either bound alter its integrated norm?

Solution

The minimum is $-1/(4V)$ and the maximum is $9/(4V)$. Their spatial pattern moves with the interference phase while the cosine integral remains zero. The conserved norm stays one.

3. A neutral state contains one particle of charge $q$ and its antiparticle. Which of total charge and total excitation number distinguishes it from the vacuum in a free Fock-space description?

Solution

Both have total charge zero. The pair state has excitation number two and positive excitation energy, whereas the vacuum has zero excitations. Charge alone cannot identify the particle content.

References

• J. D. Bjorken and S. D. Drell, Relativistic Quantum Mechanics, McGraw–Hill, 1964 — KG and Dirac density interpretation.
• W. Greiner, Relativistic Quantum Mechanics: Wave Equations, 3rd ed., Springer, 2000 — scalar currents, positive-frequency restrictions, and spinor probability.
• S. Weinberg, The Quantum Theory of Fields, Volume I: Foundations, Cambridge University Press, 1995 — particle states, charge, and field interpretation.