Negative-Energy Solutions

Negative-energy solutions are modes on the negative branch of a relativistic wave equation’s time-translation spectrum. They are neither negative probabilities nor ordinary positive-energy particles viewed from another inertial frame. Free positive-energy sectors can be defined consistently; the deeper question is whether a chosen sector remains closed under the interactions and localization operations one wants to describe.

Required background. The Klein–Gordon Equation and Covariant Dirac Equation supply the free frequency branches and their different currents.

Helpful background. The Square-Root Hamiltonian constructs an explicit free positive-energy theory.

Energy eigenvalues and frequency conventions

With the convention $i\hbar\partial_t\phi=E\phi$, a factor $e^{-iE_{\mathbf p}t/\hbar}$ has positive energy and $e^{+iE_{\mathbf p}t/\hbar}$ has negative energy, where $E_{\mathbf p}=\sqrt{c^2\mathbf p^2+m^2c^4}>0$. A general free scalar solution can be written

$$\phi(t,\mathbf x)=\int d^3p\, \left[a(\mathbf p)e^{-iE_{\mathbf p}t/\hbar} +b(\mathbf p)e^{+iE_{\mathbf p}t/\hbar}\right] e^{i\mathbf p\cdot\mathbf x/\hbar},$$

with any fixed consistent Fourier normalization absorbed into $a,b$. Both coefficients are needed for general independent initial values of $\phi$ and $\partial_t\phi$.

The free Dirac equation also has energies $\pm E_{\mathbf p}$, but its first-order initial data live in a four-component spinor. At each momentum and for positive mass, two spin states belong to each energy branch. The positive local Dirac density is a separate fact: a negative-energy Dirac mode still has $\psi^\dagger\psi\geq0$.

Proper orthochronous Lorentz transformations preserve the sign of energy on each mass-shell sheet. A boost never converts the negative sheet to the positive sheet. A constant shift of a Hamiltonian’s energy reference is also a different operation from exchanging these branches. In a background potential, branch labels must be tied to the chosen asymptotic or stationary reference, not to a gauge-dependent absolute potential offset.

When a positive-energy restriction is consistent

In the free Dirac Hilbert space, let $P_+$ be the positive-energy projector of the self-adjoint Hamiltonian $H_0$. Since $[P_+,H_0]=0$, a state initially in its range remains there under free evolution. This is a valid invariant subspace. Nothing in free relativistic kinematics forces a single isolated particle to produce additional particles.

For a prescribed perturbation $V(t)$, the same free subspace remains closed only if the evolution preserves it. A sufficient instantaneous condition is

$$P_-V(t)P_+=0,\qquad P_-=I-P_+,$$

for all times, with the complementary block absent by Hermiticity. An arbitrary spatially varying background does not satisfy this condition. A static interacting Hamiltonian may instead admit its own separated spectral subspaces. Their existence and interpretation require analysis of that Hamiltonian; they need not coincide with the free projectors.

A two-level mixing diagnostic

In a truncated model of one state from each free sector, take

$$H=\begin{pmatrix}E&g\\g&-E\end{pmatrix}, \qquad E>0,\quad g\in\mathbb R.$$

Since $H^2=(E^2+g^2)I$, define $\Omega=\sqrt{E^2+g^2}$ and obtain

$$e^{-iHt/\hbar}=I\cos(\Omega t/\hbar) -i\frac{H}{\Omega}\sin(\Omega t/\hbar).$$

Starting in the upper basis vector gives lower-sector weight

$$P_-(t)=\frac{g^2}{E^2+g^2}\sin^2(\Omega t/\hbar).$$

Weak coupling $|g|\ll E$ suppresses this weight. A resonant time-dependent drive can accumulate transitions and needs a different estimate. This finite-dimensional calculation diagnoses failure of a chosen free-sector restriction; it is not a pair-production probability, because its Hilbert space still contains no variable particle number or vacuum.

Why discarding modes is not field quantization

A free complex scalar field is schematically expanded as

$$\widehat\phi(x)=\int d\Pi_p\, \left[\widehat a(\mathbf p)e^{-ip\cdot x/\hbar} +\widehat b^\dagger(\mathbf p)e^{+ip\cdot x/\hbar}\right], \qquad p^0=E_{\mathbf p}/c>0,$$

where normalization factors depend on the field convention. The negative frequency factor is paired with a creation operator. The resulting antiparticle excitation has positive physical energy; it is not a state with energy $-E_{\mathbf p}$ in the physical Fock-space spectrum. The Dirac field has the corresponding particle-annihilation and antiparticle-creation structure, with fermionic anticommutation relations.

The operator algebra, vacuum, and Hamiltonian are additional input to this interpretation. Replacing the letter $b$ by $b^\dagger$ in a classical mode expansion is not a derivation of them. Both frequency sectors participate in local field commutators or anticommutators, even when the physical excitation spectrum is bounded below. This is the bridge described in Weinberg’s and Tong’s free-field treatments.

Common pitfalls

Equating negative energy and negative norm. KG’s standard bilinear form has opposite signs on opposite frequency sectors. The Dirac Hilbert-space norm is positive on both. The Hamiltonian spectrum and the norm are different mathematical objects.

Claiming every external field creates pairs. Background-induced mixing, available energy, and the vacuum pair-production rate are distinct. Static magnetic backgrounds, adiabatic limits, and selection rules demonstrate why a single energy-scale slogan cannot decide the process.

Exercises

1. For a spatially constant KG mode, initial data satisfy $\phi(0)=f$ and $\partial_t\phi(0)=h$. Find its positive- and negative-frequency amplitudes.

Solution

With $\omega=mc^2/\hbar$, $f=a+b$ and $h=-i\omega a+i\omega b$. Thus $a=(f+ih/\omega)/2$ and $b=(f-ih/\omega)/2$. The positive branch requires $h=-i\omega f$; it is a restriction on the initial data.

2. In the mixing model let $g/E=0.1$. What is the largest lower-sector weight? Does the exact Hamiltonian lose its two real eigenvalues?

Solution

The maximum is $0.01/1.01\simeq0.00990099$. The eigenvalues are still the real pair $\pm\sqrt{E^2+g^2}$. Mixing relative to the free basis does not mean loss of unitary evolution.

3. In the field expansion, apply $i\hbar\partial_t$ and $-i\hbar\nabla$ to $e^{+ip\cdot x/\hbar}$. Why must the mode label be distinguished from the created antiparticle’s physical momentum?

Solution

The c-number factor has differential eigenvalues $-E_{\mathbf p}$ and $-\mathbf p$. The operator $b^\dagger(\mathbf p)$ creates an excitation labeled by positive energy and momentum $\mathbf p$ under the quantized translation generators. The field term combines the mode and operator; one cannot infer the excitation spectrum from the c-number factor alone.

References

• J. D. Bjorken and S. D. Drell, Relativistic Quantum Mechanics, McGraw–Hill, 1964 — the scalar and Dirac branches and their one-particle interpretation.
• D. Tong, Lectures on Quantum Field Theory, University of Cambridge, 2006, sections 2 and 5 — scalar and fermion mode quantization.
• S. Weinberg, The Quantum Theory of Fields, Volume I: Foundations, Cambridge University Press, 1995, chapters 2 and 5 — positive-energy particle states and covariant fields.