Pair Creation

Pair creation is a statement about particle counts in a quantum state. A classical wave equation supplies the mode conversion needed for that calculation, but a negative-frequency component alone is not a pair probability. One must also specify the quantized field, the initial state, and the definitions of incoming and outgoing particles. For an incoming vacuum and a normalized independent pair channel, the outgoing mean number of pairs is $N=|\beta|^2$, where $\beta$ is a Bogoliubov mixing coefficient. Its allowed magnitude and probability interpretation depend on statistics.

Required background. Antiparticles fixes charges and species; Why Fock Space Is Necessary organizes variable-number states; Dirac in Electromagnetic Fields provides the background-field mode evolution.

Helpful background. Pair-Creation Thresholds supplies reaction kinematics; Klein–Gordon Inner Product explains the signed scalar mode normalization.

Incoming and outgoing particle modes

Use $\hbar=c=1$ and quantize the matter field in a prescribed classical electromagnetic background. The source of the field can exchange energy with the matter. Dynamical photons, depletion of the source, and backreaction are excluded from this approximation. Take $m>0$ and a background that admits well-defined early and late positive-frequency subspaces. A finite box and suitable switching make the mode bookkeeping explicit before continuum limits.

For example, in temporal gauge let $\mathbf A(t)=A_z(t)\hat{\mathbf z}$, with constant limits as $t\to\pm\infty$, and $\Phi=0$. Then $E_z=-\dot A_z$. Spatial translation invariance conserves canonical momentum $\mathbf p$, while the kinetic momentum is

$$\boldsymbol\pi(t)=\mathbf p-q\mathbf A(t).$$

For a complex scalar, the temporal part of a Fourier mode obeys

$$\ddot f_{\mathbf p}+\omega_{\mathbf p}^2(t)f_{\mathbf p}=0,$$ $$\omega_{\mathbf p}^2(t) =m^2+p_\perp^2+[p_z-qA_z(t)]^2.$$

An incoming positive-frequency solution approaches $e^{-i\omega_{\rm in}t}/\sqrt{2\omega_{\rm in}}$ and has Wronskian

$$f\dot f^*-f^*\dot f=i.$$

At late times it generally contains both frequency signs. The Dirac problem instead evolves a spinor by the Hermitian momentum-mode Hamiltonian; its conserved positive norm gives a different mixing constraint. Neither calculation by itself specifies an occupation number.

The asymptotic energies use kinetic, not canonical, momentum. Under $A_z\mapsto A_z+C$, the corresponding mode label changes as $p_z\mapsto p_z+qC$. Keeping $\pi_z$ fixed leaves the physical prediction unchanged. Comparing the same numerical canonical label in both gauges would compare different modes. This gauge relabeling assumes continuum momentum labels or consistently transformed boundary conditions; an arbitrary constant shift in a fixed periodic box can change its holonomy. In the homogeneous pair channels, a particle with canonical momentum $\mathbf p$ pairs with an antiparticle of canonical momentum $-\mathbf p$. Their opposite charges also make their kinetic momenta opposite.

Bogoliubov transformations and statistics

Expand the same quantum field in incoming or outgoing normalized modes. Equality of the two expansions relates their operators. For one independent particle–antiparticle channel, choose phases so that the fermionic transformation is

$$\begin{pmatrix}a_{\rm out}\\ b_{\rm out}^\dagger\end{pmatrix} = \begin{pmatrix}\alpha&\beta\\-\beta^*&\alpha^*\end{pmatrix} \begin{pmatrix}a_{\rm in}\\ b_{\rm in}^\dagger\end{pmatrix}.$$

Preservation of the canonical anticommutators requires

$$|\alpha|^2+|\beta|^2=1.$$

For bosons, a compatible convention is

$$\begin{pmatrix}a_{\rm out}\\ b_{\rm out}^\dagger\end{pmatrix} = \begin{pmatrix}\alpha&\beta\\\beta^*&\alpha^*\end{pmatrix} \begin{pmatrix}a_{\rm in}\\ b_{\rm in}^\dagger\end{pmatrix},$$ $$|\alpha|^2-|\beta|^2=1.$$

Indeed, $[a_{\rm out},a_{\rm out}^\dagger]$ contains $|\beta|^2[b_{\rm in}^\dagger,b_{\rm in}]=-|\beta|^2$; the corresponding fermionic anticommutator contains $+|\beta|^2$. The off-diagonal bracket vanishes with the displayed relative signs. Mode-function coefficients may use conjugated or oppositely signed conventions; the operator equations above define $\alpha$ and $\beta$ here.

Let the initial state be $|0,{\rm in}\rangle$, annihilated by $a_{\rm in}$ and $b_{\rm in}$. Direct substitution gives

$$\begin{aligned} \langle a_{\rm out}^\dagger a_{\rm out}\rangle_{\rm in} &=|\beta|^2,\\ \langle b_{\rm out}^\dagger b_{\rm out}\rangle_{\rm in} &=|\beta|^2. \end{aligned}$$

Thus the channel contributes $N=|\beta|^2$ mean pairs, $2N$ mean quanta, and zero net charge. Fermions satisfy $0\leq N\leq1$ per channel; bosons have no such upper bound. Summing pair channels once, including each independent spin or other degeneracy, gives $\langle N_{\rm pairs}\rangle=\sum_\lambda N_\lambda$. Counting both members as separate pairs would double this result.

In a general nonuniform background, $\alpha$ and $\beta$ are matrices between mode spaces. The vacuum particle count is $\operatorname{Tr}(\beta\beta^\dagger)$. Independent scalar channel formulas require an appropriate paired mode basis; an arbitrary matrix entry is not a separate statistically independent event. In an infinite system, unitary implementation on a common Fock space requires additional conditions, including a Hilbert–Schmidt mixing operator. Finite-volume calculations do not establish that global property automatically.

Initial particles: blocking and enhancement

Suppose the initial density operator is diagonal in incoming occupation numbers for this channel. Write $n_a=\langle a_{\rm in}^\dagger a_{\rm in}\rangle$ and $n_b=\langle b_{\rm in}^\dagger b_{\rm in}\rangle$. There are then no anomalous averages such as $\langle a_{\rm in}b_{\rm in}\rangle$. Using the same transformations yields

$$\begin{aligned} \text{fermions:}\qquad n_{a,{\rm out}}&=n_a+N(1-n_a-n_b),\\[2pt] \text{bosons:}\qquad n_{a,{\rm out}}&=n_a+N(1+n_a+n_b). \end{aligned}$$

The equation for $b$ interchanges $a$ and $b$. Consequently $n_{a,{\rm out}}-n_{b,{\rm out}}=n_a-n_b$: charge is conserved even when total number changes.

For fermions, one occupied member and one empty member give no net change in this isolated channel. Both occupied members allow pair annihilation and give a negative number change. For bosons, initial occupation enhances the change above the vacuum value. These are statements about the specified linear mode transformation, not a universal kinetic rate equation for an interacting plasma. Initial coherent pair correlations add interference terms; occupations alone then do not determine the answer. For a nonvacuum initial state, a net occupation change should not automatically be called the number of newly created pairs.

A solvable scalar frequency change

A single oscillator mode gives an explicit check on normalization. Let its frequency jump from $\omega_{\rm in}>0$ to $\omega_{\rm out}>0$ at $t=0$. For the normalized incoming solution, write the late-time form as

$$f(t>0)= \frac{A e^{-i\omega_{\rm out}t} + B e^{i\omega_{\rm out}t}}{\sqrt{2\omega_{\rm out}}}.$$

Continuity of $f$ and $\dot f$ gives

$$\begin{aligned} A&=\frac12\left( \sqrt{\frac{\omega_{\rm out}}{\omega_{\rm in}}} +\sqrt{\frac{\omega_{\rm in}}{\omega_{\rm out}}}\right),\\ B&=\frac12\left( \sqrt{\frac{\omega_{\rm out}}{\omega_{\rm in}}} -\sqrt{\frac{\omega_{\rm in}}{\omega_{\rm out}}}\right). \end{aligned}$$

These real coefficients satisfy $A^2-B^2=1$, and the corresponding bosonic incoming-vacuum occupation is

$$N=B^2 =\frac{(\omega_{\rm out}-\omega_{\rm in})^2} {4\omega_{\rm out}\omega_{\rm in}}.$$

No frequency change gives $N=0$; doubling the frequency gives $N=1/8$. A ratio of nine gives $N=16/9$, which is a valid bosonic mean occupation, not a probability exceeding one. The Wronskian remains conserved throughout.

This is a single-mode sudden-quench model. It does not prove ultraviolet convergence of a field’s total particle number or energy after an arbitrarily abrupt change. A physical pulse requires its full momentum spectrum and switching behavior to be checked.

Energy supply and the meaning of a pair count

In an isolated reaction, energy–momentum conservation fixes the kinematic threshold. In the present calculation the prescribed source supplies work, and a time-dependent background need not conserve the matter Hamiltonian’s instantaneous energy. These are compatible descriptions with different systems held dynamical.

The coefficients above depend on the whole field history, not only its peak strength. Adiabatic evolution in a gapped problem can strongly suppress mixing, whereas a fast or sufficiently extended pulse can produce it. A decomposition at an intermediate time may depend on the chosen instantaneous particle basis; well-defined asymptotic detectors provide the in/out interpretation used here.

Finally, a mean pair count does not determine the vacuum survival probability by the rule $P_{\rm vac}=1-\sum N_\lambda$. Independent-channel probabilities and the constant-electric-field limit are developed in Vacuum Instability in Strong Fields. The present result is the counting input to that calculation.

Exercises

Fermionic occupation bookkeeping. For $N=1/4$, calculate the outgoing mean occupations from incoming occupations $(0,0)$, $(1,0)$, and $(1,1)$. Check the charge difference in each case.

Solution

The results are $(1/4,1/4)$, $(1,0)$, and $(3/4,3/4)$. The differences remain $0,1,0$. The last case has a mean loss of one half of a quantum in total, corresponding to a probability $1/4$ for annihilating the occupied pair in this two-mode model.

Small scalar quench. Set $\omega_{\rm out}=\omega_{\rm in}(1+\epsilon)$ with $|\epsilon|\ll1$. Find the leading pair occupation and explain why a first-order mode change need not give a first-order count.

Solution

$N=\epsilon^2/[4(1+\epsilon)] =\epsilon^2/4+O(\epsilon^3)$. The negative-frequency amplitude is $B=\epsilon/2+O(\epsilon^2)$; the vacuum occupation is its squared magnitude.

Bosonic stimulation. For the frequency doubling example, take incoming occupations $n_a=2$, $n_b=1$ without anomalous correlations. Find both outgoing occupations and the net charge.

Solution

With $N=1/8$, each occupation increases by $(1/8)(1+2+1)=1/2$. Thus the results are $5/2$ and $3/2$. The net charge remains $q(5/2-3/2)=q$.

References

• Gavrilov, S. P., and D. M. Gitman. “Vacuum Instability in External Fields.” Physical Review D 53, 7162–7175 (1996). doi:10.1103/PhysRevD.53.7162; author manuscript. In/out quantization, mean occupations, and vacuum probabilities in time-dependent backgrounds.
• Gavrilov, S. P., and D. M. Gitman. “Quantization of Charged Fields in the Presence of Critical Potential Steps.” Physical Review D 93, 045002 (2016). doi:10.1103/PhysRevD.93.045002; author manuscript. Distinguishes spatial-step scattering data, particle numbers, and the choice of asymptotic states.
• Greiner, Walter. Relativistic Quantum Mechanics: Wave Equations. 3rd ed. Springer (2000). doi:10.1007/978-3-662-04275-5. Background for relativistic wave equations and their particle-interpretation limits.