Bogoliubov Coefficients and Pair Creation

The previous page separated two real-time questions. The in–out formalism computes transition amplitudes between a specified past vacuum and a specified future vacuum. The in–in formalism computes expectation values in the state actually prepared in the past. This page makes that distinction concrete by following a system whose Hamiltonian changes with time.

The model is deliberately simple: a harmonic oscillator whose frequency changes with time. In field theory, every momentum mode of a free field in a spatially uniform time-dependent background is such an oscillator. If the background changes, a mode that was positive-frequency in the past need not remain purely positive-frequency in the future. The coefficient of the negative-frequency component is the Bogoliubov coefficient $\beta$. Its absolute square is the number of produced quanta in that mode.

This is the mechanism behind particle production in external fields, cosmological particle creation, parametric amplification, and the mode mixing that later appears near horizons. When the Hamiltonian is stationary in the remote past and future, the two asymptotic particle bases have an operational meaning and the comparison is sharp:

$$\text{positive frequency in the past}\quad\ne\quad\text{positive frequency in the future}.$$

The particle concept remains basis-dependent, but the asymptotic mismatch is measurable: prepare the past vacuum and count quanta with detectors calibrated to the future Hamiltonian.

Required background. In–out and in–in functionals supplies the boundary-condition distinction used throughout, while free scalar mode quantization supplies the normalized oscillator expansion.

Helpful background. WKB, eikonal approximation, and turning points reviews complex turning points, and vacuum decay explains why an imaginary in–out effective action measures loss from the vacuum channel.

Time-dependent oscillator and conserved norm

Mode and correlator conventions. For the oscillator part of this page, positive frequency means $e^{-i\omega t}$. A normalized positive-frequency mode is

$$f(t)={e^{-i\omega t}\over\sqrt{2\omega}},$$

so that

$$i(f^*\dot f-\dot f^*f)=1.$$

The formulas below use $a_{\rm in}$ and $a_{\rm out}$ for annihilation operators associated with the past and future positive-frequency bases. Correlators are written without a leading $-i$, so the oscillator Feynman function obeys

$$\left(\partial_t^2+\omega^2(t)\right)G_F(t,t')=-i\delta(t-t').$$

Keeping this convention explicit prevents a common factor-of-$i$ error in the in–out Green function.

Consider a single real oscillator obeying

$$\left({d^2\over dt^2}+\omega^2(t)\right)f(t)=0,$$

with asymptotically constant frequency,

$$\omega(t)\longrightarrow \omega_{\rm in}\quad(t\to-\infty), \qquad \omega(t)\longrightarrow \omega_{\rm out}\quad(t\to+\infty).$$

A concrete example is

$$\omega^2(t)=\omega_0^2+U(t), \qquad U(t)\to0\quad(t\to\pm\infty),$$

where $U(t)$ is a finite disturbance. The equation is mathematically a one-dimensional scattering problem, except that the scattering coordinate is time. The conserved Wronskian is the analogue of flux conservation.

For any two solutions $f$ and $g$, define the Klein–Gordon product

$$(f,g)=i(f^*\dot g-\dot f^*g).$$

It is time independent. Indeed,

$${d\over dt}(f,g) =i(f^*\ddot g-\ddot f^*g) =i\left[-\omega^2(t)f^*g+\omega^2(t)f^*g\right]=0.$$

Choose the in-mode $f_{\rm in}$ by its past behavior,

$$f_{\rm in}(t)\underset{t\to-\infty}{\longrightarrow} {e^{-i\omega_{\rm in}t}\over\sqrt{2\omega_{\rm in}}},$$

and choose the out-mode $f_{\rm out}$ by its future behavior,

$$f_{\rm out}(t)\underset{t\to+\infty}{\longrightarrow} {e^{-i\omega_{\rm out}t}\over\sqrt{2\omega_{\rm out}}}.$$

The pair $f_{\rm out},f_{\rm out}^*$ is a basis of solutions. Therefore

$$\boxed{ f_{\rm in}=\alpha f_{\rm out}+\beta f_{\rm out}^*. }$$

The coefficients $\alpha$ and $\beta$ are the Bogoliubov coefficients.

Taking the product with itself gives

$$(f_{\rm in},f_{\rm in})=1,$$

while

$$(f_{\rm out},f_{\rm out})=1, \qquad (f_{\rm out}^*,f_{\rm out}^*)=-1, \qquad (f_{\rm out},f_{\rm out}^*)=0.$$

Hence

$$\boxed{ |\alpha|^2-|\beta|^2=1. }$$

This is not the ordinary conservation law $|T|^2+|R|^2=1$ of Schrödinger scattering. The negative-frequency solution has negative Klein–Gordon norm, so the group of transformations is hyperbolic rather than unitary. The same minus sign is responsible for amplification and pair creation.

A positive-frequency in-mode passes through a time-dependent background and becomes a mixture of positive- and negative-frequency out-modes. Wronskian conservation gives $|\alpha|^2-|\beta|^2=1$.

Operators and particle number

Quantize the oscillator by expanding the same Heisenberg operator in either basis:

$$\phi(t)=a_{\rm in}f_{\rm in}(t)+a_{\rm in}^\dagger f_{\rm in}^*(t) =a_{\rm out}f_{\rm out}(t)+a_{\rm out}^\dagger f_{\rm out}^*(t).$$

Substitute

$$f_{\rm in}=\alpha f_{\rm out}+\beta f_{\rm out}^*$$

and compare coefficients of $f_{\rm out}$ and $f_{\rm out}^*$. One obtains

$$\boxed{ a_{\rm out}=\alpha a_{\rm in}+\beta^*a_{\rm in}^\dagger, \qquad a_{\rm out}^\dagger=\alpha^*a_{\rm in}^\dagger+\beta a_{\rm in}. }$$

The commutator is preserved precisely because

$$[a_{\rm out},a_{\rm out}^\dagger] =|\alpha|^2-|\beta|^2=1.$$

The in-vacuum and out-vacuum are defined by

$$a_{\rm in}|0\rangle_{\rm in}=0, \qquad a_{\rm out}|0\rangle_{\rm out}=0.$$

The number of future quanta in the past vacuum is

$${}_{\rm in}\langle0|a_{\rm out}^\dagger a_{\rm out}|0\rangle_{\rm in} =|\beta|^2.$$

This is the first central result:

$$\boxed{ N_{\rm out\ particles\ in\ the\ in\ vacuum}=|\beta|^2. }$$

The result is not obtained by saying that the vacuum “contains particles” in an absolute sense. It says something operational: prepare the state with no particles in the past, let the background act, and count particles using detectors calibrated to the future Hamiltonian.

Bosons versus fermions

This page uses bosonic oscillators. For fermionic modes the algebra changes the normalization condition. A fermionic Bogoliubov transformation preserves anticommutators and obeys

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

not $|\alpha|^2-|\beta|^2=1$. The produced occupation number is still $|\beta|^2$, but now $0\le |\beta|^2\le1$ because of Pauli exclusion. In Schwinger pair creation this difference changes the multi-pair statistics and the signs in the vacuum persistence formula, while the leading exponential is controlled by the same complex turning points.

The in-vacuum as a squeezed out-state

The relation between the two vacua is especially transparent. Invert the Bogoliubov transformation:

$$a_{\rm in}=\alpha^*a_{\rm out}-\beta^*a_{\rm out}^\dagger.$$

The in-vacuum obeys

$$(\alpha^*a_{\rm out}-\beta^*a_{\rm out}^\dagger)|0\rangle_{\rm in}=0.$$

Now use the identity

$$a\exp\left({\zeta\over2}a^{\dagger 2}\right)|0\rangle =\zeta a^\dagger\exp\left({\zeta\over2}a^{\dagger 2}\right)|0\rangle.$$

It follows that

$$\boxed{ |0\rangle_{\rm in} =N\exp\left({1\over2}{\beta^*\over\alpha^*}a_{\rm out}^{\dagger 2}\right)|0\rangle_{\rm out}. }$$

The normalization is fixed by ${}_{\rm in}\langle0|0\rangle_{\rm in}=1$. Since

$$\left|{\beta\over\alpha}\right|^2=1-{1\over|\alpha|^2},$$

one finds

$$|N|^2={1\over|\alpha|}.$$

Thus the vacuum persistence probability for one real oscillator is

$$\boxed{ P_{0\to0}=|{}_{\rm out}\langle0|0\rangle_{\rm in}|^2={1\over|\alpha|}. }$$

The exponential contains only even powers of $a_{\rm out}^\dagger$. A centered time-dependent quadratic Hamiltonian preserves number parity and therefore creates quanta in pairs. Expanding the squeezed state gives

$$|0\rangle_{\rm in} =N\sum_{n=0}^{\infty}{1\over n!} \left({\beta^*\over2\alpha^*}\right)^n (a_{\rm out}^\dagger)^{2n}|0\rangle_{\rm out}.$$

Since

$$|2n\rangle_{\rm out}={(a_{\rm out}^\dagger)^{2n}\over\sqrt{(2n)!}}|0\rangle_{\rm out},$$

the amplitude ratio is

$$\boxed{ {A_{0\to2n}\over A_{0\to0}} ={\sqrt{(2n)!}\over2^n n!} \left({\beta^*\over\alpha^*}\right)^n ={ (2n-1)!!\over\sqrt{(2n)!}} \left({\beta^*\over\alpha^*}\right)^n. }$$

Odd-particle amplitudes vanish in this simple quadratic problem. The probability distribution is

$$\boxed{ P_{2n} ={ (2n)!\over 2^{2n}(n!)^2} { |\beta/\alpha|^{2n}\over |\alpha|}, \qquad P_{2n+1}=0. }$$

The probabilities sum to one because

$$\sum_{n=0}^{\infty}{(2n)!\over2^{2n}(n!)^2}x^n={1\over\sqrt{1-x}}, \qquad x=\left|{\beta\over\alpha}\right|^2.$$

The mean occupation number is

$$\sum_{n=0}^{\infty}2nP_{2n}=|\beta|^2,$$

as it must be.

The past vacuum is a squeezed state in the future basis. For a single real oscillator, only even occupation numbers occur, and the distribution is controlled by $\zeta=\beta^*/\alpha^*$.

In–out propagator and pair amplitudes

The ordinary Feynman Green function with in–out boundary conditions is

$$G_{\rm in\text{-}out}(t,t') ={{}_{\rm out}\langle0|T\phi(t)\phi(t')|0\rangle_{\rm in} \over {}_{\rm out}\langle0|0\rangle_{\rm in}}.$$

It is the inverse of the differential operator with Feynman boundary conditions: negative frequency toward the far past and positive frequency toward the far future. With the mode conventions above,

$$\boxed{ G_{\rm in\text{-}out}(t,t') ={1\over\alpha^*} f_{\rm in}^*(t_<)f_{\rm out}(t_>), \qquad t_<\equiv\min(t,t'),\quad t_>\equiv\max(t,t'). }$$

The factor $1/\alpha^*$ is not decorative. It normalizes the derivative jump at $t=t'$ so that

$$\left({d^2\over dt^2}+\omega^2(t)\right)G_{\rm in\text{-}out}(t,t')=-i\delta(t-t').$$

To see the pair amplitude inside the Green function, take both times late. Then

$$f_{\rm in}^*(t_<) \underset{t_<\to+\infty}{\longrightarrow} \alpha^*f_{\rm out}^*(t_<)+\beta^*f_{\rm out}(t_<),$$

and therefore

$$G_{\rm in\text{-}out}(t,t') \longrightarrow f_{\rm out}(t_>)f_{\rm out}^*(t_<) +{\beta^*\over\alpha^*}f_{\rm out}(t_>)f_{\rm out}(t_<).$$

For equal future frequency $\omega_{\rm out}$, this is

$$G_{\rm in\text{-}out}(t,t') \longrightarrow {1\over2\omega_{\rm out}}e^{-i\omega_{\rm out}|t-t'|} +{1\over2\omega_{\rm out}}{\beta^*\over\alpha^*}e^{-i\omega_{\rm out}(t+t')}.$$

The first term is ordinary propagation in the out-vacuum. The second term has the time dependence of two future annihilation operators acting on the in-vacuum:

$${}_{\rm out}\langle0|a_{\rm out}a_{\rm out}|0\rangle_{\rm in} =\sqrt2\,A_{0\to2}.$$

Comparing coefficients gives

$${A_{0\to2}\over A_{0\to0}}={1\over\sqrt2}{\beta^*\over\alpha^*},$$

which is the $n=1$ case of the squeezed-state result above.

The in–out Green function is fixed by a negative-frequency condition in the past and a positive-frequency condition in the future. When both insertions are late, it contains both ordinary propagation and a pair-production amplitude.

This is why in–out correlators are excellent for extracting amplitudes. They are not, however, expectation values in the produced state. For actual particle number, current, energy density, or backreaction, one should use in–in quantities.

Field theory: one oscillator per momentum mode

For a real scalar field in a spatially uniform time-dependent background,

$$\mathcal L={1\over2}\dot\phi^2-{1\over2}(\nabla\phi)^2 -{1\over2}\left[m^2+U(t)\right]\phi^2,$$

For literal number operators and normalized vacuum vectors, first use a periodic box of spatial volume $V$ and a finite, inversion-symmetric momentum cutoff. Each retained oscillator must have positive asymptotic frequency; a massless zero mode needs separate treatment. Box operators obey $[a_{\mathbf k},a_{\mathbf k'}^\dagger]=\delta_{\mathbf k\mathbf k'}$ and multiply normalized spatial waves $V^{-1/2}e^{i\mathbf k\cdot\mathbf x}$. The corresponding continuum Fourier notation is

$$\phi(\mathbf x,t)=\int{d^{d-1}\mathbf k\over(2\pi)^{d-1}} e^{i\mathbf k\cdot\mathbf x}\phi_{\mathbf k}(t),$$

with

$$\ddot f_{\mathbf k}(t)+\omega_{\mathbf k}^2(t)f_{\mathbf k}(t)=0, \qquad \omega_{\mathbf k}^2(t)=\mathbf k^2+m^2+U(t).$$

It is safest to include the spatial plane wave in the mode label. With

$$u_{\mathbf k}(x)=e^{i\mathbf k\cdot\mathbf x}f_{\mathbf k}(t),$$

complex conjugation reverses momentum, and the normalized modes satisfy

$$u_{{\mathbf k},{\rm in}} =\alpha_{\mathbf k}u_{{\mathbf k},{\rm out}} +\beta_{\mathbf k}u_{-{\mathbf k},{\rm out}}^*.$$

Consequently,

$$a_{{\mathbf k},{\rm out}} =\alpha_{\mathbf k}a_{{\mathbf k},{\rm in}} +\beta_{\mathbf k}^*a_{-{\mathbf k},{\rm in}}^\dagger,$$

and momentum conservation makes quanta appear in opposite-momentum pairs. For an isotropic background, $\alpha_{-\mathbf k}=\alpha_{\mathbf k}$ and $\beta_{-\mathbf k}=\beta_{\mathbf k}$. In the regulated box, the normalized squeezed state has the form

$$|0\rangle_{\rm in} \propto \exp\left[{1\over2}\sum_{\mathbf k} {\beta_{\mathbf k}^*\over\alpha_{\mathbf k}^*} a_{\mathbf k,\rm out}^\dagger a_{-\mathbf k,\rm out}^\dagger \right]|0\rangle_{\rm out},$$

with an overall normalization given by the finite product over independent modes. The future occupation in one normalized box mode is

$${}_{\rm in}\langle0|a_{\mathbf k,\rm out}^\dagger a_{\mathbf k,\rm out}|0\rangle_{\rm in} =|\beta_{\mathbf k}|^2.$$

The factor $1/2$ in the exponent compensates for summing over both $\mathbf k$ and $-\mathbf k$. One may instead sum over one representative of each distinct pair and omit that factor. A self-paired zero mode retains the single-oscillator factor.

In continuum normalization, $[a_{\mathbf k},a_{\mathbf k'}^\dagger]=(2\pi)^{d-1}\delta^{(d-1)}(\mathbf k-\mathbf k')$, and the distributional statement is

$$\langle a_{\mathbf k,\rm out}^\dagger a_{\mathbf k',\rm out}\rangle_{\rm in} =(2\pi)^{d-1}\delta^{(d-1)}(\mathbf k-\mathbf k')\,n_{\mathbf k}, \qquad n_{\mathbf k}=|\beta_{\mathbf k}|^2.$$

Thus $n_{\mathbf k}$ is an occupation per normalized mode, while the total number and its thermodynamic density are

$$\langle N\rangle_V=\sum_{\mathbf k}n_{\mathbf k}, \qquad \rho_N=\lim_{V\to\infty}{\langle N\rangle_V\over V} =\int{d^{d-1}\mathbf k\over(2\pi)^{d-1}}\,n_{\mathbf k},$$

when the ultraviolet and infrared limits converge. Setting the continuum delta function to one would lose the volume factor.

Removing the mode cutoff at fixed volume requires $\sum_{\mathbf k}|\beta_{\mathbf k}|^2<\infty$ for a bounded canonical map to be unitarily implementable. Finite density in infinite volume is a different statement: a nonzero homogeneous continuum mixing need not define a vacuum vector in the original Fock space. Bogoliubov transformations and unitary implementability explains this distinction and the precise Hilbert–Schmidt test. The squeezed vector and vacuum product here are regulated statements, not an assumed global unitary in the thermodynamic limit.

The vacuum persistence probability is a product over modes. Equivalently,

$$|{}_{\rm out}\langle0|0\rangle_{\rm in}|^2 =e^{-2\operatorname{Im}W},$$

where $W$ is the in–out effective action. A nonzero imaginary part of $W$ is therefore the many-mode version of the statement $|\alpha|>1$: the vacuum-to-vacuum channel has lost probability to states with particles.

For bookkeeping, distinguish two related squeeze problems. One real oscillator has

$$P_{0\to0}={1\over|\alpha|},$$

whereas one independent particle–antiparticle or $\mathbf k,-\mathbf k$ two-mode pair has

$$P_{0\to0}^{(\mathrm{pair})}={1\over|\alpha_{\mathbf k}|^2}.$$

The momentum product must count independent pairs only; any self-paired real zero mode is counted as one real oscillator. For a complex charged field and nonzero $\mathbf k$, the pair $(a_{\mathbf k},b_{-\mathbf k})$ and the pair $(a_{-\mathbf k},b_{\mathbf k})$ are distinct. At zero momentum, $a_0$ and $b_0$ still form a two-species pair. Thus all particle momenta label independent particle–antiparticle pairs. The volume factor and the distinction between mean multiplicity and vacuum survival are explicit in Gelis and Tanji 2015, §3.3, Eqs. (71)–(75), PDF pp. 22–23. In particular, a vanishing infinite-volume vacuum overlap does not by itself imply a divergent particle density.

Perturbative and adiabatic estimates

The coefficient $\beta$ is a reflection amplitude in time. If

$$\omega^2(t)=\omega_0^2+U(t), \qquad |U(t)|\ll \omega_0^2,$$

then the first Born approximation gives

$$\boxed{ \beta\simeq {i\over2\omega_0} \int_{-\infty}^{\infty}dt\,U(t)e^{-2i\omega_0t}. }$$

The phase $e^{-2i\omega_0t}$ is the energy cost of producing two oscillator quanta. Slow backgrounds have little Fourier support at frequency $2\omega_0$, so production is small.

A more geometric estimate comes from WKB. Write

$$f(t)\sim {1\over\sqrt{2\omega(t)}} \exp\left[-i\int^t dt'\,\omega(t')\right]$$

where the adiabaticity parameter is

$$\epsilon(t)={|\dot\omega(t)|\over\omega^2(t)}.$$

If $\epsilon\ll1$ on the real axis and $\omega(t)$ is analytic, pair production is controlled by complex turning points $t_\pm$ where

$$\omega(t_\pm)=0, \qquad t_-=t_+^*$$

for a real analytic profile. When a single conjugate pair dominates, the probability has the schematic form

$$|\beta|^2\sim \exp\left[-2\left|\operatorname{Im}\int_{t_-}^{t_+}dt\,\omega(t)\right|\right].$$

The contour, Stokes sector, and possible interference among several turning-point pairs are part of the approximation; the displayed formula records only the leading isolated-pair exponent. It makes precise the old intuition that a particle can be created only by a sufficiently nonadiabatic background. If the background changes abruptly, $\beta$ is often only power-law suppressed. In the analytic, slowly varying regime with the isolated turning points specified above, production is exponentially suppressed.

A useful special case is a very slow change of mass. If the frequency evolves from $\omega_{\rm in}$ to $\omega_{\rm out}$ over a time scale $T$ with $\omega T\gg1$, the action variable of the oscillator is adiabatically conserved and $\beta\to0$. The energy changes because the Hamiltonian changes, but the occupation number does not.

Pair creation in a spatially uniform electric field

Now consider a charged scalar field in a classical electric field pointing in the $z$ direction. Use the gauge

$$A_0=0, \qquad A_z=A_z(t), \qquad E_z(t)=-\dot A_z(t).$$

Here $A_z$ denotes the $z$-component of the spatial vector $\mathbf A$. With the mostly-minus metric, the covariant four-potential is $A_\mu=(A_0,-\mathbf A)$. Thus $D_\mu=\partial_\mu-iqA_\mu$ gives the kinetic momentum $k_z+qA_z(t)$.

Here $q$ is the signed coefficient in the covariant derivative. The positive-frequency particle sector has Maxwell-source charge $Q=-q$: with the inherited action convention, the current obtained by varying the matter action is the negative of the source in $\partial_\mu F^{\mu\nu}=J^\nu$. The current derivation below fixes this sign explicitly. Thus $\dot\pi_z=-qE_z=QE_z$, and a positive source charge accelerates along the electric field. The scalar action and derivative convention agree with Gelis and Tanji 2015, §3.1, Eq. (54), PDF pp. 18–19, with their $g=q$; positive-charge acceleration must still be assigned to the stated particle sector.

For derivative coefficient $q$, a Fourier mode

$$\phi(\mathbf x,t)=e^{i\mathbf k\cdot\mathbf x}f_{\mathbf k}(t)$$

obeys

$$\boxed{ \ddot f_{\mathbf k}(t)+\omega_{\mathbf k}^2(t)f_{\mathbf k}(t)=0, \qquad \omega_{\mathbf k}^2(t)=m^2+k_\perp^2+\left(k_z+qA_z(t)\right)^2. }$$

This is again a time-dependent oscillator. The canonical momentum $k_z$ labels the Fourier mode. The physical kinetic momentum is

$$\pi_z(t)=k_z+qA_z(t).$$

For a constant electric field, choose

$$A_z(t)=-Et.$$

Then

$$\omega_{\mathbf k}^2(t)=m_\perp^2+(k_z-qEt)^2, \qquad m_\perp^2=m^2+k_\perp^2.$$

The mode is most nonadiabatic near the time when the kinetic longitudinal momentum crosses zero,

$$k_z-qEt=0.$$

The complex turning points are

$$t_\pm={k_z\pm i m_\perp\over qE}.$$

The WKB exponent is

$$2\left|\operatorname{Im}\int_{t_-}^{t_+}dt\,\omega_{\mathbf k}(t)\right| ={\pi m_\perp^2\over |qE|}.$$

Therefore

$$\boxed{ |\beta_{\mathbf k}|^2 =\exp\left[-{\pi(m^2+k_\perp^2)\over |qE|}\right] }$$

for a constant field. This is the exact mean occupation number per scalar mode in an eternal uniform field, and it is also the locally constant result for a long pulse away from its switching regions. Because an eternal field never becomes field-free, its in/out states are defined by the positive-frequency WKB branches as $t\to\pm\infty$, not by free plane waves with a fixed kinetic momentum.

The mode exponent also gives the leading pair-production rate in a long constant field. During a time interval $T$ in a box of length $L_z$, the kinetic momentum $k_z-qEt$ sweeps through the nonadiabatic region for

$${L_z\over2\pi}|qE|T$$

longitudinal modes. Therefore, in $3+1$ dimensions and in the dilute scalar limit, the mean number $N$ of produced pairs satisfies

$${N\over VT} \simeq {|qE|\over2\pi} \int {d^2k_\perp\over(2\pi)^2} \exp\left[-{\pi(m^2+k_\perp^2)\over |qE|}\right] ={ (qE)^2\over8\pi^3} \exp\left[-{\pi m^2\over |qE|}\right].$$

This is the leading term in the scalar Schwinger rate. The full vacuum persistence probability resums multiple production events and, for spinor QED, includes spin degeneracy and Fermi statistics.

In the figure, distinguish the mode label $q$ from the physical source charges $Q$ on the separating particles. The upper panel shows the minimum real gap; the lower panel shows which charge accelerates in each direction.

A spatially uniform electric field gives $\omega_{\mathbf k}^2=m_\perp^2+\pi_z^2$, with $\pi_z=k_z+qA_z$. The minimum real gap and adjacent complex turning points fix the constant-field production exponent. For the illustrated $q>0$ and $E>0$, the particle has source charge $Q=-q$ and accelerates left, while its antiparticle accelerates right. The geometry is schematic, with no numerical axis scale; the displayed constant-field exponent is exact for the prescribed scalar background.

There is also a simple spacetime estimate behind the exponent. To create a pair of rest mass $m$, the field must do work of order $2m$. If the particles separate by a distance $\Delta z$, the work is roughly

$$|qE|\Delta z\sim2m.$$

Quantum mechanics allows such a process through tunneling over a distance of order

$$\Delta z\sim {2m\over |qE|}.$$

The precise calculation replaces this estimate by the exponential $e^{-\pi m^2/|qE|}$, but the physical content is the same: strong fields shorten the tunneling distance and enhance pair production.

In–in current and backreaction

The in–out effective action tells us whether the vacuum remains the vacuum. The electric current produced by the pairs is an in–in observable. For $\mathcal L_{\rm m}=(D_\mu\phi)^*D^\mu\phi-m^2|\phi|^2$, variation with respect to the covariant potential gives

$$j_q^\mu\equiv{\delta S_{\rm m}\over\delta A_\mu} =iq\left[\phi^*D^\mu\phi-(D^\mu\phi)^*\phi\right].$$

With the Maxwell term $-F_{\mu\nu}F^{\mu\nu}/4$, its equation and the physical source current are

$$\partial_\mu F^{\mu\nu}+j_q^\nu=0, \qquad J^\nu=-j_q^\nu.$$

These signs also fix the particle charge. For a unit-normalized positive-frequency plane wave, the action-current charge is $q$, so the Maxwell-source charge is $Q=-q$. Its antiparticle has charge $-Q=q$. A spatial lower index introduces an additional minus sign: $j_{q,z}=-j_q^z$ and $J_z=-J^z$.

To fix the mode normalization, expand the complex field in a spatially homogeneous background as

$$\phi(\mathbf x,t)=\int{d^3\mathbf k\over(2\pi)^3}e^{i\mathbf k\cdot\mathbf x} \left[a_{\mathbf k,\rm in}f_{\mathbf k}(t) +b_{-\mathbf k,\rm in}^\dagger f_{\mathbf k}^*(t)\right].$$

Both time functions solve the same real oscillator equation for this Fourier label, and $i(f_{\mathbf k}^*\dot f_{\mathbf k}-\dot f_{\mathbf k}^*f_{\mathbf k})=1$. The $a$ and $b$ operators have the continuum commutator given above and annihilate the in-vacuum. Since $D^z=-i\pi_z$ on each spatial plane wave, the vacuum contraction in the $b b^\dagger$ sector gives

$$\begin{aligned} \langle j_q^z(t)\rangle_{\rm ren} &=2q\int{d^3\mathbf k\over(2\pi)^3}\, \pi_z(t)|f_{\mathbf k}(t)|^2_{\rm sub},\\ \langle J^z(t)\rangle_{\rm ren} &=-2q\int{d^3\mathbf k\over(2\pi)^3}\, \pi_z(t)|f_{\mathbf k}(t)|^2_{\rm sub}. \end{aligned}$$

The subscript denotes subtraction of the local ultraviolet terms with a consistent gauge-invariant current prescription, including charge renormalization, for example the appropriate adiabatic subtraction. It does not mean that arbitrary mode-by-mode vacuum removal is gauge invariant. One may first derive the contractions with regulated sums; a fixed canonical-momentum cutoff alone is not the final gauge-invariant prescription.

After a pulse has ended, let $\omega_{\mathbf k,\rm out}$ be the constant future frequency and write $\theta_{\mathbf k}=\omega_{\mathbf k,\rm out}t$. Subtracting the out-vacuum term leaves

$$|f_{\mathbf k}|^2-{1\over2\omega_{\mathbf k,\rm out}} ={|\beta_{\mathbf k}|^2+\operatorname{Re} [\alpha_{\mathbf k}\beta_{\mathbf k}^*e^{-2i\theta_{\mathbf k}}] \over\omega_{\mathbf k,\rm out}}.$$

If the oscillatory coherence dephases under momentum integration or an explicitly coarse time resolution, the conduction part is

$$\langle J^z\rangle_{\rm cond} =-2q\int{d^3\mathbf k\over(2\pi)^3} {\pi_{z,\rm out}\over\omega_{\mathbf k,\rm out}}|\beta_{\mathbf k}|^2.$$

The factor two counts the two charges in each pair: $Q(\pi_z/\omega)$ and $(-Q)(-\pi_z/\omega)$ have the same sign. The full transient current also contains coherence and vacuum polarization; it is not determined by particle number alone or by the instantaneous value of $E(t)$.

For an electric-field pulse, the vector potential changes by

$$A_z(t)-A_z(-\infty)=-\int_{-\infty}^t dt'\,E_z(t').$$

Thus the signed kinetic-momentum change is

$$\Delta\pi_z(t)=-q\int_{-\infty}^t dt'\,E_z(t'),$$

and a late-time current can depend on this accumulated impulse. This is not a violation of gauge invariance. A constant shift of $A_z$ is compensated by relabeling the canonical momentum $k_z$, while the difference $A_z(t)-A_z(-\infty)$ is fixed by the physical field history.

The next figure isolates a possible late conduction current after a positive pulse. Its curves illustrate retained production and impulse, not a computed transient current or a claim that coherence always decays monotonically.

Schematic collisionless response for $q>0$ and a positive pulse: the particle’s signed impulse is $\Delta\pi_z=-q\int E\,dt$, while its source charge is $Q=-q$. The particle and antiparticle contributions to $J^z$ can therefore add to a positive conduction current after the pulse. The late curve assumes produced pairs and dephased coherence; it is not a numerical solution for the full renormalized transient current. All curves have arbitrary vertical scales.

If the electric field is externally prescribed, this current is an output. For a homogeneous dynamical field, Maxwell’s equation is $\dot E_z=-\langle J^z\rangle_{\rm ren}$ in the normalization above. A positive conduction current drains a positive field; energy balance gives $\partial_t(E_z^2/2)=-E_z\langle J^z\rangle_{\rm ren}$. Continuous production eventually invalidates the fixed-background approximation. Coherent transients and plasma oscillations can occur, so screening need not be monotonic.

Periodic driving and moving-frame instabilities

Next contrast nonperturbative production with ordinary thresholds. Suppose a weak perturbation contains a Fourier component of frequency $\Omega$. At first order, creating two massive quanta requires

$$\Omega\ge 2m.$$

At $n$th order, $n$ quanta of the drive can combine, and the threshold becomes

$$n\Omega\ge2m.$$

At any fixed perturbative order, sufficiently small $\Omega$ lies below that order’s threshold. This does not forbid a strong slowly varying field from creating pairs through complex turning points. At nonzero $\Omega$, sufficiently high multiphoton orders can also cross the threshold. The constant-field Schwinger factor is the distinct zero-frequency example: it is nonanalytic in the field strength at $E=0$ and therefore invisible at every finite order in a power series in $E$.

The same energy-accounting logic diagnoses a moving body’s energetic instability. Suppose a heavy impurity moves at $\mathbf v=v\hat{\mathbf v}$ through a homogeneous medium at zero temperature. For a branch with $\epsilon(\mathbf p)>0$ at nonzero momentum, the leading net energy cost of transferring momentum $\mathbf p$ from the body into one excitation is

$$\Delta E(\mathbf p)=\epsilon(\mathbf p)-\mathbf v\cdot\mathbf p.$$

The neglected nonrelativistic recoil term is $|\mathbf p|^2/(2M)$ for impurity mass $M$. In the infinite-mass limit, a negative-cost excitation exists precisely when

$$\boxed{ v>v_c(\hat{\mathbf v}), \qquad v_c(\hat{\mathbf v})=\inf_{\hat{\mathbf v}\cdot\mathbf p>0} {\epsilon(\mathbf p)\over\hat{\mathbf v}\cdot\mathbf p}. }$$

The denominator must retain the direction of motion when the dispersion is anisotropic Yu 2017, Eqs. (1)–(3), PDF p. 1. For an isotropic branch $\epsilon=\epsilon(p)$, choosing momentum parallel to the velocity minimizes the cost at fixed $p$, and this reduces to $v_c=\inf_{p>0}\epsilon(p)/p$. A minimum may fail to be attained; equality $v=v_c$ does not give a negative-cost excitation.

For example, $\epsilon=\sqrt{4p_x^2+p_y^2}$ gives $v_c=2$ for motion along $x$, since $\epsilon\ge2|p_x|$. Minimizing $\epsilon/|\mathbf p|$ instead gives 1 along $y$, where the body transfers no energy. For the isotropic linear branch $\epsilon=cp$, the criterion is $v>c$.

This is an energetic threshold for the specified branch and heavy-impurity problem. An actual emission rate also requires an allowed coupling, nonzero matrix elements, and energy-conserving phase space. Other excitation channels can set a lower threshold. It does not explain Rindler thermality: an accelerated observer in vacuum instead changes the time generator and has access to only one causal wedge.

In a compact isolated system, a permanently driven state need not approach a steady state. Energy has nowhere to escape, recurrences can matter, and backreaction must eventually be included.

Summary

A time-dependent quadratic background mixes positive and negative frequencies. The phase of the Bogoliubov coefficient $\beta$ depends on mode conventions, while $|\beta|$ measures the mixing invariantly and $|\beta|^2$ is the number of produced quanta in the corresponding out-mode. The identity

$$|\alpha|^2-|\beta|^2=1$$

is Wronskian conservation, or equivalently preservation of canonical commutation relations.

The in-vacuum is a squeezed state in the out-basis. For a single real oscillator,

$$|0\rangle_{\rm in} =N\exp\left({\beta^*\over2\alpha^*}a_{\rm out}^{\dagger 2}\right)|0\rangle_{\rm out}, \qquad |N|^2={1\over|\alpha|}.$$

In–out propagators contain transition amplitudes such as $A_{0\to2n}$. In–in expectation values contain actual observables such as produced particle number, current, energy density, and backreaction.

A spatially uniform electric field turns each charged momentum mode into a time-dependent oscillator with

$$\omega_{\mathbf k}^2(t)=m^2+k_\perp^2+(k_z+qA_z(t))^2.$$

For a constant field,

$$|\beta_{\mathbf k}|^2=\exp\left[-{\pi(m^2+k_\perp^2)\over |qE|}\right],$$

which is the basic Schwinger exponent. The next page will use the same positive/negative-frequency logic in a different setting: mode functions adapted to accelerated observers and Rindler wedges.

Common pitfalls

The most common mistake is to call $\beta$ a reflection coefficient and then impose $|\alpha|^2+|\beta|^2=1$. That is the wrong conservation law. Negative-frequency modes have negative Klein–Gordon norm, so the correct identity is $|\alpha|^2-|\beta|^2=1$.

Another common mistake is to treat particle number as absolute. In time-dependent backgrounds, particle number is defined with respect to a choice of positive-frequency modes. It is unambiguous when the background becomes time independent in the past and future. If the background never switches off, one must use an adiabatic, detector-based, or otherwise physical prescription.

Finally, an in–out current is not the produced current. The in–out effective action diagnoses vacuum persistence and transition amplitudes. Backreaction is driven by the in–in Maxwell-source current $J^\mu=-j_q^\mu$. Do not identify a spatial lower component with the upper component, or the derivative coefficient $q$ with the source charge $Q=-q$ of the chosen positive-frequency mode.

A further trap is to identify every exponentially small production process with a perturbative threshold. Multiphoton thresholds depend on Fourier support and order in the driving amplitude; the Schwinger process is nonanalytic at $E=0$ and is controlled by complex turning points.

Exercises

Exercise 1: Wronskian normalization

Let $f_{\rm in}=\alpha f_{\rm out}+\beta f_{\rm out}^*$ and assume

$$(f_{\rm out},f_{\rm out})=1, \qquad (f_{\rm out}^*,f_{\rm out}^*)=-1, \qquad (f_{\rm out},f_{\rm out}^*)=0.$$

Show that $(f_{\rm in},f_{\rm in})=1$ implies $|\alpha|^2-|\beta|^2=1$.

Solution

Use sesquilinearity of the Klein–Gordon product:

$$(f_{\rm in},f_{\rm in}) =(\alpha f_{\rm out}+\beta f_{\rm out}^*,\alpha f_{\rm out}+\beta f_{\rm out}^*).$$

The cross terms vanish because

$$(f_{\rm out},f_{\rm out}^*)=(f_{\rm out}^*,f_{\rm out})=0.$$

Thus

$$(f_{\rm in},f_{\rm in}) =|\alpha|^2(f_{\rm out},f_{\rm out})+|\beta|^2(f_{\rm out}^*,f_{\rm out}^*) =|\alpha|^2-|\beta|^2.$$

Since $f_{\rm in}$ is normalized to one,

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

Exercise 2: squeezed-state probabilities

Starting from

$$a_{\rm in}=\alpha^*a_{\rm out}-\beta^*a_{\rm out}^\dagger,$$

derive the squeezed-state expression for $|0\rangle_{\rm in}$ and the probability

$$P_{2n} ={ (2n)!\over 2^{2n}(n!)^2} { |\beta/\alpha|^{2n}\over |\alpha|}.$$

Solution

The condition $a_{\rm in}|0\rangle_{\rm in}=0$ becomes

$$(\alpha^*a_{\rm out}-\beta^*a_{\rm out}^\dagger)|0\rangle_{\rm in}=0.$$

Try

$$|0\rangle_{\rm in}=N\exp\left({\zeta\over2}a_{\rm out}^{\dagger 2}\right)|0\rangle_{\rm out}.$$

Since

$$a\exp\left({\zeta\over2}a^{\dagger 2}\right)|0\rangle =\zeta a^\dagger\exp\left({\zeta\over2}a^{\dagger 2}\right)|0\rangle,$$

the vacuum condition gives

$$\alpha^*\zeta-\beta^*=0, \qquad \zeta={\beta^*\over\alpha^*}.$$

Expanding the exponential,

$$|0\rangle_{\rm in} =N\sum_{n=0}^\infty{1\over n!} \left({\zeta\over2}\right)^n(a^\dagger)^{2n}|0\rangle.$$

Using

$$(a^\dagger)^{2n}|0\rangle=\sqrt{(2n)!}|2n\rangle,$$

we get

$$A_{0\to2n}=N{\sqrt{(2n)!}\over2^n n!}\zeta^n.$$

The normalization is

$$1=|N|^2\sum_{n=0}^\infty{(2n)!\over2^{2n}(n!)^2}|\zeta|^{2n} =|N|^2{1\over\sqrt{1-|\zeta|^2}}.$$

Because

$$1-|\zeta|^2=1-{|\beta|^2\over|\alpha|^2}={1\over|\alpha|^2},$$

we find

$$|N|^2={1\over|\alpha|}.$$

Therefore

$$P_{2n}=|A_{0\to2n}|^2 ={ (2n)!\over 2^{2n}(n!)^2} { |\beta/\alpha|^{2n}\over |\alpha|}.$$

Exercise 3: Born approximation for mode mixing

For

$$\omega^2(t)=\omega_0^2+U(t), \qquad |U(t)|\ll\omega_0^2,$$

show that the first Born approximation has the form

$$\beta\simeq {i\over2\omega_0}\int_{-\infty}^{\infty}dt\,U(t)e^{-2i\omega_0t}$$

up to a convention-dependent overall phase.

Solution

Write the solution as a slowly corrected superposition of free modes,

$$f(t)={1\over\sqrt{2\omega_0}} \left[A(t)e^{-i\omega_0t}+B(t)e^{+i\omega_0t}\right],$$

with $A(-\infty)=1$ and $B(-\infty)=0$. To first order in $U$, one may compute the negative-frequency amplitude generated by the perturbation using the free retarded Green function of $d^2/dt^2+\omega_0^2$:

$$G_R(t-t')={\theta(t-t')\over\omega_0}\sin\omega_0(t-t').$$

The equation is

$$(\partial_t^2+\omega_0^2)f(t)=-U(t)f(t).$$

Insert the zeroth-order solution

$$f^{(0)}(t)={e^{-i\omega_0t}\over\sqrt{2\omega_0}}$$

on the right-hand side. The late-time correction is

$$\delta f(t)=-\int dt'\,G_R(t-t')U(t'){e^{-i\omega_0t'}\over\sqrt{2\omega_0}}.$$

For $t$ later than the support of $U$,

$$\sin\omega_0(t-t')={1\over2i} \left(e^{i\omega_0(t-t')}-e^{-i\omega_0(t-t')}\right).$$

The coefficient of $e^{+i\omega_0t}/\sqrt{2\omega_0}$ is therefore

$$\beta\simeq -{1\over2i\omega_0}\int dt'\,U(t')e^{-2i\omega_0t'} ={i\over2\omega_0}\int dt'\,U(t')e^{-2i\omega_0t'}.$$

Changing the phase convention for the negative-frequency mode changes the overall phase of $\beta$, but not $|\beta|^2$.

Exercise 4: constant-field Schwinger exponent

For a charged scalar in a constant electric field, use

$$\omega_{\mathbf k}^2(t)=m_\perp^2+(k_z-qEt)^2$$

and the complex turning points

$$t_\pm={k_z\pm i m_\perp\over qE}$$

to show that

$$|\beta_{\mathbf k}|^2=\exp\left[-{\pi m_\perp^2\over |qE|}\right].$$

Solution

The answer depends only on $|qE|$, so take $qE>0$ while evaluating the oriented contour; reversing the field interchanges the two turning points. The WKB exponent is

$$2\operatorname{Im}\int_{t_-}^{t_+}dt\,\omega_{\mathbf k}(t).$$

Set

$$\nu=qEt-k_z, \qquad dt={d\nu\over qE}.$$

The turning points become

$$\nu_-=-i m_\perp, \qquad \nu_+=+i m_\perp.$$

Thus

$$\int_{t_-}^{t_+}dt\,\omega_{\mathbf k}(t) ={1\over qE}\int_{-im_\perp}^{+im_\perp}d\nu\,\sqrt{m_\perp^2+\nu^2}.$$

Put $\nu=i y$, with $y$ running from $-m_\perp$ to $m_\perp$. Then $d\nu=i dy$ and

$$\sqrt{m_\perp^2+\nu^2}=\sqrt{m_\perp^2-y^2}.$$

Therefore

$$\int_{t_-}^{t_+}dt\,\omega_{\mathbf k}(t) ={i\over qE}\int_{-m_\perp}^{m_\perp}dy\,\sqrt{m_\perp^2-y^2}.$$

The remaining integral is the area of a semicircle of radius $m_\perp$:

$$\int_{-m_\perp}^{m_\perp}dy\,\sqrt{m_\perp^2-y^2} ={\pi m_\perp^2\over2}.$$

Hence

$$2\operatorname{Im}\int_{t_-}^{t_+}dt\,\omega_{\mathbf k}(t) ={\pi m_\perp^2\over |qE|}.$$

The tunneling probability is the exponential of minus this quantity:

$$|\beta_{\mathbf k}|^2=\exp\left[-{\pi m_\perp^2\over |qE|}\right].$$

Exercise 5: produced occupation number

Show directly from

$$a_{\rm out}=\alpha a_{\rm in}+\beta^*a_{\rm in}^\dagger$$

that the in-vacuum contains $|\beta|^2$ out-quanta on average.

Solution

Compute

$$N_{\rm out}=a_{\rm out}^\dagger a_{\rm out}.$$

Using

$$a_{\rm out}^\dagger=\alpha^*a_{\rm in}^\dagger+\beta a_{\rm in},$$

we get

$$a_{\rm out}^\dagger a_{\rm out} =(\alpha^*a^\dagger+\beta a)(\alpha a+\beta^*a^\dagger),$$

where $a=a_{\rm in}$. Expanding,

$$N_{\rm out}=|\alpha|^2a^\dagger a+ \alpha^*\beta^*a^\dagger a^\dagger+ \alpha\beta aa+|\beta|^2aa^\dagger.$$

Taking the expectation value in the in-vacuum kills all terms except the last one:

$${}_{\rm in}\langle0|aa^\dagger|0\rangle_{\rm in}=1.$$

Therefore

$${}_{\rm in}\langle0|N_{\rm out}|0\rangle_{\rm in}=|\beta|^2.$$

Exercise 6: dilute scalar production rate

Use the mode probability

$$|\beta_{\mathbf k}|^2= \exp\left[-{\pi(m^2+k_\perp^2)\over |qE|}\right]$$

for a constant electric field in $3+1$ dimensions to derive the dilute scalar estimate

$${N\over VT}\simeq { (qE)^2\over8\pi^3} \exp\left[-{\pi m^2\over |qE|}\right].$$

Solution

In a time $T$, the longitudinal kinetic momentum changes by

$$|\Delta \pi_z|=|qE|T.$$

With box normalization, the density of longitudinal modes is $L_z/(2\pi)$, so the number of longitudinal modes that pass through the nonadiabatic region is

$${L_z\over2\pi}|qE|T.$$

The number of produced pairs is therefore

$$N\simeq {L_z|qE|T\over2\pi} L_xL_y\int {d^2k_\perp\over(2\pi)^2} \exp\left[-{\pi(m^2+k_\perp^2)\over |qE|}\right].$$

Divide by $V=L_xL_yL_z$ and by $T$:

$${N\over VT} \simeq {|qE|\over2\pi}e^{-\pi m^2/|qE|} \int {d^2k_\perp\over(2\pi)^2} e^{-\pi k_\perp^2/|qE|}.$$

The Gaussian integral is

$$\int {d^2k_\perp\over(2\pi)^2} e^{-\pi k_\perp^2/|qE|} ={|qE|\over4\pi^2}.$$

Thus

$${N\over VT}\simeq { (qE)^2\over8\pi^3} \exp\left[-{\pi m^2\over |qE|}\right].$$

This calculation gives the leading dilute production rate, not the full multi-pair vacuum persistence formula.

Exercise 7: Landau critical velocity

An excitation branch has $\epsilon(\mathbf p)>0$ for nonzero momentum in the medium rest frame. In the infinite-mass impurity limit, show that a negative-cost excitation exists for motion along $\hat{\mathbf v}$ precisely when

$$v>v_c(\hat{\mathbf v}),\qquad v_c(\hat{\mathbf v})=\inf_{\hat{\mathbf v}\cdot\mathbf p>0} {\epsilon(\mathbf p)\over\hat{\mathbf v}\cdot\mathbf p}.$$

Recover the isotropic result, and evaluate the threshold for $\epsilon=\sqrt{4p_x^2+p_y^2}$ with motion along $x$. Explain why this criterion alone does not determine an emission rate.

Solution

In the heavy-body limit, transferring momentum $\mathbf p$ changes the body’s kinetic energy by $-\mathbf v\cdot\mathbf p$ to leading order. The net energy cost is therefore

$$\Delta E=\epsilon(\mathbf p)-\mathbf v\cdot\mathbf p.$$

If $\hat{\mathbf v}\cdot\mathbf p\le0$, the cost is positive. Otherwise, dividing by the positive longitudinal projection gives

$$\Delta E<0\quad\Longleftrightarrow\quad v>{\epsilon(\mathbf p)\over\hat{\mathbf v}\cdot\mathbf p}.$$

Some momentum satisfies this inequality precisely when $v$ exceeds the infimum. At equality no ratio lies below $v$. If $\epsilon$ depends only on $p=|\mathbf p|$, alignment with $\hat{\mathbf v}$ maximizes the denominator without changing the numerator, giving

$$v_c=\inf_{p>0}{\epsilon(p)\over p}.$$

For a linear isotropic branch $\epsilon=cp$, this is $c$. For the anisotropic example, $p_x>0$ gives $\epsilon/p_x=\sqrt{4+(p_y/p_x)^2}\ge2$, with equality at $p_y=0$, so $v_c(\hat x)=2$. The unrelated radial minimum 1 lies along $y$ and cannot describe motion along $x$. A nonzero rate further requires a coupling to the mode and energy-conserving phase space; a forbidden matrix element is not repaired by a negative energy cost.

References

• Gelis, François, and Naoto Tanji. “Schwinger Mechanism Revisited.” Progress in Particle and Nuclear Physics 87 (2016): 1–49. DOI; Open PDF, 2015 preprint v1.
• Yu, Zeng-Qiang. “Landau Criterion for an Anisotropic Bose–Einstein Condensate.” Physical Review A 95 (2017): 033618. DOI; Open PDF, preprint v1.

Further reading

• Birrell, N. D., and P. C. W. Davies. Quantum Fields in Curved Space. Cambridge Monographs on Mathematical Physics. Cambridge University Press, 1982.
• Dunne, Gerald V. “Heisenberg–Euler Effective Lagrangians: Basics and Extensions.” In From Fields to Strings: Circumnavigating Theoretical Physics, edited by M. Shifman, A. Vainshtein, and J. Wheater, vol. 1, 445–522. World Scientific, 2005.
• Kim, Sang Pyo, and Don N. Page. “Schwinger Pair Production in Electric and Magnetic Fields.” Physical Review D 73 (2006): 065020.
• Parker, Leonard, and David Toms. Quantum Field Theory in Curved Spacetime: Quantized Fields and Gravity. Cambridge University Press, 2009.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014, chapters 33–34.
• Schwinger, Julian. “On Gauge Invariance and Vacuum Polarization.” Physical Review 82 (1951): 664–679.
• Weinberg, Steven. The Quantum Theory of Fields. Vol. 1, Foundations. Cambridge University Press, 1995.