Fakultät für Physik

Solutions of the Dirac equation inspacetime-dependent electric fields

Master’s thesis

by

Johannes Oertel

from

Freiberg

September 2014

Abstract

Solving the Dirac equation with spacetime-dependent potentials is crucial for theunderstanding of pair creation due to the Sauter-Schwinger effect. Only few ex-act solutions are known today, but no solutions have been established where theelectric field does depend on more than one spacetime coordinate. In this the-sis, a method for generating solutions of the Dirac equation in the presence ofspacetime-dependent electromagnetic fields is developed. By swapping the rolesof known and unknown quantities in the Dirac equation, we are able to choosearbitrary solutions and calculate the corresponding electromagnetic field. Usingthe presented method, solutions with an electromagnetic field depending on ei-ther one of the light cone coordinates or both can be found in (1 + 1) and (2 + 1)dimensions. Additionally, choosing a specific ansatz, it is possible to formulateconditions for solutions that correspond to non-perturbative pair creation.

Lösungen der Dirac-Gleichung inraumzeitabhängigen elektrischen Feldern

Zusammenfassung

Für ein vertieftes Verständnis der Paarerzeugung durch den Sauter-Schwinger-Effekt ist es notwendig, die Dirac-Gleichung mit raumzeitabhängigen Potentia-len zu lösen. Bis heute sind nur wenige exakte Lösungen bekannt. Zudem sindbisher keine Lösungen für elektrische Felder gefunden worden, die von mehr alseiner Raumzeitkoordinate abhängen. Eine Methode zur Erzeugung von Lösungender Dirac-Gleichung mit raumzeitabhängigen elektromagnetischen Felder wird indieser Arbeit entwickelt. Durch das Vertauschen der Rollen von bekannten undunbekannten Größen in der Dirac-Gleichung ist es möglich aus der Wahl einerbeliebigen Lösung das entsprechende elektromagnetische Feld zu berechnen. Mitdieser Methode können Lösungen in (1 + 1) und (2 + 1) Dimensionen gefundenwerden, bei denen das elektromagnetische Feld von einer oder sogar beiden Licht-kegelkoordinaten abhängt. Zusätzlich können durch Wahl eines bestimmten An-satz Bedingungen an solche Lösungen formuliert werden, die nicht-perturbativerPaarerzeugung entsprechen.

iii

Contents

1. Introduction 1

2. Dirac equation 32.1. Motivation and derivation . . . . . . . . . . . . . . . . . . . . . . 32.2. Gamma matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3. Minimal coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4. Light cone coordinates . . . . . . . . . . . . . . . . . . . . . . . . 102.5. Lorentz transformations . . . . . . . . . . . . . . . . . . . . . . . 122.6. Free solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.7. Pair creation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3. Known exact solutions 213.1. Constant electric field . . . . . . . . . . . . . . . . . . . . . . . . . 213.2. Constant magnetic field . . . . . . . . . . . . . . . . . . . . . . . 223.3. Constant orthogonal electric and magnetic field . . . . . . . . . . 233.4. Sauter pulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.5. Lightfront field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4. Inverse approach to the Dirac equation 274.1. Basic formalism . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.2. Plane wave solutions . . . . . . . . . . . . . . . . . . . . . . . . . 304.3. Single moving pulses . . . . . . . . . . . . . . . . . . . . . . . . . 334.4. Two moving pulses . . . . . . . . . . . . . . . . . . . . . . . . . . 364.5. Emerging pulses . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.6. Perturbed solution . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4.6.1. Expansion in powers of the mass parameter . . . . . . . . 414.6.2. Expansion of the mixing square root . . . . . . . . . . . . 42

5. Inverse approach in (2+1) dimensions 475.1. Basic formalism . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475.2. Landau levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525.3. Plane wave solutions . . . . . . . . . . . . . . . . . . . . . . . . . 535.4. Single wavefront . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545.5. Two wavefronts . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

6. Conclusion and outlook 63

v

Contents

A. Expansion in powers of the mass parameter 65

Bibliography 67

vi

1. Introduction

When Dirac formulated his famous equation in 1928, he had been looking for arelativistic wave equation that could explain certain “duplexity phenonema” thatappeared when calculating the stationary states for an electron in an atom [Dir28]:The number of states measured in the experiment is twice the number from calcula-tions using the (non-relativistic) Schrödinger equation. As Dirac states, Goudsmitand Uhlenbeck introduced the idea of the electron being a spin-1

2particle. Indeed

Pauli was able to show that the number of states comes out right if one introducesa spin angular momentum which can only take on the values ±1

2~ [Pau27]. Dirac

was not completely satisfied with this spin model as it is based on the assumptionthat the electron is not simply a point charge, but also has got the spin as an ad-ditional property. He guessed that the quantum mechanical theory of the electronis incomplete and assumed that this is due to the fact that Schrödinger’s quantummechanics is a non-relativistic theory. As it turns out, Dirac could reproduce theresults of Pauli’s spin model under the assumption that the electron is a pointcharge which should be described by a relativistic wave equation that is linearin ∂t. Dirac also found an interpretation for the negative-energy solutions of hisrelativistic wave equation which finally led to the discovery of the positron [seeDir30; Dir31].By connecting quantum mechanics and special relativity, Dirac laid the ground-

work for modern quantum field theory. Today the Dirac equation is known as thewave equation for all spin-1

2particles, for example electrons, positrons, neutrinos

and quarks. To study the behaviour of these particles in electromagnetic fields,one has to solve the Dirac equation with a (possibly spacetime-dependent) poten-tial term. Although Dirac formulated his equation more than eighty years ago,even today few exact analytic solutions to the Dirac equation in arbitrary electricfields are known (see chapter 3 for some examples). However, the electromagneticpotential Aµ for these solutions depends on only one spatial coordinate (e.g. x),on the time t or on the linear combination t ± x. No solutions are known formore general potentials. In this thesis a method to generate solutions of the Diracequation in arbitrary potentials is developed and applied.Chapter 2 contains a brief introduction to the Dirac equation, its properties

and basic predictions. Furthermore, light cone coordinates are introduced as theyare used in the formulation of the method or inverse approach in (1 + 1)-dimen-sional spacetime described in chapter 4. Chapter 5 extends the method to (2 + 1)dimensions.

1

2. Dirac equation

In this chapter the Dirac equation and its basic properties will be described to-gether with an introduction to light cone coordinates and pair creation. As in allother chapters natural units with ~ = c = 1 are used unless otherwise noted. Vec-tors are printed bold (e.g. p), quantum mechanical operators are denoted by a hat(e.g. H) and the Einstein summation convention is used (e.g. γµpµ =

∑µ γ

µpµ).For partial derivatives the usual notation is used

∂µ =∂

∂xµ, ∂t =

∂

∂t, ∂x =

∂

∂x, . . . ,

where xµ denotes the position four-vector with components (t,x).

2.1. Motivation and derivation

The free Schrödinger equation for a particle with mass m and momentum p canbe written in Dirac notation as follows

i ∂t |ψ〉 = HS |ψ〉 =p2

2m|ψ〉 .

The Schrödinger equation is manifestly not Lorentz invariant as the Hamiltonoperator does not fulfil the relativistic energy-momentum relation which is givenby

ε2 = p2 +m2.

In contrast the Klein-Gordon equation

− ∂2t |ψ〉 = H2

KG |ψ〉 =(p2 +m2

)|ψ〉

−m

+m

ε

p = 0

p = 0

Figure 2.1.: The relativistic energy spectrum consists of two continua (grey areas)which are separated by a forbidden region between ε = −m and ε =+m.

3

2. Dirac equation

is a relativistic wave equation. Basically there are two major problems whentrying to interpret the solutions of the Klein-Gordon equation in the same wayas the solutions of the Schrödinger equation. First, due to the quadratic formof the relativistic energy-momentum relation the solutions can have positive ornegative energy. In the relativistic theory of classical particles the positive energysolutions are taken as the physical solutions while negative energy solutions areneglected. This can be done because the dynamical variables are only allowedto vary continuously in classical mechanics. Because the solutions with positiveand negative energy are separated from each other by the gap energy 2m2, asolution that has positive energy initially can never reach the negative continuum(see fig. 2.1). In the quantum theory though it is possible that a positive-energysolution evolves into a negative-energy solution because of quantum fluctuationsor scattering. However, this problem of negative energy solutions is not specific tothe Klein-Gordon equation in particular, but instead occurs with every relativisticwave equation.The second problem is related to the interpretation of the wave function ψ

itself. As Noether’s theorem states, every continuous symmetry of a physicalsystem has a corresponding conserved quantity. The conserved probability densityin Schrödinger’s quantum mechanics due to the global U(1) symmetry of theSchrödinger equation is given by |ψ|2 which upon correct normalisation meets allconditions for a probability density function. Therefore,

∫V d3r |ψ|2 is interpreted

as the probability of finding an electron inside the region V . When deriving acontinuity equation from the global U(1) symmetry of the Klein-Gordon equationsimilar to how it is done in the non-relativistic Schrödinger case, one finds thatthe conserved density is given by

i (ψ∗ ∂t ψ − ψ ∂t ψ∗) , (2.1)

which is not necessarily positive. Hence an intepretation of the conserved densityas a probability density is not feasible.Dirac tried to overcome these problems by deriving a new relativistic wave

equation with an ansatz that only contained first-order derivatives

i ∂t |ψ〉 = H |ψ〉 = (α · p+ βm) |ψ〉 , (2.2)

where α and β are arbitrary coefficients. These are going to be fixed by the con-dition that H2 should have the form of the relativistic energy-momentum relation

p2 +m2 != (α · p+ βm)2

=∑i

α2i p

2i +

∑i,ji>j

(αiαj + αjαi) pipj +m∑i

(αiβ + βαi) pi + β2m2.

4

2.1. Motivation and derivation

For this equation to hold the following conditions must be true

α2i = 1, β2 = 1, (2.3a)

αi, αj = αiαj + αjαi = 0 for i 6= j, (2.3b)αi, β = αiβ + βαi = 0, (2.3c)

where ·, · denotes the anticommutator. As ordinary numbers can not anti-commute with each other, the αi and β have to be other mathematical objectsthat can meet the above conditions, for example matrices. Therefore, becauseα and β act on |ψ〉 in (2.2), the state |ψ〉 has to be an object with multiplecomponents – a spinor. To arrive at the common version of the Dirac equation,we multiply (2.2) by β (

iβ ∂t−βα · p− β2m)|ψ〉 = 0,

and define another set of matrices

γ0 = β, γi = βαi, i = 1, 2, 3.

We now can rewrite (2.1) using these so-called gamma matrices and the fact thatp in position space is represented by −i ∂µ to get the equation that is known asthe Dirac equation (

iγµ ∂µ−m)ψ = 0. (2.4)

The conditions given in (2.3) can be transformed into equations for the gammamatrices

1 = β2 (2.3a)=(γ0)2

=1

2

γ0, γ0

,

1 = α2i

(2.3a)= αiββαi

(2.3b)= − (βαi)

2 = −(γi)2

=−1

2

γi, γi

,

0 = αiαj + αjαi(2.3a)= αiββαj + αjββαi

(2.3b)= −βαiβαj − βαjβαi = −

γi, γj

,

0 = αiβ + βαi ⇒ 0 = βαiβ + ββαi =γi, γ0

.

These conditions can be unified into a single anticommutator relation which isknown as the defining relation for a Clifford algebra

γµ, γν = 2gµν , (2.5)

where gµν = diag(+1,−1,−1,−1) is the metric tensor for the flat Minkowskispacetime in cartesian coordinates. When changing the coordinate system, thistensor has to be replaced by the corresponding metric tensor for the new coor-dinates. The anticommutator relation can be verbalised as: When multipliedby themselves, the gamma matrices give the identity matrix (up to a sign) anddifferent gamma matrices anticommute with each other.

5

2. Dirac equation

2.2. Gamma matrices

The anticommutator relation does only constrain the choice of the gamma matri-ces, but does not uniquely determine them. However, the relation can be used toderive several properties of the gamma matrices. For example, using the followingidentity, which can be read off from (2.5) immediately,

1

gµµγµγµ = 1,

where 1 denotes the identity matrix, we can prove that the gamma matrices aretraceless

tr(γν) =1

gµµtr(γνγµγµ) = − 1

gµµtr(γµγνγµ) = − 1

gµµtr(γνγµγµ) = − tr(γν) = 0,

where we used that γµ and γµ with µ 6= ν anticommute and that the trace isinvariant under cyclic permutations.If uµ is a eigenvector of γµ to the eigenvalue λµ, the following equations hold

1u0 = (γ0)2u0 = (λ0)2u0,

−1uj = (γj)2uj = (λi)2uj,

where the index j runs over the spatial components only. From these relations itsimmediately clear that the eigenvalues of γ0 are ±1 and the eigenvalues of the γjare ±i. Together with the condition that the gamma matrices have to be traceless,we can therefore conclude that the gamma matrices must have even dimension.Thus, the smallest gamma matrices we could choose are traceless 2×2 matrices.

As there are three linear independent traceless 2 × 2 matrices, it suffices to use2 × 2 matrices in (1 + 1) and (2 + 1) dimensions, where we need two and threegamma matrices respectively. However, there are four gamma matrices in (3 + 1)dimensions. Therefore, we need at least 4× 4 matrices there.Indeed, it can be shown (see for example Appendix L of [Cor89]) that the

dimension of the smallest irreducible representation of the Clifford algebra in (d+1) dimensions is given by 2b(d+1)/2c, where b·c denotes the floor function

bxc = maxn ∈ Z|n ≤ x,

which rounds any real number x to the largest previous integer.The standard (Dirac) choice for the γµ in (3 + 1) dimensions is given by

γ0 =

[1 O

O −1

], γ1 =

[O σ1

−σ1 O

], γ2 =

[O σ2

−σ2 O

], γ3 =

[O σ3

−σ3 O

], (2.6)

whereO =

(0 00 0

), 1 =

(1 00 1

),

6

2.2. Gamma matrices

and σi are the Pauli matrices

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

).

In (1 + 1) and (2 + 1) dimensions however, the formula given above predicts theexistence of 2 × 2 gamma matrices. Indeed, a set of such matrices can be con-structed easily: A set of 2× 2 matrices which when multiplied by themselves givethe identity matrix 1 is given by σ1, σ2, σ3. Additionally, these three matricesall anticommute with each other. In (1 + 1) dimensions there are only 2 gammamatrices. Therefore, we can choose any two of the three Pauli matrices, multiplyone of them by i, call this matrix γ1 and the other one γ0. For example

γ0 = σ1, γ1 = iσ2. (2.7)

In the same way one can construct gamma matrices in (2 + 1) dimensions usingall three Pauli matrices, e.g.

γ0 = σ1, γ1 = iσ2, γ2 = iσ3. (2.8)

Using the gamma matrices from (2.6), it can be shown that

(γµ)† = γ0γµγ0. (2.9)

Because the anticommutator relation (2.5) is invariant under unitary transforma-tions of the gamma matrices

γµ 7→ U †γµU,

the γµ are only determined up to such transformations. The expression for (γµ)†,which can be verified in a specific representation, for example in the one given in(2.6), is invariant under unitary transformations γµ 7→ γµ = U †γµU and thereforeindependent of the chosen representation, as the following calculation shows

(γµ)† 7→ (γµ)† =(U †γµU

)†= U † (γµ)† U = U †γ0γµγ0U

= U †γ0UU †γµUU †γ0U = γ0γµγ0

We now want to derive a continuity equation from the Dirac equation (2.4). Mul-tiplying the Dirac equation by γ0 and solving for ∂t ψ gives

∂t ψ = −γ0γk ∂k ψ − imγ0ψ,

where j is understood to take values 1, 2, 3. Using the last equation, one canwrite down an expression for the time derivative of the adjoint spinor

∂t ψ† = − ∂k ψ†

(γk)† (

γ0)†

+ imψ†(γ0)†,

where the right-hand side can be further simplified using (2.9)

∂t ψ† = − ∂k ψ†γ0γk + imψ†γ0.

7

2. Dirac equation

eiχ(xµ)

Figure 2.2.: Visualisation of a local U(1) transformation: The rotation of the com-plex plane is different at every point. Note however that the functionχ(xµ) has to be differentiable.

Calculating the time derivative of ψ†ψ gives

∂t(ψ†ψ

)= −ψ†γ0γk ∂k ψ− imψ†γ0ψ−∂k ψ†γ0γkψ+ imψ†γ0ψ = − ∂k

(ψ†γ0γkψ

),

which can be written as a continuity equation for the four-current jµ

∂µ jµ = 0,

withj0 = ψ†ψ, jk = ψ†γ0γkψ. (2.10)

In contrast to the conserved density of the Klein-Gordon equation given in (2.1),j0 = ψ†ψ is always positive and therefore could be interpreted as a probabilitydensity function similar to ψ∗ψ in non-relativistic quantum mechanics.

2.3. Minimal coupling

This section contains a brief review of minimal coupling in the context of theDirac equation. Like the Schrödinger and Klein-Gordon equation, the free Diracequation (2.4) is globally U(1) invariant: The form of the equation does not changewhen applying a global U(1) gauge transformation ψ 7→ ψ′ = eiχψ where χ is areal constant. This transformation is equivalent to rotating the complex planein which ψ lives. From a physical point of view this change should not makeany measurable difference. Therefore, the invariance of the Dirac equation underthis transformation makes perfect sense. However, there is not any measurabledifference when we change the rotation of the complex plane at every point inspace and time either – the Dirac equation should be invariant under a local U(1)transformation ψ 7→ ψ′ = eiχ(xµ)ψ, too. But because of the partial derivative term

8

2.4. Light cone coordinates

iγµ ∂µ acting not only on the spinor ψ but also on the phase factor eiχ(xµ), extraterms are created so that the Dirac equation is not locally U(1) invariant. Tomake it invariant, we introduce a gauge field Aµ and change the partial derivative∂µ in the Dirac equation to the covariant derivative ∂µ +iqAµ:(

iγµ[∂µ +iqAµ

]−m

)ψ = 0. (2.11)

Inserting ψ = e−iχ(xµ)ψ′ leads to

0 =(iγµ[∂µ +iqAµ

]−m

)e−iχψ′

= e−iχ(iγµ[∂µ−i

(∂µ χ

)+ iqAµ

]−m

)ψ′

⇒ 0 =(iγµ[∂µ +iqA′µ

]−m

)ψ′.

The (covariant) Dirac equation (2.11) is therefore locally U(1) invariant if Aµtransforms under a local U(1) transformation as

Aµ 7→ A′µ = Aµ −1

q∂µ χ.

When introducing charged particles in classical Hamiltonian mechanics, every oc-currence of pµ has to be replaced with pµ + qAµ where Aµ is the electromagneticfour-potential with components (φ,−A). We did exactly the same by introduc-ing the gauge field Aµ when we made the Dirac equation locally gauge invariant(i.e. invariant under a local U(1) transformation). Therefore, we conclude thatthe gauge field actually is the same as the electromagnetic four-potential. Thecovariant Dirac equation (2.11) then describes an electron in an electromagneticfield given by Aµ. The electric field E and the magnetic field B are connectedto the electromagnetic four-potential Aµ by the antisymmetric electromagneticfield tensor Fµν = ∂µAν−∂ν Aµ whose components in Minkowski spacetime usingcartesian coordinates are given by

F (3+1)µν =

0 Ex Ey Ez−Ex 0 −Bz By

−Ey Bz 0 −Bx

−Ez −By Bx 0

. (2.12)

In (1 + 1) and (2 + 1) dimensions the field tensor reduces to the upper-left 2× 2and 3× 3 part respectively:

F (1+1)µν =

(0 Ex−Ex 0

), F (2+1)

µν =

0 Ex Ey−Ex 0 −Bz

−Ey Bz 0

(2.13)

It should be noted that the conserved four-current corresponding to the local U(1)symmetry of the covariant Dirac equation is the same as (2.10) if and only if Aµis real.

9

2. Dirac equation

2.4. Light cone coordinates

All former calculations were done using cartesian coordinates. However, as we willsee in chapter 4, another set of coordinates is better suited to the problem we aretrying to solve. We introduce light-cone coordinates in (1 + 1) dimensions by

x+ =t+ x√

2, x− =

t− x√2.

The inverse transformation has exactly the same form because of the factor 1/√

2:

t =x+ + x−√

2, x =

x+ − x−√2

.

Hence, the Jacobian matrix of the coordinate transformation from the standardMinkowski coordinates (t, x) to light-cone coordinates (x+, x−) given by

Jµν =∂ (x+, x−)

∂ (t, x)=

1√2

(1 11 −1

)is equal to its own inverse (J−1)µν . Using the Jacobian matrix, we can transformthe components of any tensor from cartesian to light cone coordinates. For exam-ple, if we have the gradient in cartesian coordinates ∂µ = (∂t, ∂x), we can calculateits light cone components as follows

∂µ′ = (J−1)νµ ∂µ =

1√2

(∂t + ∂x∂t− ∂x

)=

(∂x+∂x−

)=..

(∂+

∂−

).

It is also possible to transform tensors of higher rank to light cone coordinates,for example the electromagnetic field tensor

F ′µν = (J−1)λµ(J−1)ρνFλρ = (J−1)λµFλρ((J−1)T) ρ

ν

=1

2

(1 11 −1

)(0 Ex−Ex 0

)(1 11 −1

)=

(0 −ExEx 0

).

(2.14)

x

t x+x−

Figure 2.3.: The coordinate axes of the light cone coordinates are the light lines.

10

2.4. Light cone coordinates

Similarly, the inverse Minkowski metric tensor ηµν = diag(+1,−1) can be trans-formed to light-cone coordinates

gµν = JµλJνρη

λρ = Jµληλρ(JT) ν

ρ =

(0 11 0

). (2.15)

The metric tensor itself is therefore given by

gµν =

(0 11 0

).

Using this metric tensor, we see that the length of any two-vector vµ does nolonger consist of purely quadratic terms as in cartesian coordinates (v0, v1)

vµvµ = ηµνvµvν = (v0)2 − (v1)2,

but instead mixes the two light-cone components (v+, v−)

vµvµ = gµνvµvν = 2v+v−.

If we take for example the momentum pµ, the following relation holds

pµpµ = gµνpµpν = 2p+p−

!= m2,

which immediately reveals that none of the light cone components of pµ can bezero (as opposed to the cartesian components).Due to the change of the metric tensor, the anticommutation relation of the

Clifford algebra (2.5) has also been changed: When multiplied with themselves,the gamma matrices now should give zero and their anticommutator with eachother should be the identity matrix up to a factor. A choice that satisfies theserelations for the inverse metric tensor given in (2.15) is

γ+ =

(0√

20 0

), γ− =

(0 0√2 0

), (2.16)

which can also be found by calculating

γ± =γ0 ± γ1

√2

with the gamma matrices from (2.7). Using the light cone gamma matrices givenin (2.16), we can calculate

γµ ∂µ = γ+ ∂+ +γ− ∂− =

(0

√2 ∂+√

2 ∂− 0

).

This shows that light cone coordinates separate the partial derivatives in the Diracequation nicely into the two components of the equation, a fact that will be usedin chapter 4.

11

2. Dirac equation

2.5. Lorentz transformations

Light cone coordinates have another nice feature besides the separation of thecomponents of the gradient in the Dirac equation: They diagonalise Lorentz trans-formations.In (1 + 1) dimensions the only possible Lorentz transformation is a boost in x-

direction which can be written as

t′ = t cosh(φ)− x sinh(φ),

x′ = −t sinh(φ) + x cosh(φ),

where the rapidity parameter φ is determined by the boost velocity v = tanh(φ).If we transform all coordinates to light cone coordinates we get

x′± =t′ ± x′√

2=

1√2

[t cosh(φ)− x sinh(φ)∓ t sinh(φ)± x cosh(φ)]

= x± [cosh(φ)∓ sinh(φ)]

= x±e∓φ.

This means that Lorentz transformations in light-cone coordinates just rescale thecoordinate axes. We can also rewrite the last equation as an equation betweentwo-vectors (

x′+x′−

)︸ ︷︷ ︸x′µ

=

(e−φ 00 eφ

)︸ ︷︷ ︸

Λµν

(x+

x−

)︸ ︷︷ ︸xν

. (2.17)

The matrix Λµν is a vector representation of the Lorentz transformation, i.e. a

matrix that when applied to a vector in a specific coordinate system acts as howthe group element, that it is representing, would act on the coordinate-independentvector. We now want to find a spinor representation of the Lorentz transformationwhich we will denote by S(Λ) so that a spinor ψ transforms according to ψ′ =S(Λ)ψ. Applying a Lorentz transformation to the Dirac equation (2.4) results inthe following

0 =(iγµ∂µ

′ −m)ψ′

=[iγµ(Λ−1

)νµ∂ν −m

]S(Λ)ψ,

where we used the transformation law for covectors to write the partial derivative∂µ′ in the unprimed frame and the transformation law for spinors with the still

unknown matrix S(Λ) to write ψ′ in the unprimed frame. Inserting the identitymatrix 1 = S(Λ)S(Λ)−1 and reordering the factors gives

= S(Λ)S(Λ)−1[iγµ(Λ−1

)νµ∂ν −m

]S(Λ)ψ

= S(Λ)[iS(Λ)−1γµS(Λ)

(Λ−1

)νµ∂ν −m

]ψ

= S(Λ)[iγµ ∂µ−m

]ψ,

12

2.5. Lorentz transformations

where the last step is only possible if

S−1(Λ)γµS(Λ) = Λµνγ

ν . (2.18)

The left-hand side transforms the spinor indices of the gamma matrix γµ, theright-hand side transforms the vector index; the result has to be the same forthe Dirac equation to be Lorentz invariant. Equation (2.18) relates the spinorrepresentation S(Λ) to the vector representation Λµ

ν . Using the Λµν from (2.17)

and the gamma matrices given in (2.16), we can write down an expression forS(Λ) quite easily

S(Λ) =

(eφ/2 0

0 e−φ/2

), (2.19)

which indeed fulfills the relation given in (2.18).In non-relativistic quantum mechanics, the wave function ψ is normalized using

the probability density ψ∗ψ which, as mentioned before, is also a conserved density.As was shown in section 2.2, ψ†ψ is a conserved density in the Dirac theory andcould be used in the same way as ψ∗ψ in Schrödinger’s theory. However, ψ†ψ isnot Lorentz invariant as one can easily verify

ψ′†ψ′ = ψ†S(Λ)†S(Λ)ψ 6= ψ†ψ,

because S(Λ) is not unitary (see (2.19)). Thus, we try to construct a ψ such thatψψ transfroms as a Lorentz scalar, ψγµψ as a Lorentz vector, and ψγµγνψ asa Lorentz rank-2 tensor according to the number of their indices. We make anansatz for ψ

ψ = ψ†M,

where M is a 2× 2 matrix that has to be determined. To turn ψψ into a Lorentzscalar we have to require that ψ transforms according to

ψ′!

= ψS−1(Λ).

Inserting our ansatz for ψ we get the following relation

ψ†S†(Λ)M!

= ψ†MS−1(Λ) ⇔ S†(Λ)MS(Λ)!

= M.

Setting

M =

(m11 m12

m21 m22

),

and inserting (2.19), we get(eφ/2 0

0 e−φ/2

)(m11 m12

m21 m22

)(eφ/2 0

0 e−φ/2

)=

(eφm11 m12

m21 e−φm22

)!

=

(m11 m12

m21 m22

).

For this condition to hold for any parameter φ, m11 and m22 have to be zero

M!

=

(0 m12

m21 0

).

13

2. Dirac equation

Additionally, we require ψψ to be real

ψψ =(ψ∗1 ψ∗2

)( 0 m12

m21 0

)(ψ1

ψ2

)= m12ψ

∗1ψ2 +m21 (ψ∗1ψ2)∗

= < (ψ∗1ψ2) (m12 +m21) + i= (ψ∗1ψ2) (m12 −m21) ,

which is only the case if m12 = m21. If we set the only remaining coefficient of M ,m12 = 1, we get

M =

(0 11 0

)=γ+ + γ−√

2. (2.20)

Because M does only swap the two components of a spinor, we have ψ =(ψ∗2 ψ∗1

)for a spinor ψ =

(ψ1

ψ2

). We can check that using this definition of ψ the

bilinear ψγµψ indeed transforms as a Lorentz vector

ψ′γµψ′ = ψS−1(Λ)γµS(Λ)ψ = Λµνψγ

νψ,

using the relation given in (2.18). Analogously, it can be proven that ψγµγνψtransforms as a Lorentz rank-2 tensor.

2.6. Free solutions

To derive solutions of the (free) Dirac equation (2.4), we make a plane-wave ansatzbecause the Dirac equation is a first order linear differential equation with onlyconstant coefficients

ψ = ueipµxµ

,

where u is a spinor independent of xµ. Inserting this into the Dirac equation leadsto

(γµpµ +m)u = 0.

Multiplying the last equation by γνpν −m gives(γνγµpνpµ −m2

)u = 0.

Using the Clifford algebra’s anticommutation relation and the fact that pνpµ issymmetric in ν and µ, we can simplify the first term:

γνγµpνpµ =1

2(γνγµ + γµγν)︸ ︷︷ ︸

=γµ,γν=2gµν

pνpµ = pµpµ,

so that we can diagonalise the former equation(pµpµ −m2

)u = 0.

Thus, the plane-wave ansatz can only solve the Dirac equation if pµ is the four-momentum that satisfies the relativistic energy-momentum relation pµpµ = m2.

14

2.7. Pair creation

If we use pµ = (ε,p), we see that for every fixed momentum p two solutions exist– one with positive and one with negative energy

pµpµ = m2 ⇔ ε = ±√m2 + p2.

Therefore, there are at least two linearly independent solutions to the free Diracequation with either positive or negative energy ψ(±)

ψ(+) = ue−ipµxµ

, ψ(−) = veipµxµ

.

Inserting ψ(±) into the Dirac equation leads to equations for the spinors u and v

(γµpµ −m)u = 0,

(γµpµ +m)v = 0.

There are two independent solutions to each of these two equations in (3 + 1)dimensions. This can be easily seen using the gamma matrices from (2.6) in therest frame where p = 0 and thus ε = m

0 = (γ0 − 1)u = −2

[O O

O 1

]u,

0 = (γ0 + 1)v = 2

[1 O

O O

]v.

Therefore, there are four independent solutions – two with positive energy andtwo with negative energy. These two additional degrees of freedom correspond tothose two associated to the spin of the electron.In (1 + 1) and (2 + 1) dimensions however, there are no additional degrees of

freedom when using 2× 2 gamma matrices. For example, for those given in (2.7),we get the following two equations in the rest frame

0 = (γ0 − 1)u = −2

(−1 11 −1

)u,

0 = (γ0 + 1)v = 2

(1 11 1

)v.

Therefore, when using two-dimensional representations of the gamma matrices,the spinor ψ does not have any degrees of freedom associated with the particle’sspin. Solving the above equations gives the following spinors:

u =1√2

(11

), v =

1√2

(1−1

),

where we haven chosen the normalization factor so that uu = 1 and vv = 1. Notethat these normalization conditions are Lorentz invariant in contrast to the formof the spinors which have been calculated explicitly in the rest frame.

15

2. Dirac equation

ε

+m

−m

(a) Vacuum state

ε

+m

−m

(b) Excited state

ε

(c) Sauter-Schwingertunneling

Figure 2.4.: Visualization of the Dirac sea: Each circle represents a state, blackcircles are filled states, white ones are unoccupied ones. In the groundstate all states with negative energy are occupied.

2.7. Pair creation

This chapter started with a motivation of the Dirac equation. As it turns out,Dirac finally found solutions to the problems we discussed in section 2.1. We didalready address the second problem of negative probability densities when usingthe Klein-Gordon equation at the end of section 2.2: the conserved density in theDirac theory is manifestly positive.The first problem of negative energy solutions can be resolved by a clever inter-

pretation of the solutions. Let us assume that these negative energy solutions areindeed physical solutions and there are particles associated with such solutions.Then, however, the total energy will be unbounded below: The total energy canalways be further minimized by creating particles with negative energy. Dirac’sfamous idea to overcome this disastrous situation was the so-called Dirac sea in-terpretation [Dir30]. Because electrons (which the Dirac equation describes) arefermions, they obey the Pauli principle stating that no two particles can be inthe same state. Therefore, we can define the vacuum state as the state where allnegative energy states are filled (see fig. 2.4a).Of course, this means that our vacuum contains an infinite charge, but, accord-

ing to Dirac, as long as the charge of the Dirac sea is uniformly distributed, itwill not be observable – only charge differences would be measurable. Thus, apositive state can only be filled by exciting a negative energy state to a positiveenergy state, thereby leaving a hole in the Dirac sea (see fig. 2.4b). Consequently,such a hole will have positive energy but also positive charge as a hole basicallycan be understood as the absence of a negative energy electron. Because of thehole having positive energy, it can be interpreted as a particle. This particle, thathas exactly the same properties as the electron excluding the positive instead ofnegative charge, is the anti-particle of the electron, namely the positron. Thus, anexcitation from the Dirac sea always creates electron-positron pairs and thereforeconserves overall charge. For this pair creation to happen, there has to be enoughenergy (e.g. in the form of radiation, i.e. photons) to bridge the gap between thenegative and the positive continuum.

16

2.7. Pair creation

However, there is another mechanism for pair production which is called theSauter-Schwinger effect. If we add a potential to the relativistic energy-momentumrelation, we get:

ε = V ±√p2 +m2.

For example, the potential energy for a homogeneous electric field in x-direction isgiven by V = −qEx. Because V depends linearly on x, the level spectrum shownin fig. 2.4a is tilted, allowing for negative energy states to tunnel into the positivecontinuum and thereby creating electron-positron pairs(see fig. 2.4c).To create an electron-positron pair as depicted in fig. 2.4b, an oscillating electric

field with enough energy (i.e. a frequency that is high enough) can be used. Underthe assumption that the electric field is weak, the probability for pair productioncan be calculated using perturbation theory. In contrast, for pair production dueto the Sauter-Schwinger effect a strong electric field has to be used to minimizethe distance to tunnel (see fig. 2.4c). Therefore, perturbation theory cannot beused to calculate the probability in this case – the Sauter-Schwinger effect is anon-perturbative effect.Nevertheless, the pair creation rate for the Sauter-Schwinger effect can be ap-

proximated using a quantum mechanical tunneling picture. In non-relativisticquantum mechanics the probability for an electron to tunnel through a barrier ofheight ∆V and length L is given by

Ptunneling ∼ exp(−2L√

2m∆V).

In the case of a constant electric field, the length of the barrier is L = 2m/(qE)and the height is of order m. Inserting this into the tunneling probability givenabove leads to the following result

Ptunneling ∼ exp

(−4√

2m2

qE

),

which is notably close to the exact result [Sau31]

P ∼ exp

(−πm

2

qE

).

This shows again that the Sauter-Schwinger effect is a non-perturbative effect asthis expression does not permit a Taylor expansion for small q or E as it is donein standard perturbation theory using Feynman diagrams.To obtain the pair creation probability or rate for arbitrary electric fields, the

covariant Dirac equation has to be solved for these fields. Fritz Sauter then deter-mined the pair creation rate by calculating the tunneling probability as it is donein Schrödinger’s quantum mechanics, but instead of the Schrödinger equation heused the Dirac equation as wave equation.However, the interpretation of the Dirac equation’s solutions as single particle

wave functions is flawed as the Dirac sea interpretation relies crucially on the

17

2. Dirac equation

fact that electrons are fermions. Today it is a well-known fact that bosons haveanti-particles, too. A cleaner approach is to interpret the Dirac equation as theequation of motion of a classical field and quantize it to get particles and anti-particles. As it turns out, the Hamiltonian of the Dirac quantum field theorycan be made positive definite, so that the problem of negative energy solutions issolved naturally.Nevertheless, the solutions to the Dirac equation can be used in the quantized

theory as the Dirac field operator ψ has to solve the Dirac equation, too. Thesingle particle solutions of the Dirac equation are used as a basis for the fieldoperator. For example, the solutions of the free Dirac equation can be used towrite down the free Dirac field operator

Ψ =

∫d3p

(2π)3/2√p0/m

∑s

[a(p, s)u(p, s)e−ipµx

µ

+ b†(p, s)v(p, s)eipµxµ],

where p0 =√m2 + p2, s is an index for the spin of a solution, a(p, s) and a†(p, s)

are the annihilation and creation operators for electrons, and b(p, s) and b†(p, s)are those for positrons. These operators fulfil the following anticommutator rela-tions

a(p, s), a†(p′, s′)

= δss′δ(3)(p− p′),

b(p, s), b†(p′, s′)

= δss′δ(3)(p− p′),

and all other anticommutators vanish. The vacuum state |0〉 is defined by

a(p, s) |0〉 = b(p, s) |0〉 = 0.

To calculate the pair creation rate for a specific electric field, we assume thatthe electric field vanishes for t→ ±∞. We then have to find two sets of solutionsto the covariant Dirac equation: Solutions ψ(±,s)

in (p;xµ) that initially (i.e. fort → −∞) behave like plane wave solutions of either positive or negative energyand solutions ψ(±,s)

out (k;xµ) that finally (i.e. for t → ∞) behave like plane wavesolutions of either positive or negative energy. The Dirac field operator can beexpanded into both sets independently

Ψ =

∫d3p

(2π)3/2√p0/m

∑s

[ain(p, s)ψ

(+,s)in (p;xµ) + b†in(p, s)ψ

(−,s)in (p;xµ)

](2.21)

=

∫d3k

(2π)3/2√k0/m

∑s′

[aout(k, s

′)ψ(+,s′)out (k;xµ) + b†out(k, s

′)ψ(−,s′)out (k;xµ)

],

(2.22)

where the in- and out-vacuum states are defined by

ain(p, s) |0in〉 = bin(p, s) |0in〉 = 0,

aout(p, s) |0out〉 = bout(p, s) |0out〉 = 0.

18

2.7. Pair creation

As these two sets are both bases of the solutions of the Dirac equation in thechosen electric field, they have to be connected by a unitary basis transformation

ψ(+,s)in (p;xµ) =

∫d3k

√p0

k0

∑s′

[αp,k;s,s′ψ

(+,s′)out (k;xµ) + βp,k;s,s′ψ

(−,s′)out (k;xµ)

],

ψ(−,s)in (p;xµ) =

∫d3k

√p0

k0

∑s′

[β∗p,k;s,s′ψ

(+,s′)out (k;xµ) + α∗p,k;s,s′ψ

(−,s′)out (k;xµ)

].

This means that solutions that have either positive or negative energy initiallyevolve into mixtures of both positive and negative energy solutions of arbitrarymomentum, which is exactly what happens when a solution tunnels through theforbidden region of the relativistic level spectrum as shown in fig. 2.4c.Moreover, inserting this transformation into the Dirac field operator given in

(2.21) and reordering terms leads to the following

Ψ =

∫d3k d3p

(2π)3/2√k0/m

∑s′,s

[ψ

(+,s′)out (k;xµ)

(αp,k;s,s′ ain(p, s) + β∗p,k;s,s′ b

†in(p, s)

)+ψ

(−,s′)out (k;xµ)

(βp,k;s,s′ ain(p, s) + α∗p,k;s,s′ b

†in(p, s)

)].

Comparing this to (2.22), we can read off the Bogoliubov transformation betweenthe creation and annihilation operators

aout(k, s′) =

∫d3p

∑s

(αp,k;s,s′ ain(p, s) + β∗p,k;s,s′ b

†in(p, s)

),

b†out(k, s′) =

∫d3p

∑s

(βp,k;s,s′ ain(p, s) + α∗p,k;s,s′ b

†in(p, s)

).

The pair production rate is equal to the number of positrons in the out-vacuum

〈0in| b†out(k, s′)bout(k, s

′) |0in〉 =∫d3p d3p′

∑s,s

〈0in| βp,k;s,s′β∗p′,k;s,s′ ain(p, s)a†in(p′, s) |0in〉

=

∫d3p

∑s

|βp,k;s,s′ |2.

It should be noted that in order to investigate the Sauter-Schwinger effect in greatdetail, the contribution of non-perturbative pair creation to the pair creation rateshould be minimized by choosing slowly varying electric fields. In this case, thefrequency and thus the energy of the electric field is too low to bridge the gap (seefig. 2.4b) directly without tunneling.In summary, the pair creation probability can be calculated by solving the co-

variant Dirac equation and relating the in- and out-states. The following chapterswill focus on finding solutions of the Dirac equation in spacetime-dependent elec-tric fields.

19

3. Known exact solutions

So far, only solutions of the free Dirac equation (2.4) were discussed. However,for the purpose of studying the behaviour of electrons in arbitrary electromag-netic fields, solutions of the covariant Dirac equation (2.11) have to be found.This chapter contains a description of some known exact analytic solutions to thecovariant Dirac equation for different electromagnetic fields. All these solutionsare found by reducing the Dirac equation to an ordinary differential equation forwhich the solutions are well-known.

3.1. Constant electric field

In 1931, Fritz Sauter published an article with the derivation of solutions of thecovariant Dirac equation in a constant electric field [Sau31]. To understand whichproblems occur when trying to solve the Dirac equation in the presence of aspacetime-dependent electric field, this section contains an overview of the stepsneeded to arrive at the solution.A possible choice of a four-potential Aµ that generates a constant electric field

of strength E pointing in x-direction is given by

A0 = −Ex, A1 = A2 = A3 = 0.

This can be inserted into the covariant Dirac equation (2.11), and after separatingthe t-, y-, and z-dependence by using

ψ = ξ(x)e−iωt+ikyy+ikzz

and multiplying the remaining equation by iγ1, we get[d

dx+ iγ1γ0 (ω + qEx)− iγ1γ2ky − iγ1γ3kz − iγ1m

]ξ(x) = 0.

Differentiating this equation with respect to x and manipulating a bit gives thefollowing equation, where we set m2

eff = m2 + k2y + k2

z ,[d2

dx2+ iγ1γ0qE + (ω + qEx)2 −m2

eff

]ξ(x) = 0.

Because (γ1γ0)2

= 1, the eigenvalues of γ1γ0 are ±1. Therefore, solutions tothis equation are given by ξ±(x) = v±f±(x), where v± is an eigenvector to the

21

3. Known exact solutions

-5 0 5

-4

-2

0

2

4

x

(a) f(1)+

-5 0 5-6

-4

-2

0

2

4

x

(b) f(2)+

Figure 3.1.: Real (solid) and imaginary (dashed) part of the two independent so-lutions to (3.1) with qE/m2

eff = 1, and ω/meff = 1

eigenvector ±1 of the operator γ1γ0 and f±(x) is a scalar function solving theWeber equation [

d2

dx2+ (ω + qEx)2 −m2

eff ± iqE]f±(x) = 0. (3.1)

The solutions to this equation are given by parabolic cylinder functions, whichare plotted for certain parameters in fig. 3.1. Note that the parabolic cylinderfunctions can be also expressed using the Whittaker function or the confluenthypergeometric function of the first kind [AS72].

3.2. Constant magnetic field

The next case, the Dirac equation with a constant magnetic field was first solvedby Rabi in 1928 [Rab28]. Assuming that the constant magnetic field of strengthB is pointing in z-direction, a possible choice of the gauge field Aµ is

A0 = 0, A1 = 0, A2 = −Bx, A3 = 0.

As in the previous case, these components of Aµ do only depend on x so that wecan perform exactly the same steps (separating, differentiating) to arrive at thefinal second order differential equation for ξ(x)[

d2

dx2+ ω2 − (ky − qBx)2 −m2

eff + iγ1γ2qB

]ξ(x) = 0,

where m2eff

..= m2 + k2z . The eigenvalues of γ1γ2 are ±i because (γ1γ2)

2= −1.

Thus, solutions to this equation are ξ±(x) = v±f±(x) where v± is eigenvector tothe eigenvalue ±i of γ1γ2 and f±(x) is again a solution of a Weber equation[

d2

dx2+ ω2 − (ky − qBx)2 −m2

eff ∓ qB]f±(x) = 0.

22

3.3. Constant orthogonal electric and magnetic field

Using a coordinate transformation u = x + kyqB

, this equation can be broughtinto the same form as the differential equation for a one-dimensional harmonicoscillator in quantum mechanics

d2

du2f±(u) +

[ω2 −m2

eff ∓ qB︸ ︷︷ ︸∧=2mE

− (qB)2︸ ︷︷ ︸∧=m2ω2

0

u2

]f±(u) = 0.

Thus, the unnormalised solutions f±(x) are given by

f±(x) = Hn

(√qB

[x− ky

qB

])exp

(−1

2qB

[x− ky

qB

]2), (3.2)

where Hn(x) are the Hermite polynomials and

n =ω2 −m2

eff ∓ qB2qB

− 1

2.

3.3. Constant orthogonal electric and magneticfield

The previous two cases can be combined. The four-potential components

A0 = −Ex, A1 = 0, A2 = −Bx, A3 = 0

generate a constant electric field pointing in x-direction and a constant magneticfield pointing, orthogonal to the electric field, in z-direction. In this case, thesecond order differential equation for ξ(x) is given by[

d2

dx2+ (ω + qEx)2 − (ky − qBx)2 −m2

eff + iγ1γ0qE + iγ1γ2qB

]ξ(x) = 0,

where m2eff = m2 + k2

z . A solution to this equation is ξλ(x) = vλfλ(x), where(γ1γ0 +

B

Eγ1γ2

)vλ = λvλ (3.3)

and fλ(x) is a solution to the following Weber equation[d2

dx2+ (ω + qEx)2 − (ky − qBx)2 −m2

eff + iλqE

]fλ(x) = 0.

Calculating the eigenvalues λ± of the operator given in (3.3) gives

λ± = ± 1

E

√E2 −B2.

Therefore, if E > B, the eigenvalues are real as in the case of a constant electricfield, thereby leading to the same differential equations as in that case. In the othercase E < B, the eigenvalues are purely imaginary as in the case of a constantmagnetic field, thus giving the differential equation of the quantum harmonicoscillator as in that case.

23

3. Known exact solutions

3.4. Sauter pulse

A non-constant electric field for which the covariant Dirac equation can be solvedexactly was found by Sauter in 1932 and is therefore often called Sauter pulsenowadays [Sau32]

E(x) =E

cosh2(Ωx)ex.

To derive solutions for this case, the same steps as in the previous solutions can beapplied (separating and differentiating). However, the resulting ordinary differen-tial equation is a bit more sophisticated than before. A coordinate transformationcan be used to bring the equation into a known form (see [Sau32] or [Heb11] fordetails); the solutions to this equation can then again be expressed using hyperge-ometric functions. Note that it is also possible to derive solutions for a temporalSauter pulse

E(t) =E0

cosh2(Ωt)ex.

The solutions for a temporal Sauter pulse in the (1 + 1)-dimensional case will begiven here as they are used as a reference in section 4.6. After separating the x-dependence, the components of the spinor ξ(t) are given by

ξ1(τ) = τ ν(1− τ)µ 2F1(ν + µ+ λ, ν + µ− λ+ 1; 2ν + 1; τ),

ξ2(τ) = τ ν(1− τ)µ 2F1(ν + µ− λ, ν + µ+ λ+ 1; 2ν + 1; τ),(3.4)

where 2F1(a, b; c;x) denotes the hypergeometric function,

τ =1

1 + e−2ωt, λ = −iqE0

ω2,

and

−ν2 =

(1

2ω

)2[(

k − qE0

ω

)2

+m2

],

−µ2 =

(1

2ω

)2[(

k +qE0

ω

)2

+m2

].

3.5. Lightfront field

The electric fields mentioned so far were dependent on one coordinate at most.This fact was crucial for reducing the covariant Dirac equation, which is a partialdifferential equation, to an ordinary differential equation. However, it is also possi-ble to derive solutions for electric fields that depend on a special linear combinationof two cartesian coordinates by using light cone coordinates. After transforming

24

3.5. Lightfront field

-3 -2 -1 0 1 2 3

0.0

0.2

0.4

0.6

0.8

1.0

x

Figure 3.2.: Sauter pulse 1/ cosh2(x).

to light cone coordinates, the covariant Dirac equation can be solved for an elec-tromagnetic potential depending only on x+ = t+x√

2using similar methods as for

the previously discussed fields (see for example [TTW00], [TTW01], or [HIM11]for actual calculations).

25

4. Inverse approach to the Diracequation

To derive solutions of the covariant Dirac equation,(iγµ[∂µ +iqAµ

]−m

)ψ = 0, (4.1)

usually a specific potential Aµ is fixed and one tries to find the spinor solutions ψto the coupled partial differential equations using standard methods as describedin the previous chapter. Unfortunately this only works for a few special choices forAµ. The approach described in this chapter swaps known and unknown quantities:Instead of fixing Aµ, one chooses a specific ψ and instead of solving the Diracequation for the spinor ψ, one tries to solve it for Aµ. That this is feasible canbe seen from (4.1): In contrast to ψ, only the gauge potential Aµ itself and noderivatives of it appear in the Dirac equation. Using this method, new solutionsof the Dirac equation in electric fields can be generated. Furthermore, it allowsfor more insight into the relationship between a spinor and its associated electricfield.This chapter contains a description of the inverse approach in (1+1)-dimensional

spacetime including solutions that were generated using this method.

4.1. Basic formalism

Using ψ =(ψ1

ψ2

)and the gamma matrices given in (2.7), we can write down the

two components of the Dirac equation in cartesian coordinates(−m i [∂t +iqA0] + i [∂x +iqA1]

i[∂−+iqA−

]− i [∂x +iqA1] −m

)(ψ1

ψ2

)= 0.

We now assume that we know the solution ψ for a given two-potential Aµ. Thenit is possible to solve the two components of the above equation for the twocomponents of Aµ. However, the terms A0±A1 and ∂t± ∂x look familiar. Indeed,as we have seen in section 2.4, ∂± = (∂t± ∂x)/

√2. So it is reasonable to write

down the Dirac equation in light cone coordinates using the gamma matrices from(2.16) (

−m i√

2[∂+ +iqA+

]i√

2[∂−+iqA−

]−m

)(ψ1

ψ2

)= 0. (4.2)

27

4. Inverse approach to the Dirac equation

Due to the form of the Dirac equation in light cone coordinates, it is triviallypossible to solve the two equations in (4.2) for the light-cone components of theelectromagnetic two-potential

qA+ = i∂+ ψ2

ψ2

− m√2

ψ1

ψ2

,

qA− = i∂− ψ1

ψ1

− m√2

ψ2

ψ1

.

(4.3)

In general, the covariant Dirac equation is only invariant under local U(1) trans-formations if and only if the gauge field Aµ is real. Furthermore, the four-currentgiven in (2.10) is the corresponding conserved quantity to the local U(1) symmetryof the covariant Dirac equation. Therefore, it is desirable that the electromagneticpotential Aµ is real. Nevertheless, the light cone components given in (4.3) are notnecessarily real for arbitrary (complex) ψ1 and ψ2. Thus, we require the imaginaryparts of A+ and A− to vanish

<(∂+ ψ2

ψ2

)− m√

2=(ψ1

ψ2

)= 0,

<(∂− ψ1

ψ1

)− m√

2=(ψ2

ψ1

)= 0.

Using polar coordinates for the spinor components ψi = rieiϕi leads to

<(∂+ r2

r2

+ i ∂+ ϕ2

)− m√

2=(r1

r2

eiϕ1−iϕ2

)= 0,

<(∂− r1

r1

+ i ∂− ϕ1

)− m√

2=(r2

r1

eiϕ2−iϕ1

)= 0.

After computing the real and imaginary parts and multiplying the first equationby r2 and the second one by r1, we have

∂+ r2 −m√

2r1 sin (ϕ1 − ϕ2) = 0,

∂− r1 +m√

2r2 sin (ϕ1 − ϕ2) = 0.

Multiplying the second equation by r1r2

and adding it to the first finally gives thefollowing equations

∂− r21 = − ∂+ r

22, (4.4a)

∂− r1 =m√

2r2 sin (ϕ2 − ϕ1) . (4.4b)

We integrate (4.4a) once to solve for r2

r2 =

√c(x−)−

∫∂− r

21 dx+ , (4.5)

28

4.1. Basic formalism

where c(x−) is a integration constant for the x+-integration that may dependon x−. This step shows again how convenient light cone coordinates are for thiscalculation: If we had used cartesian coordinates, we could not have integrated thisequation easily. Derivatives of r2 with respect to t and x would have appeared in(4.4a), which we then would have to treat as a more complicated partial differentialequation for r2.Additionally, we can solve (4.4b) for the phase difference ϕ2 − ϕ1 and obtain

the following two solutions

ϕ2 − ϕ1 = arcsin

(√2

m

∂− r1

r2

), ϕ2 − ϕ1 = π − arcsin

(√2

m

∂− r1

r2

), (4.6)

where we neglected 2π-periodic constants as those terms will eventually drop outwhen we reinsert the phases into the spinor components. We insert r2 from (4.5)into (4.6) and solve for ϕ2. Thus, we are able to express r2 and ϕ2 solely using r1

and ϕ1, thereby reducing the number of real degrees of freedom of the spinor ψfrom four to two. As r2 and ϕ2 are obsolete now, we set r ..= r1 and ϕ ..= ϕ1 sothat we finally have for the spinor ψ:

ψ =

(ψ1

ψ2

)= eiϕ

r

±√c−

∫∂− r

2 dx+ − 2m2 (∂− r)

2 + i√

2m∂− r

.

Gauge freedom allows us to always eliminate the phase factor eiϕ by a gaugetransformation ψ 7→ ψ′ = e−iϕψ:

ψ′ =

(ψ′1ψ′2

)=

r

±√c−

∫∂− r

2 dx+ − 2m2 (∂− r)

2 + i√

2m∂− r

. (4.7)

Thus, we are able to express any solution to the Dirac equation in (1 + 1) dimen-sions using a spinor of the form given in (4.7) that does only depend on r(x+, x−),a real-valued function of the coordinates, and the integration constant for the x+

integration c(x−).As explained in section 2.3, the gauge transformation ψ 7→ e−iϕψ does also

change the components of the electromagnetic potential given in (4.3)

qA+ = i∂+ ψ2

ψ2

− m√2

ψ1

ψ2

+ ∂+ ϕ = i∂+ ψ

′2

ψ′2− m√

2

ψ′1ψ′2,

qA− = i∂− ψ1

ψ1

− m√2

ψ2

ψ1

+ ∂− ϕ = i∂− ψ

′1

ψ′1− m√

2

ψ′2ψ′1.

Using the spinor components from (4.7), the components of the electromagnetic

29

4. Inverse approach to the Dirac equation

two-potential finally are

qA+ = ∓ m√2

r

s∓√

2

m

∂+ ∂− r

s,

qA− = ∓ m√2

s

r,

(4.8)

where we used s ..=√c−

∫∂− r

2 dx+ − 2m2 (∂− r)

2 as an abbreviation. Note that2 ∂+ ∂− represents the d’Alembert operator in light-cone coordinates

2 ∂+ ∂− = gµν ∂µ ∂ν = ∂µ ∂µ .

Using (2.14), we can calculate the electric field corresponding to the electromag-netic two-potential given above

qE = q ∂−A+ − q ∂+A− =m√

2

[∂+

(sr

)− ∂−

(r

s+

2

m2

∂+ ∂− r

s

)]. (4.9)

For a solution ψ to be well-defined, the functions r(x+, x−) and c(x−) have tomeet some conditions. First, r and s should not be zero, as their inverse valueappears in the expression for the electric field (4.9). Second, s2 should be positiveas s is a real number. To ensure this, the function c(x−) has to be large enough

c(x−)!>

∫∂− r

2 dx+ +2

m2

(∂− r

)2.

In conclusion, choosing a real-valued function r(x+, x−) and adjusting an inte-gration constant c(x−) allows us to calculate a well-defined solution to the Diracequation in an electric field. The rest of this chapter is dedicated to specific solu-tions that can be found using this method. As can be seen in (4.7) and (4.8), theform of the spinor and the two-potential is significantly simplified if r does notdepend on x−. Because all derivatives with respect to x− are zero, the imaginarypart of the second spinor component vanishes and s reduces to

√c(x−). We will

start by looking at solutions where this is the case.

4.2. Plane wave solutions

The simplest solutions are those where both r and s =√c are constant

ψ =

(r±s

).

In this case Aµ is constant, too, as can be calculated using (4.8)

qA+ = ∓ m√2

r

s= const,

qA− = ∓ m√2

s

r= const,

30

4.2. Plane wave solutions

which implies that the electric field associated with these solutions is zero. There-fore, the solutions defined as

u =

(rs

), v =

(r−s

)are essentially solutions to the free Dirac equation. We impose the followingLorentz invariant normalization conditions, that we also used in section 2.6, onthem

uu = 2rs!

= 1, vv = −2rs!

= −1,

so that the spinors have the following form

u =

(r

(2r)−1

), v =

(r

−(2r)−1

).

Using these spinors, the components of the (constant) electromagnetic two-poten-tial change to

qA+ = ∓ m√2

2r2,

qA− = ∓ m√2

1

2r2,

which we can set to zero using the gauge transformation

ψ 7→ ψ exp

[∓i(m√

22r2x+ +

m√2

1

2r2x−

)].

The phase factor can be shortened to exp (∓ipµxµ) if we define

pµ ..=

(p+

p−

)=

m√2

(2r2

(2r2)−1

).

The two-vector pµ indeed is the relativistic two-momentum of a particle with massm as the following calculation shows

pµpµ = gµνpµpν = 2p+p− = m2.

Thus, ue−ipµxµ is a positive energy solution of the free Dirac equation (2.4),whereas veipµxµ is a negative energy solution. Note that we can also expressthe spinors u and v in terms of the momentum using the relationship between thecomponents of the two-momentum and the spinor components

u = 2−1/4m−1/2

(√p+√p−

), v = 2−1/4m−1/2

( √p+

−√p−

).

If we have a solution that is a linear combination of positive and negative energysolutions

ψ = αu e−ipµxµ

+ β v eipµxµ

,

31

4. Inverse approach to the Dirac equation

Figure 4.1.: The function r as given in (4.11) with p+/m = p−/m = 1/√

2, α2 =0.99, β = 0.1 and argα− arg β = 0.

where α and β are arbitrary (complex) constants, we should also be able to expressψ as

ψ =

(r1

r2ei(ϕ2−ϕ1)

)eiϕ1

to make contact with the spinors in our formalism. A rather tedious but simplecalculation leads to

ψ = 2−1/4m−1/2

√p+

√|α|2 + |β|2 + 2|α||β| cos (argα− arg β − 2pµxµ)

√p−|α|2 − |β|2 + i 2|α||β| sin (argα− arg β − 2pµx

µ)√|α|2 + |β|2 + 2|α||β| cos (argα− arg β − 2pµxµ)

eiϕ1 ,

(4.10)where we can get rid of the global phase eiϕ1 by using a gauge transformation aswe did before. Indeed if we choose

r = 2−1/4m−1/2√p+

√|α|2 + |β|2 + 2|α||β| cos (argα− arg β − 2pµxµ), (4.11)

c = 2−1/2m−1p−(|α|2 + |β|2

),

we can show that the resulting spinor according to (4.7) is exactly the same as theone given in (4.10) when neglecting the global phase. Additionally, we can checkthat the electric field that belongs to this solution vanishes indeed.

32

4.3. Single moving pulses

4.3. Single moving pulses

Solutions that are a bit more sophisticated than the constant ones are those wherer does only depend on x+. We first look at such solutions where additionally cand therefore also s is constant

ψ =

(r(x+)±s

).

In this case, the light cone components of the electromagnetic potential Aµ aregiven by

qA+ = ∓ m√2

r(x+)

s,

qA− = ∓ m√2

s

r(x+).

Depending on the sign of the second spinor component, if r(x+) is initially con-stant, the solution initially behaves like a plane wave solution with either positiveor negative energy. As the components of Aµ do only depend on x+, the electricfield given in (4.9) can be simplified to

qE = q ∂−A+︸ ︷︷ ︸=0

−q ∂+A− = ± m√2s ∂+

1

r(x+).

In fact, this is a first order ordinary differential equation for r(x+) that can beintegrated easily

q

∫ x+

−∞E(x+) dx+ = ± m√

2s

[1

r(x+)− 1

rin

],

where rin = r(x+ → −∞). We can invert this equation so that we can calculater(x+) from a field profile E(x+)

r(x+) =rin

1±√

2m

rinsq∫ x+−∞E(x+) dx+

. (4.12)

If the electric field vanishes for x+ → ±∞, the field can be imagined as a singlepulse that moves along the x−-axis of the light cone. Similarly, if we choose r tobe constant and let c and therefore also s depend on x−, we can construct singlepulses moving along the x+-axis of the light cone

ψ =

(r

±s(x−)

).

Using (4.8), we can again calculate A+ and A−

qA+ = ∓ m√2

r

s(x−),

qA− = ∓ m√2

s(x−)

r.

33

4. Inverse approach to the Dirac equation

Thus, the electric field associated with this spinor does only depend on x−

qE = q ∂−A+ − q ∂+A−︸ ︷︷ ︸=0

= ∓ m√2r ∂−

1

s(x−). (4.13)

This equation can be integrated in the same way as above, so that we finally get

s(x−) =sin

1∓√

2m

sinrq∫ x−−∞E(x−) dx−

, (4.14)

where sin = s(x− → −∞).We want to investigate whether these solutions correspond to electron-positron

pairs. As we discussed in section 2.7, pair creation occurs when a solution thatinitially behaves like a plane wave with either positive or definite energy evolvesinto a solution that is a mix of positive and negative energy solutions. As can beseen in (4.11), such a mix of solutions does depend both on x+ and x− (rememberthat p+ and p− cannot be zero). Because the function r given in (4.12) does at notime depend on x−, the solution given here can not evolve into a mix of positiveand negative energy solutions. Hence, pairs can only be created if initially ψ haspositive energy and evolves into a solution with negative energy (or vice versa).This can only happen if r(x+) changes its sign somewhere in time. We define rout

as the value of r(x+) for x+ →∞

rout..=

rin

1±√

2m

rinsq∫∞−∞E(x+) dx+

.

We introduce the initial (x+ → −∞) and final (x+ → ∞) momentum named pµand kµ respectively

pµ =

(p+

p−

)=

m√2

(rinssrin

), kµ =

(k+

k−

)=

m√2

(routssrout

)Using these definitions, r(x+) and rout can be written in terms of the momentumcomponents

r(x+) =

√2

ms

p+

1± p−1− q

∫ x+−∞E(x+) dx+

, rout..=

√2

ms

p+

1± p−1− q

∫∞−∞E(x+) dx+

.

We can calculate a relation between the initial and final momentum from the lastexpression

k− = p− ± q∫ ∞−∞

E(x+) dx+ .

If ψ has positive energy for t→ −∞, the final solution must have negative energyfor pair creation to happen. Thus, k− has to be negative

p− < −q∫ ∞−∞

E(x+) dx+ .

34

4.3. Single moving pulses

(a) E(x+) (b) r(x+)

Figure 4.2.: Visulization of Sauter pulse solution with rin = 1, qE0/m2 = 1, and

ω/m = 2.

If ψ has negative energy initially, pair production occurs only if k− is positive

p− < q

∫ ∞−∞

E(x+) dx+ .

For definiteness, let us assume that E(x+) > 0. As p− was defined as a positivenumber, pair creation then can only happen for solutions that have negative energyinitially. The number of modes for which pair creation occurs is large if the integralq∫∞−∞E(x+) dx+ is large. This is the case for slow and strong pulses E(x+).In both cases, if pairs are created, r(x+) changes its sign somewhere, so that

r(x+) will diverge at this point. To understand this divergence physically let usexamine the behaviour of a relativistic charged particle in the same electric field.The energy of the particle is given by

ε =√m2 + (p1 + qA1)2 − qA0

=

√m2 +

1

2(p+ − p− + qA+ − qA−)2 − 1√

2(qA+ + qA−).

As qA+ diverges for r(x+)→∞, the energy will diverge at this point, too. Thus,the particle will be accelerated to the speed of light and move parallel to thelight cone. Note that the divergence of A+ does not influence the electric fieldbecause only the derivative with respect to x− of A+, which vanishes nevertheless,contributes to the electric field.To visualise the above solutions, we choose a specific electric field, namely a

Sauter pulse (see fig. 4.2a)

E(x+) =E0

cosh2(ωx+), (4.15)

35

4. Inverse approach to the Dirac equation

so that the integral over E(x+) can be calculated as follows∫ x+

−∞E(x+) dx+ = E0

∫ x+

−∞

1

cosh2(ωx+)dx+ =

E0

ω[tanh(ωx+)− tanh(−∞)]

= E01 + tanh(ωx+)

ω.

Using this result, we can write down the expression for r(x+) immediately

r(x+) =rin

1±√

2m

rinsqE0

1+tanh(ωx+)ω

. (4.16)

A plot of this function r(x+) for specific parameters can be seen in fig. 4.2b.

4.4. Two moving pulses

Using the solutions from the previous section, we can construct another type ofsolutions where two pulses move along both axes of the light cone. Basically, wejust let r depend on x+ and s on x−:

ψ =

(r(x+)±s(x−)

).

The electric field associated with this type of solution can be calculated using thefollowing formula

qE = q ∂−A+ − q ∂+A− = ∓ m√2r ∂−

1

s± m√

2s ∂+

1

r,

where, as we will see later, the first term corresponds to a pulse moving along thex−-axis while the second term describes a pulse travelling along the x+-axis.If we use the general form of the single pulses

r(x+) =rin

1±√

2m

rinsinq∫ x+−∞E+(x+) dx+

,

s(x−) =sin

1∓√

2m

sinrinq∫ x−−∞E−(x−) dx−

,

we can calculate the electric field using the formula given above, so that we finallyget

qE =r

rin

qE−(x−) +s

sin

qE+(x+)

=qE−(x−)

1±√

2m

rinsinq∫ x+−∞E+(x+) dx+

+qE+(x+)

1∓√

2m

sinrinq∫ x−−∞E−(x−) dx−

, (4.17)

36

4.5. Emerging pulses

For t → −∞, the electric field simply consists of a superposition of the singlepulses E+(x+) and E−(x−). As long as the integrals stay small, i.e. as long asthe pulses are far away from each other, they move along the axes of the lightcone undisturbedly. When the pulses get near to each other, the integrals start tocontribute and increase or decrease the amplitude of each pulse. After the pulseshave passed each other, the integrals stay constant again, so that the pulses areagain moving undisturbedly for t → ∞. However, their amplitude might bedifferent than for t→ −∞.As in the previous section, the sign of the second spinor component determines

whether the solution initially has positive or negative energy. In contrast to thesingle pulse solutions however, the initially positive and negative energy solutionscorrespond to different electric fields, as can be seen in (4.17).We will now give a condition for pair creation if the second spinor component

is positive initially. We define the initial and final momentum as before

pµ =

(p+

p−

)=

m√2

( rinsinsinrin

), kµ =

(k+

k−

)=

m√2

(routsoutsoutrout

),

where

rout..= r(x+ →∞) =

rin

1 +√

2m

rinsinq∫∞−∞E+(x+) dx+

,

sout..= s(x− →∞) =

sin

1−√

2m

sinrinq∫∞−∞E−(x−) dx−

.

Using these definitions, we can write down a relation between the initial and finalmomentum

k− =p− + q

∫∞−∞E+(x+) dx+

1− p−1+ q∫∞−∞E−(x−) dx−

If pµ is positive initially, k− has to be negative for pair creation to occur. This isthe case, if either

p− < −q∫ ∞−∞

E+(x+) dx+ and p+ > q

∫ ∞−∞

E−(x−) dx−

or

p− > −q∫ ∞−∞

E+(x+) dx+ and p+ < q

∫ ∞−∞

E−(x−) dx−

hold.An example for this type of solution can be seen in fig. 4.3. The electric field was

calculated using eq. (4.17), where E+ and E− were chosen to be Sauter pulses.

37

4. Inverse approach to the Dirac equation

Figure 4.3.: Electric field consisting of two Sauter pulses E+ = E(+)0 sech2(ω+x+)

and E− = E(−)0 sech2(ω−x−). The parameters chosen were rin = 1,

sin = 1, qE(+)o /m2 = 1, qE(−)

0 /m2 = 1, ω+/m = 4.5, and ω−/m = 4.5.The two pulses are moving along the axes of the light cone and areamplified after interfering with each other in the origin.

4.5. Emerging pulses

We now choose a function r(x+, x−) that depends both on x+ and x−

r(x+, x−) = rin +ξ

1 + e−γx+ + e−γx−(4.18)

For non-vanishing ξ and γ > 0, the chosen r(x+, x−) will be constant almosteverywhere except in the vicinity of the forward light cone. The value of r insidethe future light cone is different from the value outside this area; the difference isgiven by ξ (see fig. 4.4). As r does depend on x−, we have to calculate the integraland the derivative in the expression for s

s2 = c(x−)−∫∂− r

2 dx+ −2

m2

(∂− r

)2.

Calculating the integral gives the following result∫∂− r

2 dx+ = 2

∫ [rin +

ξ

1 + e−γx+ + e−γx−

] [ξγe−γx−

(1 + e−γx+ + e−γx−)2

]dx+

=2(a− 1)ξrine

γx+

a (1 + aeγx+)2 +(a− 1)ξ

a3 (1 + aeγx+)2 (2a (rin + 2ξeγx+) + 3ξ)

+ 2(a− 1)ξ

a3(arin + ξ) ln (1 + aeγx+) ,

38

4.5. Emerging pulses

Figure 4.4.: Plot of r(x+, x−) as given in (4.18) with rin = 1, ξ = 0.2, and γ =1.2/m.

where a = 1 + e−γx− was used as an abbreviation. By carefully choosing c(x−),we can achieve that s = sin = const for x+ → −∞ so that s is constant almosteverywhere except along the x+-axis: As s does only depend on the derivativeof r with respect to x−, we expect s to be non-constant only where ∂− r 6= 0which is only the case along the positive x+-axis. The integral

∫∂− r

2 dx+ in theexpression for s is responsible for a linear growth of s along that axis. In addition,if s is constant intially (i.e. for x+ → −∞), the whole spinor is constant initially.Thus, the solution initially is a plane wave.For choosing an appropiate c(x−), we calculate the limit of

∫∂− r

2 dx+ +2m2

(∂− r

)2 as x+ approaches −∞ using the result from above

limx+→−∞

(∫∂− r

2 dx+ +2

m2

(∂− r

)2︸ ︷︷ ︸→0

)=

(a− 1)ξ

a3(2arin + 3ξ) =.. c(x−)− s2

in.

The final function s(x+, x−) can be seen in fig. 4.5.As r and s are constant everywhere except in the vicinity of the forward light

cone, we expect the electric field to vanish everywhere else. Indeed the field canbe imagined as two pulses emerging from the origin and moving on the forwardlight cone (see fig. 4.6). If we choose a negative γ the field will be reversed in time;we then have two pulses moving along the negative axes of the light cone towardsthe origin where they annihilate each other.These pulses can only create electron-positron pairs if we choose ξ in a way that

r is negative inside the future light cone and positive outside (so that the solutioninitially behaves like a plane wave with positive energy and finally ends up as anegative energy solution). In this case, a contour line with r = 0 has to exist.Note that this line lies in the vicinity of the forward light cone. By inspecting the

39

4. Inverse approach to the Dirac equation

Figure 4.5.: Plot of s(x+, x−) with rin = 1, s2in = 2, ξ = 0.2, and γ = 1.2/m.

expression for qE in terms of r and s in (4.9) we find a term proportional to

∂+

s

r=∂+ s

r− s

r2∂+ r. (4.19)

The derivative ∂+ r vanishes along the positive x+-axis but it is finite alongthe positive x−-axis so that the second term in (4.19) diverges for r = 0. Be-cause ∂+ s does not vanish around the positive x+-axis the first term in (4.19)diverges there for r = 0, too. This means that it is impossible to produceelectron-positron pairs using these pulse as the electric field would have to di-verge. Note that the argument given above also applies to other choices ofr(x+, x−) that are qualitatively equal to the one used in this section (for exampler(x+, x−) = 1 + ξ/4[tanh(γx+) + 1][tanh(γx−) + 1]) as we did not make use of theexact form of r(x+, x−).

4.6. Perturbed solution

The solutions discussed so far have one thing in common: The functions r and sare both constant initially (i.e. for t → −∞) and finally (i.e. for t → ∞). Thatmeans (see section 4.2) that they are initially and finally plane wave solutions witheither positive or negative energy. In this situation, electron-positron pairs canonly be created if a initial positive energy solution finally evolves into a negativeenergy solution (or the other way round). In general, pair production occursif a plane wave solution evolves into a mixture of positive and negative energysolutions. Such a mixture can be described using the function given in (4.11)which is an oscillating function of the coordinates. However, (4.11) is not a veryrealistic choice for a solution corresponding to pairs, as the momentum of thepositive and negative energy solutions that make up the mixing solution are equal(up to the sign). To generate a more general solution that evolves into a mix of

40

4.6. Perturbed solution

Figure 4.6.: Electric field qE belonging to r(x+, x−) = 1 + ξ

1+e−γx++e−γx−with

rin = 1, s2in = 2, ξ = 0.2 and γ = 1.2/m

positive and negative energy solutions, we start with an ansatz that also containsan oscillating term

r = α + β sin(mγ), (4.20)

where α, β, and γ are all functions of the light cone coordinates. The functions αand β should not be confused with the identically named constants in (4.11). Theidea behind this ansatz is that α is a known exact solution, for example one of thesolutions described in the previous sections, and β sin(mγ) is a small perturbation.Further justification for this ansatz is provided by fig. 4.7 which shows the

function r for a temporal Sauter pulse that does indeed produce electron-positronpairs. The function r shown there is initially constant and oscillates finally.

4.6.1. Expansion in powers of the mass parameter

We make the assumption that α, β, and γ are of order O (m0). We expand theelectric field

qE =m√

2

[∂+ s

2

2sr− s

r2∂+ r −

∂− r

s+

r

2s3∂− s

2

]−√

2

m

[∂+ ∂

2− r

s−∂+ ∂− r

2s3∂− s

2

]in powers of m and keep only the highest order term ∼ m2. The integrals withoscillating terms can be calculated using integration by parts like in the followingexample∫

f(x+, x−) cos(mγ) dx+ =1

m

∫f(x+, x−)

∂+ γm∂+ γ cos(mγ) dx+

=1

m

f(x+, x−)

∂+ γsin(mγ)− 1

m

∫∂+

(f(x+, x−)

∂+ γ

)sin(mγ) dx+ ,

41

4. Inverse approach to the Dirac equation

-10 -5 0 5 10

0.7

0.8

0.9

1.0

t

1 /m

r

Figure 4.7.: Plot of the function r for the temporal Sauter pulse E(t) =ex sech2(ωt) obtained by calculating the absolute value of the (un-normalised) first spinor component of the solution given in (3.4). Pa-rameters chosen are k/m = 0, ω/m = 0.6, and qE0/m

2 = 0.6.

where the first term is of higher order in m than the second one. Using thismethod, such oscillating integral terms can be pushed to lower order in m.The quite lengthy result for the electric field can be found in appendix A.

4.6.2. Expansion of the mixing square root

The square root in the mixing solution given in (4.11) is of the form

r ∼√a+ b cos(c),

where b is small if the coefficient of the negative energy solution β is small. If thatis the case, we can expand the square root in powers of b

√a+ b cos(c) =

√a+

cos(c)

2√ab+O

(b2),

which has the same form as the ansatz given in (4.20). Thus, that ansatz canbe regarded as an expansion of the mixing square root for small β. Therefore,when calculating the associated electric field, we can neglect terms of order β2

and higher. Note that all derivatives of β will be treated as if they have the sameorder as β.Using the aforementioned ansatz, we can calculate the real part of the second

42

4.6. Perturbed solution

spinor component up to order β

s2 =

=..s2α︷ ︸︸ ︷c−

∫∂− α

2 dx+ −2

m2(∂− α)2−2

∫∂− (αβ sin(mγ)) dx+

− 4

m2(∂− α) ∂−(β sin(mγ)) +O

(β2).

Therefore, the first light cone component of the electromagnetic potential can becalculated as follows

qA+ = −m√

2r +

√2m∂+ ∂− r

s

= −m√

2α +

√2m∂+ ∂− α

sα

[1 +

m√2β sin(mγ) +

√2m∂+ ∂−(β sin(mγ))

m√2α +

√2m∂+ ∂− α

+1

s2α

(∫∂− (αβ sin(mγ)) dx+ +

2

m2(∂− α) ∂−(β sin(mγ))

)]+O

(β2),

while the result for the second one is

qA− = − m√2

s

r

= − m√2

sαα

[1− β sin(mγ)

α− 1

s2α

∫∂−(αβ sin(mγ)

)dx+

+2

m2(∂− α) ∂−

(β sin(mγ)

)]+O

(β2).

The prefactors in these expressions are the unperturbed components of the two-potential Aµ

qA(α)+ = − m√

2

α

sα−√

2

m

∂+ ∂− α

sα, qA

(α)− = − m√

2

sαα.

Using these abbreviations, the electric field can be written as

qE = q ∂−A+ − q ∂+A−

= qE(α) + ∂−

[qA

(α)+

m√2β sin(mγ) +

√2m∂+ ∂−(β sin(mγ))

m√2α +

√2m∂+ ∂− α

+1

s2α

(∫∂−(αβ sin(mγ)

)dx+ +

2

m2(∂− α) ∂−

(β sin(mγ)

))]

+ ∂+

[qA

(α)−

β sin(mγ)

α+

1

s2α

(∫∂−(αβ sin(mγ)

)dx+

43

4. Inverse approach to the Dirac equation

+2

m2(∂− α) ∂−

(β sin(mγ)

))]+O

(β2),

where the unperturbed electric field qE(α) ..= q ∂−A(α)+ − q ∂+A

(α)− was used.

Expanding qE(β), the β1 term of the electric field, in powers of m and keepingonly the highest order term gives the following result

qE(β) =√

2β cos(mγ)

sα ∂+ γ

[m2(∂+ γ)2(∂− γ)2 −

(m√

2

α

sα︸ ︷︷ ︸≈−qA(α)

+

∂− γ +m√

2

sαα︸ ︷︷ ︸

=−qA(α)−

∂+ γ

)2].

(4.21)The eikonal or Hamilton-Jacobi equation in covariant form for the Dirac theoryis given by

m2 =(∂µ S + qAµ

)(∂µS + qAµ) ,

where S is the Hamilton-Jacobi action. Writing down the Hamilton-Jacobi equa-tion using light cone coordinates gives

m2

2=(∂+ S + qA+

) (∂− S + qA−

)= (∂+ S)(∂− S) + qA+qA− + qA+ ∂− S + qA− ∂+ S.

(4.22)

If S = mγ solves the eikonal equation with qA+ = m√2αsα

and qA− = m√2sαα, the

term in brackets in (4.21) vanishes. Thus, the electric field does not oscillaterapidly up to order m2. That means, that only non-perturbative pair creationdue to the Sauter-Schwinger effect can occur. Moreover, if α is a slowly varyingfunction, the electric field does not contain any term of order m2.So for a given function α, the function γ can be calculated in leading order

from the eikonal equation, if we do not want the electric field to oscillate rapidlyat highest order. For example, if we choose the x+-dependent single pulse from(4.12),

α(x+) =αin

1±√

2m

αin

sαq∫ x+−∞E

(α)(x+) dx+

, sα = const, (4.23)

the eikonal equation (4.22) reads

0 = (∂+ S)(∂− S)∓ m√2

α(x+)

sα∂− S ∓

m√2

sαα(x+)

∂+ S.

Using the ansatz S = f(x+) + g(x−), the eikonal equation separates into twoordinary differential equations

f ′(x+) = −

[√2

m+

1

ω−

(sα

α(x+)

)2]−1

, g′(x−) = ω−,

44

4.6. Perturbed solution

-20 -10 0 10 20 30

0.2

0.4

0.6

0.8

1.0

t

1 /m

r

Figure 4.8.: Plot of r(x+, x−) = α(x+) +β(x+, x−) sin [mγ(x+, x−)] with the func-tions from (4.23), (4.24), and (4.25) for x = 0, αin = 1, ωα/m = 0.6,qE

(α)0 /m2 = 0.6, βout = 0.05, ωβ/m = 0.2, ω−/m = 1, and χ = 0.

where ω− is the separation constant. After integrating both equations a solutionof the eikonal equation is given by

S(x+, x−) = f(x+) + g(x−) = ω−x− −∫ [√

2

m+

1

ω−

(sα

α(x+)

)2]−1

dx+ + χ,

(4.24)with χ as an integration constant. Remember that S(x+, x−) = mγ(x+, x−).Choosing

β(x+, x−) =βout

2

[tanh

(ωβx+ + x−√

2

)+ 1

](4.25)

to slowly turn on the perturbation we can plot the function r = α+β sin(mγ) fora fixed position x (see fig. 4.9), resembling fig. 4.7. The full spacetime-dependenceof r(x+, x−) can be seen in fig. 4.9. The electric field for this solution is byconstruction equal to qE(α) to lowest order of β and highest order of m.

45

4. Inverse approach to the Dirac equation

Figure 4.9.: Plot of r(x+, x−) = α(x+) +β(x+, x−) sin [mγ(x+, x−)] with the func-tions from (4.23), (4.24), and (4.25) for αin = 1, ωα/m = 0.6,qE

(α)0 /m2 = 0.6, βout = 0.05, ωβ/m = 0.2, ω−/m = 1, and χ = 0.

46

5. Inverse approach in (2+1)dimensions

The method described in the previous chapter can be extended to (2 + 1)-di-mensional spacetimes. In contrast to (1 + 1) dimensions, there exists a magneticfield in (2 + 1) dimensions. Additionally, the electric field has two dimensions sothat electric pulses with a transverse component are possible. However, althoughthe (2 + 1)-dimensional spacetime has more physically interesting properties thanin the (1 + 1)-dimensional case, a Dirac spinor still only has two components(see section 2.2) which reduces the complexity of the problem (in contrast to a(3+1)-dimensional calculation, where one would have to deal with four-componentspinors).The chapter starts with some remarks about light cone coordinates in (2 + 1)

dimensions and then continues along the lines of chapter 4 by giving a derivationof the inverse approach in (2 + 1) dimensions, followed by solutions that can befound using the method.

5.1. Basic formalism

In (2 + 1)-dimensional Minkowski spacetime one can use cartesian coordinates t,x, and y. When using light cone coordinates, one has to choose a direction forthe light cone – x or y. We select the x-direction here (see fig. 5.1), so that thetransformation to light cone coordinates is

x+ =t+ x√

2, x− =

t− x√2, y = y,

i.e. t and x are transformed as in (1+1) dimensions while y does not change underthe transformation. Therefore, many results from section 2.4 can be adaptedeasily. For example, the inverse transformation is given by

t =x+ + x−√

2, x =

x+ − x−√2

, y = y,

and the Jacobian matrix can be written down easily, too,

Jµν =1√2

1 1 01 −1 0

0 0√

2

. (5.1)

47

5. Inverse approach in (2+1) dimensions

t

x+

x−

x

y

Figure 5.1.: The coordinate axes of light cone coordinates in (2 + 1) dimensions.

Thus, the metric tensor of (2 + 1)-dimensional Minkowski spacetime in light conecoordinates is given by

gµν = JµλJνρη

λρ =

0 1 01 0 00 0 −1

,

where ηλρ = diag(+1,−1,−1).The first two gamma matrices have to behave like the (1 + 1)-dimensional light

cone gamma matrices while the third gamma matrix is the same as with cartesiancoordinates. Therefore a straightforward choice combining (2.16) and (2.8) is

γ+ =

(0√

20 0

), γ− =

(0 0√2 0

), γ2 =

(i 00 −i

).

Using these matrices, the covariant Dirac equation can be written as follows(−m−

[∂y +iqAy

]i√

2[∂+ +iqA+

]i√

2[∂−+iqA−

]−m+

[∂y +iqAy

])(ψ1

ψ2

)= 0.

As in chapter 4, we can solve these two equations easily for the light cone com-ponents of the electromagnetic potential. However, in contrast to the (1 + 1)-dimensional case, these do still depend on the third component Ay

qA+ = i∂+ ψ2

ψ2

− m√2

ψ1

ψ2

− 1√2

∂y ψ1

ψ2

− i√2qAy

ψ1

ψ2

,

qA− = i∂− ψ1

ψ1

− m√2

ψ2

ψ1

+1√2

∂y ψ2

ψ1

+i√2qAy

ψ2

ψ1

,

(5.2)

48

5.1. Basic formalism

which are partially the same expressions as in (4.3); only the terms where ∂y andAy appear are new. Again, we require the imaginary parts of the components ofAµ to vanish. Therefore, using the polar form of ψ1 = r1e

iϕ1 and ψ2 = r2eiϕ2 we

can write down two conditions

∂+ r2

r2

− 1√2

[mr1

r2

+∂y r1

r2

]sin (ϕ1 − ϕ2)− 1√

2

r1

r2

[∂y ϕ1 + qAy

]cos (ϕ1 − ϕ2) = 0,

∂+ r1

r1

+1√2

[mr2

r1

−∂y r2

r1

]sin (ϕ1 − ϕ2) +

1√2

r2

r1

[∂y ϕ2 + qAy

]cos (ϕ1 − ϕ2) = 0.

(5.3)The sum of the first equation multiplied by r2

2 and the second multiplied by r21

gives the following condition

0 = r1 ∂− r1 + r2 ∂+ r2 −1√2

[r1 ∂y r2 + r2 ∂y r1

]sin (ϕ1 − ϕ2)

− 1√2r1r2

[∂y ϕ1 − ∂y ϕ2

]cos (ϕ1 − ϕ2)

=1

2∂− r

21 +

1

2∂+ r

22 −

1√2∂y [r1r2 sin (ϕ1 − ϕ2)] .

This equation can be integrated and solved for the sine of the phase differenceeasily

sin (ϕ1 − ϕ2) =1√

2r1r2

[c(x+, x−) +

∫ (∂− r

21 + ∂+ r

22

)dy

]=

u

r2

, (5.4)

where c(x+, x−) is the integration constant for the y-integration and u is definedas

u ..=1√2r1

[c(x+, x−) +

∫ (∂− r

21 + ∂+ r

22

)dy

].

As in chapter 4, (5.4) can be used to simplify the form of the spinor ψ by solvingfor ϕ2. Again, because of the arcus sine, there are two solutions

ϕ2 = ϕ1 − arcsin

(u

r2

), ϕ2 = ϕ1 + arcsin

(u

r2

)− π.

Using these solutions, one can easily derive an identity for the cosine of the phasedifference

cos (ϕ1 − ϕ2) = ± cos

[arcsin

(u

r2

)]= ±

√1− u2

r22

=s

r2

, (5.5)

where s ..= ±√r2

2 − u2. Note that s was defined with a sign here in contrastto the function s defined in section 4.1. This makes the following expressions

49

5. Inverse approach in (2+1) dimensions

more readable. Equation (5.5) and (5.4) can be used to write the second spinorcomponent as

ψ2 = r2eiϕ2 = r2e

iϕ1e−i(ϕ1−ϕ2) = eiϕ1 (s− iu) .

As one could guess from the the expression for ψ2, it should again be possibleto get rid of the phase function eiϕ1 by applying a gauge transformation. To dothis, one has to calculate the components of Aµ first to evaluate how they wouldchange under a gauge transformation.The difference of the first equation in (5.3) multiplied by r2

2 and the secondmultiplied by r2

1 gives the following condition

0 = −r1 ∂− r1 + r2 ∂+ r2 −1√2

[2mr1r2 − r1 ∂y r2 + r2 ∂y r1

]sin (ϕ1 − ϕ2)

− 1√2r1r2

[∂y ϕ1 + ∂y ϕ2 + 2qAy

]cos (ϕ1 − ϕ2) ,

which can be used to calculate Ay

qAy = −1

2

(∂y ϕ1 + ∂y ϕ2

)+

1

cos (ϕ1 − ϕ2)

[− 1√

2

∂− r1

r2

+1√2

∂+ r2

r1

−(m− 1

2

∂y r2

r2

+1

2

∂y r1

r1

)sin (ϕ1 − ϕ2)

].

(5.6)

In order to further simplify this expression, we derive an useful identity by differ-entiating (5.4) with respect to xµ

∂µ (ϕ1 − ϕ2) =1

cos (ϕ1 − ϕ2)∂µ

u

r2

. (5.7)

In the special case where µ = 2, this expression can be simplified even furtherusing the definition of u and (5.4)

∂y (ϕ1 − ϕ2) =1

cos (ϕ1 − ϕ2)

[√2

(∂− r1

r2

+∂+ r2

r1

)− sin (ϕ1 − ϕ2)

(∂y r1

r1

+∂y r2

r2

)].

Solving this equation for ∂y ϕ2 and inserting the result into (5.6) finally gives

qAy =1

cos (ϕ1 − ϕ2)

[√2∂+ r2

r1

− sin (ϕ1 − ϕ2)

(m+

∂y r1

r1

)]− ∂y ϕ1

=1√2

∂+ r22

r1s− u

mr1 + ∂y r1

r1s− ∂y ϕ1.

50

5.1. Basic formalism

To calculate the light cone components of Aµ, we insert the spinor componentsinto (5.2) and use (5.4), (5.5), and (5.7) to eliminate all occurrences of ϕ2, finallygiving

qA+ = − m√2

r1

s− 1√

2

∂y r1

s+∂+ u

s− ∂+ ϕ1,

qA− = − m√2

r22

r1s+

1√2

r2 ∂y r2

r1s−u ∂− r1

r1s− ∂− ϕ1.

In every component of the electromagnetic potential Aµ there is a term − ∂µ ϕ1

present which can be eliminated by the gauge transformation ψ 7→ e−iϕ1ψ, therebyalso removing the global phase factor from the spinor components.In conclusion, a spinor of the form

ψ =

r1

s− iu

, s = ±√r2

2 − u2, u =1√2r1

[c(x+, x−) +

∫ (∂− r

21 + ∂+ r

22

)dy

]

is a solution to the Dirac equation in (2 + 1) dimensions with the electromagneticpotential

qA+ = − m√2

r1

s− 1√

2

∂y r1

s+∂+ u

s,

qA− = − m√2

r22

r1s+

1√2

r2 ∂y r2

r1s−u ∂− r1

r1s,

qAy = −mu

s−u ∂y r1

r1s+

1√2

∂+ r22

r1s.

(5.8)

Using the Jacobian matrix given in (5.1), we can transform the cartesian compo-nents of the electromagnetic field tensor as given in (2.13) to light cone coordinates

Fµν =

0 −Ex Ey−Bz√2

Ex 0 Ey+Bz√2

−Ey−Bz√2−Ey+Bz√

20

.

Thus, the components of the electric and magnetic field can be calculated asfollows

Ex = F10 = ∂−A+ − ∂+A−,

Ey =F12 + F02√

2=

1√2

(∂−Ay − ∂y A− + ∂+Ay − ∂y A+

),

Bz =F12 − F02√

2=

1√2

(∂−Ay − ∂y A− − ∂+Ay + ∂y A+

).

(5.9)

The inverse approach to generate solutions of the Dirac equation works in (2 + 1)dimensions, too. In contrast to the (1 + 1)-dimensional case, we have to choosetwo (instead of one) real functions r1 and r2 of the coordinates x+, x−, and y in

51

5. Inverse approach in (2+1) dimensions

addition to the integration constant c(x+, x−) to fully determine a solution. Whenr1 and r2 are independent of y, we can set

r22(x+, x−) = c(x−)−

∫∂− r

21 dx+ , c(x+, x−) =

2

mr1 ∂− r1,

so that the spinor has exactly the same form as in (1 + 1) dimensions. It can bechecked easily, that the components of Aµ given in (5.8) reduce to the (1 + 1)-dimensional components given in (4.8) with Ay = 0. Thus, any of the solu-tions discussed in the previous chapter can be embedded into (2 + 1)-dimensionalMinkowski spacetime.However, the full expressions for the components of Aµ look more complicated

than in section 4.1. They simplify a lot when neither r1 nor r2 depend on y. In thiscase, the integral in u vanishes, effectively reducing u to c√

2r1, and all derivatives

with respect to y in (5.8) vanish. The next sections will discuss solutions wherethis is the case.

5.2. Landau levels

However, before we start generating new solutions, we want to convince ourselvesthat the inverse approach can also reproduce solutions known beforehand. As in(2+1) dimensions it is possible to examine the Dirac equation in the presence of amagnetic field, we use the solutions given in section 3.2 for the constant magneticfield to verify that inserting the (known) solutions into (5.8) leads to the correctmagnetic field. For simplicity, we use the ground state function with n = 0 andchoose the f− solution given in (3.2).In (2 + 1) dimensions, an eigenvector to the eigenvalue −i of γ1γ2 using the

gamma matrices from (2.8) is given by

v− =

(11

),

so that the full spinor ξ−(x) for the ground state up to normalization can bewritten as

ξ−(x) = exp

(−1

2qB

[x− ky

qB

]2)(

11

).

In the formalism of the inverse approach, the same spinor can be represented using

r1(x+, x−) = exp

(−1

2qB

[x+ − x−√

2− kyqB

]2),

r2(x+, x−) = r1(x+, x−), c = 0,

52

5.3. Plane wave solutions

so that s = r2(x+, x−) = r1(x+, x−). Inserting these functions into (5.8) gives

qA+ = − m√2

r1

s= − m√

2= const,

qA− = − m√2

r22

r1s= − m√

2= const,

qAy =1√2

∂+ r22

r1s=√

2∂+ r1

r1

= −qBx+ − x−√2

+ ky.

Thus, the components of the electric and magnetic field can be calculated asfollows

Ex = ∂−A+ − ∂+A− = 0,

Ey =1√2

(∂−Ay + ∂+Ay

)= 0,

Bz =1√2

(∂−Ay − ∂+Ay

)= qB,

which is the expected result.

5.3. Plane wave solutions

We will now generate new solutions similar to the ones we found in chapter 4.The simplest solutions are again those where r1, r2, and c are constant. In thiscase, the spinor has the following form

ψ =

r1

±s− iu

, s =√r2

2 − u2 = const, u =c√2r1

= const.

Note that this time s was defined without a sign so that subsequent expressionlook more familiar to the ones in the (1 + 1)-dimensional case. Calculating thecomponents of the three-potential Aµ gives

qA+ = ∓ m√2

r1

s,

qA− = ∓ m√2

r22

r1s= ∓ m√

2

s

r1

∓ m√2

u2

r1s,

qAy = ∓mu

s,

where r22 was written as s2 +u2 so that the expression for qA− is similar to the one

in the (1 + 1)-dimensional case. Because the components of Aµ are all constant,the solutions’s associated electric field is zero. Therefore, a gauge transformation

ψ 7→ ψ exp [∓ipµxµ]

53

5. Inverse approach in (2+1) dimensions

with

pµ =

p+

p−py

=m√

2

r1

ss

r1

+u2

r1s√

2u

s

can be used to set the components of Aµ to zero. Again, it can be proven that pµis the relativistic three-momentum of a particle with mass m

pµpµ = gµνpµpν = 2p+p− − p2

y = m2.

Hence, we have found plane wave solutions with positive and negative energy in(2 + 1) dimensions in our framework (note that this time s was defined with asign). Naming the positive energy solution u and the negative one v, we imposethe same normalisation conditions on them as we used before

uu =(s+ iu r1

)( r1

s− iu

)= 2r1s

!= +1,

vv =(−s+ iu r1

)( r1

−s− iu

)= −2r1s

!= −1,

so that we get the following normalised spinors

u =

(r1

(2r1)−1 − iu

), v =

(r1

−(2r1)−1 − iu

)Using the relationship between the spinor components and the momentum com-ponents, we can rewrite these solutions in terms of the momentum pµ

u = 2−1/4

√p+/m√

p−/m

1 + (py/m)2

(1− ipy

m

) , v = 2−1/4

√p+/m

−

√p−/m

1 + (py/m)2

(1− ipy

m

) .

5.4. Single wavefront

Solutions similar to the ones discussed in section 4.3 can be found in the (2 + 1)-dimensional formulation, too. If r1 and c are both constant, and r2(x−) does onlydepend on x−, i.e. we have a spinor

ψ =

(r1

±s(x−) + iu

)

54

5.4. Single wavefront

with r1 and u constant, the components of Aµ can be written as follows

qA+ = ∓ m√2

r1

s(x−),

qA− = ∓ m√2

s(x−)

r1

∓ m√2

u2

r1s(x−),

qAy = ∓m u

s(x−).

As these depend only on x−, the components of the electric and magnetic fieldcan be calculated easily

qEx = q ∂−A+ = ∓ m√2r1 ∂−

1

s(x−),

qEy =1√2q ∂−Ay = ∓ m√

2u ∂−

1

s(x−)=

u

r1

Ex,

qBz =1√2q ∂−Ay = qEy.

(5.10)

The expression for Ex is exactly the same as in (4.13); integrating this equationgives (4.14). The other components Ey and Bz have the same form but theiramplitude might be different as it is determined by u/r1. Because the electricfield depends only on x− = (t−x)/

√2, it can be imagined as a wavefront parallel

to y moving in x-direction. If u is non-zero, the electric field has a transversecomponent, too. A vector plot of the electric field at different times can be seen infig. 5.2. The parameter to control the transverse component was set to u/r1 = 0.5,so that the electric has a component in y-direction. Note that it is impossibleto create an electric field with no longitudinal component but only a transversecomponent using the ansatz given here: To achieve this, one would have to chooser1 = 0 so that Ey would diverge (see (5.10)).Solutions where the electric field does only depend on x+ can be created sim-

ilarly. If we choose r2 and c to be constant, and r1(x+) to only depend on x+,the components of the three-potential will also only depend on x+. Thus, thecomponents of the electric and magnetic field are given by

qEx = −q ∂+A− = ± m√2r2

2 ∂+

1

r1(x+)s(x+),

qEy =1√2q ∂+Ay = ∓ m√

2∂+

u(x+)

s(x+),

qBz =− 1√2q ∂+Ay = −qEy.

The y-component can be simplified even further

qEy = ∓ m√2∂+

u(x+)

s(x+)= ∓ m√

2

c√2∂+

1

r1(x+)s(x+)= − c√

2r22

qEx.

55

5. Inverse approach in (2+1) dimensions

-2 -1 0 1 2

-2

-1

0

1

2

x

1 /m

y

1 /m

(a)t

1/m= −3

-2 -1 0 1 2

-2

-1

0

1

2

x

1 /m

y

1 /m

(b)t

1/m= 0

-2 -1 0 1 2

-2

-1

0

1

2

x

1 /m

y

1 /m

(c)t

1/m= 3

Figure 5.2.: Vector plot of the electric field with Ex = E0/ cosh2(ωx−) and u/r1 =0.5, qE0/m

2 = 1, and ω/m = 2 at different times.

Using

r1(x+)s(x+) =

√r2

2r21(x+)− c2

2,

the equation for Ex can be integrated as we did before

r1(x+) =

√√√√√√√√√(rin

1 )2 − c2

2r22(

1±√

2

m

√(rin

1 )2

r22

− c2

2r42

q∫ x+−∞Ex(x+) dx+

)2 +c2

2r22

,

which is exactly of the same form as (4.12) if c = 0.Another possibility to generate a solution where the associated electric field

only depends on x+ is to let r1 and r2 be constant while c(x+) depends on x+.Again, for calculating the electromagnetic field’s components only the terms withthe derivatives with respect to x+ contribute

qEx = −q ∂+A− = ± m√2r2

2 ∂+

1

r1s(x+),

qEy =1√2q ∂+Ay = ∓ m√

2∂+

u(x+)

s(x+),

qBz =− 1√2q ∂+Ay = −qEy.

The equation for Ex can be integrated much in the same way as above (using

r1s(x+) =√r2

1r22 −

c(x+)2

2)

c(x+) =

√√√√√√c2

in − 2r21r

22(

1±√

2

m

√r2

1

r22

− c2in

2r42

q∫ x+−∞Ex(x+) dx+

)2 + 2r21r

22.

56

5.5. Two wavefronts

However, the y-component of the electric field is more complicated than in thecase discussed above. Calculating u(x+)/s(x+) using the above c(x+) gives

u(x+)

s(x+)=

c(x+)√2r2

1r22 − c2(x+)

=

√√√√√√√√√(

1±√

2

m

√r2

1

r22

− c2in

2r42

q∫ x+−∞Ex(x+) dx+

)2

1− c2in

2r21r

22

− 1.

Differentiating this expression with respect to x+ allows one to calculate Ey

qEy = −r1

r2

qEx

1−1− c2

in

2r21r

22

1±√

2

m

√r2

1

r22

− c2in

2r42

q∫ x+−∞Ex(x+) dx+

−1/2

.

5.5. Two wavefronts

Similarly to how the solutions with two moving pulses in (1 + 1) dimensions wereconstructed in section 4.4, we can find solutions where the electric field consists oftwo moving wavefronts in (2 + 1) dimensions. We start by choosing the followingfunctions for the spinor ψ

r1(x+) =

√√√√√√√√√(rin

1 )2 − c2

2(rin2 )2(

1 +

√2

m

1

rin2

√(rin

1 )2 − c2

2(rin2 )2

q∫ x+−∞E+(x+) dx+

)2 +c2

2(rin2 )2

r2(x−) =

√√√√√√√√√(rin

2 )2 − c2

2(rin1 )2(

1−√

2

m

1

rin1

√(rin

2 )2 − c2

2(rin1 )2

q∫ x−−∞E−(x−) dx−

)2 +c2

2(rin1 )2

c = const,

so that

s(x+, x−) =

√r2

2(x−) +c2

2r21(x+)

57

5. Inverse approach in (2+1) dimensions

Figure 5.3.: Electric field component Ex consisting of two sauter pulses calculatedusing (5.11) with rin = 1, sin = 1, qE(+)

o /m2 = 1, qE(−)0 /m2 = 1,

ω+/m = 4.5, ω−/m = 4.5 and c = 1.

is a function of both light cone coordinates. In this case, the light cone componentsof the three-potential are

qA+ =

[− m√

2r1(x+) +

c√2∂+

1

r1(x+)

] [r2

2(x−)− c2

2r21(x+)

]−1/2

,

qA− = − m√2r2(x−)

[r2

1(x+)− c2

2r22(x−)

]−1/2

.

Calculating the x-component of the electric field gives

qEx = q ∂−A+ − q ∂+A−

= qE−r1(x+)

rin1

r22(x−)− c2

2(rin1 )2

r22(x−)− c2

2r21(x+)

3/2

+ qE+r2(x−)

rin2

r21(x+)− c2

2(rin2 )2

r21(x+)− c2

2r22(x−)

3/2

− c√2

2

m2qE+qE−

1

rin1 r

in2

(1− c2

2r21(x+)(rin

2 )2

) r22(x−)− c2

2(rin1 )2

r22(x−)− c2

2r21(x+)

3/2

.

(5.11)This expression has the same form as (4.17) if c = 0. Otherwise, there is a newterm proportional to qE+qE−, directly coupling the two wavefronts. However, forsubcritical fields E+ and E− its contribution to the total electric field is supressedby the m−2 factor.

58

5.5. Two wavefronts

Figure 5.4.: Electric field component Ey consisting of two sauter pulses calculatedusing (5.12) with rin = 1, sin = 1, qE(+)

o = 1, qE(−)0 = 1, ω+ = 4.5,

ω− = 4.5 and c = 1.

To visualise this type of solution, we use two Sauter pulses E+ = E(+)0 sech2(ω+x+)

and E− = E(−)0 sech2(ω−x−). The x-component of the electric field for a specific

choice of parameters can be seen in fig. 5.3.The y-component of the electric field Ey and the magnetic field Bz can be cal-

culated using the derivatives of the y-component of the electromagnetic potentialAµ

qAy = −m u(x+)

s(x+, x−)= − m√

2

c√r2

1(x+)r22(x−)− c2/2

.

Thus, the y-component of the electric field is given by

qEy =1√2

(q ∂−Ay + q ∂+Ay

)

=c√2

qE−rin

1 r1(x+)

r22(x−)− c2

2(rin1 )2

r22(x−)− c2

2r21(x+)

3/2

− qE+

rin2 r2(x−)

r21(x+)− c2

2(rin2 )2

r21(x+)− c2

2r22(x−)

3/2 .

(5.12)A plot of qEy for the two Sauter pulses and using the same parameters as for thex-component of the electric field given in fig. 5.3 can be seen in fig. 5.4. The pulsemoving along the x+-axis has the same form as the same pulse in fig. 5.3, but theother pulse, which moves along the x−-axis, has a different sign than the one infig. 5.3.

59

5. Inverse approach in (2+1) dimensions

Figure 5.5.: Magnetic field BZ consisting of two sauter pulses calculated using(5.13) with rin = 1, sin = 1, qE(+)

o = 1, qE(−)0 = 1, ω+ = 4.5, ω− = 4.5

and c = 1.

The magnetic field can be calculated easily, too

qBz =1√2

(q ∂−Ay − q ∂+Ay

)

=c√2

qE−rin

1 r1(x+)

r22(x−)− c2

2(rin1 )2

r22(x−)− c2

2r21(x+)

3/2

+qE+

rin2 r2(x−)

r21(x+)− c2

2(rin2 )2

r21(x+)− c2

2r22(x−)

3/2 .

(5.13)The plot of qBz in fig. 5.5 reveals that qBZ is qualitatively the same as qEz,but has different amplitudes which can be controlled by the choice of c. Notethat Ey and Bz vanish completely for c = 0. As in the case of single wavefrontsdiscussed in the previous section, the parameter c controls the strength of thetransverse component of the electric field and the strength of the magnetic field.Figure 5.6 shows the time evolution of the two pulses as a vector plot for c = 1.Initially, the transverse components of the two wavefronts are equal (up to asign), but after they have interferred with each other, the transverse componentof the wavefront moving in positive x-direction is greater than before whereas thetransverse component of the other wavefront is smaller than before.

60

5.5. Two wavefronts

-2 -1 0 1 2

-2

-1

0

1

2

x

1 /m

y

1 /m

(a) t = −2

-2 -1 0 1 2

-2

-1

0

1

2

x

1 /m

y

1 /m

(b) t = −1

-2 -1 0 1 2

-2

-1

0

1

2

x

1 /m

y

1 /m

(c) t = −0.5

-2 -1 0 1 2

-2

-1

0

1

2

x

1 /m

y

1 /m

(d) t = −0.25

-2 -1 0 1 2

-2

-1

0

1

2

x

1 /m

y

1 /m

(e) t = 0

-2 -1 0 1 2

-2

-1

0

1

2

x

1 /m

y

1 /m

(f) t = 0.25

-2 -1 0 1 2

-2

-1

0

1

2

x

1 /m

y

1 /m

(g) t = 0.5

-2 -1 0 1 2

-2

-1

0

1

2

x

1 /m

y

1 /m

(h) t = 1

-2 -1 0 1 2

-2

-1

0

1

2

x

1 /m

y

1 /m

(i) t = 2

Figure 5.6.: Vector plot of the electric field consisting of two sauter pulses cal-culated using (5.13) with rin = 1, sin = 1, qE(+)

o = 1, qE(−)0 = 1,

ω+ = 4.5, ω− = 4.5 and c = 1.

61

6. Conclusion and outlook

Using the inverse approach developed and described in this thesis, it is possibleto generate new solutions of the covariant Dirac equation in spacetime-dependentelectric and magnetic fields. The method is significantly different from the tradi-tional ways of solving the Dirac equation. Instead of fixing an electric field andcalculating the solutions of the covariant Dirac equation for it, we guess an ar-bitrary function and calculate which electric field yields the same function as asolution to the Dirac equation.In (1+1) dimensions this means, by choosing a real-valued function r(x+, x−) of

the light cone coordinates x+ and x−, and a real-valued function c(x−) constrainedto only depend on x−, a well-defined solution to the covariant Dirac equation in(1+1) dimensions can be found. The electric field associated with the solution canbe calculated directly using the chosen functions r(x+, x−) and c(x−). If r doesnot depend on x−, the calculation of the electric field is considerably simplified. In(2 + 1) dimensions, two functions of all coordinates r1(x+, x−, y) and r2(x+, x,y),and a function c(x+, x−) that does not depend on y have to be chosen to fullydetermine the solution. The expressions for the components of the electric andmagnetic field are a bit more complicated than in the (1 + 1)-dimensional case,but in principle computable.The inverse approach has several advantages over the traditional method. First,

solutions for arbitrary electromagnetic fields can be found. The traditional restric-tion of only being able to calculate solutions for electromagnetic fields that dependon no more than one coordinate does not apply here. Indeed as was shown in sec-tion 4.4, section 4.5, and section 5.5, rather simple functions could be used togenerate solutions for electromagnetic fields depending on both light cone coordi-nates. By using more sophisticated functions, one could find solutions for a varietyof electric fields.Another advantage of the method presented in this thesis is that it is possible to

generate solutions with definite properties. For example, the perturbation ansatzin section 4.6 could be chosen so that the solution initially behaves like a planewave solution (i.e. the function r is constant) and finally consists of a mixture ofplane wave solutions with positive and negative energies (due r oscillating finally),thus resembling the solutions for the temporal Sauter pulse, that does createelectron-positron pairs from the vacuum. Then the chosen solution can be used tocalculate the corresponding electric field to examine under which circumstancesit is possible to create electron-positron pairs. This is completely different fromthe traditional method or the experimental access where one has to simply trydifferent electromagnetic fields until a solution that corresponds to pair creation

63

6. Conclusion and outlook

is found.Therefore, the inverse approach allows for more insight into the nature of the

spinor field. The relationship between the spinor and the electromagnetic fieldcan be examined from another point of view. As can be seen from the expressionsfor the electromagnetic field (4.9) and (5.9), the electromagnetic field is a highlynonlinear function of the spinor components. Thus, a small change in the spinormight change the resulting electromagnetic field strongly.The highly nonlinear nature of the electric field in terms of the spinor com-

ponents complicates the calculation of the electric field for the perturbed ansatzgiven in section 4.6. Therefore, only the leading terms (for β small and m large) inthe electric field were considered. However, the terms of lower order in m cannotbe neglected completely. For example, the term of order m1 is the leading term inthe electric field together with the undisturbed field E(α) if α is slowly varying andshould therefore be considered in the expansion. Furthermore, we have chosen thefunction β at will, but it is imaginable that requiring that the electric field shouldnot oscillate rapidly to order m1, too, might impose conditions on the form ofβ. Moreover, there might be a relationship between this ansatz and WKB-typesolutionsof the Dirac equation which should be analysed in detail.The solutions discussed so far in (2 + 1)-dimensional spacetime were rather

limited as Ey and Bz were almost always of exactly the same form as Ex. Solutionsthat also depend on y should probably be used to generate more general forms ofelectromagnetic fields. In particular, if r2 does depend on y, then there is a uniquecontribution to Ay which does not appear in the other components of the three-potential and could accomplish to generate a distinct form of the magnetic field.Finally, the approach should be generalized to (3 + 1) spacetime dimensions.

This should be doable in principle as the Dirac equation then must have fourspinor components and the electromagnetic potential Aµ has four componentsin (3 + 1) dimensions, too. Therefore, the Dirac equation can then again besolved for the components of the potential. Requiring that Aµ has to be real,the eight real functions needed to describe a four-component spinor can be cutdown to four real functions. Moreover, Gauge freedom then should allow foreliminating one of these, reducing the number of real-valued functions needed tofully determine a solution to three. However, the derivation of the method in (3+1)dimensions is expected to be still more complex than in (2+1) dimensions and theresulting expressions for the four-potential Aµ might be more complicated thanbefore. Nevertheless, it would be interesting how the lower-dimensional solutionsdescribed in this thesis would appear in the (3 + 1)-dimensional formalism.

64

A. Expansion in powers of themass parameter

The result of the expansion of the electric field in powers of m including only thehighest order term:

qE =m2

√2s3r2

[sin(2mγ)

(αβ2c(∂− γ)(∂+ γ) + 2αβ2c(∂− γ)2(∂+ γ)

− αβ2(∂− γ)(∂+ γ)

∫∂− α

2 dx+ − 2αβ2(∂− γ)2(∂+ γ)

∫∂− α

2 dx+

+1

2αβ4 +

1

2αβ4(∂− γ)3 − αβ4(∂− γ)2(∂+ γ)− 4αβ4(∂− γ)3(∂+ γ)

+ α3β2(∂− γ)2 − α3β2(∂− γ)3

)+ sin(4mγ)

(1

4αβ4(∂− γ)3 − 1

2αβ4(∂− γ)2(∂+ γ)

)+ cos(mγ)

(− β(∂+ γ)

[ ∫∂− α

2 dx+

]2+ 2βc(∂+ γ)

∫∂− α

2 dx+

− βc2(∂+ γ)− 2α2βc(∂− γ) + 2α2βc(∂− γ)2(∂+ γ)

+ 2α2β(∂− γ)

∫∂− α

2 dx+ − 2α2β(∂− γ)2(∂+ γ)

∫∂− α

2 dx+

− β3c(∂− γ) +7

2β3c(∂− γ)(∂+ γ) +

1

2β3c(∂− γ)2(∂+ γ)

+ β3(∂− γ)

∫∂− α

2 dx+ −7

2β3(∂− γ)(∂+ γ)

∫∂− α

2 dx+

− 1

2β3(∂− γ)2(∂+ γ)

∫∂− α

2 dx+ − α2β3 (∂− γ)2

∂+ γ+

7

2α2β3(∂− γ)2

− 4α2β3(∂− γ)3(∂+ γ)− α4β(∂− γ)2

∂+ γ− 1

4β5 (∂− γ)2

∂+ γ+

7

4β5(∂− γ)2

− 3β5(∂− γ)2(∂+ γ) +1

4β5(∂− γ)3 − β5(∂− γ)3(∂+ γ)

)cos(3mγ)

(1

2β3c(∂− γ)(∂+ γ)− 1

2β3c(∂− γ)2(∂+ γ)

− 1

2β3(∂− γ)(∂+ γ)

∫∂− α

2 dx+ +1

2β3(∂− γ)2(∂+ γ)

∫∂− α

2 dx+

65

A. Expansion in powers of the mass parameter

+1

2α2β3(∂− γ)2 + α2β3(∂− γ)3 +

1

4β5(∂− γ)2 − β5(∂− γ)2(∂+ γ)

− 1

4β5(∂− γ)3 + β5(∂− γ)3(∂+ γ)

)]+O

(m1)

66

Bibliography

[AS72] Milton Abramowitz and Irene A. Stegun. Handbook of mathematicalfunctions. 10th Pr. 1972. url: http://people.math.sfu.ca/~cbm/aands/abramowitz%5C_and%5C_stegun.pdf.

[Cor89] John F. Cornwell. Group Theory in Physics - Volume 3. Ed. by N. H.March. Academic Press, 1989. isbn: 0121898067.

[Dir28] P. A. M. Dirac. “The Quantum Theory of the Electron”. In: Proceed-ings of the Royal Society A: Mathematical, Physical and Engineer-ing Sciences 117.778 (Feb. 1928), pp. 610–624. issn: 1364-5021. doi:10.1098/rspa.1928.0023.

[Dir30] P. A. M. Dirac. “A Theory of Electrons and Protons”. In: Proceedings ofthe Royal Society A: Mathematical, Physical and Engineering Sciences126.801 (Jan. 1930), pp. 360–365. issn: 1364-5021. doi: 10.1098/rspa.1930.0013.

[Dir31] P. A. M. Dirac. “Quantised Singularities in the Electromagnetic Field”.In: Proceedings of the Royal Society A: Mathematical, Physical andEngineering Sciences 133.821 (Sept. 1931), pp. 60–72. issn: 1364-5021.doi: 10.1098/rspa.1931.0130.

[Heb11] Florian Hebenstreit. “Schwinger effect in inhomogeneous electricfields”. Dissertation. Karl-Franzens-Universität Graz, 2011. url:http://arxiv.org/abs/1106.5965.

[HIM11] Florian Hebenstreit, Anton Ilderton, and Mattias Marklund. “Pair pro-duction: The view from the lightfront”. In: Physical Review D 84.12(Dec. 2011), p. 125022. issn: 1550-7998. doi: 10.1103/PhysRevD.84.125022.

[Pau27] Wolfgang Pauli. “Zur Quantenmechanik des magnetischen Elektrons”.In: Zeitschrift für Physik 43.9-10 (Sept. 1927), pp. 601–623. issn: 1434-6001. doi: 10.1007/BF01397326.

[Rab28] I. I. Rabi. “Das freie Elektron im homogenen Magnetfeld nach derDiracschen Theorie”. In: Zeitschrift für Physik 49.7-8 (1928), pp. 507–511.

[Sau31] Fritz Sauter. “Über das Verhalten eines Elektrons im homogenen elek-trischen Feld nach der relativistischen Theorie Diracs”. In: Zeitschriftfür Physik 69.11-12 (1931), pp. 742–764.

67

Bibliography

[Sau32] Fritz Sauter. “Zum „Kleinschen Paradoxon“.” In: Zeitschrift für Physik73.7-8 (1932), pp. 547–552.

[TTW00] T. N. Tomaras, N. C. Tsamis, and R. P. Woodard. “Back reaction inlight cone QED”. In: Physical Review D 62.12 (Nov. 2000), p. 125005.issn: 0556-2821. doi: 10.1103/PhysRevD.62.125005.

[TTW01] T. N. Tomaras, N. C. Tsamis, and R. P. Woodard. “Pair creation andaxial anomaly in light-cone QED2”. In: Journal of High Energy Physics11.008 (2001).

68

An dieser Stelle möchte ich mich für die freundliche Betreuung durch HerrnProf. Dr. Schützhold während der Bearbeitungszeit bedanken. Weiterhin dankeich der kompletten Arbeitsgruppe Schützhold für das angenehme Arbeitsklimaund insbesondere Christian Schneider für die vielen interessanten Diskussionen.Zudem gilt mein Dank meiner Familie und meinen Freunden für die fortwäh-

rende Unterstützung in dieser Zeit.

Erklärung:

Hiermit versichere ich, dass ich diese Arbeit selbständig verfasst, keine anderen alsdie angegebenen Quellen und Hilfsmittel benutzt sowie Zitate kenntlich gemachthabe.

Duisburg, den 30. September 2014