Lecture Notes: Relativistic Quantum Mechanics, Study Guides, Projects, Research of Quantum Mechanics

Download Lecture Notes: Relativistic Quantum Mechanics and more Study Guides, Projects, Research Quantum Mechanics in PDF only on Docsity! Lecture Notes: Relativistic Quantum Mechanics Frank Krauss1 1Institute for Particle Physics Phenomenology, Physics Department, Durham University, Durham DH1 3LE, UK Last Updated: May 4, 2021 1 Disclaimer These lecture notes accompany the final-year undergraduate lecture course on “Relativistic Quantum Mechanics”, consisting of 12 lectures, delivered during the 2020/2021 academic year in an online format. The notes are by no means original. Instead I shamelessly borrowed from a multitude of resources, and tried to put material into a coherent form. I also try and cite online links to books etc., where possible – I am aware that the library holds some of them, also in electronic form. The lectures mainly deal with second quantisation, a topic that has been excellently covered in the literature and on the web. There are many truly excellent textbooks on the topic, often named “Introduction to Quantum field Theory” or similar, for example the books by Peskin & Schröder [1], Griffiths [2], Schwartz, Zee, or Hatfield [3], in addition to a multitude of freely available lectures notes on the web: ˆ Mark Srednicki’s notes on Quantum Field Theory [4], which have since been published as a book; ˆ David Tong’s lectures on Quantum Field Theory [5]; ˆ lecture notes of the great Sidney Coleman on Quantum Field The- ory [6]; ˆ Jeff Dror’s summary of practically all relevant relations worked out in this course, and, in fact, many more beyond it, can be found in [7]. The lecture notes will be continuously updated over the course of the year - please check the date on the front page to keep track of changes. When you compare the notes with books you will realise that notation and conventions differ between different resources. However, quite often these differences boil down to trivial normalisations. I’ve tried, hopefully successfully, to be at least self-consistent. The notes are supplemented with worked examples and problems through- out, and I cannot overemphasise how important it is to actually calculate things on your own. Tougher, expert-level problems are identified with an asterisk. They are outside the scope of examinable material and are solely geared to helping interested students to develop a deeper understanding of the subject and to contextualising the material in a wider perspective. I have also added “extremely unbelievably hard” questions, indicated with two asterisks. They cover material that is entirely beyond the scope of the course, but may trigger some further reading and digging by students with a soft spot for the abyss that is Quantum Field Theory. Over the course of six weeks we will work through the analogue of 12 lectures - I will try to highlight and explain crucial concepts in short movies with me working through things on a white board - however, these movies are i 8 Interacting Fields 164 8.1 Perturbative Expansion: Born Series . . . . . . . . . . . . . . 164 8.2 Interacting Field Theory: General Thoughts . . . . . . . . . . 165 8.3 Interacting Field Theory: λφ4 . . . . . . . . . . . . . . . . . . 168 9 List of Problems 178 iv 1 Introduction In this course, “Relativistic Quantum Mechanics”, we combine Quantum Mechanics with Special Relativity and develop a formalism to quantise fields in a Lorentz-invariant way. We will recapitulate the Lagrange and Hamilton formalism for the treat- ment of classical point particles as well as the quantisation of the harmonic oscillator through creation and annihilation operators. Building on the for- mer, we will briefly analyse the Lagrange formalism and the derivation of Euler-Lagrange equations of motion for a discrete system, before taking the continuum limit, resulting in the Lagrange formulation of field dynamics. We will analyse free real and complex scalar fields in this formalism, and for the latter, we will find a symmetry – phase shifts of the fields – that leaves the Lagrangian invariant. We will see that such invariances result in conserved currents and charges. We will further exemplify the power of the formalism by constructing a Lagrange density for the electromagnetic fields and deriving Maxwell’s equations from it. To quantise fields we will copy the steps known from single-particle systems, in particular the harmonic oscillator, and adapt it to the case of fields. In so doing we effectively replace the role of position and momentum of the particle, and the corresponding operators, with the field and its conjugate momentum. The resulting logic is to replace the functions describing fields and their conjugate momenta with field and momentum operators, and to de- mand suitable commutator relations for them. This is called second quanti- sation. As a consequence of relativistic invariance, encoded in the quadratic energy-momentum relation of E2− p2 = m2, solutions with negative energy become possible. Demanding a Hamiltonian with an energy spectrum that is bounded from below, i.e. a physically meaningful ground state or vac- uum, necessitates their interpretation as anti-particles. It also immediately implies we have arrived at a multi-particle theory, because pairs of parti- cles and anti-particles with short lifetimes can be produced. We will check, by explicit calculation, that the resulting theory maintains causality at a microscopic level, by asserting that commutators of causally disconnected fields always vanish and that they therefore cannot impact onto each other. After second quantisation of the simplest possible theory, a single free real scalar field, we analyse the structure of a free complex scalar field. We will recover the current and charge stemming from the phase invariance of the Lagrangian and we will by explicit calculation show that the charge and the Hamilton operators commute, making charge conservation of the theory manifest. After analysing the free scalar or Klein-Gordon fields we will turn our atten- tion to the treatment of spin-1/2 particles in the celebrated Dirac equation. We will analyse its structure and ingredients – γ matrices and spinors – and their properties before second quantisation of the theory. Reflecting 1 the fermionic nature of the particles, we will use anti-commutators {·, ·} instead of commutators [·, ·] for the quantisation conditions. Similar to the case of the complex scalar field, also the Lagrangian for free spinors enjoys invariance under phase transformations of the fields, and again this leads to a conserved charge. We then turn our attention to the quantisation of electrodynamics and the free electromagnetic fields. There, we will encounter an interesting problem: the vector potential Aµ, on which we build the theory, naively speaking, has four degrees of freedom in its four-components, but the physical field has only two degrees of freedom, the well-known linear or circular polarisation states of the photons, the quanta of electromagnetism. This necessitates the imposition of additional conditions onto the theory, to correctly reflect its physical content. In more formalised language, this problem is a result of the gauge invariance of the underlying theory, electromagnetism, which results in identical physical fields for different vector potentials. It will become clear that the problem of the additional content will be fixed by fixing the gauge of the theory, and we will see how this is shapes the additional conditions we will impose on the theory. Having quantised various free field theories and discussing some of their properties, we will start with developing a framework to analyse their dy- namical behaviour. To this end we will build on the concept of Green’s functions and construct the Green’s functions of our quantised theories. It will turn out that these “propagators” are the vacuum expectation values of time-ordered products of the field operators. 2 Einstein Convention When not stated otherwise we will use Einstein’s convention of summing over repeated indices, for example p2 = pipi = p2 x + p2 y + p2 z. (8) Metric Tensor For four-vectors this is a meaningful operation only when combining contravariant objects (where the index is a superscript) with co- variant objects (where the index is a subscript). The two sets of four-objects – contravariant and covariant – are connected through the metric tensor gµν , pµ = gµνp ν and pµ = gµνpν , (9) where the Minkowski metric is given by gµν = gµν = diag(1, −1) =  1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1  . (10) In other words, if pµ = (E, p), pµ = (E, −p). From pµ = g ν µ pν we can easily infer that g ν µ = gµν = diag(1, 1) =  1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1  . (11) Raising and Lowering Indices We have already seen, how the metric tensor is used to raise or lower indices of four-vectors, e.g., xµ = gµνx ν and xµ = gµνxν , (12) which introduces a sign flip in the spatial coordinates: if xµ = (t, x) then xµ = (t, −x) . (13) For tensors with n indices, one metric tensor is necessary to raise or lower one index. For example, for a tensor Fµν of rank two, two metric tensors are necessary to lower both indices. As an example, consider the field-strength tensor of electromagnetism, given by Fµν =  0 −Ex −Ey −Ez Ex 0 −Bz By Ey Bz 0 −Bx Ez −By Bx 0  . (14) 5 Therefore, Fµν = gµµ′gν′νF µ′ν′ , (15) where, making the sequence of matrix multiplications explicit Fµν = gνν′F µ′ν′ = Fµ ′ν′gν′ν =  0 −Ex −Ey −Ez Ex 0 −Bz By Ey Bz 0 −Bx Ez −By Bx 0   1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1  =  0 Ex Ey Ez Ex 0 Bz −By Ey −Bz 0 Bx Ez By −Bx 0  (16) and Fµν = gµµ′F µ′ ν′ =  1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1   0 Ex Ey Ez Ex 0 Bz −By Ey −Bz 0 Bx Ez By −Bx 0  =  0 Ex Ey Ez −Ex 0 −Bz By −Ey Bz 0 −Bx −Ez −By Bx 0  ; (17) in other words, lowering both indices changed the sign in the 0-row and 0- column of the tensor – the mixed temporal-spatial entries – and left the temporal-temporal and spatial-spatial entries unchanged. Scalar Product Scalar products of two four-vectors are then given by x · p = xµp µ = xµpµ = x0p0 − x · p = x0p0 − x1p1 − x2p2 − x3p3 (18) Derivatives Derivatives of a scalar or scalar product with respect to a vector are given by ∂a · b ∂aµ = ∂aµ · bµ ∂aµ = bµ , (19) i.e. derivatives of a scalar quantity w.r.t a covariant vector yield a con- travariant vector. In particular it is customary to define ∂µ = ∂ ∂xµ = (∂/∂t, ∇) 6 ∂µ = ∂ ∂xµ = (∂/∂t, −∇) . (20) Note that the derivatives have a relative sign in the spatial coordinates! Relativistic Energy-Momentum Relation In particular, the energy- momentum relation for a physical particle of (rest) mass m can be written as p2 = E2 − p2 = m2 (21) Kronecker-δ and Levi-Civita Tensor Two important tensors in three dimensions are the Kronecker-δ, δij = δij = { 1 if i = j 0 otherwise, (22) and the anti-symmetric Levi-Civita Tensor, given by εijk = εijk =  1 if {ijk} = cyclical permutation of 123 −1 if {ijk} = anti-cyclical permeation of 123 0 otherwise. (23) The latter is generalised to the totally anti-symmetric tensor in four dimen- sions, εµνρσ with εµνρσ = −εµνρσ =  1 if {µνρσ} = cyclical permutation of 0123 −1 if {µνρσ} = anti-cyclical permutation of 0123 0 otherwise. (24) 2.4 Lorentz Transformations General idea Lorentz transformations, xµ −→ x′µ = Λµνx ν (25) are linear transformations that connect four-vectors with each other. The Λµν are usually divided into active transformations where the four-vector in question is moved while the reference system is fixed, and passive transfor- mations, where the four-vector is fixed, but the reference system is changed. The difference between active and passive transformations is encoded in a relative sign of the defining parameters. In the context of this lecture, the idea of Lorentz transformations is gener- alised such that they contain both boosts Bµ ν and rotations Rµν , where the former are defined by three velocities and the latter defined by three angles. In fact, the rotation are the Galilei transformations, which are superseded by the Lorentz transformations. 7 This implies that (Λ−1)µµ′ = (ΛT )µµ′) = Λ µ′ µ , (36) i.e. transposition is the inverse of a Lorentz transformation. 2.5 Lagrange and Hamilton Formalism for Point Particles Lagrange Function Consider a point particle with kinetic energy T and a set of generalised coordinates qi(t) and velocities q̇i(t) = dqi/dt that are suitable to describe its motion in a potential V . The Lagrange function is given by L(qi(t), q̇i(t), t) = T − V (37) and gives rise to the action S(t1, t0) = t1∫ t0 dtL(qi(t), q̇i(t), t) . (38) Principle of Least Action Minimising the action by employing virtual small perturbations of the particle’s path εi and ε̇i, taken to be zero at the endpoints t0 and t1, will yield the Euler-Lagrange Equations of Motion (E.o.M.). This is also known as Hamilton’s Principle or Principle of Least Action. Under the usual assumption of an explicitly time-independent La- grange function and suppressing for a moment the time dependence of the generalised coordinates and velocities, this yields δS = t1∫ t0 dt [L(qi + εi, q̇i + ε̇i)− L(qi, q̇i)] = t1∫ t0 dt [ εi ∂L ∂qi + ε̇i ∂L ∂q̇i ] = t1∫ t0 dt [ εi ∂L ∂qi − εi d dt ∂L ∂q̇i ] + [ εi ∂L ∂q̇i ]t1 t0 , (39) where in the last step the term with ε̇i has been partially integrated. Euler-Lagrange Equations of Motion Since the variations εi are as- sumed to vanish at the path’s endpoint the last term vanishes, demanding that the integral reduces to zero for arbitrary perturbations yields the Euler- Lagrange E.o.M. for systems without explicit time dependence: ∂L ∂qi − d dt ∂L ∂q̇i = 0 . (40) 10 Canonical Momentum and Hamilton Function Introducing the canon- ical momenta pi = ∂L ∂q̇i (41) and expressing the generalised velocities through the canonical momenta pi allows to construct the Hamilton function as H(pi, qi) = q̇ipi − L(qi, q̇i) = q̇i ∂L ∂q̇i − L(qi, q̇i) = T + V , (42) identical to the energy of the system if it is not explicitly time-dependent. Hamilton Equations of Motion The Hamilton equations of motions are given by the two sets of coupled partial differential equations ṗi = dpi dt = −∂H ∂qi q̇i = dqi dt = + ∂H ∂pi . (43) Poisson Brackets Poisson brackets are another possibility to express the Hamilton E.o.M. they are defined by {f, g} = ∂f ∂qi ∂g ∂pi − ∂f ∂pi ∂g ∂qi . (44) They have some interesting properties, for example ˆ anti-commutativity: {f, g} = −{g, f} (45) ˆ bilinearity (a and b constants): {af + bg, h} = a {f, h}+ b {g, h} (46) ˆ Jacobi identity: {f, {g, h}}+ {g, {h, f}}+ {h, {f, g}} = 0 (47) In particular, the Poisson brackets for the canonical coordinates (positions and momenta) enjoy the following simple properties: {qi, qj} = {pi, pj} = 0 11 {qi, pj} = δij . (48) Equations of motion can therefore be expressed as ṗi = − ∂H ∂qi = {pi, H} q̇i = + ∂H ∂pi = {qi, H} . (49) The time evolution of any function f(pi, qi, t) can be evaluated using the chain rule, df dt = ∂f ∂qi q̇i + ∂f ∂pi ṗi + ∂f ∂t = {f, H}+ ∂f ∂t . (50) This translates into explicitly time-independent f are constant of motion, if their Poisson bracket with the Hamilton function vanishes2. 2.6 First Quantisation of the Harmonic Oscillator Hamilton operator In a first step, the Hamilton function is written in terms of the usual canonical position and momentum, and position, momen- tum, and Hamilton function are promoted to operators 3, resulting in Ĥ = 1 2m p̂2 + mω2 2 x̂2 . (51) Note that we have used natural units, setting ~ = 1, and in the following we will also set m = 1 to ease the notation. Commutator of Position and Momentum Quantisation is achieved by demanding that the position and momentum operators have a non-vanishing commutator 4, namely [x̂, p̂] ≡ x̂p̂− p̂x̂ = i . (52) Creation and Annihilation Operators To cast the Hamilton operator into a form better suited for analysis, creation and annihilation operators ↠and â are introduced as â = 1√ 2 (√ ω x̂+ i√ ω p̂ ) 2Note the similarity of the Poisson brackets to the commutator in Quantum Mechanics. It is, however, important to stress that the functions here are not operators acting on a Hilbert space, but just functions. 3Throughout the lecture we will denote operators through aˆsymbol. 4Remember the Poisson brackets? Of course, as functions, the sequence of their product is irrelevant, but as operators this is not the case anymore. In this respect the commutator, although not connected to any derivative, behaves quite similarly to the Poisson brackets. 12 2.7 Problems & Solutions 1. Levi-Civita symbol (a) Show that for the Levi-Civita symbol in three dimensions, εijkεilm = δjlδkm − δjmδkl εijkεijl = 2δkl εijkεijk = 6 , (b) and that for the Levi-Civita symbol in four dimensions, εµνρσεµν ρ′σ′ = − 2(gρρ ′ gσσ ′ − gσρ′gρσ′) εµνρσεµνρ σ′ = − 6gσσ ′ εµνρσεµνρσ′ = − 24 . Solution (a) Levi-Civita in three dimensions: First of all, it is improtant to stress here that we use Einstein’s convention over repeated indices throughout. To have a non-vanishing εijk the three indices must be different. Without any loss of generality this implies that j 6= k and l 6= m must be fulfilled for the product εijkεilm to be different from 0. In addition, i has to be different to both j and k and to l and m, and the sum collapses to only one term (three dimensions, so i, j, . . . are numbers in {1, 2, 3}), where j and k are identical to l and m, so either j = l and k = m, or j = m and k = l. These are the two δ-terms. The first term, with the positive sign, emerges from ijk and ilm being both either cyclical (εijk = 1) or anticyclical (εijk = −1), with {ijk} = {ilm}, while the second term, the one with the negative sign comes from {ijk} = {iml} and one of the two being cyclical implies that the other is anti-cyclical. This proves εijkεilm = ∑ i εijkεilm = δjlδkm − δjmδkl . For the product εijkεijl, similar considerations apply. Demanding that i, j 6= k and i, j 6= l means that k = l must be fulfilled and i 6= j means that for a fixed k, there are two combinations possible for ij, either cyclical or anti-cyclical. Therefore εijkεijl = 2δkl . Finally, for εijkεijk, it suffices to count how many permutations of {ijk} = {123} exist to arrive at εijkεijk = 6. 15 (b) Levi-Civita in four dimensions: The identities for the totally antisymmetric tensor in four dimen- sions follow from using the same logic as before. The realitve minus sign in front of the expressions is relatively easy to ex- plain with the signs in the metric tensor, since for the spatial components gρρ ′ = −δρρ′ . 2. Boosts and Rotations (a) Calculate the effect of two consecutive boosts in z-direction, given by rapidities η1 and η2. Do the two operations commute (i.e. what happens if you reverse the order)? Hint: Use that coshα coshβ ± sinhα sinhβ = cosh(α± β) coshα sinhβ ± sinhα coshβ = sinh(α± β) cosα cosβ ± sinα sinβ = cosh(α∓ β) cosα sinβ ± sinα cosβ = sinh(α± β) (b) Repeat the exercise for two consecutive rotations around the z- axis with angles θ1 and θ2. Solution (a) Recalling the boost matrices B12 =  cosh η1,2 0 0 − sinh η1,2 0 1 0 0 0 0 1 0 − sinh η1,2 0 0 cosh η1,2  and therefore, consecutively applying boost 2 after boost 1 B2B1 = ( cosh η2 0 0 − sinh η2 0 1 0 0 0 0 1 0 − sinh η2 0 0 cosh η2 )( cosh η1 0 0 − sinh η1 0 1 0 0 0 0 1 0 − sinh η1 0 0 cosh η1 ) = ( cosh(η1 + η2) 0 0 − sinh(η1 + η2) 0 1 0 0 0 0 1 0 − sinh(η1 + η2) 0 0 cosh(η1 + η2) ) = B1B2 . (b) Similarly, the rotation matrices R12 =  1 0 0 0 0 cos θ1,2 − sin θ1,2 0 0 sin θ1,2 cos θ1,2 0 0 0 0 1  16 and therefore, consecutively applying boost 2 after boost 1 R2R1 = ( 1 0 0 0 0 cos θ2 − sin θ2 0 0 sin θ2 cos θ2 0 0 0 0 1 )( 1 0 0 0 0 cos θ1 − sin θ1 0 0 sin θ1 cos θ1 0 0 0 0 1 ) = ( 1 0 0 0 0 cos(θ1 + θ2) − sin(θ1 + θ2) 0 0 sin(θ1 + θ2) cos(θ1 + θ2) 0 0 0 0 1 ) = R1R2 . 3. Inverse Lorentz transformation (a) use the invariance of distances under Lorentz transformations x′ 2 = [Λµνx ν ]2 to show that [Λµν ]−1 = Λ µ ν (b) what is the form of inverse Lorentz boosts and rotations along or around the z-axis? Solution (a) x2 = gµνx µxν = x′ 2 = gρσx ′ρx′σ = gρσΛραx αΛσβx β =⇒ gµν = gρσΛρµΛσν =⇒ δγµ = gµνg νγ = gρσg νγΛρµΛσν =⇒ δγµ = gρσΛσγΛρµ =⇒ δγµ = Iγµ = Λ γ ρ Λρµ Written in matrix notation this implies that Λ−1 = gΛT g (b) Using the metric tensor to raise and lower the two indices, yields Λ µ ν = gρνΛρσg µσ = ( 1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1 )( coshu 0 0 − sinhu 0 1 0 0 0 0 1 0 − sinhu 0 0 coshu )( 1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1 ) = ( 1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1 )( coshu 0 0 sinhu 0 −1 0 0 0 0 −1 0 − sinhu 0 0 − coshu ) = ( coshu 0 0 sinhu 0 1 0 0 0 0 1 0 sinhu 0 0 coshu ) 17 (b) Let us consider a number of cases, namely the commutator of two two boosts, of two rotations, and of a boost and a rotation.[ M̂01, M̂02 ] = ig00M̂12 = iM̂12[ M̂12, M̂13 ] = −ig11M̂23 = iM̂23[ M̂01, M̂12 ] = ig11M̂02 = −iM̂02 We then have, for the vanishing commutator,[ X̂±i , X̂ ∓ j ] = 1 4 [ 1 2 εiklM̂kl ± iM̂i0, 1 2 εjmnM̂mn ∓ iM̂j0 ] = 1 4 {εiklεjmn 4 [ M̂kl, M̂mn ] ± iεjmn 2 [ M̂i0, M̂mn ] ∓ iεikl 2 [ M̂kl, M̂j0 ] + [ M̂i0, M̂j0 ]} = i 4 {εiklεjmn 4 ( gknM̂lm + glmM̂kn − gkmM̂ln − glnM̂km ) ± iεjmn 2 ( ginM̂0m + g0mM̂in − gimM̂0n − g0nM̂im ) ∓ iεikl 2 ( gk0M̂lj + gljM̂k0 − gkjM̂l0 − gl0M̂kj ) − g00M̂ij } = i 4 { −1 4 [ (δljδim − δlmδij)M̂lm + (δinδjk − δijδkn)M̂kn − (δlnδij − δljδin)M̂ln − (δijδkm − δimδkj)M̂km ] ∓ i 2 [ εijmM̂0m − εjinM̂0n ] ∓ i 2 [ εikjM̂k0 − εijlM̂l0 ] − M̂ij } = i 4 { −1 4 [ M̂ji + M̂ji + M̂ji + M̂ji ] − M̂ij ∓ i 2 [ εijmM̂0m − εjimM̂0m − εimjM̂m0 + εijmM̂m0 ]} = i 4 { −M̂ji − M̂ij ∓ i 2 [ εijmM̂0m(1 + 1− 1− 1) ]} = 0 , where, in the last step, we have used the anti-symmetry of the Levi-Civita Tensor and that M̂0m = −M̂m0. In the treatment of the terms proportional to only one of the Levi-Civita Tensors we also realised that terms of the form εijkgk0 vanish – after all εijk is only defined for permutations of the spatial dimensions, i.e. {i, j, k} ∈ perm({1, 2, 3}). We also took into account the sign for space-like components, i.e. gik = −δik. 20 The non-vanising commutators in contrast are given by[ X̂±i , X̂ ± j ] = 1 4 [ 1 2 εiklM̂kl ± iM̂i0, 1 2 εjmnM̂mn ± iM̂j0 ] = 1 4 {εiklεjmn 4 [ M̂kl, M̂mn ] ± iεjmn 2 [ M̂i0, M̂mn ] ± iεikl 2 [ M̂kl, M̂j0 ] − [ M̂i0, M̂j0 ]} = i 4 {εiklεjmn 4 ( gknM̂lm + glmM̂kn − gkmM̂ln − glnM̂km ) ± iεjmn 2 ( ginM̂0m + g0mM̂in − gimM̂0n − g0nM̂im ) ± iεikl 2 ( gk0M̂lj + gljM̂k0 − gkjM̂l0 − gl0M̂kj ) + g00M̂ij } = i 4 { −1 4 [ (δljδim − δlmδij)M̂lm + (δinδjk − δijδkn)M̂kn − (δlnδij − δljδin)M̂ln − (δijδkm − δimδkj)M̂km ] ∓ i 2 [ εijmM̂0m − εjinM̂0n ] ± i 2 [ εikjM̂k0 − εijlM̂l0 ] + M̂ij } = i 4 { −1 4 [ M̂ji + M̂ji + M̂ji + M̂ji ] + M̂ij ∓ i 2 [ εijmM̂0m − εjimM̂0m − εimjM̂m0 + εijmM̂m0 ]} = i 4 { −M̂ji + M̂ij ∓ i 2 [ εijmM̂0m(1 + 1 + 1 + 1) ]} = i 2 [ M̂ij ± iεijkM̂k0 ] and direct comparison with the definition of X̂±m, iεijmX̂ ± m = iεijm 2 (εmrs 2 M̂rs ± iM̂m0 ) = i 2 [εijmεrsm 2 M̂rs ± iM̂m0 ] = i 2 [ δirδjs − δisδjr 2 M̂rs ± iM̂m0 ] = i 2 [ 1 2 ( M̂ij − M̂ji ) ± iM̂m0 ] = i 2 [ M̂ij ± iM̂m0 ] = yields the desired result. This proves that the six generators of the Lorentz group fac- torise into two groups of three generators, where each group has a commutator structure that is identical to the one enjoyed by the 21 generators of the angular momentum group, and where the gener- ators of the two groups do commute. In other words, the Lorentz group SO(3, 1) decomposes as SO(3, 1) = SU(2)⊗SU(2), hinting at a deep connenction between the Lorentz group and spin. 5. ∗Poincare transformations The Poincare transformation U(Λ, a) is defined by the combination of a Lorentz transformation, Λµν , and a shift in space-time, aµ, as xµ → x′µ = Λµνx ν + aµ . (a) Determine the product, the unit and inverse of the resulting group. (b) Verify that U−1(Λ, 0)U(1, ε)U(Λ, 0) = U(1, Λ−1ε) and show that this implies that U−1(Λ, 0)P̂µU(Λ, 0) = (Λ−1)νµP̂ν . Use this to determine the commutator [M̂µν , P̂ρ] of the generators of the Lorentz group and the momentum operator. (c) Show that U−1(Λ, 0)U(Λ̃, 0)U(Λ, 0) = U(Λ−1Λ̃Λ, 0) and use this to prove the commutator relation of the generators M̂µν from the previous problem, i.e.[ M̂µν , M̂ρσ ] = i ( gµρM̂νσ − gµσM̂νρ − gνρM̂µσ + gνσM̂µρ ) . Solution (a) Let us start by the product of two transformations, U(Λ, a) ⊗ U(Λ̃, ã) i.e. xµ → x′µ = Λµν ( Λ̃νρx ρ + ãν ) + aµ , and we can read off that U(Λ, a)⊗ U(Λ̃, ã) = U(ΛΛ̃, Λã+ a) the product is given by a product of the Lorentz-transformation with a shift given by the sum of the Lorentz-transformed first shift and the second shift. 22 (a) write down the Lagrange function; (b) derive and solve the Euler-Lagrange E.o.M.; (c) construct the canonical momenta; (d) find the Hamilton function; (e) derive the Hamilton E.o.M.; (f) try to directly infer constants of motion where possible. Solution (a) (i) : L = m 2 ẋ2 (ii) : L = ml2 2 θ̇2 −mglθ2 (iii) : L = m 2 ( ṙ2 + r2θ̇ 2 ) − V (r) (b) For each coordinate q we have 0 = d dt ∂L ∂q̇ − ∂L ∂q and therefore (i) : 0 = mẍ (ii) : 0 = ml2θ̈ +mglθ (iii) : 0 = m(r2θ̈ + 2rṙθ̇) 0 = mr̈ −mrθ̇2 + ∂V ∂r (c) For each coordinate q we have p = ∂L ∂q̇ and therefore (i) : px = mẋ (ii) : pθ = ml2θ̇ (iii) : pθ = mr2θ̇ pr = mṙ 25 (d) Summing over all coordinates and momenta i H = ∑ i q̇ipi − L and therefore, replacing generalised velocities with the momenta, (i) : H = mẋẋ− L = 1 2m p2 (ii) : H = ml2θ̇θ̇ − L = 1 2ml2 p2 θ +mglθ2 (iii) : H = mr2θ̇θ̇ +mṙṙ − L = 1 2m p2 r + 1 2mr2 p2 θ + V (r) (e) Using ṗi = −∂H ∂qi and q̇i = + ∂H ∂pi we have (i) : ṗ = 0 and ẋ = p m (ii) : ṗθ = −mglθ and θ̇ = pθ ml2 (iii) : ṗr = − p2 θ mr3 − ∂V ∂r and ṙ = pr 2m ṗθ = 0 and θ̇ = pθ mr2 . (f) In (i), the momentum is conserved, component-by-component, i.e. ṗ = 0, and in (iii) the angular momentum pθ is conserved. 7. Conserved energy from Hamilton function Show that the energy is conserved of a system described by a classical Hamilton function without explicit time dependence. Solution Hamilton equations of motion: q̇i = ∂H ∂pi , ṗi = −∂H ∂qi , ∂tH = −∂tL = 0 if L is not explicitly dependent on t. 0 = −∂tL = ∂L ∂q q̇ + ∂L ∂q̇ q̈ − dL dt = ( ∂t ∂L ∂q̇ ) q̇ + ∂L ∂q̇ q̈ − dL dt 26 = ∂t ( ∂L ∂q̇ q̇ ) − ∂L ∂q̇ q̈ + ∂L ∂q̇ q̈ − dL dt = d dt ( ∂L ∂q̇ q̇ − L ) = d dt ( ∂L ∂q̇ q̇ − T + V ) , where Lagrange E.o.M. and L = T − V have been used with T and V are the kinetic and potential energy. Using Euler’s theorem for homogeneous functions, ∂L ∂q̇ q̇ = 2T if the kinetic energy is a quadratic function of generalised velocities q (which is usually the case), and if the potential does - as usual - only depend on generalised coordinates. This shows that d(T + V ) dt = dE dt = 0 Alternatively: dH dt = ∂H ∂p dp dt + ∂H ∂q dq dt + ∂H ∂t = q̇ṗ− ṗq̇ + ∂H ∂t (using Hamilton’s equations) = ∂H ∂t = 0 if H not explicitly dependent on t. 8. Quantum Mechanical Harmonic Oscillator (a) Starting with the Lagrange function L = mẋ2 2 − mω2 x2 2 , calculate the conjugate momenta and show that the Hamilton function reads H = 1 2m p2 + mω2 2 x2 . (b) Promote the Hamilton function to the Hamilton operator of Eq. (51) and use the commutation relation of the position and momentum operator to show that the commutator relations [â, â] = [â†, â†] = 0 [â, â†] = 1 . of Eq. (54) hold true, where the creation and annihilation oper- ators are defined through Eq. (53). 27 (e) Writing, in full analogy, the Hamiltonian as Ĥ = ω ( N̂ − 1 2 ) = ω ( â†â− 1 2 ) will lead to the same ground state, â|0〉 = 0 −→ E0 = −ω/2 Repeatedly applying the creation operator ↠will result in |1〉 = â†|0〉 but because of â†â† = 1 2 { â†, ↠} = 0 the application of creation operator on the first excited state will annihilate it: â†|1〉 = â†â†|0〉 = 0 and therefore the fermionic harmonic oscillator has only two states with energies E0,1 = ∓ω/2. 30 3 Classical Fields In this section we re-derive the Lagrange functions for classical fields. For a more exhaustive explanation of how to make the transition from a discrete to a continuous system, chapter 13 of Goldstein [9] may be helpful. There, you will also find a good derivation of the Euler-Lagrange Equations of Motion for fields. If you are mainly interested in using the formalism in the context of the course, you may want to consult Sec. 2.2 of Peskin & Schröder [1]. 3.1 One-Dimensional Lattice Setup Consider a system of massive particles with identical mass m, ar- ranged in a one-dimensional lattice with positions ξi, and with their motion confined along the lattice direction. The kinetic energy of the system is given by T = m 2 ∑ i ξ̇2 i (t) (67) Coupling the particles with springs with constants k yields the potential energy V = k 2 ∑ i (ξi+1(t)− ξi(t))2 , (68) and the Lagrange function L = 1 2 ∑ i [ mξ̇2 i (t)− k (ξi+1(t)− ξi(t))2 ] = a2 2 ∑ i m( ξ̇i(t) a )2 − k ( ξi+1(t)− ξi(t) a )2  , (69) where a is the equilibrium separation between the particles. Euler-Lagrange E.o.M. To arrive at the E.o.M. for a specific ηi we have to take into account that for the same index i the displacement ηi shows up twice in the sum over the differences, and therefore 0 =ma2 ξ̈(t) a2 − ka2 ( ξi+1(t)− ξi(t) a2 − ξi(t)− ξi−1(t) a2 ) = a [ µξ̈(t)− Y ( ξi+1(t)− ξi(t) a2 − ξi(t)− ξi−1(t) a2 )] (70) where the second line was obtained after factoring out one power of a, and by identifying µ = m/a as the mass density per unit length, and Y = ka as Young’s modulus of the continuous rod. 31 Continuum Limit Going from discrete lattice distances to a continuum can be understood as replacing the index i with a position x, ξi(t)→ ξ(x, t), and by taking the limit a → 0 for the lattice spacing. The ξ differences become lim a→0 ξi+1(t)− ξi(t) a = lim a→0 ξ(x+ a, t)− ξ(x, t) a = ∂ξ(x, t) ∂x (71) Summation over i translates into an integral over x, a ∑ i → dx (72) and the discrete Lagrange function of Eq. (69) turns into the Lagrangian L = 1 2 ∫ dx [ µξ̇2 − Y ( ∂ξ ∂x )] (73) for the continuous rod; from now on we suppress the arguments of the ξ. Going back to the equation of motion, Eq. (69), and taking a closer look at the second term in the limit of vanishing spacing a( ξi+1 − ξi a2 − ξi − ξi−1 a2 ) −→ ( ξ(x+ a)− ξ(x) a2 − ξ(x)− ξ(x− a) a2 ) a→0−→ lim a→0 ∂ξ(x+ a)/∂x− ∂ξ(x)/∂x a = ∂2ξ(x) ∂x2 , (74) it is clear that this is a second derivative, and the E.o.M. for the continuous elastic rod therefore is given by µ ∂2ξ ∂t2 − Y ∂ 2ξ ∂x2 = 0 , (75) with longitudinal waves as solution. Lagrange Density The previous considerations suggest that it is sensible to introduce a Lagrange density L(ξ, ∂ξ/∂t, ∂ξ/∂t, x, t) = µ 2 ( ∂ξ ∂t )2 − Y 2 ( ∂ξ ∂x )2 (76) which becomes the Lagrange function through integration over the (one- dimensional) space, L = ∫ dxL(ξ, ∂ξ/∂t, ∂ξ/∂x, x, t) (77) 32 = xµ1∫ xµ0 d4x [ ∂ξ ∂α ( ∂L ∂ξ − ∂µ ∂L ∂(∂µξ) ) + ∂µ ( ∂µ ∂L ∂(∂µξ) ∂ξ ∂α )] . (88) The last term is a four-dimensional volume integral over a four-dimensional divergence, which vanishes with the vanishing virtual variations of the fields, and we are left with the Euler-Lagrange E.o.M. for relativistic fields ∂µ ∂L ∂(∂µξ) − ∂L ∂ξ = 0 . (89) 3.2 Scalar Fields: Real Scalars Know Thy Equation of Motion! The desired equations of motion are a good starting point to construct Lagrange densities for realistic and physi- cally relevant examples of relativistic field theories. We will first consider the probably simplest case of a free real scalar field φ(x), i.e. a field that does not interact with other fields or with an external potential 5. To see how this works, let us start with the well-known Schrödinger equation, where the starting point is the kinetic energy, given by E = p2/2m. Substituting derivatives for energy and momentum, E → i∂t and p→ −i∇ or pj → −i∂j , we arrive at the E.o.M., i ∂ ∂t φ(x) + 1 2 ∂2 ∂x2 j φ(x) = 0 . (90) In the same vein, we start with the relativistic energy-momentum relation, E2 = p2 +m2 and find the Klein-Gordon Equation( ∂2 ∂t2 −∇2 +m2 ) φ(x) = ( ∂µ∂ µ +m2 ) φ(x) = 0 . (91) Solutions to the Klein-Gordon Equation The solution to the Klein- Gordon Equation, Eq. (91) for a fixed momentum is given by φ(x) = a(k) e−ik·x + a∗(k) eik·x , (92) where a(k) and a∗(k) are the (complex) amplitudes for the plane-wave so- lution for a fixed wave four-vector k, which satisfies the implicit “on-shell” condition k2 = k2 0 − k2 = m2. Of course, we could also sum over many such waves and we arrive at φ(x) = ∫ d3k (2π)3(2k0) [ a(k) e−ik·x + a∗(k) eik·x ] . (93) 5Note that application of external potentials would introduce an explicit dependence of the Lagrange density on the space-time coordinates x. 35 A few comments are in order here: 1. In Eq. (142) we have directly used the continuum limit. This neces- sitates the integration over all momenta instead of a summation over a discrete set of eigenvalues for the momentum. The latter would be the case for example when second quantising on a lattice with lattice spacing a, where the eigenvalues for the momentum are discrete and behave like kn = n/a. 2. The measure of integration, that sums over the different wave vector, should better be Lorentz-invariant. It is not trivial to see immedi- ately that d3k/(2k0) fulfils this criterion. To realise that this is indeed the case, let us start with a manifestly Lorentz-invariant integration measure,∫ d4k (2π)4 (2π)δ(k2 −m2)Θ(k0) = ∫ d3k (2π)3 [ dk0δ(k 2 0 − k2 −m2)Θ(k0) ] = ∫ d3k (2π)3 ∑ k0=± √ k2+m2 [ 1 2k0 Θ(k0) ] = ∫ d3k (2π)3(2k0) , (94) where the d4k obviously is a boost and rotation-invariant quantity, the factor δ(k2 −m2) encodes the (Lorentz-invariant) relativistic energy- momentum relation necessary to ensure that the quanta behave in a physically sensible way, and Θ(k0) projects on positive-energy solu- tions. In performing the k0-integration we have used a property of the δ-function, namely∫ dx δ(f(x)) = ∑ xi: f(xi)=0 δ(x− xi) f ′(xi) , (95) which replaces the integral over the δ-function of a function f(x) with an integral over a sum of its zeroes xi (given by f(xi) = 0), normalised by the first derivative of the function at the zero. Klein-Gordon Lagrange Density It is simple to show that this equation of motion, cf. Eq. (91), can be obtained from the Lagrange density L(∂µφ, φ) = 1 2 (∂µφ)(∂µφ)− m2 2 φ2 . (96) Note that, wherever the dependence is self-evident, we will ignore the ar- guments of the fields from now on. To see this, let us plug this Lagrange 36 density into Eq. (89), with the obvious replacement ξ → φ. 0 = ∂µ ∂L ∂(∂µφ) − ∂L ∂φ = ∂µ ∂[1 2(∂µφ)(∂µφ)] ∂(∂µφ) − ∂[m 2 2 φ 2] ∂φ , (97) where we have replaced the Lagrange density in the first line with the rel- evant parts of Eq. (96) in the second one. The first expression looks a bit tricky and, naively, it seems as if derivation w.r.t. ∂µφ would only deliver 1 2∂ µφ – this however is wrong, and it is easy to see why. Rewriting this part component by component we would arrive at terms like 1 2 ∂ ∂t ∂φ̇2 ∂φ̇ = 2 · 1 2 ∂φ̇ ∂t = ∂2φ ∂t2 (98) and similar for the spatial components. Another way to see this is to rewrite the Lorentz-scalar of the derivatives with other indices – replacing the µ’s with ν’s in the Lagrangian (it doesn’t matter, they get contracted anyway, so I can sum over µ’s, ν’s or any other symbol I chose as Lorentz index) ∂ ∂(∂µφ) [ 1 2 (∂νφ)(∂νφ) ] = ∂ ∂(∂µφ) [ gνρ 2 (∂νφ)(∂ρφ) ] = gνρ 2 [ (∂νφ) ∂(∂ρφ) ∂(∂µφ) + (∂ρφ) ∂(∂νφ) ∂(∂µφ) ] = gνρ 2 [ (∂νφ)δµρ + (∂ρφ)δµν ] = ∂µφ (99) Taking into account of this insight, we ultimately arrive at 0 = ∂µ∂ µφ+m2φ , (100) as requested. 3.3 Scalar Fields: Complex Scalars Two Real Scalars = One Complex Scalar Consider now two such free real scalar fields, φ1 and φ2. The Lagrange density reads L = 2∑ i=1 1 2 (∂µφi)(∂ µφi)− m2 i 2 φ2 i (101) If both masses are equal, m1 = m2, the two real fields can be re-arranged into one complex one, φ = φ1 + iφ2√ 2 and φ∗ = φ1 − iφ2√ 2 , (102) 37 3.4 Vector Fields: Maxwell’s Equations A Little Game of Symmetry Assume you want to introduce two differ- ent three-vector fields. From a (classical) symmetry point of view, they can be distinguished through parity, i.e. one of them is parity-odd – a “proper” vector – while the other one is parity-even – an axial-vector. We call the parity odd fields (or 1− in spin-parity notation) E, and the parity even ones (or 1+) B. Now let us assume that you only want to allow first derivatives of the fields, ∂t and ∇ and scalar and pseudo-scalar charge densities ρE,B and corresponding currents j E,B . Then you can sort resulting quantities by spin and parity as in Table 3. name JP allowed terms scalars 0+ ∇ · E, ρE pseudo-scalars 0− ∇ ·B, ρB vectors 1− ∂tE, ∇×B, j E axial-vectors 1− ∇× E, ∂tB, j B Table 3: Terms in Maxwell’s equations, by spin and parity Symmetry to Dynamics Each of the four rows in Tab. 3 collects possi- ble terms in one of the four equations defining the system, and this is where we will introduce data to the game. First of all we identify E and B with electric and magnetic fields, respectively. Then we realise that to date no magnetic monopoles have been found, and therefore there is no magnetic charge density of current, ρB = 0 and j B = 0. Adding lastly that electrody- namics is a theory of light, and thereby fixing prefactors and signs we arrive at Maxwell’s equations ∇ · E = 4πρE ∇ ·B = 0 ∇×B − ∂tE = 4πj E ∇× E + ∂tB = 0 (118) Note that we absorbed the usual factors of ε0 and µ0 into the definition of the charge and current, and we have used natural units with c = 1. 40 The Vector Potential The left column in Eq.(118) suggest to use a scalar potential Φ, which we denote as A0, and a vector potential A and write E = −∇A0 − ∂tA and B = ∇×A . (119) Of course this now forms a four-vector potential Aµ = (A0, A), and we will continue the analysis of electrodynamics mainly based on this object. Gauge Transformation and Gauge Invariance One of the first bene- fits of introducing the vector potential is that it is relatively easy to formulate gauge transformations. To this end we introduce an arbitrary scalar gauge function, Λ(x), under which A(x) transforms as 6 Aµ → A′µ = Aµ − ∂µΛ , (120) and therefore E → E′ = −∇(A0 − ∂0Λ)− ∂t(A+∇ · Λ) = −∇A0 − ∂tA = E B → B′ = ∇× (A+∇ · Λ) = ∇×A = B , (121) where we have used that rot ·grad of a scalar function vanishes. This suggest that it would be beneficial to express the theory in terms of gauge invariant quantities made from Aµ, to directly encode this symmetry. The Field-Strength Tensor One such gauge-invariant quantity is the anti-symmetric field-strength tensor Fµν = ∂µAν − ∂νAµ =  0 −Ex −Ey −Ez Ex 0 −Bz By Ey Bz 0 −Bx Ez −By Bx 0  , (122) and another such tensor is its dual, F̃µν = 1 2 εµνσρFρσ =  0 −Bx −By −Bz Bx 0 Ez −Ey By −Ez 0 Ex Bz Ey −Ex 0  . (123) They allows to express the inhomogeneous and homogeneous Maxwell’s equations, i.e. the left and right column of Eq. (118), as ∂µF µν = 4πjν and ∂µF̃ µν = 0 . (124) 6Remember that ∂µ = (∂t, −∇)! 41 Lagrange Density in Terms of the Fields There are various ways to express the Lagrange density; a version probably familiar from previous lectures expresses it through the electric and magnetic fields and reads L = E2 −B2 8π − ρφ+ j ·A . (125) The E.o.M. are obtained in terms of the potential φ and A, using the fact that the electromagnetic fields are expressed through their derivatives. This also fixes the two homogeneous Maxwell equations, i.e. the right column of Esq. (118). This also implies that we are left with the task to check if the Lagrange density above yields the correct inhomogenous equations – the left column of Eq. (118). For example, for φ we have: ∂L ∂φ = − ρ ∂L ∂(∂φ/∂xk) = Ek 4π ∂Ek ∂(∂φ/∂xk) = −Ek 4π , (126) where ∂Ek ∂(∂φ/∂xk) = −1 (127) follows directly from Eq. (121). Assembling all parts, and making the sum- mation over repeated indices explicit therefore yields Gauss’ law,∑ k [ ∂ ∂xk ∂L ∂(∂φ/∂xk) ] − ∂L ∂φ = −∇ · E 4π + ρ = 0 . (128) Similarly, for an arbitrary component of A, Ai we find ∂L ∂Ai = ji ∂L ∂(∂Ai/∂t) = Ei 4π ∂Ei ∂(∂Ai/∂t) = −Ei 4π ∂L ∂(∂Ai/∂xj) = − Bk 4π ∂Bk ∂(∂Ai/∂xj) = −εijk Bk 4π ∂L ∂(∂Ai/∂xk) = − Bj 4π ∂Bj ∂(∂Ai/∂xk) = −εijk Bj 4π , (129) where Eq. (121) has again been used, noting that, expressed in component notation B = ∇×A ←→ Bk = εijk∂iAj . (130) 42 3.6 Problems & Solutions 1. General Solutions for the Klein-Gordon Equation Consider a real scalar field, given by thr Klein-Gordon Lagrangian, Eq. (96). (a) Proof that the solutions to its Equation of Motion, Eq. (91), are given by the expression in Eq. (93). (b) Calculate the Hamiltonian and momentum for a free scalar field using their definitions, H = 1 2 ∫ d3x [ (∂tφ)2 + (∇φ)2 +m2φ2 ] P = − ∫ d3x [(∂tφ)(∇φ)] . Solution (a) Inserting the solution for the Klein-Gordon equation from Eq. (93) into the E.o.M. yields[  +m2 ] ∫ d3k (2π)32k0 [ a(k)e−ik·x + a∗(k)eik·x ] = ∫ d3k (2π)32k0 [ a(k)( +m2)e−ik·x + a∗(k)( +m2)eik·x ] = ∫ d3k (2π)32k0 (−k2 +m2) [ a(k)e−ik·x + a∗(k)eik·x ] = ∫ d3k (2π)32k0 (−k2 0 + k2 +m2) [ a(k)e−ik·x + a∗(k)eik·x ] = 0 because of the relativistic energy-momentum relation k2 0 = k2 + m2. (b) Inserting the solutions into the two expressions for the Hamilto- nian and the momentum results in H = 1 2 ∫ d3x [ (∂tφ)2 + (∇φ)2 +m2φ2 ] = 1 2 ∫ d3x d3k (2π)32k0 d3q (2π)32q0 { a(k)a(q)e−i(k+q)·x [−k0q0 − k · q +m2 ] + a(k)a∗(q)e−i(k−q)·x [ k0q0 − k · q +m2 ] + a∗(k)a(q)e+i(k−q)·x [k0q0 − k · q +m2 ] 45 + a∗(k)a∗(q)e+i(k+q)·x [−k0q0 − k · q +m2 ]} = 1 2 ∫ d3k (2π)32k0 d3q (2π)32q0 { a(k)a(q)(2π)3δ3(k + q)e−i(k0+q0)x0 [ −k0q0 − k · q +m2 ] + a(k)a∗(q)(2π)3δ3(k − q)e−i(k0−q0)x0 [ k0q0 + k · q +m2 ] + a∗(k)a(q)(2π)3δ3(k − q)e+i(k0−q0)x0 [ k0q0 + k · q +m2 ] + a∗(k)a∗(q)(2π)3δ3(k + q)e+i(k0+q0)x0 [ −k0q0 − k · q +m2 ]} = 1 2 ∫ d3k (2π)3(2k0)2 { a(k)a(−k)e−i(k0+k0)x0 [ −k2 0 + k2 +m2 ] + a(k)a∗(k)e−i(k0−k0)x0 [ k2 0 + k2 +m2 ] + a∗(k)a(k)e+i(k0−k0)x0 [ k2 0 + k2 +m2 ] + a∗(k)a∗(−k)e+i(k0+k0)x0 [ −k2 0 + k2 +m2 ]} = 1 2 ∫ d3k (2π)3(2k0)2 2k2 0 [ a(k)a∗(k) + a(k)a∗(k) ] = ∫ d3k (2π)3(2k0) k0a(k)a∗(k) and P = − ∫ d3x [(∂tφ)(∇φ)] = − ∫ d3x d3k (2π)32k0 d3q (2π)32q0 { a(k)a(q)e−i(k+q)·x(k0q) + a(k)a∗(q)e−i(k−q)·x(−k0q) + a∗(k)a(q)e+i(k−q)·x(−k0q) + a∗(k)a∗(q)e+i(k+q)·x(k0q) } = − ∫ d3k (2π)32k0 d3q (2π)32q0 k0q { a(k)a(q)(2π)3δ3(k + q)e−i(k0+q0)x0 46 − a(k)a∗(q)(2π)3δ3(k − q)e−i(k0−q0)x0 − a∗(k)a(q)(2π)3δ3(k − q)e+i(k0−q0)x0 + a∗(k)a∗(q)(2π)3δ3(k + q)e+i(k0+q0)x0 } = ∫ d3k (2π)3(2k0)2 k0k { a(k)a(−k)e−i(k0+k0)x0 + a(k)a∗(k)e−i(k0−k0)x0 + a∗(k)a(k)e+i(k0−k0)x0 + a∗(k)a∗(−k)e+i(k0+k0)x0 } = ∫ d3k (2π)3(2k0)2 k0k { a(k)a∗(k) + a∗(k)a(k) } = ∫ d3k (2π)32k0 k a(k)a∗(k) where the terms proportional to ka(k)a(−k) and ka∗(k)a∗(−k) vanish due to the symmetry of the integration. 2. ∗Klein-Gordon Equation in Two-Component Form Introduce a two-component form of the real scalar (Klein-Gordon) field as χ = (χ+, χ−)T , where χ± = 1 2 ( φ± i m ∂φ ∂t ) . (a) Rewrite the Klein-Gordon E.o.M. in the Schrödinger form, i.e. as i∂χ ∂t = Hχ and construct the Hamiltonian as a 2×2 matrix. Solve the energy eigenvalue equation Hχ = Eχ. (b) Consider the non-relativistic limit where the solution of the E.o.M. has the form χ(t) = e−i(m+T )tχ(0) for the time-evolution of the fields and where T is the kinetic energy. What does this imply for the relative sizes of χ+ and χ−? Expand the product Tχ+(0) to second order in T and deduce the first relativistic correction to the Hamiltonian. 47 Solution Before starting to derive E.o.M.’s let us first look at some relevant derivatives that appear more than once: ∂Fµν ∂(∂ρAσ) = ∂(∂µAν − ∂νAµ) ∂(∂ρAσ) = gρµg σ ν − gρνgσµ and therefore −1 4 ∂FµνF µν ∂(∂ρAσ) = −1 2 Fµν ∂Fµν ∂(∂ρAσ) = −1 2 Fµν ( gρµg σ ν − gρνgσµ ) = −1 2 (F ρσ − F σρ) = −F ρσ = F σρ and −1 4 ∂ρ ∂FµνF µν ∂(∂ρAσ) = −∂ρF ρσ = Aσ − ∂σ(∂ ·A) . (a) We quickly arrive at[  +m2 + λ 3! ] φ = 0 . (b) There are three terms to evaluate: 1 2 ∂(m2A2) ∂Aρ = m2Aρ ∂ρ λ 2 ∂(∂ ·A)2 ∂(∂ρAσ) = λgρµ∂ρg σµ(∂ ·A) = λ∂σ(∂ ·A) ∂ρ ∂[−(∂µA ν)(∂νA µ)] ∂(∂ρAσ) = −∂ρ [ (∂µA ν) ∂(∂νA µ) ∂(∂ρAσ) + (∂νA µ) ∂(∂µA ν) ∂(∂ρAσ) ] = −∂ρ [ (∂µA ν)gρνgσµ + (∂ρ∂νA µ)gρµgσν ] = −2∂ρ∂ σAρ = −2∂σ(∂ ·A) and therefore we arrive at (λ− 2)∂σ(∂ ·A)−m2Aσ = 0 (c) Previous results mean that we only have to put terms together and arrive at ∂σF σρ +m2Aρ = [gρσ( +m2)− ∂σ∂ρ]Aσ = 0 . 50 (d) Here we have three active fields and arrive at: (− + ∂σ∂ µ)Aσ = ie [φ∗∂µφ− φ∂µφ∗] + 2e2φ∗φAµ ( +m2)φ = 2ieAρ∂ρφ + ieφ∗∂ρA ρ − e2A2φ∗ ( +m2)φ∗ = −2ieAρ∂ρφ ∗ − ieφ∗∂ρAρ − e2A2φ∗ . 4. Massive Vector Field The Lagrangian of a massive vector field Vµ is given by L = −1 4 VµνV µν + m2 2 VµV µ , where the field strength tensor Vµν assumes the usual form Vµν = ∂µVν − ∂νVµ . (a) derive the Euler-Lagrange equations of motion from the Lagrangian. (b) show that the condition ∂µV µ = 0 is a consequence of the equations of motion. (c) use this condition and construct three linearly independent po- larisation vectors ε (λ) µ (k) that satisfy this condition, or after Fourier transformation kµε µ(k) = 0 . Solution (a) For the various derivatives we find ∂L ∂(∂ρVσ) = −1 2 V µν ∂Vµν ∂(∂ρVσ) = −1 2 V µν ( gρµg σ ν − gρνgσµ ) = −1 2 (V ρσ − V σρ) = −V ρσ ∂L ∂Vσ = mV σ , and therefore ∂ρ ∂L ∂(∂ρVσ) − ∂L ∂Vσ = −∂ρV ρσ −mV σ = 0. (b) Forming the derivative (four-dimensional divergence) of the E.o.M. of part (a) of the problem yields ∂σ (∂ρV ρσ +mV σ) = m∂ · V = 0 . 51 (c) Assuming the momentum being oriented along the z-axis, we have kµ = (ω, 0, 0, κ) with ω = √ κ2 +m2 and therefore the following three polarisation vectors are orthonor- mal: ε(1) =  0 1 0 0  , ε(2) =  0 0 1 0  , ε(3) = 1√ −m2  κ 0 0 −ω  . 5. Electrodynamics with gauge-fixing term An alternative form for the Lagrange density of the electromagnetic fields reads L = −1 4 FµνFµν − 1 2 (∂µA µ)2 − 4πjµA µ , where Fµν = ∂µAν − ∂νAµ is the field strength tensor for the vector potential Aµ. The additional term 1/2(∂ ·A)2 is also known as “gaug-fixing” term, and in this case corresponds to the Lorentz gauge. We will come back to this in Section 6 of the notes. Show that the Euler-Lagange E.o.M. lead directly to a wave equation of the form Aµ = ∂ν∂ νAµ = 4πjµ . Solution Using Aµ as the dynamical variable, the Euler-Lagrang E.o.M. read 0 = ∂ν ∂L ∂(∂νAµ) − ∂L ∂Aµ . Let us first rewrite the product of field strength tensors, −1 4 F κλFκλ = − 1 4 [( ∂κAλ − ∂λAκ ) (∂κAλ − ∂λAκ) ] = − 1 4 [ ∂κAλ∂κAλ − ∂κAλ∂λAκ − ∂λAκ∂κAλ + ∂λAκ∂λAκ ] = − 1 2 [ ∂κAλ∂κAλ − ∂κAλ∂λAκ ] , 52 = ∫ +∞ −∞ dt +L∫ −L dx [ (∂tφ)∂t(δφ)− (∂xφ)∂x(δφ)−m2φδφ ] = +L∫ −L dx (∂tφ)δφ ∣∣∣∣t=+∞ t=−∞ − +∞∫ −∞ dt (∂2 t φ)δφ  − +∞∫ −∞ dt (∂xφ)δφ ∣∣∣∣x=+L x=−L − +L∫ −L dx (∂2 xφ)δφ  − +∞∫ −∞ dt +L∫ −L dxm2φδφ = +L∫ −L dx (∂tφ)δφ ∣∣∣∣t=+∞ t=−∞ − +∞∫ −∞ dt +L∫ −L dx ( +m2)φδφ , where we have assumed, as usual, that the variations of the field vanish for t→ ±∞, δφ(t→ ±∞, x) = 0. This leaves us with the equation of motion ( +m2)φ = 0 . This however holds true only either if the (Dirichlet) boundary condi- tions δφ(t, , x = ±L) = 0 or if the (Neumann) boundary conditions ∂xφ(t, , x = ±L) = 0 are fulfilled. The latter are better suited for the solution of our prob- lem, since they are formulated as conditions on the field φ or its deriva- tive and not on its – in principle arbitrary – variation δφ. 8. Symmetry and Conserved Current Consider the Lagrangian density for two real scalars φ1, 2 given by L = 1 2 [ (∂µφ1)(∂µφ1) + (∂µφ2)(∂µφ2) ] −m 2 2 [ φ2 1 + φ2 2 ] − λ 4! [ φ2 1 + φ2 2 ]2 . ˆ Show that it invariant under the transformation φ1 → φ′1 = cos θ φ1 − sin θ φ2 φ2 → φ′2 = sin θ φ1 + cos θ φ2 . ˆ Construct the corresponding conserved current and charge, 55 Solution ˆ To see the invariance of the Lagrangian it suffices to realise 1. that the angle θ does not depend on space-time, and therefore ∂µφ ′ 1,2 = cos θ∂µφ1,2 ∓ sin θφ2,1 , 2. and that the mixed terms ∝ cos θ sin θ vanish in the sum of squares (∂µφ ′ 1)2 + (∂µφ ′ 2)2 and φ′21 + φ′22 , leaving only terms ∝ (cos2 θ + sin2 θ) = 1 behind. ˆ To construct the conserved current we use that jµ = 2∑ i=1 ∂L ∂(∂µφi) δφi = ∂µφ1δφ1 + ∂µφ2δφ2 and that to first order in the angle δφ1,2 = φ′1,2 − φ1,2 = ∓θφ2,1 . Therefore the current jµ = ∂µφ1δφ1 + ∂µφ2δφ2 = θ (φ1∂µφ2 − φ2∂µφ1) is conserved, ∂µjµ = 0 – easy to see when using the E.o.M. on the evaluation of ∂µjµ. The conserved charge is given by Q = ∫ d3x(φ1φ̇2 − φ̇1φ2) . 9. ∗A SU(2) Symmetry Consider a doublet of conplex scalars Φ = ( φ1, φ2 ) with dynamics defined by the free Lagrangian L = (∂µΦ)†(∂µΦ)−m2Φ†Φ . (a) Show that this Lagrangian is invariant under the three-parameter transformations Φ→ Φ† = exp [ i 2 θaσa ] Φ , with the three constant real angles θ1, 2, 3 and where the σa are the three Pauli matrices, σ1 = ( 0 1 1 0 ) , σ2 = ( 0 −i i 0 ) , σ3 = ( 1 0 −1 0 ) which enjoy the commutation relation [σi, σj ] = iεijkσk 56 Solution (a) Because the angles are constant, the derivatives do not act on them, and we only have to evaluate terms of the form exp [ − iθa 2 σ†a ] exp [ + iθb 2 σb ] = exp [ X ] [ Y ] = exp [ Z ] , where Z is given by the Baker–Hausdorff formula as Z = X + Y + 1 2 [X, Y ] + 1 12 [X, [X, Y ]]− 1 12 [X, [X, Y ]] + . . . . Making the summation over a and b explicit and using that σ†a = σa, the commutator is given by [X, Y ] = 3∑ a,b=1 θaθb 4 [σa, σb] = 3∑ a,b=1 iεabcθaθbσc 4 = 0, because we multiply a symmetric tensor with an anti-symmetric one. This means that we have exp [ − iθa 2 σ†a ] exp [ + iθb 2 σb ] = exp [ − 3∑ a=1 iθa 2 σ†a + 3∑ b=1 iθb 2 σb ] = 1 . Therefore, the Lagrangian density is invariant under the SU(2) symmetry. 10. Energy-Momentum Tensor The energy-momentum tensor for a Lagrangian given as a function of a set of fields {φα} and their first derivatives {∂µφα} is defined as Tµν = ∂L ∂(∂µφα) ∂νφα − gµν L . (a) Can you identify the T 00 component? (b) Can you identify the T 0j component? (c) Using the Euler-Lagrange E.o.M., show that it is a conserved quantity, i.e. ∂µT µν = 0 . (d) Show that for the free real scalar field, the energy-momentum tensor is symmetric, Tµν = T νµ. (e) Verify that the interpretations of T 00 and T 0i from questions (a) and (b) are correct for the free electromagentic field. 57 4 Second Quantisation At the beginning of this part of the course you may have read the title and asked yourself: What does second quantisation actually mean? Haven’t we already quantised the theory? The answer is that in “first quantisation” we quantised the position and momentum of point particles. This led to important properties related to our ability to measure them – the uncer- tainty principle – and to a crucial reassessment of the inner working of the world around us, replacing Laplace’s demon of deterministic physics with a determinism of probabilities. So while in this first quantisation we re- placed the real numbers x and p with operators x̂ and p̂ and constructed wave functions for the emerging states, in “second quantisation” we quantise something else, namely the fields. Consequently, x and p become “ordinary” numbers again, which serve as arguments of the field operators φ̂ and π̂. This step necessitates the introduction of a new state. While, formally speaking, the states of Quantum Mechanics constitute a Hilbert space, the field operators act on objects in a more complicated Fock space, which is not labelled by eigen-positions or momenta, but by the number of field quanta of a given momentum. We will, however, not discuss the properties of these vector spaces in the lecture. Simply put: while first quantisation quantised the point dynamics of Clas- sical Mechanics, leading to Quantum Mechanics, second quantisation pro- duces a Quantum Field Theory. If you like to do some additional reading, I would recommend to take a closer look at Chapter 3 of Hatfield’s book [3], in particular sections 3.1-3.4 or at Sections 2.3-2.4 in Peskin & Schroeder [1]. 4.1 A How-To Guide Process Summary The process of second or field quantisation follows the logic of the familiar first quantisation programme, with suitably replacing position and momentum with the field and its conjugate momentum, and by replacing the δ’s of the commutators with δ-functions of the positions. This proceeds in a relatively straight-forward “algorithmic” fashion, as outlined in Fig. 1. The crucial part in it is to demand equal-time commutator rela- tions between fields and their momenta, which also fixes a Lorentz frame in which the field quantisation is performed. Obviously, there are other choices for such a programme, for example a quantisation on the light-cone, which however is beyond the scope of the lecture here. It is, nevertheless, im- portant to stress that despite the implicit choice of a Lorentz frame during quantisation, the resulting theory has the correct causal properties. This will be shown towards the end of this section. 60 How-to: Second Quantisation 1. determine conjugate momenta of fields: π = ∂L/∂(∂tφ) = ∂L/∂φ̇ 2. construct Hamiltonian as function of fields φ and their mo- menta π H = ∫ d3x ( φ̇π − L ) 3. promote fields and momenta to operators, φ −→ φ̂, π −→ π̂ 4. demand equal time commutators of fields and momenta[ φ̂(t, x), π̂(t, y) ] = iδ3(x− y)[ φ̂(t, x), φ̂(t, y) ] = [ π̂(t, x), π̂(t, y) ] = 0 5. express fields as linear combination of plane waves and anni- hilation and creation operators (which will “inherit” com- mutator relations) φ̂(x) = ∑ k [ â(k)e−ik·x + â†(k)eik·x ] , where summation is replaced with integration for continuous momenta k. Figure 1: The steps performed during second quantisation in form of an “algorithm”. Details will be worked out and highlighted through examples during the course. 4.2 Second Quantisation of the Real Scalar Field Lagrangian: Fields and Conjugate Momenta Starting with the La- grangian of Eq. (96), L(∂µφ, φ) = 1 2 (∂µφ)(∂µφ)− m2 2 φ2 , the conjugate momentum reads π = ∂L ∂φ̇ = φ̇ . (139) 61 Hamiltonian The Hamilton function therefore is given by H = ∫ d3x [ πφ̇− L ] = ∫ d3x 1 2 [ π2 + (∇φ)2 +m2φ2 ] . (140) Field Operators and Commutators Promoting the field and its con- jugate momentum to operators, φ(x)→ φ̂(x) and π(x)→ π̂(x), we demand the equal-time commutators,[ φ̂(t, x), π̂(t, y) ] = iδ3(x− y)[ φ̂(t, x), φ̂(t, y) ] = [ π̂(t, x), π̂(t, y) ] = 0 (141) Creation and Annihilation Operators The field and the conjugate momentum is expressed through creation and annihilation operators as φ̂(x) = ∫ d3k (2π)32k0 [ â(k) e−ik·x + â†(k) eik·x ] π̂(x) = ∫ d3k (2π)32k0 [ −ik0â(k) e−ik·x + ik0â †(k) eik·x ] , (142) where we have obtained the momentum operator through straightforward calculation of the derivative π̂(x) = ∂tφ̂(x). Comparing the expression for the field operator φ̂(x) in the equation above with the solution to the Klein- Gordon equation for the classical field φ(x), Eq. (93), we recognise the same pattern of an expansion in amplitude factors and plane waves. But while the amplitude factors for the classical field are merely numbers a(k) and their complex conjugate a∗(k), they become operators for the quantised fields, and the complex conjugation turns into an Hermitean conjugate7. The in- terpretation is clear. While for classical fields every value of the amplitude is allowed, in quantised fields the amplitude is composed by adding finite quanta. This “amplitude quantisation” is reflected by using creation and annihilation operators from which the field “inherits” its quantised proper- ties. We will build on this in the following by expressing the Hamiltonian through these operators, by creating a number operator, and by analysing their inherent properties. Commutators of the Creation and Annihilation Operators To cal- culate the commutators it is necessary to express, in a first step, the creation and annihilation operators through the field and conjugate momentum op- erators. To see how this works, let us first try to combine φ̂ and π̂ in such 7As a result the field operator is Hermitean φ̂ = φ̂†, which guarantees that it has real eigenvalues – as you would expect from a real scalar field. 62 +â†(k)â(k′) [ +k0k ′ 0kk ′ +m2 ] e+i(k−k′)·x +â†(k)â†(k′) [ −k0k ′ 0 − kk′ +m2 ] e+i(k+k′)·x } = 1 2 ∫ d3k (2π)32k0 d3k′ (2π)32k′0 { (2π)3δ3(+k + k′) e−i(k0+k′0)x0 â(k)â(k′) [ −k0k ′ 0 − kk′ +m2 ] + (2π)3δ3(+k − k′) e−i(k0−k′0)x0 â(k)â†(k′) [ +k0k ′ 0 + kk′ +m2 ] + (2π)3δ3(−k + k′) e+i(k0−k′0)x0 â†(k)â(k′) [ +k0k ′ 0 + kk′ +m2 ] + (2π)3δ3(−k − k′) e+i(k0+k′0)x0 â†(k)â†(k′) [ −k0k ′ 0 − kk′ +m2 ]} = 1 2 ∫ d3k (2π)34k2 0 { â(k)â(−k)e−2ik0x0 [ −k2 0 + k2 +m2 ] +â(k)â†(k) [ +k2 0 + k2 +m2 ] + â†(k)â(k) [ +k2 0 + k2 +m2 ] +â†(k)â†(−k)e2ik0x0 [ −k2 0 + k2 +m2 ]} = 1 2 ∫ d3k (2π)34k2 0 {[ 2k2 0 ] [ â(k)â†(k) + â†(k)(̂k) ]} = 1 2 ∫ d3k (2π)32k0 k0 [ â(k)â†(k) + â†(k)(̂k) ] , (150) where the δ functions in the first step emerge from the integral over x and where we have eliminated the terms ââ and â†â† by realising that due to the relativistic energy-momentum relation k2 0 = k2 + m2. Therefore, the Hamilton operator for the real scalar field is given by Ĥ = 1 2 ∫ d3k (2π)32k0 k0 [ â(k)â†(k) + â†(k)(̂k) ] (151) It suggests that the Quantum Field Theory for a real scalar field can be in- terpreted as a continuous sum of harmonic oscillators, permeating all space. Simple States: Ground State and First Excited State Following the same logic already present in the harmonic oscillator in Quantum Mechanics we introduce a ground state – the “vacuum” – |0〉 which is annihilated by any annihilation operator, â(k) |0〉 = 0 ∀k . (152) 65 States containing fields (or particles) with momenta ki are generated by repeated application of the corresponding creation operators â†(k1) |0〉 = |k1〉 â†(k1)â†(k2) |0〉 = |k1k2〉 . . . . (153) Normalisation of States But here’s a new problem. Let us take a look at the norm of a one-field (one-particle) state |k〉: 〈k|k〉 = ∣∣|k〉 ∣∣2 = ∣∣∣â†(k) |0〉 ∣∣∣2 = [ â†(k) |0〉 ]† [ â†(k) |0〉 ] = 〈0|â(k)â†(k)|0〉 = 〈 0 ∣∣∣[â(k)â†(k)− â†(k)â(k) ]∣∣∣ 0〉 = 〈 0 ∣∣2k0(2π)3δ3(k − k) ∣∣ 0〉 = 〈 0 ∣∣2k0(2π)3δ3(0) ∣∣ 0〉 = 2k0(2π)3δ3(0) , (154) where in the second line we have subtracted a 0 – remember that â |0〉 = 0, and where in the last step we used our normalisation of the ground state, 〈0|0〉 = 1. Using that (2π)3δ(k) = ∫ d3xeik·x −→ (2π)3δ(0) = ∫ d3x (155) suggests that the normalisation of the state equals the (infinite) spatial vol- ume – a veritable divergence. This is actually not a surprising finding, after all, the uncertainty principle tells you that a particle with completely fixed momentum has no localisation. Our particle here, with its fixed momentum represents a plane wave, filling all volume. If the volume is infinite – which it is for us to have a continuous spectrum – such states have no normalisation. The solution to this conundrum is to “smear” the state with a modulating function f(k), and to define |k〉 −→ |k〉f = f(k)â†(k) |0〉 (156) which will lead to perfectly normalisable states, if∫ d3k |f(k)|2 <∞ . (157) In this respect, states obtained through application of the creation operator on the vacuum, â†(k) |0〉 are physically sensible only, if used together with a test function that smear them out. A natural question to ask is: in what space to these new states live? The answer is that they populate a Fock space, which is the sum of all n-particle Hilbert spaces plus, possibly, some symmetrisation that takes care of the fact that identical particles are indistinguishable. It also has a meaningful scalar product, again allowing the definition of a distance, which it “inherits” from the underlying Hilbert space. 66 Ground-State Energy The ground state |0〉 is an eigenstate of the Hamil- tonian; calculating its energy E0 we arrive at E0 = 〈0| Ĥ |0〉 = 〈 0 ∣∣∣∣12 ∫ d3k (2π)32k0 k0 [ â(k)â†(k) + â†(k)â(k) ]∣∣∣∣ 0〉 = 1 2 ∫ d3k (2π)32k0 k0 〈 0 ∣∣∣â(k)â†(k) ∣∣∣ 0〉 = 1 2 ∫ d3kk0δ 3(0) =∞ , (158) the product of the volume in both position and momentum space, infinity for a Quantum Field Theory in an infinite volume. A simple solution is to just subtract the ground state energy, by redefining the Hamiltonian as Ĥ −→ Ĥ ′ = Ĥ − 〈0| Ĥ |0〉 . (159) Normal Ordering Alternatively, we can define normal-ordering of the operators, indicated by colons around the operators as :â(k)â†(k): = :â†(k)â(k): = â†(k)â(k) , (160) and therefore we replace Ĥ −→ :Ĥ: = 1 2 ∫ d3k (2π)32k0 k0 : [ â(k)â†(k) + â†(k)â(k) ] : (161) and, finally, :Ĥ: = ∫ d3k (2π)32k0 k0 â †(k)â(k) . (162) This obviously cures the divergence stemming from 〈0| â(k)â†(k) |0〉 in the ground state energy and similar observables. In the remainder of this lecture we will always assume implicit normal ordering, if not stared otherwise. 4.3 A Little Detour: Causal Structure of the Theory Commutator of Field Operators To guarantee the correct causal struc- ture of the theory, we need to convince ourselves that fields in space-like distances cannot influence each other: they must commute for space-like distances. To see this, define the commutator of two field operators at arbi- trary four-positions x and y as i∆(x− y) = [ φ̂(x), φ̂(y) ] = ∫ d3k (2π)32k0 d3k′ (2π)32k′0 {[ â(k), â†(k′) ] e−ik·x+ik′·y 67 we arrive at commutator relations, for example,[ â(k), â†(k′) ] = ∫ d3xd3x′e+ik·x−ik′·x′ × [ k0φ̂(t, x) + iπ̂∗(t, x), k′0φ̂ ∗(t, x′)− iπ̂(t, x′) ] = ∫ d3xd3x′e+ik·x−ik′·x′ { −ik0 [ φ̂(t, x), π̂(t, x′) ] + ik′0 [ π̂∗(t, x), φ̂∗(t, x′) ]} = ∫ d3xd3x′e+ik·x−ik′·x′ [(k0 + k′0)δ3(x− x′) ] = ∫ d3xe+i(k−k′)·x(k0 + k′0) = 2k0(2π)3δ3(k − k′) , (170) where we have used the definition of the δ function, as usual. Therefore,[ â(k), â†(k′) ] = [ b̂(k), b̂†(k′) ] = 2k0(2π)3δ3(k − k′) (171) and all other commutators vanishing. Hamilton and Number Operators The normal-ordered Hamilton op- erator is given by :Ĥ: = ∫ d3k (2π)32k0 k0 [ â†(k)â(k) + b̂†(k)b̂(k) ] , (172) and it looks like the Hamilton operator for the sum of two free real scalar fields. This further fortifies the idea that we are presented by two kinds of particles – those created and annihilated by ↠and â, and those created and annihilated by b̂† and b̂, and that the vacuum is annihilated by both â and b̂, â(k) |0〉 = b̂(k) |0〉 = 0 . (173) It is therefore natural to introduce two number operators for the two kinds of particles, N̂a = ∫ d3k (2π)32k0 â†(k)â(k) N̂b = ∫ d3k (2π)32k0 b̂†(k)b̂(k) . (174) It is easy to check that they are indeed number operators counting the number of a and b fields in a given state |ψ〉. Denoting∣∣∣k(a) 1 k (a) 2 . . . k(a) na k (b) 1 k (b) 2 . . . k(b) nb 〉 = na∏ i=1 [ â†(ki) ] nb∏ i=1 [ b̂†(ki) ] |0〉 , (175) 70 it is easy to show that〈 k (a) 1 k (a) 2 . . . k(a) na k (b) 1 k (b) 2 . . . k(b) nb ∣∣∣ N̂a ∣∣∣k(a) 1 k (a) 2 . . . k(a) na k (b) 1 k (b) 2 . . . k(b) nb 〉 = na . (176) We leave this as part of a problem. Current and Charge As noted in Sec. 3.3, the Lagrangian for the com- plex scalar field enjoys invariance under the “gauge transformation” φ→ φ′ = exp(iθ)φ , φ∗ → φ′∗ = exp(−iθ)φ∗ , cf. Eq. (106). This leads to a conserved current given by Eq. (113) jµ = i [ φ∗(∂µφ)− (∂µφ∗)φ ] ≡ iφ∗ ←→ ∂µφ , where we added a factor i to ensure that the current is a real number. This factor, obviously, does not change the fact that ∂µj µ = 0. Of course the current can be promoted to a current operator by replacing the fields with field operators, ĵµ = iφ̂∗ ←→ ∂µ φ̂ . (177) The spatial integral over the (normal-ordered) time-component of the cur- rent is the charge, given in operator form by :Q̂: = ∫ d3x :̂j0: = i ∫ d3x :φ̂∗(∂tφ̂)− (∂tφ̂ ∗)φ̂: = ∫ d3k (2π)32k0 [ â†(k)â(k)− b̂†(k)b̂(k) ] = N̂a − N̂b . (178) This suggests that our two particle types a and b have opposite charged with qa,b = ±1 . (179) Conserved Charge To show that the charge is conserved, we need to verify that the charge operator :Q̂: = ∫ d3k (2π)32k0 [ â†(k)â(k)− b̂†(k)b̂(k) ] = N̂a − N̂b (180) commutes with the Hamiltonian, i.e. [:Ĥ:, :Q̂:] = 0. This is indeed the case, and we leave this proof for the problems below. 71 4.5 Problems & Solutions 1. States and Operators of the Real Scalar Field (a) one- and two particle states i. create one- and two-particle states of particles with momenta k1,2, |k1〉 and |k1k2〉 ii. show that the two-particle state |k1k2〉 is symmetric, i.e. |k1k2〉 = |k2k1〉. (b) calculate the energy of the two-particle state above, i.e. E12 |k1k2〉 = Ĥ |k1k2〉 . (c) show that the number operator N̂ counts the number of quanta: N̂ |k1k2 . . . kn〉 = n|k1k2 . . . kn〉 (181) Solution (a) In real scalar field theory, |k1〉 = â†(k1)|0〉 |k1k2〉 = â†(k1)|k2〉 = â†(k1)â†(k2)|0〉 = â†(k2)â†(k1)|0〉 The last equality shows the symmetry of the state. (b) With the (normal-ordered) Hamilton operator given by :Ĥ: = 1 2 ∫ d3k (2π)3(2k0) [ k0 : ( â†(k)â(k) + ââ†(k)(k) ) : ] = ∫ d3k (2π)3(2k0) [ k0 â †(k)â(k) ] , the energy of the state |k1k2〉 is given by Ek1k2 |k1k2〉 = Ĥ|k1k2〉 = 1 2 ∫ d3k (2π)3 ( â†(k)â(k) ) â†(k1)â†(k2) ∣∣∣∣ 0〉 = 1 2 ∫ d3k (2π)3 { â†(k) [ â(k), â†(k1) ] â†(k2) + â†(k)â†(k1)â(k)â†(k2) }∣∣∣∣ 0〉 = 1 2 ∫ d3k (2π)3 { â†(k)â†(k2) [ (2π)3δ(k − k2) 2 √ k2 +m2 ] 72 (b) As usual, Hamiltonian density given by H = 2∑ i=1 [ πiφ̇i ] − L = 2∑ i=1 1 2 [ (φ̇i)(φ̇i) + (∇φi)(∇φi) +m2 iφ 2 i ] (c) Commutators in the usual way: field operators and and their conjugate momentum operators do not commute, everything else does: [ φ̂i(x, t), π̂j(y, t) ] = iδijδ 3(x− y)[ φ̂i(x, t), φ̂j(y, t) ] = [ π̂i(x, t), π̂j(y, t) ] = 0 (d) Employ the usual plane-wave expansion with factors exp[±ik · x] and operators: φ̂i(x) = ∫ d3k (2π)3(2k0) [ âi(k)e−ik·x + â†i (k)eik·x ] π̂i(x) = ∫ d3k (2π)3(2k0) [ −ik0âi(k)e−ik·x + ik0â † i (k)eik·x ] and therefore∫ d3xe−ik ′·x [ k′0φ̂i(x) + iπ̂i(x) ] = ∫ d3xe−ik ′·x d3k (2π)3(2k0)[ âi(k)e−ik·x(k′0 + k0) + â†i (k)eik·x(k′0 − k0) ] = ∫ d3k (2π)3(2k0) d3x [ âi(k)e−i(k+k′)·x−ik0t(k′0 + k0) +â†i (k)ei(k−k ′)·x+ik0t(k′0 − k0) ] = ∫ d3k (2π)3(2k0) [ âi(k)(2π)3δ3(k + k′)(k′0 + k0)e−ik0t +â†i (k)(2π)3δ3(k − k′)(k′0 − k0)eik0t ] = 1 (2k′0) [ âi(k ′)(2k′0)e−ik ′ 0t + 0 ] = âi(k ′)e−ik ′ 0t , where we have used that k0 = √ k2 +m2 = √ (−k)2 +m2 = k′0 and the Fourier transform of the δ function,∫ d3xeip·x = (2π)3δ3(p) . 75 Multiplying on both sides with eik ′ 0t and replacing k′ → k yields âi(k) = ∫ d3xeik·x [ k0φ̂i(x) + iπ̂i(x) ] Taking the Hermitean conjugate and using that φ̂†i = φ̂i and π̂†i = π̂i for real fields implies that â†i (k) = ∫ d3xe−ik·x [ k0φ̂i(x)− iπ̂i(x) ] This allows us to directly calculate the commutators, for example[ âi(k), â†j(q) ] = ∫ d3xeik·x ∫ d3ye−iq·y [ k0φ̂i(x) + iπ̂i(x), q0φ̂j(y)− iπ̂j(y) ] = ∫ d3xeik·x ∫ d3ye−iq·y{ −ik0 [ φ̂i(x), π̂j(y) ] + iq0 [ π̂i(x), φ̂j(y) ]} = ∫ d3xeik·x ∫ d3ye−iq·y{ −ik0 · iδijδ3(x− y) + iq0 · (−i)δijδ3(y − x) } = ∫ d3xei(k−q)·x {(k0 + q0)δij} = (2π)32k0δijδ 3(k − q) , where we have used that we discussed equal time communtators, i.e. x0 = y0 = t. Commutators of two creation or annihilation operators will vanish, because in the last line the term (k0 + q0) will become ±(k0 − q0)→ 0. (e) Write the two complex fields as linear combinations of the two real fields: φ = 1√ 2 (φ1 + iφ2) and φ∗ = 1√ 2 (φ1 − iφ2) , therefore8 π = φ̇∗ and π∗ = φ̇ . 8Expressing the fields φ1,2 through φ and φ∗, φ1 = 1√ 2 (φ+ φ∗) and φ2 = − i√ 2 (φ− φ∗) and yields the Lagrangian L = (∂µφ)(∂µφ∗)−m2φ∗φ and the conjugate momenta are given, as before, by π = ∂L/∂φ̇ and π∗ = ∂L/∂φ̇∗. 76 Suitably combining the expansions for φ1 and φ2 yields φ̂(x) = ∫ d3k (2π)3(2k0) [ â1(k) + iâ2(k)√ 2 e−ik·x + â†1(k) + iâ†2(k)√ 2 eik·x ] = ∫ d3k (2π)3(2k0) [ â+(k)e−ik·x + â†−(k)eik·x ] φ̂∗(x) = ∫ d3k (2π)3(2k0) [ â1(k)− iâ2(k)√ 2 e−ik·x + â†1(k)− iâ†2(k)√ 2 eik·x ] = ∫ d3k (2π)3(2k0) [ â−(k)e−ik·x + â†+(k)eik·x ] , where â±(k) = â1(k)± iâ2(k)√ 2 and ↱(k) = â†1(k)∓ iâ†2(k)√ 2 Using that π̂i = ˙̂ φi and that [φ1, π2] = 0 and similar, the equal- time commutators read[ φ̂(x, t), π̂(y, t) ] = 1 2 [ φ̂1(x, t) + iφ̂2(x, t), ˙̂ φ1(x, t)− i ˙̂ φ2(x, t) ] = 1 2 {[ φ̂1(x, t), π̂1(y, t) ] + [ φ̂2(x, t), π̂2(y, t) ]} = iδ3(x− y)[ φ̂∗(x, t), π̂∗(y, t) ] = 1 2 [ φ̂1(x, t)− iφ̂2(x, t), ˙̂ φ1(x, t) + i ˙̂ φ2(x, t) ] = 1 2 {[ φ̂1(x, t), π̂1(y, t) ] + [ φ̂2(x, t), π̂2(y, t) ]} = iδ3(x− y)[ φ̂(x, t), π̂∗(y, t) ] = 1 2 [ φ̂1(x, t) + iφ̂2(x, t), ˙̂ φ1(x, t) + i ˙̂ φ2(x, t) ] = 1 2 {[ φ̂1(x, t), π̂1(y, t) ] − [ φ̂2(x, t), π̂2(y, t) ]} = 0 and similarly for all commutators of fields with fields and mo- menta with momenta. 77 The total four-momentum operator of a real scalar field is given by :P̂µ:= ∫ d3k (2π)32k0 kµ â†(k)â(k) . (a) Show that P̂µ can be expressed in terms of the field operator φ̂(x) and the conjugate momentum operator π̂(x) as :P̂µ:= ∫ d3x :π̂(x)∂µφ̂(x): (b) show that [ P̂µ, φ̂(x) ] = −i∂µφ̂(x) . Solution (a) Inserting the expansion of the field operator and its conjugate momentum through plane waves and creation and annihilation operators we have :P̂µ: = ∫ d3x d3k (2π3)2k0 d3q (2π3)2q0 : [ −ik0 ( â(k)e−ik·x − â†(k)eik·x ) iqµ ( −â(q)e−iq·x + â†(q)eiq·x )] : = ∫ d3x d3k (2π3) d3q (2π3) qµ 4q0 : [ −â(k)â(q)e−i(k+q)·x + â†(k)â†(q)ei(k+q)·x +â(k)â†(q)e−i(k−q)·x + â†(k)â(q)ei(k−q)·x ] : = ∫ d3q (2π3) qµ 4q0 : [ −â(−q)â(q)e−2iq0t + â†(−q)â†(q)e2iq0t +â(q)â†(q) + â†(q)â(q) ] : = ∫ d3q (2π3) qµ 2q0 â†(q)â(q) The first two terms in the second-to last line vanishes because â(q) commutes with â(−q), and similarly for the “daggered” operators. We can therefore replace∫ d3q qµe−2iq0t q0 â(−q)â(q) = ∫ d3q qµe−2iq0t q0 1 2 [ â(−q)â(q) + â(q)â(−q) ] , 80 showing for each component of qµ that this is an integration of an odd function over an even integration space. The same reasoning holds also true for the daggered operators. Therefore only the two last terms survive and we have shown that indeed :P̂µ: = ∫ d3k (2π)32k0 kµ â†(k)â(k) = ∫ d3x :π̂(x)∂µφ̂(x): (b) Direct calculation shows that [P̂µ, φ̂(x)] = ∫ d3k (2π)32k0 d3q (2π)32q0 kµ [ â†(k)â(k), â(q)e−iq·x + â†(q)eiq·x ] = ∫ d3k (2π)32k0 d3q (2π)32q0 kµ { e−iq·x [ â†(k), â(q) ] â(k) + eiq·x â†(k) [ â(k), â†(q) ]} = ∫ d3k (2π)32k0 kµ { −e−ik·x â(k) + eik·x â†(k) } = ∫ d3k (2π)32k0 (−i∂µ) { e−ik·x â(k) + eik·x â†(k) } = −i∂µφ̂(x) , as demanded. 5. Causality and anti-commutators (real scalars) Consider real scalar fields and define ∆1(x− y) = ∆+(x− y) + ∆−(x− y) (a) show that ∆1 is equal to the vacuum expectation value of the anti-commutator of the field operators φ̂(x) and φ̂(y) ∆1(x− y) = 〈0|{φ̂(x), φ̂(y)}|0〉 = 〈0|[φ̂(x)φ̂(y) + φ̂(y)φ̂(x)]|0〉 (b) show that ∆1(x − y) does not vanish outside the light-cone, i.e. that ∆1(x− y) 6= 0 for (x− y)2 < 0. Solution (a) Remember definitions for ∆±, expansion of fields in terms of cre- ation and annihilation operators and â(k)|0〉 = 0, then ∆+(x− y) = 〈0|φ̂(x)φ̂(y)|0〉 = ∫ d3k (2π)3(2k0) d3k′ (2π)3(2k′0) e−ikx+ik′y 〈 0 ∣∣∣â(k)â†(k′) ∣∣∣ 0〉 81 = ∫ d3k (2π)3(2k0) d3k′ (2π)3(2k′0) e−ikx+ik′y 〈 0 ∣∣∣[â(k), â†(k′) ]∣∣∣ 0〉 = ∫ d3k (2π)3(2k0) d3k′ (2π)3(2k′0) e−ikx+ik′y 〈 0 ∣∣(2π)3(2k0)δ3(k − k′) ∣∣ 0〉 = ∫ d3k (2π)3(2k0) e−ik(x−y) In a similar way, we can write ∆−(x− y) = ∫ d3k (2π)3(2k0) eik(x−y) = ∫ d3k (2π)3(2k0) d3k′ (2π)3(2k′0) e−ik(y−x) 〈 0 ∣∣(2π)3(2k0)δ3(k − k′) ∣∣ 0〉 = ∫ d3k (2π)3(2k0) d3k′ (2π)3(2k′0) e−ik ′y+ikx 〈 0 ∣∣∣[â(k′), â†(k) ]∣∣∣ 0〉 = 〈0|φ̂(y)φ̂(x)|0〉 Therefore, as demanded ∆(x−y) = ∆+(x−y)+∆−(x−y) = 〈 0 ∣∣∣[φ̂(x)φ̂(y) + φ̂(y)φ̂(x) ]∣∣∣ 0〉 (b) ∆1(x− y) = ∫ d3k (2π)3(2k0) ( e−ik(x−y) + eik(x−y) ) x0→y0−→ ∫ d3k (2π)3(2k0) ( e−ik(x−y) + eik(x−y) ) x→y −→ 2 ∫ d3k (2π)3(2k0) →∞ 6. Commutators for free reals scalar fields Calculate the equal time commutators for (a) [ P̂µ, φ̂(x) ] , where the momentum operator is given by P̂µ = ∫ d3k (2π3)2k0 kµâ†(k)â(k) = ∫ d3xπ̂(x)∂µφ̂(x) ; (b) [ Ĥ, â†(k)â(q) ] ; 82 = 1 2 ∫ d3k (2π)32k0 [ â†(k)â(k)− b̂†(k)b̂(k) ] . (b) To calculate the commutator let us first take a look at one typical term, namely [φ̂(x)π̂(x), P̂µ] = φ̂(x)π̂(x)P̂µ − P̂µφ̂(x)π̂(x) = φ̂(x)P̂µπ̂(x)− P̂µφ̂(x)π̂(x) + φ̂(x)π̂(x)P̂µ − φ̂(x)P̂µπ̂(x) = [φ̂(x), P̂µ]π̂(x) + φ̂(x)[π̂(x), P̂µ] . Using the result from one of the previous problems that [P̂µ, φ̂(x)] = −i∂µφ̂(x) and similarly (because the derivatives commute) that [P̂µ, π̂(x)] = [P̂µ, ∂tφ̂ ∗(x)] = −i∂t∂µφ̂∗(x) = −i∂µπ̂(x) we see that[ :Q:, :P̂µ: ] = − ∫ d3x {[ P̂µ, φ̂∗(x) ] π̂∗(x) + φ̂∗(x) [ P̂µ, π̂∗(x) ] − [ P̂µ, φ̂(x) ] π̂(x) + φ̂(x) [ P̂µ, π̂(x) ]} = i ∫ d3x { [∂µφ̂∗(x)]π̂∗(x) + φ̂∗(x)[∂µπ̂∗(x)] − [∂µφ̂(x)]π̂(x)− φ̂(x)[∂µπ̂(x)] } = i ∫ d3x { ∂µ [ φ̂∗(x)π̂∗(x)− φ̂(x)π̂(x) ]} = 0 8. ∗Parity of a Scalar Field The parity operator P̂ for a real scalar field is given by P̂ = exp { − iπ 2 ∫ d3k (2π)32k0 [ â†(k)â(k)− ηP â†(k)â(−k) ]} , where the phase ηP = ±1 is the intrinsic parity of the field. Fields with ηP = 1 are scalars and those with ηP = −1 are pseudoscalars. Prove that [P̂, Ĥ] = 0. Solution To prove this, we need to realise that any function of an operator Ô commutes with another operator X̂, if the operators commute, i.e. 85 [f(Ô), X̂] = 0 if [Ô, X̂] = 0, because functions of operators are defined through their series expansion. This means that we have to show that 0 = [∫ d3k (2π)32k0 { â†(k)â(k)− ηP â†(k)â(−k) } ,∫ d3q (2π)32q0 q0 â †(q)â(q) ] = ∫ d3k (2π)32k0 ∫ d3q (2π)32q0 q0 {[ â†(k)â(k), â†(q)â(q) ] − ηP [ â†(k)â(−k), â†(q)â(q) ]} = ∫ d3k (2π)32k0 ∫ d3q (2π)32q0 q0 { (2π3)2q0δ 3(k − q)â†(k)â(q)− â†(k)â†(q)â(k)â(q) −(2π3)2q0δ 3(k − q)â†(q)â(k) + â†(q)â†(k)â(q)â(k) − ηP [ (2π3)2q0δ 3(k + q)â†(k)â(q)− â†(k)â†(q)â(−k)â(q) − (2π3)2q0δ 3(k − q)â†(q)â(−k) + â†(q)â†(k)â(q)â(−k) ]} = −ηP ∫ d3k (2π)32k0 ∫ d3q (2π)32q0 q0 { â†(k)â(−k)− â†(k)â(−k) } = 0 and therefore the parity operator commutes with the Hamiltonian. This implies that parity is a conserved quantity for the free scalar field. 86 5 Fermions In this section we will get acquainted with the Dirac equation, which intro- duces not only fermions, but also provides insight into anti-particles. The Dirac equation emerges through linearisation of the Klein-Gordon equa- tion, after realising that its quadratic form yields negative energy solutions. Such a linearised form, however, only satisfies the original energy-momentum relation – essentially the kernel of the free Klein-Gordon equation – if the fields have an even number of components, at least two. This proves to be a blessing, as it allows us to describe spin-1/2 particles, and the corresponding fields are dubbed “spinors”. Insisting on maintaining that the spinors sat- isfy the Klein-Gordon equation for massive particles leads to spinors with four components - two more than necessary for spin-1/2 particles. These additional degrees of freedom are identified with negative energy solutions and interpreted as anti-particles. As before, in the case of the scalar fields, this implies that the energy spectrum of the theory is unbounded from be- low. Consequently the vacuum is not empty, and i fact it contains short- lived quantum fluctuations of particle+anti-particle with opposite energy, momentum and spin. The Dirac equation has been covered ubiquitously in the literature. Keeping in mind that we use a somewhat different (and in my opinion, more mod- ern) normalisation, it would maybe be a good idea to also take a look at Sections 4.1 and 4.2 of Hatfield [3] or Sections 3.1-3.4 and 3.6 in Peskin & Schroeder [1], the latter section more of some extended reading. It is also worthwhile to check out Chapter 2 of Itzykson & Zuber [?], if you can find it somewhere. 5.1 The Dirac Equation Short-comings of the Klein-Gordon Lagrangian Consider, again, the Klein-Gordon equations of motion, Eq. (91) in Sec. 3.2,( ∂2 ∂t2 −∇2 +m2 ) φ(x) = ( ∂µ∂ µ +m2 ) φ(x) = 0 . Fourier-transforming it into (E2 − p2 −m2)φ = 0 −→ E2 = p2 +m2 (182) we realise that, due to its quadratic form, nothing prevents us from con- structing solutions with negative energies. Assuming plane wave solutions for the fields, φ(x) ∼ exp(ikx) the charge or probability density for the complex scalar field is given by ρ = j0 = (∂tφ ∗)φ− φ∗(∂tφ) = −2ik0 (183) 87 Dirac Equation Multiplying the Dirac equation, expressed through the α and β matrices, Eq. (185), from the left with γ0 we arrive at (iγµ∂µ −m1)ηξ ψξ = (i∂/−m)ψ = 0 . (192) In the equation above, Eq. (192) the components of the Dirac equation in “spinor space” have been made explicit, indicated by the indices η and ξ. It is important to stress that this exhibits the fact that there are two spaces in the Dirac equation, namely the “normal” Minkowski space with index µ, incorporating the external Lorentz symmetry of space-time, and this spinor space. The γ matrices and the spinor ψ have multiple components in this space, and the mass term is diagonal in this space, indicated by the 1- symbol. As before, the Lorentz indices µ etc. run from 0 to 3, while the Dirac or spinor indices run from 1 to 4. Dirac Equation for ψ† The nature of the equation above suggest that the Hermitean conjugate spinor ψ† represents a second, independent field, similar to φ∗ and φ. Straightforward Hermitean conjugation of Eq. (185) results in −i∂ψ †(x, t) ∂t = i∇ψ†(x, t) · α† +mψ†(x, t)β† , (193) and multiplying from the right with β† = β = γ0 yields −iψ†(x, t) ←− ∂µγ †µ = mψ†(x, t) . (194) Using γ02 = 1 and γ† = (βα)† = αβ = β(βα)β = γ0γγ0 while defining the “barred” spinor ψ̄ = ψ†γ0 allows to find the E.o.M. for the barred spinor as ψ̄(i ←− ∂/+m) = 0 . (195) Lagrangian It is easy to check that the two E.o.M. for the spinors ψ and ψ† can be obtained from the free Dirac Lagrangian L = ψ̄(x) ( i ←→ ∂/ −m ) ψ(x) , (196) where a ←→ ∂ b = 1 2 [a(∂b)− (∂a)b] . (197) The E.o.M. for ψ (ψ̄) are obtained,as usual, by varying the Lagrangian with respect to ψ̄ (ψ): ∂L ∂ψ̄ − ∂µ ∂L ∂(∂µψ̄) = −mψ + 1 2 [i∂/ψ − ∂µ(−iγµψ)] = ( i −→ ∂/ −m ) ψ = 0 ∂L ∂ψ − ∂µ ∂L ∂(∂µψ) = −mψ̄ − 1 2 [ ψ̄(i ←− ∂/ ) + ∂µ(iψ̄γµ) ] = −ψ̄ ( i ←− ∂/ +m ) = 0 . (198) 90 Conserved Current It is relatively straightforward to construct a con- served current from the two E.o.M. Eqs (192) and (195): multiply the former from the left with ψ̄ and the latter from the right with ψ and add. This results in 0 = ψ̄ · (i −→ ∂/ −m)ψ + ψ̄ · (i ←− ∂/ +m) · ψ = iψ̄( −→ ∂/ + ←− ∂/ )ψ (199) and we arrive at the conserved current ∂µj µ = ∂µ [ iψ̄γµψ ] . (200) Solutions to the Dirac E.o.M.: Spinors at Rest To construct solu- tions for the Dirac equation, it is important to keep in mind that the ψ and ψ̄ are objects with four components9. Let us for the moment describe the ψ as a product of advanced and retarded plane wave factors and polarisation eigenstates u(p) and v(p), ψη(x) = ∫ d3p (2π)3 [ e−ip·xuη(p) + eip·xvη(p) ] , (201) where we have made explicit the spinor index η. This expansion moves the spinor index to the u and v spinors, i.e. they are objects with four entries, and the Dirac matrices act on these indices10. To construct them, it is sufficient to realise that the E.o.M. become a system of linear equations for the eigenstates u(p) and v(p). Let us first solve this equation for a particle at rest, p = 0, p0 = E = m, leading to (Eγ0 −m)u(0) = m(γ0 − 1)u(0) = 0 and, similarly, (γ0 + 1)v(0) = 0 , (202) Inserting the (diagonal) form of γ0 =  1 0 0 0 0 1 0 0 0 0 −1 0 0 0 0 −1  (203) implies that the third and fourth component of u and the first and second component of v must be zero. Both u and v therefore have two independent 9But, although they look like vectors because of the four-components, they differ from four-vectors in how they behave under Lorentz transformations. Simply put: spinor index 6= Lorentz index) 10Positive and negative energy solutions ψ± are of course related to the wave factors such that ψ+ = e−ip·xu(p) and ψ− = eip·xv(p) , and we will recycle them later when quantising the Dirac fields. 91 solutions each, and the corresponding eigenstates can be readily identified with the two spin states: u(1/2) describe positive-energy particles with spin up/down, and v(1/2) decibel negative-energy particles with spin up/down. Choosing normalised “eigenvectors” then results in u(1)(0) =  1 0 0 0  , u(2)(0) =  0 1 0 0  , v(1)(0) =  0 0 1 0  , v(2)(0) =  0 0 0 1  . (204) Solutions to the Dirac E.o.M.: General Momenta To obtain so- lutions for general momenta, we use the fact that suitable multiplication of the kernels of the E.o.M. for u and v with terms (p/ ± m) encodes the energy-momentum relation for a massive particle, (p/−m)(p/+m) = p2 −m2 = 0 . (205) This means that, including normalisation factors η(p), the transformed eigen- states u(i)(p) = ηi(p/+m)u(i)(0) v(i)(p) = ηi(−p/+m)v(i)(0) (206) will satisfy the E.o.M. (p/ −m)u = 0 and (p/ + m)v = 0. Introducing p± = px ± ipy of the momentum components the spinors and using p/±m = pµγ µ ±m =  E ±m 0 −pz −px + ipy 0 E ±m −px − ipy pz pz px − ipy −E ±m 0 px + ipy −pz 0 −E ±m  (207) we arrive at u(1)(p) = η  1 0 pz E+m p+ E+m  , u(2)(p) = η  0 1 p− E+m −pz E+m  , v(1)(p) = η  pz E+m p+ E+m 1 0  , v(2)(p) = η  p− E+m −pz E+m 0 1  , (208) 92