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Relativistic Quantum Physics and the Dirac Equation

Maurice R. TREMBLAY,PMPSuivre

Sr Project Manager at Shared Services Canada (SSC) à Shared Services CanadaPublicité

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- From First Principles January 2017 – R4.4 Maurice R. TREMBLAY PART VII – QUANTUM ELECTRODYNAMICS ∫∫ +++++−=′ 65 2 56 2 )2,6()1,5()()6,4()5,3()()2,1;4,3( ττδγγ µµ ddKKsKKeiK bababa ELECTRONS VIRTUAL QUANTUMTIME 1 2 3 4 5 6 )( 2 56s+δ )6,4(+K )1,5(+K )2,6(+K )5,3(+K µγ µγ a b Circa 1949 Chapter 2
- Contents PART VII–QUANTUM ELECTRODYNAMICS Particles and Fields Second Quantization Yukawa Potential Complex Scalar Field Noether’s Theorem Maxwell’s Equations Classical Radiation Field Quantization of Radiation Oscillators Klein-Gordon Scalar Field Charged Scalar Field Propagator Theory Dirac Spinor Field Quantizing the Spinor Field Weyl Neutrinos Relativistic Quantum Mechanics Quantizing the Maxwell Field Cross Sections and the Scattering Matrix Propagator Theory and Rutherford 2017 MRT 2 Scattering Time Evolution Operator Feynman’s Rules The Compton Effect Pair Annihilation Møller Scattering Bhabha Scattering Bremsstrahlung Radiative Corrections Anomalous Magnetic Moment Infrared Divergence Lamb Shift Overview of Renormalization in QED Brief Review of Regularization in QED Appendix I: Radiation Gauge Appendix II: Path Integrals Appendix III: Dirac Matrices References
- In order to study radiation phenomena in the quantum domain we shall first discuss in detail the properties of the vector potential A satisfying the solenoïdal condition: 3 2017 MRT 0=• A∇∇∇∇ within the framework of classical electrodynamics. The equation above is known as the transversality condition; it should not be confused with the Lorentz condition ∂µ Aµ =0. The electric of magnetic fields derivable from vector potentials satisfying ∇∇∇∇•A=0 are called transverse fields or radiation fields. Often the term transverse field or radiation field is used to refer to a vector potential itself satisfying ∇∇∇∇•A=0. Classical Radiation Field The transversality condition ∇∇∇∇•A=0 is of interest under a variety of circumstances. First, suppose Jµ =0. We can then consider a gauge transformation that eliminates the A0 component of Aµ and makes A obey ∇∇∇∇•A=0. Consider: χ χ ∇∇∇∇++++AAA =′→ ∂ ∂ −=′→ and tc AAA 1 000 such that: 0 1 A tc = ∂ ∂χ Since the A0 component of Aµ has been eliminated in the new gauge, the Lorentz condition ∂µ Aµ =0 reduces to the transversality condition ∇∇∇∇•A=0.
- Let us consider now the situation in which Jµ ≠0, as in the case of mutually interacting electrons. We may first decompose A so that: 4 2017 MRT ||AAA ++++⊥= where: ( ) ( )allongitudinandtransverse 0AA ==• ⊥ ||0 ××××∇∇∇∇∇∇∇∇ This can always be done. Here A⊥ and A| | are called respectively the transverse and the longitudinal component of A. Radiation)()( 2)()( π4 )]([ 2 1 H ee e m H ji ji ji j j j j j + − +−= ∑∑ > ⊥ xx xAp where HRadiation is the free field Hamiltonian of A⊥ only. Note that nowhere in this equation do A0 and A| | appear explicitly. In 1930, E. Fermi was able to show that A| | and A0 together give rise to the instantaneous static Coulomb interaction between the charged particles, whereas A⊥ accounts for the electromagnetic radiation of moving charged particles. The total Hamiltonian of the charged particles (treated nonrelativistically) and the electromagnetic fields generated by them can be written (apart from the magnetic moment interaction) as (c.f., Appendix I: Radiation Gauge):
- Fermi's formalism based on the Hamiltonian H=HEM +HRadiation above is called the radiation (or Coulomb) gauge method. Since this decomposition shown in the Hamiltonian is not relativistically covariant, nor is the transversality condition itself, the whole formalism appears noncovariant; each time we perform a Lorentz transformation, we must simultaneously make a gauge transformation to obtain a new set of A and A0. 5 2017 MRT tc ∂ ∂ −== A EAB 1 and××××∇∇∇∇ It is worth studying a theory of transverse electromagnetic field before we learn about more sophisticated formalisms. When the theory is quantized, it provides simple and physically transparent descriptions of a variety of processes in which real photons are emitted, absorbed, or scattered. The three basic equations we work with for the free-field case are: as well as: 0 1 2 2 2 2 = ∂ ∂ −∇ tc A A where A satisfies the transversality condition ∇∇∇∇•A=0. Note that for the time being we will include the velocity of light constant c in the equations – for old time sake! This will especially come to bear when we introduce the angular momentum ω=|k|c in the equations that will follow.
- At a given instant, say starting at t =0, we expand the vector potential A in terms of Fourier series and assume periodic boundary conditions for A enclosed in a box taken to be a cube of volume V and of side L=V 1/3. Remembering the reality of A, we have: 6 2017 MRT ∑ ∑= = += k kkkk xuxuxA 2,1 ** 0 ])()0()()0([ 1 ),( r rrrrt cc V t (N.B., ckr (0) and its complex conjugate c* kr(0) are Fourier coefficients) where: xk k εxu • = ir r e)( )( and εεεε(r), called a (linear) polarization vector, is a real unit vector whose direction depends on the propagation direction k (i.e., the wave vector). Given k, we choose εεεε(1) and εεεε(2) in such a way that {εεεε(1),εεεε(2),k/|k|} for a right-handed set of mutually orthogonal unit vectors (N.B., εεεε(1) and εεεε(2) are, in general, not along the x- and the y-axes since k is, in general, not along the z-axis). Since εεεε(r) is perpendicular to k, the transversality condition is guaranteed. The Fourier components ukr satisfy: sr sr sr srsr d V d V δδδδ kk kk kk kkkk uu uu xuux ′− ′ ′ ′′ = • • =• ∫∫ ,** 3*3 11 and where: because of the periodic boundary conditions. ( )K,2,1 π2 ,, ±±== n L n kkk zyx
- To obtain A(x,t) for t≠0, we simply replace ckr (0) and c* kr (0) by: 7 2017 MRT ti rr ti rr ctcctc ω**ω e)0()(e)0()( kkkk == − and where the angular frequency is given by: ck=ω With this replacement, both the wave equation ∇2A−(1/c2)∂2A/∂t2 =0 and the reality condition on A are satisfied. So: ∑∑ ∑∑ ∑∑ ⋅−⋅ +•−−• •−• += += += k kk k xk k xk k k xk k xk k εε εε εεxA r xkir r xkir r r tiir r tiir r r ir r ir r cc V cc V tctc V t ]e)0(e)0([ 1 ]e)0(e)0([ 1 ]e)(e)([ 1 ),( )(*)( ω)(*ω)( )(*)( where: tctxkxk kxkxk −•=−•=≡⋅ ωµ µ since: ],[]ω,[ xk txk == µµ and
- The Hamiltonian of the field is: 8 2017 MRT A typical term we must evaluate for the |B|2 integration is: ∫∫ ∂ ∂ +=−= 2 23223 1 2 1 )( 2 1 tc ddH A AxEBx ××××∇∇∇∇ sr sr srsrsr V c d ddd δδ kk kk kkkkkk uux uuxuuxuux ′ ′ ′′′ = •∇−= •+•=• ∫ ∫∫∫ 2 *23 *3*3*3 ω )]([)]([)()( ××××∇∇∇∇××××∇∇∇∇××××∇∇∇∇××××∇∇∇∇××××∇∇∇∇ where we have used the periodic boundary conditions and the identity ∇∇∇∇××××(∇∇∇∇××××X)=∇∇∇∇(∇∇∇∇•X) −−−−∇2X. Similarly, for the |E|2 integration it is useful to evaluate first: * 2 **3 ω )( 1 )( 1 srsrssrr ccV c c tc c tc d kkkkkkkk uux ′′′′ = ∂ ∂ • ∂ ∂ ∫ δδ Using these relations, we obtain: ∑∑ = k kk r rrcc c H * 2 ω 2 where ckr is a time-independent Fourier coefficient satisfying ckr =−ω2ckr.−
- This last relation with the Hamiltonian reminds us of an expression for the energy in a collection of independent and uncoupled harmonic oscillators! (N.B., This means that in order to analyze the system we decompose it into its normal modes. Denoting the normal mode coordinates by Qi we thus obtain a system of uncoupled harmonic oscillator equations of the form Qi +ωi 2Qi =0.) To make the analogy more vivid, we define: 9 2017 MRT )( ω )( 1 ** rrrrrr cc c iPcc c Q kkkkkk −−=+= and Then: ∑∑ ∑∑ += + − = k kk k kkkk r rr r rrrr QP PiQcPiQc c H )ω( 2 1 ω2 )ω( ω2 )ω(ω 2 222 2 The Pkr and Qkr are now seen as canonical variables: r r r r Q P H P Q H k k k k && += ∂ ∂ −= ∂ ∂ and Thus, the radiation field can be regarded as a collection of independent harmonic oscillators each of which is characterized by {k,r} and whose dynamical variables are orthogonal linear combinations of the Fourier coefficients. ⋅⋅
- As we saw towards the end of PART II – MODERN PHYSICS, at the end of the nineteenth century it was recognized that the space-time development of the radiation field satisfying the wave equation ∇2A−(1/c2)∂2A/∂t2 =0 (c.f., Classical Radiation Field chapter) resembles the dynamical behavior of a collection of harmonic oscillators.* By assigning an average energy kBT to each radiation oscillator, Lord-Rayleigh (a.k.a., J. W. Strutt) and J. H. Jeans wrote an expression for the energy distribution of the radiation field as a function of angular frequency ω in an ideal situation where the radiation field is enclosed by perfectly absorbing walls. The expression they obtained was in satisfactory agreement with observation for low values of ω at sufficiently high temperatures, but in marked disagreement for high values of ω. This difficulty led M. Planck to take one of the most revolutionary steps ever taken in the history of science. He proposed that: The energy of each radiation oscillator is not an arbitrary quantity but must be an integral multiple of hω, where h is a new fundamental constant in nature. This he did in 1901. Four years later, in order to explain the photoelectric effect, A. Einstein proposed that an electromagnetic wave λ=2πc/ω be regarded as a collection of massless particles each of which has energy hω and ω can also be written ω=|k|c with standard momentum k. 10 2017 MRT We can now do better than Planck and Einstein, but only because we know nonrelati- vistic quantum mechanics. Indeed, no sooner was nonrelativistic quantum mechanics fully developed that P. A. M. Dirac proposed that the canonical dynamical variables of a radiation oscillator be treated as noncommutable operators, just as x and p of a one- dimensional harmonic oscillator are treated in nonrelativistic quantum mechanics. Quantization of Radiation Oscillators * The whole of quantum mechanics is based on this idea… i.e., until a different approximation than ‘harmonic oscillators’ becomes available.
- We postulate that P and Q of the radiation oscillators are no longer mere numbers but are operators satisfying: 11 2017 MRT 0],[0],[],[ === ′′′′ srsrsrsr PPQQiPQ kkkkkkkk and,δδh We next consider linear combinations of Pkr and Qkr given by: )ω( ω2 1 )ω( ω2 1 † rrrrrr PiQaPiQa kkkkkk −=+= hh and The akr and a† kr are seen to be the operator analogs of the Fourier coefficients ckr and c† kr when we insert a factor to make akr and a† kr dimensionless: rr acc kk ω2 h → They satisfy the commutation relations: srsrsrsr QP i PQ i aa δδ kkkkkkkk ′′′′ =+−= ],[ 2 ],[ 2 ],[ † hh 0],[],[ †† == ′′ srsr aaaa kkkk and These commutation relations are to be evaluated for operators taken at equal times (e.g., [akr,a† k′s] actually stands for [akr(t),a† k′s(t)] in the Heisenberg Picture where operators incorporate a dependency on time, but the state vectors are time- independent).
- Before we begin the physical interpretations of akr and a† kr, it is instructive to study the properties of the operator defined by: 12 2017 MRT We have: rrr aaN kkk † = rsr rssssr rssssrsr a aaaaaa aaaaaaNa kkk kkkkkk kkkkkkkk δδ ′ ′′′′ ′′′′′ = −= −= ],[],[ ],[ †† †† Similarly: †† ],[ rsrsr aNa kkkkk δδ ′′ −= For the time being, we will suppress the indices k and r.
- Unlike a and a†, the operator N is Hermitian. The Hermiticity of N encourages us to consider a normalized eigenvector of the operator N denoted by |n〉 such that: 13 2017 MRT nnnN = where n is the eigenvalue of N. Because N is Hermitian, n must be real. Now: nannaNanaN †††† )1()( +=+= where we have used [a†,N]=−a† above. This can be viewed as a new eigenvalue equation in which the eigenvalue a†|n〉 is shown to have eigenvalue n+1. Similarly: nannaN )1( −= The roles of a† and a are now clear; a† (a) acting on |n〉 gives a new eigenvector with eigenvalue increased (decreased) by one. So: 11† −=+= −+ ncnancna and where c+ and c− are constants. To determine c± we evaluate: 1],[11 ††††22 +=+===++= ++ nnaaNnnaannananncc and The phases of c± are indeterminate; they may be chosen to be zero at t=0 by convention. nnaannanac ===− †2
- We thus have, at t=0: 14 2017 MRT Meanwhile: 111† −=++= nnnannna and 0† ≥== naannNnn since the norm of a|n〉 must be positive. This immediately tells us that n cannot be a noninteger; otherwise, the eigenvalue of |n〉 could be made to decrease indefinitely as we successfully apply a, and n would eventually take negative values, in contradiction to n≥0 above. On the other hand, if n is a positive integer, successive applications of the operator a proceed as: 000211221 ===−= aaannna ,,,L Note that we obtain a null vector 0 (to be distinguished from |0〉) when a is applied to |0〉. By applying a to such a null vector, we again obtain a null vector. Hence n=0 is the lowest possible eigenvalue of the operator N.
- Explicit matrix representations of a, a†, and N consistent with the commutation relations [a,a†]=1 & [a,a]=[a†,a†]=0, [a,N]=a and [a†,N]=−a† above* can be written as follows: 15 2017 MRT = = OMMMMM L L L L OMMMMM L L L L 00300 00020 00001 00000 40000 03000 00200 00010 † aa & and = OMMMM L L L L 3000 0200 0010 0000 N * Technically, these were written above in the form: ††††† ],[],[0],[],[],[ rsrsrrsrsrsrsrsrsr aNaaNaaaaaaa kkkkkkkkkkkkkkkkkk δδδδδδ ′′′′′′′′ −===== and,&
- They are assumed to act on a column vector represented by: 16 2017 MRT where only the (n+1)-entry is different than zero and where the superscript T means ‘transpose’ (i.e., a transposed matrix is a new matrix whose rows are the columns of the original). [ ]T LL M M 01000 0 1 0 0 0 ≡ =n
- The algebra developed so far can be applied to a physical situation in which the number of photons with given momentum and polarization is increased or decreased. The wave vector k is identified with the photon momentum p divided by h, and r represents the polarization state of the photon. We interpret an eigenvector of Nkr as the state vector for a state with a definite number of photons in state (k,r). To represent a situation in which there are many types of photons with different sets of (k,r), we consider the direct product of eigenvectors as follows: 17 2017 MRT LLKK iiii rrrrrr nnnnnn kkkkkk 22112211 ,,,, = This state vector corresponds to the physical situation in which there are Nk1r1 photons present in state (k1,r1), Nk2r2 photons present in state (k2,r2), &c. The number nkr is the occupation number for state (k,r).
- As an example, the state represented by: 18 2017 MRT since the eigenvalue of Nkr is one (c.f., Nkra† kr |n〉=(n+1)a† kr |n〉 above). A two-photon state is represented by the normalized eigenvector: LL ii rrr kkk 0000 2211 = has the property that if akr is applied to it, then we obtain a null vector for any (k,r). Hence the eigenvalue of Nkr =a† kr akr is zero for any (k,r). Therefore the state corresponding to |0〉 above is called the vacuum state. A single-photon state with definite (k,r) is represented by: 0† rak 0 2 1 †† rr aa kk when the two photons are in the same state, and: 0†† rr aa kk when the two photons are in different states. More generally: 0)( 1 ,, † 2211 ∏= ii iri ii iir n r r rr a n nn k k k kk k K
- Note that when a† ki ri is applied to the most general state vector, we obtain: 19 2017 MRT Thus a† ki ri has the property of creating an additional photon in state (ki,ri), leaving the occupation numbers of states other than (ki,ri) unchanged. For this reason a† kr is called the creation operator for a photon in state (k,r). Similarly, akr can be interpreted as the annihilation operator for a photon in state (k,r). In contrast, Nkr, being diagonal, does not change the occupation number of photons; it simply gives as its eigenvalue the number of photons in state (k,r). KKKK ,1,,,1,,,, 22112211 † ++= iiiiiiii rrrrrrrr nnnnnnna kkkkkkkk Our formalism is capable of describing a physical situation in which the number of photons in a given state is unrestricted. Moreover, any many-photon systems we can construct is necessarily symmetric under interchange of any pair of labels. For instance, the two-photon state a† kr a† kr |0〉 above is evidently symmetric under interchange (k,r)↔ (k′,s) since: The quantization of the radiation field is shown, although using a somewhat different notation, in the Appendix of PART V – THE HYDROGEN ATOM. 00 †††† ssrr aaaa kkkk ′′= because of the commutation rule. Thus, the state vectors we obtain by applying creation operators to |0〉 are automatically consistent with Bose-Einstein statistics. With essentially no modifications our formalism can be applied to physical states made up of identical particle other than photons as long as they obey Bose-Einstein statistics.
- Let us begin our discussion by quantizing the simplest possible relativistic field theory, the free scalar field. The theory was proposed in various papers by six different physicists in 1926-27: W. Gordon, O. Klein, V. Fock, E. Schrödinger, Th. de Donder & H. van den Dungen, and J. Kudar. The Lagrangian (density) is given by: 20 2017 MRT Klein-Gordon Scalar Field Historically, the quantization of the Klein-Gordon equation caused much confusion. Schrodinger,even before he postulated his celebrated nonrelativistic equation,i∂Ψ(x,t)/∂t =HΨ(x,t), considered this relativistic equation but ultimately discarded it because of problems with negative probabilities and negative energy states. In any fully relativistic equation, we must obey the mass-shell condition pµ 2= E2 −p2 =m2. This means that the energy is given by: 22 2 1 2 1 φφφ µ µ m−∂∂=L 22 mE +±= p The energy can be negative, which is quite disturbing. Even if we banish the negative energy states by fiat, we find that interactions with other particles will reduce the energy and create negative energy states. This means that all positive energy states will eventually collapse into negative energy states, destabilizing the entire theory! However, the solution of these problems with negative probability and negative energy can be resolved once one quantizes the theory… where φ =φ(x,t). From this Lagrangian we get the equation of motion (∂2/∂t2 −∇2 +m2)φ(x,t) =0 (or (∂µ∂µ +m2)φ(x)=0) which is the well-known Klein-Gordon (wave) equation.
- The canonical quantization program begins with fields φ and their conjugate momen- tum fields π, which satisfy equal time commutation relations among themselves.Then the time evolution of these quantized fields is governed by a HamiltonianH=∫d3x H. Thus, we closely mimic the dynamics found in ordinary quantum mechanics. We begin by sin- gling out time as a spatial coordinate and then defining the canonical conjugate field to φ: We first introduce the Hamiltonian (density) as: ),( ),( ),( t t t x x x φ φδ δ π & & == L 21 2017 MRT ])([ 2 1 2222 φφπφπ m++=−= ∇∇∇∇LH & Then the transition from classical mechanics to quantum field theory begins when we postulate the commutation relations between the field and its conjugate momentum: )()],(,),([ 3 yxyx −−−−δπφ itt = with the right-hand side proportional to h which we set to unity. Also, [φ,φ]=[π ,π]=0. Much of what follows is a direct consequence of this commutation relation above. There are an infinite number of ways in which to satisfy this relationship, but our strategy will be to find a specific Fourier representation of this commutation relation in terms of plane waves. When these plane-wave solutions are quantized in terms of harmonic oscillators, we will be able to construct the multiparticle Hilbert space and also find a specific operator representation of the Lorentz group in terms of oscillators.
- We first define the quantity: We want a decomposition of the scalar field where the energy k0 =E is positive, and where the Klein-Gordon equation is explicitly obeyed. In momentum space, the operator ∂µ 2 +m2 (c.f., (∂µ∂µ +m2)φ (x)=0 above) becomes k2 −m2. Therefore, we choose: xp•−=≡⋅ tExkxk µ µ 22 2017 MRT ∫ ⋅⋅− +−= ]e)(e)()[()( )π2( 1 )( † 0 224 23 xkixki kAkAkmkkdx θδφ where θ is a step function (i.e., θ(k0)=+1 if k0 >0 and θ(k0)=0 otherwise), and where A(k) are operator-valued Fourier coefficients. It is now obvious that this field satisfies the Klein-Gorgon equation (i.e., if we hit this expression with (∂µ 2+m2), then this pulls down a factor of k2 −m2, which then cancels against the delta function δ (k2 −m2)). We can simplify this expression by integrating out dk0 (which also breaks manifest Lorentz invariance). To perform the integration, we need to re-express the delta function. We note that a function f (x), which satisfies f (a)=0, obeys the relation: )( )( )]([ af ax xf ′ − = δ δ for x near a. Since k2 =m2 has two roots, we find: 0 220 0 220 22 2 )( 2 )( )( k mk k mk mk ++ + +− =− kk δδ δ
- Putting this back into the integral, and using only the positive value of k0, we find: 23 2017 MRT ∫∫ ∫∫ ∫∫ =−=+−=− ∞ k k k kkk ω2 )ω( 2 1 )( 2 1 )()( 3 0 00 0322 00 03 0 224 d k k kddmk k kddkmkkd δδθδ where ωk ≡√(k2 +m2) and d4k ≡d3kdk0. Now let us insert this expression back into the Fourier decomposition of φ(x). We now find: ∫∫ +=+= ⋅⋅− )]()()()([]e)(e)([ ω2)π2( 1 )( *†3† 3 23 xekaxekadkaka d x kk xkixki k k k φ with a(k) ≡ A(k)/√(2ωk). Furthermore: where: ∫ +−== )]()()()([ω)()( *†3 xekaxekadixx kkkkφπ & kω2 e )π2( 1 )( 23 xki k xe ⋅− = and note that with the k0 appearing in k⋅x it is now equal to ωk. We can also invert these relations, solving for the Fourier modes a(k) in terms of the original scalar field: with the operator: ∫∫ ∂=∂= )()()()()()( 0 3† 0 *3 xxedikaxxedika kk φφ tt xx and BABABA )(∂−∂≡∂ t + −
- Because the fields satisfy equal-time canonical commutation relations, the Fourier modes must also satisfy commutation relations: 24 )()](,)([ 3† kk ′=′ −−−−δkaka and all other commutators (i.e., between a† and itself and a and itself) are zero. Now we can calculate the Hamiltonian H in terms of these Fourier modes using the Hamiltonian density H =½[π2 +(∇∇∇∇φ)2 +m2φ2] introduced above with ∇∇∇∇φ ≡∂iφ : ∫ ∫ ∫∫ += += +∂∂+== )()(ω )]()()()([ω 2 1 ][ 2 1 †3 ††3 22233 kakad kakakakad mddH ii k k k k xx φφφπH Similarly, we can calculate the momentum: ∫ ∫∫ += +=−= )()( )]()()()([ 2 1 †3 ††33 kakad kakakakadd kk kkxP φπ ∇∇∇∇ 2017 MRT 1 ‒ 2 + 1 ‒ 2 + (N.B.,Both H and P are divergent because of the ½ factor appearing in the infinite sum).
- With these expressions, the operators Pµ and Mµν generate translations and Lorentz rotations: as they should. If we exponiate these generators of translation and Lorentz rotations, we can calculate how the field φ(x) transforms under the Poincaré group. Let us define: φφφφ µννµµν µµ )(],[],[ ∂−∂=∂= xxMiPi and 25 2017 MRT µ µ µν µν PaiMi aU −− =Λ ω 2 1 e),( where Λµν =ηµν +ωµν +…. Then it can be shown that: )(),()(),( 1 axaUxaU +Λ=ΛΛ − φφ This demonstrates that φ(x) transforms as a scalar field under the Poincaré group. Now that we have successfully shown how to quantize the Klein-Gordon field, we must now calculate the eigenstates of the Hamiltonian to find the spectrum of states. Let us define the vacuum states as follows: 00)( =ka By convention, we call a(k) an annihilation operator. We define a one-particle state via the creation operator a†(k) as a Fock space (i.e., an algebraic construction used in quantum mechanics to construct the quantum states space of a variable or unknown number of identical particles from a single particle Hilbert space HHHH ): kka =0)(†
- The problem with this construction, however, is that the energy associated with the vacuum state is formally infinite because of the presence of ½ in the sum in the equation H=∫d3k ωk[a†(k)a(k)+½]=∫d3k ωka†(k)a(k) +½∫d3k ωk above. We will drop this infinite term, since infinite shifts in the Hamiltonian cannot be measured. Dropping the zero point energy in the expression for harmonic oscillators has a simple counterpart in x space. The zero-point energy emerged when we commuted creation and annihilation operators past each other. Dropping the zero-point energy is therefore equivalent to moving all creation operators to the left and annihilation operators to the right. This operation, in x space, can be accomplished by normal ordering. Since the product of two or more fields at the same point is formally divergent (i.e., limx→y φ(x)φ(y)=∞), we can remove this divergence by the normal ordering operation, which corresponds to moving the part containing the creation operators to the left and the annihilation operators to the right. For example, if we decompose φ(x)=φ†(x)+φ∅(x), where †(∅) represent the creation (annihilation) part of an operator with negative (positive) frequency, then we define: Then, by applying the normal ordering to the definition of the Hamiltonian, we can simply drop the ½ appearing in H above. From now on, we assume that when two fields are multiplied at the same point in space-time, they are automatically normal ordered. 26 2017 MRT Once we have normal ordered the operators, we now have an explicitly positive Hamiltonian. In this fashion, we have been able to handle the question of negative energy states for the Klein-Gordon theory. )()()()()()()()(:)()(: yxxyyxyxyx φφφφφφφφφφ +++≡ † † † †∅ ∅ ∅ ∅
- One essential point in introducing these creation and annihilation operators is that we can write down the N-particle Fock space: This is the chief distinguishing feature between first and second quantization. In first quantization, we quantized the xi corresponding to a single particle. First quantized systems were hence inherently based on single-particle dynamics. In the second quantized formalism, by contrast, we quantize multiparticle states. To count how many particles we have of a certain momentum, we introduce the number operator: ∫= )()(†3 kakadN k 27 2017 MRT 0)()()(,,, † 2 † 1 † 21 NN kakakakkk LK = The advantage of this number operator is that we can now calculate how many particles there are of a certain momentum.
- Forexample,let |n(k)〉 equal a state consistentof n(k) identical particles with momentumk: Then the number operator N acting on this multiparticle state just counts the number of particles present: 28 2017 MRT 0)]([ !)( 1 )( )(† kn ka kn kn = By commuting a(k) to the right, until they annihilate on the vacuum, we see that: )()()( knknknN = that is, N simply counts the number of state there are at momentum k. Not surprisingly, a multiparticle state, consisting of many particles of different momenta, can be represented as a Fock space: 0)]([ !)( 1 )()()( 1 )(† 21 ∏= = m i kn i i m i ka kn knknkn L )()()()()()()( 21 1 21 m m i im knknknknknknknN LL = ∑= Finally, it is essential to notice that the norm of these multiparticle states is positive. If we define 〈k |≡〈0 |a(k) and set 〈0|0〉=1, then the norm is given by 〈k|k〉=+δ 3(k−−−−k′). The norm is positive because the appropriate sign appears in the above commutation relation [a(k),a†(k)]=δ 3(k−−−−k′).
- We can generalize our discussion of the Klein-Gordon field by postulating the existence of several scalar fields. In particular, we can arrange two independent scalar field into a single complex field: The action then becomes: )( 2 1 21 φφφ += 29 2017 MRT Charged Scalar Field φφφφ µ µ †2† m−∂∂=L Nevertheless, if we insert this decomposition of φ into the action S, then we find the sum of two independent actions for φ1 and φ2, since it is the case for the Lagrangian (density). The quantization of this action proceed as before by calculating the conjugate field and postulating the canonical commutation relations. The conjugate field is given by: † φ φδ δ π & & == L The commutation relations now read: )()],(,),([ 3 yxyx −−−−δπφ itt = which will be a habitual rough-in coinage of the term since the action is actually given by: ∫= LxdS 4
- Since the decomposition of the scalar field (c.f., Klein-Gordon Scalar Field chapter) was: where i=1,2. Then the canonical commutation can be satisfied if the Fourier compo- nents obey the following commutation relations: 30 2017 MRT ∫ ⋅⋅− += ]e)(e)([ ω2)π2( 1 )( † 3 23 xki i xki ii kaka d x k k φ jiji kaka δδ )()](,)([ 3† kk ′=′ −−−− All other commutators vanish. Similarly, we can choose the decomposition: )]()([ 2 1 )()]()([ 2 1 )( † 2 † 1 † 21 kaikakakaikaka −=+= and as well as: )]()([ 2 1 )()]()([ 2 1 )( † 2 † 1 † 21 kaikakbkaikakb +=−= and For these operators, the new commutation relations read: )()](,)([)](,)([ 3†† kk ′=′=′ −−−−δkbkbkaka All other commutators are zero. ∫ ⋅⋅− += ]e)(e)([ ω2)π2( 1 )( † 3 23 xkixki kaka d x k k φ we can always decompose the φ = (1/√2)(φ1 +φ2) field in terms of the Fourier components:
- Now let us construct the symmetries of the action and the corresponding Noether currents. The action is symmetric under the following transformation: which generates a U(1) symmetry. Written out in components, we find the following SO(2) transformation (c.f., PART VI – GROUP THEORY): †† ee φφφφ θθ ii − →→ and 31 2017 MRT − = ′ ′ 2 1 2 1 cossin sincos φ φ θθ θθ φ φ This symmetry generates a Noether current (e.g., Jµ =(δ L/δ ∂µφ)(δφ/δθ)) which equals: φφφφ µµµ †† ∂−∂= iiJ Now let us calculate the charge Q corresponding to this current in terms of the quantized operators: ba NNkbkbkakadidQ −=−=−= ∫∫ )]()()()([)( ††3††3 xx φφφφ && where the number operator for the a and b oscillators is given by: ∫∫ == )()()()( †3†3 kbkbdNkakadN ba xx and Historically, this conserved current caused a certain amount of confusion. If J 0 is considered to be the probability density of the wave function, then it can be negative, and hence negative probabilities creep into the theory.
- In 1934, W. Pauli and V. Weisskopf finally gave the correct quantum interpretation of these negative probabilities: Second, we will interpret the b† oscillators as the creation operator for a new state of matter, antimatter. It was Dirac who originally grappled with these new states found in any relativistic theory and deduced the fact that a new form of matter, with opposite charge, must be given serious physical consideration. The discovery of the antielectron (i.e., the positron) gave graphic experimental proof of this conjecture. 32 2017 MRT First, because of the crucial minus sign appearing in front of the b oscillators, we will find it convenient to redefine the current Jµ as the current corresponding to the electric charge, rather than probability density, so that the a oscillators correspond to a positively charged particle and the b oscillators correspond to a negatively charged one. In this way, we can construct the quantum theory of charged scalar particles, where the minus sign appearing in the current is a desirable feature, rather than a fatal illness of the theory. Third, we no longer have a simple, single-particle interpretation of the φ(x), which now contains both matter and antimatter fields. Thus, we must abandon the strict single- particle interpretation for φ and reinterpret it as a field.
- Now that we have defined the canonical commutation relations among the particle fields, we are interested in how these particles actually move in space-time. To solve this equation, we use the standard theory of Green functions. We first define a propagator that satisfies: 33 2017 MRT Propagator Theory )()()( 22 xJxm =−∂ φµ )()()( 422 yxyxm F −−=−∆−∂ δµ Then the solution of the interacting φ field is given by: ∫ −∆−= )()()()( 4 0 yJyxxdxx Fφφ where φ0(x) is any function that satisfies the Klein-Gordon equation without any source term (i.e., (∂µ∂µ +m2)φ0(x)=0). If we hit both sides of this expression with (∂µ 2 +m2), then we find that it satisfies the original Klein-Gordon equation in the presence of a source term. To define a propagator, and also anticipate interactions, let us modify the Klein-Gordon equation to include a source term J(x):
- The way to solve this last equation is to take the Fourier transform: If we hit both sides of this equation with (∂µ 2 +m2), then we can solve for ∆F(k): 34 2017 MRT ∫ ∆=−∆ −⋅− )(e )π2( )( )( 4 4 k xd yx F yxki F 22 1 )( mk kF − =∆ At this point, however, we realize that there is an ambiguity in this equation. The integral over d4k cannot be performed on the real axis, because the denominator diverges at kµ 2 =m2. This same ambiguity occurs even in the classical theory of wave equations, and is not specific to the Lorentz covariant theory. The origin of this ambiguity lies not in the mathematics, but in the physics, in the fact that we have yet to fix our boundary conditions.
- For example, consider the Green function for the Schrödinger equation: If we take the Fourier transform of this equation and solve for the Green function, we get: 35 2017 MRT )()( 4 00 xxxxGH t i ′−=′− − ∂ ∂ δ ∫ ′−⋅− − =′− )( 24 4 0 e 2ω 1 )π2( )( xxpi m xd xxG pk where pµ =[ωk,p]. This expression also suffers from an ambiguity, because the integration over the energy, ωk, is divergent. Let us take the convention that we integrate over the real axis (see Figure), so that we integrate above the singularity. This can be accomplished by inserting a factor of iε into the denominator, replacing ωk −p2/2m with ωk −p2/2m+iε. Then the ωk integration can be performed. Contour integration for the Green function. (Left) The contour gives us the nonrelativistic retarded Green function. (Right) Contour giving the Feynman prescription for a relativistic (ωk ≡√(k2 +m2)) Green function. p2/2m Im k0Im ωk Re k0Re ωk−ε −ε ωk+ε
- We simply convert the integration over the real axis into a contour integration over the complex variable ωk. We are going to add the contour integral over the upper half plane (which vanishes) such that: where the θ function has been written as: xpi p x ⋅− = e )π2( 1 )( 23 φ 36 2017 MRT ∫∫ ∫ ′−−= +− =′− ′−−′• ∞+ ∞− ′−− ′• )(e )π2(2ω e π2 ω e )π2( )( ))(2()( 3 3 2 )(ω )( 3 3 0 2 tt d i εim dd xxG ttmii tti i θpxxp k kxxp p p p k −−−−−−−− ′> = + −=′− ∫ ∞+ ∞− ′−− → otherwise if 0 1 ω e ω π2 1 lim)( )(ω 0 tt εi d i tt tti ε k k k θ which equals 1 for t >t′, and vanishes otherwise. To see this a bit more explicitly, let us define: so that the Green function can be written as: ∫ ′′−−=′− )()()()( *3 0 xxdttixxG pp φφθ p where φp(x) above correspond to a formulation based on the plane wave approximation. In this way, the +iε insertion has selected out the retarded Green function, which obeys the usual concept of causality. Taking the −iε prescription would have given us the advanced Green function, which would violate causality.
- Finally, taking the d3p integration (which is Gaussian integral), we find the final result for the Green function (c.f., Appendix II: Path Integrals): which is just the Green function found in ordinary quantum mechanics – a result we obtained in PART III – QUANTUM MECHANICS using other means. 37 2017 MRT )(e )(π2 )( 2 2 23 0 tt tti m ixxG tt m i ′− ′− −=′− ′− ′ − θ xx−−−−
- Now that we have seen how various prescriptions for iε give us various boundary conditions, let us apply this knowledge to the relativistic case and choose the following, unorthodox prescription: To see how this iε prescription modifies the boundary conditions, we will find it useful to decompose this as: 38 2017 MRT εimk kF +− =∆ 22 1 )( −++ + ++− = +− εimkεimkkεimk 220220022 11 2 11 kk The integral over k0 now picks up contributions from both terms. Performing the integration as before (c.f., Klein-Gordon Scalar Field chapter), we find: ∫∫ ∫∫ ′−′−′′−−= −′−′−−=′−∆ ′−⋅′−⋅− )()( ω2)π2( )()()( ω2)π2( )( e ω2)π2( )(e ω2)π2( )()( * 3 3 * 3 3 )( 3 3 )( 3 3 xx d ttixx d tti d tti d ttixx pppp xxkixxki F φφθφφθ θθ kk kk kk kk We see the rather unusual feature of this prescription: positive energy solutions are carried forward in time, but negative energy solutions are carried backwards in time!
- In classical physics, the usual solutions of the Maxwell theory give us retarded and advanced waves, and we eliminate the advanced waves by a choice of boundary conditions. However, in the quantum theory we are encountering a new type of propagator that, classically, makes no sense, with negative energy solutions going backwards in time. This propagator never appears in classical physics because it is complex and hence forbidden classically. Quantum mechanically though, negative energy solutions are an inherent problem with any relativistic theory. Even if we ban them at the beginning, quantum interactions will inevitably re-create them later! However, as we saw in the previous chapter (c.f., the Pauli-Weisskopf interpretation of negative probabilities in the Charged Scalar chapter), these negative energy states can also be reinterpreted. Feynman’s approach to this problem was to assume that these negative energy states, because they are going back in time, appear as a new form of matter with positive energy going forward in time, antimatter! Although matter going backwards in time seems to contradict causality, this poses no problem because one can show that, experimentally, a system where matter is going backwards in time is indistinguishable from antimatter going forward in time. For example, an electron placed in an electric field may move to the right; however, if it is moving backwards in time, it appears to move to the left. However, this is indistinguishable experimentally from a positively charged electron moving forwards in time to the left. In this way, we can interpret this theory as one in which everything (i.e., matter plus antimatter) has positive energy. 39 2017 MRT
- Now, the previous expression for the Green function was written in terms of plane waves φp. However, we can replace the plane wave φp by the quantum field φ(x) if we take the vacuum expectation value (i.e., 〈0|…|0〉) of the product of fields (i.e., φ(x)φ(x′)). From the previous equation for the propagator ∆F(x−x′), we find: where T is called the time-ordered operator, defined as: 40 2017 MRT 0)]()([0)( xxTxxi F ′=′−∆ φφ >′′ ′>′ =′ ttxx ttxx xxT if if )()( )()( )]()([ φφ φφ φφ So, the operator T makes sure that the operator with the latest time component always appears to the left (mnemonic: lt-lt) This equation for ∆F is our most important result for propagators. It gives us a bridge between the theory of propagators, in which scattering amplitudes are written in terms of ∆F(x−x′), and the theory of operators, where everything is written in terms of the quantum field φ(x).
- Finally, we remark that our theory must obey the laws of causality. For our purposes, we will define microscopic causality as the statement that information cannot travel faster than the speed of light. For field theory, this means that φ(x) and φ(y) cannot interact with each other if they are separated by space-like distances. Mathematically, this means that the commutator between these two fields must vanish for space-like separations. Repeating the earlier steps, we can show that this commutator equals: 41 2017 MRT −<−− <<− >− ∂ ∂ = −= −∆= ∫ −⋅− rtrtmJ rtr rtrtmJ rr i kmkkd yxiyx yxki if if if )( 0 )( π4 e)()( )π2( 1 )()](,)([ 22 0 22 0 )( 0 224 23 εδ φφ where ε (k) equals +1 (−1) for positive (negative) k, t=x0 −y0, r=|x−−−−y|, and J0 is the Bessel function. With this explicit form for the commutator, we can easily show that, for space- like separations, we have: ( )0)(0)( 2 <−=−∆ yxyx if This shows that our construction obeys microscopic causality.
- After Dirac considered the relativistic theory of radiation in 1927, he set out the next year to construct the relativistic theory of electrons. One severe limitation was the problem of negative probabilities. He started with the observation that the nonrelativistic Schrödinger equation did not have negative probabilities because it was linear in time, while the Klein-Gordon equation, being quadratic in time, did have negative probabilities. Therefore Dirac tried to find a wave equation that was linear in time but still satisfied the relativistic mass-shell constraint: 2222 mEppp =−=≡ pµ µ Dirac Spinor Field 42 2017 MRT Dirac’s original idea was to take the square-root of the energy equation. In this way, he stumbled onto the spinorial representation of the Lorentz group… ψ ψ H t i = ∂ ∂ where H is the Hamiltonian (e.g., H=T=p2/2m=−(1/2m)∇2 with p=−i∇∇∇∇, the quantum momentum operator in h≡1 units). Schrödinger’s time-dependent equation for a free particle of mass m is: ψβα ψ )( mi t i i i +∇−= ∂ ∂ where αi (with summation implied over i=1,2,3) and β are now constant matrices, not ordinary c-numbers, which act on ψ, a column matrix. Dirac began with a first-order equation:
- By squaring the operator in front of the ψ field, we want to recover the mass-shell condition p2 = m2: This is only possible if we demand that the matrices satisfy: 43 2017 MRT ψ ψβ ψ )( )( 22 2 2 2 m mi t +−∇= +•−= ∂ ∂ − ∇∇∇∇αααα 0},{ 2},{ = =+≡ βα δαααααα i kiikkiki , To make the equations more symmetrical, we can then define γ 0 =β and γ i =βαi. Multiplying the wave equation by β, we then have the celebrated Dirac equation: 0)( =−∂ ψγ µ µ mi where the γ µ matrices satisfy: νµνµ ηγγ 2},{ = It is no accident that such a relativistic construction is possible since this is precisely the Clifford algebra which are formed by the γ µ. What we are really constructing is the spin-½ representation of the Lorentz group, that is, the spinors. and 122 == βαi
- To calculate the behavior of this equation under the Lorentz group, let us define how a Dirac spinor transforms under some (similarity) representation S(Λ) of the Lorentz group: Then the Dirac equation transforms as follows: )()()( xSx ψψ Λ=′′ 44 2017 MRT 0)(])()([ 1 =−∂Λ=−∂ΛΛ− ψγψγ ν ν µ µ µ µ mimSSi where we have multiplied the transformed Dirac equation by S−1(Λ) on the left, and we have taken into account the transformed ∂′µ=Λµ ν ∂ν . In order for the above equation to be Lorentz covariant, we must therefore have the following relation: νµ ν µ γγ ][)()( 11 −− Λ=ΛΛ SS To find an explicit representation for S(Λ), let us introduce the following matrix: ],[ 2 νµνµ γγ i =Σ The ½Σµν are the generators of the Lorentz group in this representation. Thus, we can write a new Lorentz group generator that is the sum of the old generator Jµν =xµ pν −xν pµ =i(xµ ∂ν −xν ∂µ) (which acts on the space-time coordinates xµ) plus a new piece that also generates the Lorentz group but in the spinorial representation: νµνµνµ Σ+= 2 1 JM The Σµν also obey the [γ µ,Σρσ ]=2i(δρ µγσ −δσ µγρ) relation which shows that the Dirac matrices transform as vectors under the spinor representation of the Lorentz group.
- In terms of this new Mµν matrix, we can find an explicit representation of the S(Λ) matrix: νµ νµ Σ− =Λ ω)4( e)( i S 45 2017 MRT For example, consider a rotation R around the z-axis through an angle θ. Then: − = θθ θθ θ cos00sin 0100 0010 sin00cos )(zR and, for an infinitesimal transformation, θ <<1: 2112 ωω −= and all other ωµν =0. Since M12 =−M21 =½γ1γ2, we obtain: + ==Λ 2 sin 2 cose 21 )2( 21 θ γγ θθγγ ii and − ==Λ −− 2 sin 2 cose 21 )2(1 21 θ γγ θθγγ ii With the help of Pauli matrices (e.g., σz), these can be written as Λ=exp(iσzθ/2). In general, for a rotation through and angle θ around the direction n, Λ=exp(iσσσσ•nθ/2).ˆ ˆ
- Let us look at a specific example of an infinitesimal expression for S. Consider a boost in the x-direction with v=tanhζ. Then, we have: − − =Λ=Λ 1000 0100 00coshsinh 00sinhcosh ]01[ ζζ ζζ ν µ 46 2017 MRT which, using Λµ ν =δ µ ν +εωµ ν , implies that: ν µ ν µ ζεζεε M= − − = 0000 0000 0001 0010 ω where the matrix actually represents M01, one of the generators of the Poincaré group. In general though, if M=Mµ ν , then M2 =diag(1,1,0,0), M3 =M, M4 =M2, &c. Thus, we have: µν µ ζµ α µ ν α µν µ ν ζδζδ xxM N M N xx M N N ]e[ 11 lim 1 = + +=Λ=′ ∞→ L
- Now this last result means that: 47 2017 MRT Note that the series contain only terms proportional to 1, M, and M2 and that it cuts thereafter, which is due to the special form of the matrix M. Now, we have for a boost in the x-direction: − =Λ 2 sinh 2 cosh)]([ 10 ]01[ ζ γγ ζ ζ 1S and since Σ01=½i[γ0,γ1]=iγ0γ1 (i.e., with (−γ0γ1)2 =γ0γ1γ0γ1 =−γ0 2γ1 2 =1 since γ0γ1 =−γ0γ1, γ0 2 =1 and γ1 2 =−1), so that: ]01[22 1 sinhcosh ! e Λ=++−=+= ∑ ∞ = ζζ ζζ MMM N M N NN M 11 01)2(]01[ e)]([ Σ =Λ ζ ζ i S 10)2(]01[ e)]([ γγζ ζ − =ΛS Therefore, using exp(ζ M)=1− M2 + M2coshζ + Msinhζ above, we can write the expression for S as: In general, for a boost in the xi-direction, we have: − =Λ 2 sinh 2 cosh)]([ 0 ]0[ ζ γγ ζ ζ i i S 1
- Now, instead of boosts, we could have rotations such as a rotation R around the z-axis through an angle θ just like before, which has: 48 2017 MRT Thus, in this case, we obtain: 12)2( e)]([ Σ = θ θ i zRS and since Σ12=½i[γ1,γ2]=iγ1γ2 (i.e., with (iγ1γ2)2 =−γ1γ2γ1γ2 =γ1 2γ2 2 =1), so that: 21)2(]01[ e)]([ γγθ ζ ii S =Λ − = − = 2 sin 2 cos 2 sinh 2 cosh)]([ 2121 θ γγ θθ γγ θ θ 11 iiiRS z where we used the identities cosh(ix)=cosx and sinh(ix)=isinx (for x real) in the last step. Similarly, for a rotation about the xk-axis, we find (for cyclic permutations of i, j,k=1,2,3): − = 2 sin 2 cos)]([ θ γγ θ θ jikRS 1 so that: − = − = 2 sin 2 cos)]([ 2 sin 2 cos)]([ 132321 θ γγ θ θ θ γγ θ θ RSRS , and − = 2 sin 2 cos)]([ 213 θ γγ θ θRS
- 49 2017 MRT 0)( †† =+∂ mi µ µ γψ s We will show shortly that there exists a representation of the Dirac matrices that satisfies: ii γγγγ −== †0†0 )()( and where γ i is anti-Hermitian and γ 0 is Hermitian. This can also be written as: Now that we know how spinors transform under the Lorentz group, we would like next to construct invariant under the group. Let us take the Hermitian conjugate of the Dirac equation: µ µ γγ =† Now, let us define: If we hit the Hermitian conjugated Dirac equation above with γ 0, we can replace the γ † with γ matrices, leaving us with: 0† γψψ ≡ 0)( =+∂ mi µ µ γψ s
- Under a Lorentz transformation, this new field ψ obeys: 50 2017 MRT − )()()]([)()( 10†0 Λ=Λ=′′ − SxSxx ψγγψψ This is just what we need to form invariants and covariant tensors. For example, notice that ψψ is an invariant under the Lorentz group:− )()()()()()()()()()( 1 xxxSSxxxxx ψψψψψψψψ =ΛΛ=′′′′=′′′′ − 1 since S −1(Λ)S(Λ)=1. Similarly, ψ γ µψ is a genuine vector under the Lorentz group. We find: − )()()()()()()()( 1 xxxSSxxx ψγψψγψψγψ νµ ν µµ Λ=ΛΛ=′′′′ − where we have used the fact that S −1(Λ)γ µ S(Λ)=Λµ ν γ ν which is nothing but the statement that the γ µ transform as a vector under the spinor representation of the Lorentz group. In the same manner, it can also be shown that ψ Σµνψ transforms as a genuine antisymmetric second-rank tensor under the Lorentz group. −
- To find other Lorentz tensors that can be represented as bilinears in the spinors, let us introduce the matrix: where ε µνρσ =−εµνρσ and ε 0123 =+1. Because γ5 transforms like ε µνρσ , it is a pseudoscalar (i.e., it changes sign under a parity transformation). Thus, ψ γ5ψ is a pseudoscalar. σρνµ σρνµ γγγγεγγγγγγ !4 32105 5 i i −=≡= 51 2017 MRT − In fact, the complete set of bilinears, their transformation properties, and the number of elements within each tensor are given by: ]6[ ]4[ ]4[ ]4[ ]1[ 5 5 ψψ ψγγψ ψγψ ψγψ ψψψψ νµ µ µ Σ = :Tensor :vector-Axial :Vector :arPseudoscal :Scalar 1 There is a total of 16 independent components in the list.
- The following set Γ of 16 matrices are linearly independent: 52 2017 MRT },,,,{ 55 νµµµ γγγγ Σ=Γ 1A where A=1,2,…,16 and note that: 321 15 210 14 130 13 320 12 30 11 20 10 10 9 21 8 13 7 32 6 3 5 2 4 1 3 0 2 441 γγγ γγγγγγγγγ γγγγγγ γγγγγγ γγγ γ i iii iii iii I =Γ =Γ=Γ=Γ =Γ=Γ=Γ =Γ=Γ=Γ =Γ=Γ=Γ =Γ ≡=Γ × ,,, ,,, ,,, ,,, , ,1 and 53210 16 γγγγγ ≡=Γ i and Specifically, these Γ matrices are: 1≡Γ 2 )( A ABBA ΓΓ±=ΓΓ
- Because γ µ transforms as a vector under the Lorentz group, the following Lagrangian is invariant under the Lorentz group: This, in turn, is the Lagrangian corresponding to the Dirac equation, LDirac. Variations of this equation by ψ or by ψ will generate the two versions of the Dirac equation. µνσ ρ σνµρνµ ρ νµρµ ρ µρ µ µ γγγγγγγγηγγγγγγγγγγ 2424 −==−== and,, 53 2017 MRT ψγψ µ µ )(Dirac mi −∂=L − Up to now, we have not said anything specific about the representation of the Dirac matrices themselves. In fact, a considerable number of identities can be derived for these matrices in four dimensions, without even mentioning a specific representation, such as: Some trace operations can also be defined: σρνµσρνµ ρνσµσνρµσρνµσρνµ νµνµνµνµµ εγγγγγ ηηηηηηγγγγ ηγγγγγγγ i4)(Tr )(4)(Tr 4)(Tr0)(Tr)(Tr)(Tr 5 55 = +−= ===Σ= and ,, In particular, this means: where the Feynman slash notation is used: µ µ γ aa ≡/ )])(())(())([(4)(Tr cbdadbcadcbadcba ⋅⋅+⋅⋅−⋅⋅=////
- It is often convenient to find an explicit representation of the Dirac matrices. The most common representation of these matrices is the Dirac representation: where σ i are the familiar Pauli spin matrices (c.f., Appendix III: Dirac Matrices). Then the spinor ψ is a complex-valued field with four components describinga massive spin-½ field! = − = − = − = 0 0 10 01 0 0 10 010 i i i i i i σ σ αβ σ σ γγ and,, 54 2017 MRT Now let us try to decompose ψ(x) into plane waves in order to begin the canonical quantization process… To do this, we need to find a set of independent basis spinors for ψ. We will make the obvious (i.e., a ‘1’ in a column vector, a unit spinor, &c.) choice: = = = = 1 0 0 0 )0( 0 1 0 0 )0( 0 0 1 0 )0( 0 0 0 1 )0( 2121 vvuu ,and, The trick is to act upon these basis spinors with S(Λ) in order to boost them up from zero (i.e., p=k=0) to arbitrary momentum p. The momentum-dependent spinors are given by: )0()()()0()()( αααα vSpvuSpu Λ=Λ= and with the spinor indices α =1,2. This last set of equations can be shown to obey: 0)()(0)()(0)()(0)()( =+⋅=−⋅=+⋅=−⋅ mppvmppupvmppump γγγγ &,,
- The Lorentz transformation (similarity) matrix S(Λ) is not difficult to construct if we set all rotations to zero, leaving us with only Lorentz boosts. Then the only generators we have are the K generators: where: + • + • + = • • =Λ 1 p p 1 p p mE mE m mE S σσσσ σσσσ σσσσ σσσσ 2 2 cosh 2 sinhˆ 2 sinhˆ 2 cosh )( ζζ ζζ 55 2017 MRT m mE m mE 22 sinh 22 cosh − = + = ζζ and which in turn are proportional to σ i ≡σσσσ. Specifically, we have: === k k kjiiii i MK σ σ εγγ 0 0 2 1 ],[ 4 00 and p≡p/|p| is a unit vector in the direction of the 3-momentum p.ˆ
- Applying S(Λ) to the independent spinor basis, we find: 56 2017 MRT ,, + − + −+ = + + + + = mE p mE pip m mE pu mE pip mE p m mE pu z yx yx z 1 0 2 )( 0 1 2 )( 21 So, because of the particular decomposition we have chosen, the u spinors correspond to electrons with positive energy (i.e., moving forward in time), while the v spinors correspond to electrons with negative energy (i.e., moving backwards in time). That being said, electrons with negative energy are known as positrons! and + − + − + = + + + + = 1 0 2 )( 0 1 2 )( 21 mE p mE pip m mE pvmE pip mE p m mE pv z yx yx z ,
- Next, we would like to describe spinors of definite spin. In many experiments, we can produce polarized beams of electrons; so it becomes important to understand how to incorporate projection operators that can select definite spin. This is not as simple as one might suspect, since the intuitive concept of spin is rooted in our notion of the rotation group (i.e.,O(3)), which is only a subgroup of the Lorentz group (i.e.,SO(1,3)). In the rest frame, however, we know that the spin of a system can be described by a three-vector s that points in a certain direction; so we may introduce the four-vector sµ which, in the rest frame, reduces to sµ =[0,s]. Then, by demanding that this transform as a four-vector, we can boost this spin vector by a Lorentz transformation Λ. Since we define s2 =1, this means that sµ 2 = −1. In the rest frame, we have pµ =[m,0] (N.B., m=mo, the rest mass) so we also have pµ sµ = 0, which must also hold in any boosted frame by Lorentz invariance. Thus, we now have two Lorentz-invariant conditions on the spin four- vector: 012 =−= µ µµ sps and 57 2017 MRT
- Now we would like to define a projection operator that selects out states of definite spin. Again, we will define the Lorentz invariant projection operator by first examining the rest frame. At rest, we know that the operator σσσσ •s serves as an operator that determines the spin of a system: where the + refers to the u spinor, and − refers to the v spinor. 2 1 )( s s •± = σσσσ P 58 2017 MRT )0()0()()0()0()( αααα vvuu −=•=• ss σσσσσσσσ and For a spin-½ system, the projection operator at rest can be written as: 2 1 )( 5 γµ + ≡sP where γ 5 = iγ 0γ 1γ 2γ 3. •− •+ =→ s s s σσσσ σσσσ 10 01 2 1 )()( PsP µ Therefore, this operator reduces to the previous one, so this is the desired expression. In the rest frame, this projection reduces to: Our goal is to write a boosted version of this expression. Let us define the projection operator :
- The new eigenfunctions now have a spin s (a quantum number) associated with them: These spinors are quite useful for practical calculations because they satisfy certain completeness relations. Any four-spinor can be written in terms of linear combinations of the four uα(0) and vα(0) because they span the space of four-spinors. If we boost these spinors with S(Λ), then uα(p) and vα(p) span the space of all four-spinors satisfying the Dirac equation. Likewise, uα T(0)vβ (0), &c. have 16 independent elements, which in turn span the entire space of 4×4 matrices. Thus, uα T(p)vβ (p), &c. span the space of all 4×4 matrices that also satisfy the Dirac equation. 0),()(),()(),(),()(),(),()( =−=−== skvsPskusPskvskvsPskuskusP and, 59 2017 MRT − We normalize our spinors with the following convention: 1),(),(1),(),( −== spvspvspuspu and With these normalizations, these spinors obey certain completeness relations: βαβαβα δ=−∑s spvspvspuspu )],(),(),(),([ For the particular representation we have chosen, we find: βα βα βα βα γγ /+−/−= /++/= 2 1 2 ),(),( 2 1 2 ),(),( 55 s m mp spvspv s m mp spuspu and They satisfy: ),()( skuku →
- If we sum over the helicity s, we have two projection operators: These projection operators satisfy: 102 =Λ+Λ=ΛΛΛ=Λ −+−+±± and, 60 2017 MRT βα βαβα +/==Λ ∑± + m mp spuspup s 2 ),(),()]([ βα βαβα +/− =−=Λ ∑± − m mp spvspvp s 2 ),(),()]([ and: Because of the completeness relations, Λ± has a simple interpretation: It projects out the positive or negative energy solution!
- So far, we have only discussed the classical theory. To ‘second quantize’ the Dirac field, we first calculate the momentum canonically conjugate to the spinor fieldψ : Quantizing the Spinor Field 61 2017 MRT since LDirac =ψ (iγ µ∂µ −m)ψ andψ =ψ †γ 0 the adjointofψ (N.B.,ψ † is the conjugate trans- pose of ψ). Let us decomposethe spinor fieldψ (and its adjointψ )into Fourier moments: †Dirac )( )( ψ ψδ δ π i x x == & L ∫ ∑= ⋅⋅− += 2,1 † 3 3 0 ]e)()(e)()([ )π2( )( α ααααψ xkixki kvkdkukb d k m x k and ∫ ∑= ⋅−⋅ += 2,1 † 3 3 0 ]e)()(e)()([ )π2( )( α ααααψ xkixki kvkdkukb d k m x k In terms of particles and antiparticles, this particular decomposition gives the following physical interpretation: = ⋅+ ⋅− positronenergypositiveCreates electronenergypositivesAnnihilate xpi xpi pvpd pupb x e)()( e)()( )( † ψ (N.B., Having d† create a positive energy positron can be viewed as annihilating a negative energy electron since if you recall, the u spinors correspond to electrons with positive energy while the v spinors correspond to electrons with negative energy).
- Repeating the steps we took for the scalar particle (c.f., Klein-Gordon Scalar Field chapter), we insert these equations and solve for the Fourier moments in terms of the fields themselves: where: ∫∫ == )()()()()()( 03†03 xUxdkbxxUdkb kk α α α α γψψγ kk , 62 2017 MRT and ∫∫ == )()()()()()( 03†03 xxVdkdxVxdkd kk ψγγψ α α α α kk , xki k xki k kv k m xVku k m xU ⋅⋅− ≡≡ e)( )π2( )(e)( )π2( )( 0 3 0 3 and Now let us insert the Fourier decomposition back into the expression for the Hamiltonian: ∫ ∑ ∫ ∫∫ = −= ∂= −== 2,1 ††3 0 0 03 33 )]()()()([ )( )( α αααα ψγψ ψπ kdkdkbkbdk id ddH x x xx LH & Here we encounter a serious problem. We find that the energy of the Hamiltonian can be negative. There is, however, an important way in which this minus sign can be banished…
- For starters, let us define the canonical equal-time commutation relations of the fields and conjugate fields with ‘anticommutators’, instead of commutators: In order to satisfy the canonical anticommutation relations, the Fourier moments must themselves obey anticommutation relations given by: jiji tt δδψψ )(}),(,),({ 3† yxyx −−−−= 63 2017 MRT )(})(,)({)(})(,)({ 3 , †3 , † kkkk ′=′′=′ ′′′′ −−−−−−−− δδδδ αααααααα kdkdkbkb and Now, if we ‘normal order’ the Hamiltonian, we must also drop the infinite zero point energy, and hence: ∫ ∑= += 2,1 †† 0 3 )]()()()([ α αααα kdkdkbkbkdH x and ∫ ∑= += 2,1 ††3 )]()()()([ α αααα kdkdkbkbd kxP Thus, the use of anticommutation relations and normal ordering nicely solves the problem of the Hamiltonian with negative energy eigenvalues! Furthermore, the d† operators can be interpreted as creation operators for antimatter (or annihilation operators for negative energy electrons). In fact, this was Dirac’s original motivation for postulating antimatter in the first place!
- To see how this interpretation of these new states emerges, we first notice that the Dirac Lagrangian is invariant under : Therefore, there should be a conserved current associated with this symmetry. A direct application of Noether’s theorem yields: λλ ψψψψ ii − →→ ee and 64 2017 MRT 0=∂= µ µ µµ ψγψ JJ and which is conserved, if we use the Dirac equation. Classically, the conserved charge is positive definite since it is proportional to ψ †ψ . This was, in fact, an improvement over the classical Klein-Gordon equation, where the charge could be negative. However, once we quantize the system, the Dirac charge can also be negative. The quantized charge associated with this current is given by: ∫ ∑∫∫ = −=== 2,1 ††3†303 ])()()()([:: α ααααψψ kdkdkbkbddJdQ xxx This quantity can be negative, and hence cannot be associated with the probability density. However, we can, as in the Klein-Gordon case, interpret this as the current associated with the coupling to electromagnetism; so Q corresponds to the electric charge. In this case, the minus sign in Q is a desirable feature, because it means that d† is the creation operator of antimatter, that is, a positron with opposite charge to the electron.
- Again, this also means that we have to abandon the simple-minded interpretation of ψ as a single-electron wave function, since it now describes both the electron and the antielectron. with only one particle in any given quantum state. This is the first example of the spin- statistics theorem, that field theories defined with integer spin are quantized with commutators are called bosons, while theories with half-integer spins are quantized with anticommutators and are called fermions. 0)()( 1 † 1 † j M j i N i kbkd ∏∏ == αα 65 2017 MRT Now, the anticommutation relations also reproduce the Pauli Exclusion Principle found in quantum mechanics. Because dα †(k)dα †(k)= 0, only one particle can occupy a distinct energy state with definite spin. Thus, a multiparticle state is given by: The existence of two types of statistics, one based on commutators (i.e., Bose- Einstein statistics) and one based on anticommutators (i.e., Fermi-Dirac statistics), has been experimentally observed in a wide variety of physical situations and has been applied to explain the behavior of low-temperature systems and even which dwarf stars.
- Repeating the same steps that we used for the Klein-Gordon field (c.f., Klein-Gordon Scalar Field chapter), we can also compute the energy-momentum tensor, Tµν, and the angular momentum tensor, Mλµν, from Noether’s theorem: The conserved angular momentum tensor is therefore: ψγψψγψψγψ νµµννµλνµµνλνµλνµνµ Σ−∂−∂= Σ−=∂= 22 i xxi i JiiT Mand 66 2017 MRT ∫∫ Σ−∂−∂== ψψ νµµννµνµνµ 2 †303 i xxiddM xx M The crucial difference between these equations and the Klein-Gordon case is that the angular momentum tensor contains an extra piece, proportional to Σµν, which represents the fact that the theory has nontrivial spin-½. We now calculate how a quantized spinor field transforms under the Poincaré group: ψψψψ νµµννµνµ µµ Σ−∂−∂=∂= 2 ],[],[ i xxMiPi and With these operator identities, we can confirm that the quantized spinor field transforms as a spin-½ field under the Poincaré group: )()]([),(),( 11 axSaUaU +ΛΛ=ΛΛ −− ββαα ψψ which we have written in PART VI – GROUP THEORY as U(Λ,a)Ψα(x)U−1(Λ,a)= D[Λ−1]α β Ψβ(Λx+a). This closes the loop on the S(Λ) representation as being D[Λ].
- As we mentioned earlier, one of the fundamental assumptions about quantum field theory is that it must be causal. Not surprisingly, the spin-statistics theorem is also intimately tied to the question of microcausality, that is, that no signal can propagate faster than the speed of light. From our field theory perspective, microcausality can be interpreted to mean that the commutator (anticommutator) of two boson (fermion) fields vanishes for spacelike separations: To demonstrate the spin-statistics theorem, let us quantize bosons with anticommutators and arrive at a contradiction. Repeating earlier steps, we find, for large separations: ( )0)(0])(,)([ 2 <−= yxyx forφφ 67 2017 MRT 2002 )( )()( 23 3 )( e ~ ]ee[ ω2)π2( 0)}(,)({0 2002 yx d yx yxm yxkiyxki −− += −−− −⋅−⋅− ∫ yx k yx k −−−− −−−− φφ which clearly violates our original assumption of microcausality. and ( )0)(0])(,)([ 2 <−= yxyx forψψ
- Historically, anticommutators and antimatter were introduced by Dirac, who was troubled that his theory seemed riddled with negative energy states. Since physical sys- tems prefer the state of lowest energy, there is a finite probability that all the electrons in nature would decay into these negative energy states, thereby creating a catastrophe. To solve the problem of negative energy states, Dirac was led to postulate a radically new interpretation of the vacuum. He postulated that the vacuum consisted of an infinite sea of filled negative energy states. Ordinary matter did not suddenly radiate an infinite amount of energy and cascade down to negative energy because the negative energy sea was completely filled. By the anticommutation relations, only one electron can occu- py a negative energy state at a time; so an electron could not decay into the negative energy sea if it was already filled. In this way, electrons of positive energy could not cascade in energy down to negative energy states. However, once in a while an electron may be knocked out of the negative energy sea, creating a hole. This hole would act as if it were a particle. Dirac noticed that the absence of an electron of charge −|e| and negative energy −E is equivalent to the presence of a particle of positive charge +|e| and positive energy +E. This hole then has positive charge and the same mass as the elec- tron. All particles therefore had positive energy: Both the original positive energy electron as well as the absence of a negative energy electron possessed charge e (see Figure). 68 2017 MRT γ − e 2 mc 2 mc−
- The interpretation that we have chosen earlier (i.e., that of negative electrons going backwards in time are equivalent to positive energy antielectrons going forward in time) is equivalent to Dirac’s infinite negative energy sea. In fact, when we subtract off an infinite constant in the Hamiltonian, this can be interpreted as subtracting off the energy of Dirac’s infinite sea of negative energy states. To see how these positive and negative energy states move in time, let us define the evolution of a wave function via a source as follows: )()()( xJxmi =−∂ ψγ µ µ 69 2017 MRT To solve this inhomogeneousdifferential equation,we introduce the Dirac propagatorSF by: )()()( 4 yxyxSmi F −=−−∂ δγ µ µ Then the solution to the inhomogeneous wave equation is given by: ∫ −+= )()()()( 4 0 yJyxSydxx Fψψ where ψ0 solves the homogeneous Dirac equation (i.e., (iγ µ∂µ −m)ψ0(x)= 0). An explicit representation of the Dirac propagator can be obtained by using the Fourier transform: where ∆F (x−y) is the Klein-Gordon propagator. which satisfies: ∫ +− + =− −⋅− ε γ µ µ imp mppd yxS yxpi F 22 )( 4 4 e )π2( )( )()()( yxmiyxS FF −∆−∂=− µ µ γ
- As before, we can solve for the propagator by integrating over k0. The integration is identical to the one found earlier for the Klein-Gordon equation, except now we have additional γ µpµ + m factor in the numerator. Integrating out the energy, we can write the Green function in terms of plane waves. The result is almost identical to the expression found for the Klein-Gordon propagator, except for the insertion of γ matrices: where: 70 2017 MRT ∫ ∑∑ ∫ ′−′+′′−−= −′Λ+′−Λ−=′− == ′−⋅ − ′−⋅− + 4 3 2 1 3 )()( 3 3 )()()()()()( )](e)()(e)([ )π2( )( r pp r pp xxpixxpi F xxttixxttid ttpttp d E m ixxS αααα ψψθψψθ θθ p p xpir p w E m x ⋅− ≡ α ε αψ e)( )π2( 1 )( 23 p with ε α =(1,1,−1,−1) and w1 =u1, w2 =u2, w3 =v1, and w4 =v2. Written in this fashion, the states with positive energy propagate forward in time, while the states with negative energy propagate backwards in time. 0)]()([0)]([ yxTyxSi F βαβα ψψ=− As in the Klein-Gordon case, we can now replace the plane waves ψp with the quantized spinor ψ (x) by taking the vacuum expectation value of the spinor fields: which is one of the most important results of this chapter.
- For example, we can take an imaginary representation of the γ matrices, which gives the Majorana spinors (c.f., Appendix III: Dirac Matrices). For our purpose, what is more interesting is taking the Weyl representation, which gives us a representation of neutrinos. If we take the representation: Weyl Neutrinos 71 2017 MRT − = − = − − = 10 01 0 0 01 10 5 0 γ σ σ γγ and, i i i then in this representation, we can write down two chiral operators: = + ≡ 00 01 2 1 5γ RP and = − ≡ 10 00 2 1 5γ LP The spinor representation of the Lorentz group is reducible if we introduce projection operators PL and PR. Normally, we are not concerned with this because this spinorial representation is irreducible under the full Poincaré group for massive states. However, there is a situation when the spinorial representation becomes reducible even under the Poincaré group, and that is when the fermion is massless.
- To see how these projection operators affect the electron field, let us split ψ as follows: Then we have: = L R ψ ψ ψ 72 2017 MRT = − = + L R ψ ψ γψ ψ γ 0 2 1 02 1 55 and Because PL and PR commute with the Lorentz generators (i.e., [PL,R,Σµν ]=0) it means that the four-component spinor ψ actually spans a reducible representation of the Lorentz group. Contained within the complex Dirac representation are tow distinct chiral representations of the Lorentz group. Although ψL and ψR separately form an irreducible representation of the Lorentz group, they do not form a representation of the Poincaré group for massive particles. To find an irreducible representation of the Poincaré group, we must impose m=0! The reason that these chiral fermions must be massless is because the mass term mψψ in the Lagrangian is not invariant under these two separate Lorentz transformations. Because ψψ =ψLψR +ψRψL mass terms in the action necessarily mix these two distinct representations of the Lorentz group. Thus, this representation forces us to have massless fermions; that is, this is a theory of massless neutrinos. − − − −
- The problem facing us now is to write down the action for the Dirac theory coupled to the Maxwell theory, creating the quantum theory of electrodynamics, the subject of the rest of this work. The most convenient way is to use the electron current as the source for the Maxwell field. The electron current is given by J µ =ψ γ µψ (with ∂µ J µ =0 ‒ c.f., Quantizing the Spinor Field chapter), and hence we propose the coupling: Relativistic Quantum Mechanics 73 2017 MRT where e is the charge. We saw earlier that this current emerged because of the invariance of the Dirac action under the symmetryψ ′→exp(iλ)ψ. Let us promote λ into a local gauge parameter, so that λ is a function of space-time, λ(x). We want an action invariant under: ψγψ µ µAe − )()(e)( )( xAAxx xei λψψ µµµ λ ∂−→→ and The problem with this transformation is that λ(x) is a function of space-time; therefore, the derivative of a spinor ∂µψ is not a covariant object. It picks up an extraneous term ∂µdλ(x) in its transformation. To eliminate this extraneous term, we introduce the covariant derivative: µµµµ AeiD +∂≡→∂ The advantage of introducing the covariant derivative is that it transforms covariantly under a gauge transformation: )(e)()()(e)( )()( xDxeixeixDxD xeixei ψλλψψ µ λ µ λ µ →−+→
- 2017 MRT which we obtain by simply replacing ∂µ by Dµ . 74 This means that the following action is gauge invariant: νµ νµµ µνµ νµµ µ ψγψψγψ FFmDiFFmi 4 1 )( 4 1 )(EMDiracQED −−→−−∂=+= LLL The coupling of the electron to the Maxwell field reproduces the coupling proposed earlier. In the limit of velocities small compared to the speed of light, the Dirac equation should reduce to a modified version of the Schrödinger equation. We are hence interested in checking the correctness of the Dirac equation to lowest order, to see if we can reproduce the nonrelativistic results and corrections to them. The Dirac equation of motion, in the presence of an electromagnetic potential, now becomes: To find solutions of this equation, let us decompose the four-spinor ψ into two smaller two-spinors φ and χ: ψβ ψ ])([ 0 Aemei t i ++−•= ∂ ∂ A−−−−∇∇∇∇αααα Ψ Φ = ≡ − − tmi tmi e e χ φ ψ Then the Dirac equation can be decomposed as the sum of two two-spinor equations: χχφ χ φφχ φ mAe t imAe t i −+•= ∂ ∂ ++•= ∂ ∂ 00 )()( ππππσσσσππππσσσσ and where ππππ = p−eA is the canonical momentum and σσσσ represents the Pauli matrices.
- Next, we will eliminate χ. For small fields, we can make the (first-order) approximation that e A0 <<2m. Then we can solve the second equation as: In this approximation, the Dirac equation can be expressed as a Schrödinger-like equation, but with crucial spin-dependent corrections (i.e., the (e/2m)σσσσ •B term): 75 2017 MRT φφχ << • m2 ~ ππππσσσσ Φ +•−=Φ + • = ∂ Φ∂ 0 2 0 2 22 )( 2 )( Ae m e m e Ae mt i B Ap σσσσ −−−−ππππσσσσ where we have used the fact that: B•−=• σσσσππππππππσσσσ e22 )( The previous equation gives the first corrections to the Schrödinger equation in the presence of a magnetic and electric field.
- Classically, we know that the energy of a magnetic dipole in a magnetic field is given by the dot product of the magnetic moment with the field: the magnetic moment of the electron (with rest mass m=me) is therefore: σσσσ 2 h =S 76 2017 MRT SS B cm e cm e µ2 2 2 2 ee = = ≡ σσσσµµµµ h Thus, the Dirac theory predicted that the electron should have a magnetic moment twice what one might normally expect, that is, twice the Bohr magneton µB. This was perhaps the first major success of the Dirac theory of the electron. Historically, it gave confidence to physicists that the Dirac theory was correct, even if it seemed to have problems with negative energy states. From the Stern-Gerlach experiment (1922), we know that (N.B., this is the meaning of the ubiquitous and casual use of the term ‘spin-½’): BBB •−=•−=•−= σσσσσσσσµµµµ cm e cm he E 2π4 h with the velocity of light c, the Planck quantum constant h, and the Dirac h-bar constant: restored. π2 h ≡h
- Yet another classic result that gave credibility to this relativistic formulation was the splitting of the spectral lines of the hydrogen atom. in Gaussian (CGS) and MKS units, respectively, is the fine structure constant, and: 04.137 1 π4 2 ≅= c e h α 77 2017 MRT ψψ α β ψ ψ H r Z cmci t iE = −+•−= ∂ ∂ = 2 e∇∇∇∇ααααhh with Z the atomic number (i.e., the number of protons) and where: −−• •− = r Z cm r Z cm H α α 2 e 2 e p p σσσσ σσσσ To solve for corrections to the Schrödinger atom, we set B to zero and take A0 to be the Coulomb potential. The Dirac equation is now written as: using Heaviside-Lorentz units or: c e c e hh o 22 επ4 == αα and is the Hamiltonian for a bound electron in a relativistically correct description of the hygrogen atom.
- In hindsight, it turn out to be convenient to introduce a judicious ansatz (i.e., an educated guess that is verified later by its results) for the solution of the Dirac equation. First, as in the Schrödinger case, we want to separate variables, so that we take: where the radial function is explicitly separated off. Second, because ∇2 contains the (Casimir) operator L2 in the usual Schrödinger formalism, we choose Y(θ,ϕ) to be the standard spherical harmonics (i.e., eigenfunctions of the angular momentum operators). For the spinning case, however, what appears is: ),()(~ ϕθψ Yrf 78 2017 MRT 2 σσσσ ++++++++ LSLJ == where J is the total spin, that is, the combination of the orbital spin L and intrinsic spin S. Thus, our eigenfunctions for the Dirac case must be labeled by eigenvalues j, l, and ml.
- Based on these arguments, we choose as our ansatz (i.e., an educated guess!): Inserting this ansatz into the Dirac equation above and factoring out the angular part, we find that the radial part of our eigenfunctions obey: r rF j rd rFd rG r Z mE jj j )( 2 1)( )(e ll l m +−= +− α 79 2017 MRT • = ll ll l l l l mj j mj j mj r rF r rG i φ φ ψ )ˆ( )( )( rσσσσ r rG j rd rGd rF r Z mE jj j )( 2 1)( )(e ll l m += ++ α and where we use the (E+… ((E−…) sign for j=l +½ ( j=l −½). For the interested reader who would consult the References, you will see that by power expanding this equation in r, this series of equations can be solved (although with quite some amount of difficulty unless you’re a genius like Lev Landau) in terms of the confluent hypergeo- metric function 1F1(a,b;z) (or Kummer’s function of the first kind).
- These equations, when power expanded, give us the energy eigenstates of the hydrogen atom (N.B., recall that the fine structure constant is given by α =e2/4π εohc): Experimentally, this formula correctly gave spin-dependentcorrections to the Bohr formula (i.e., En =−½mec2Z2α 2/n2) to lowest order. In contrast to the usual Bohr formula, the energy is now a function of both the principal quantum number n as well as the total spin j. By power expanding, we can recover the usual nonrelativistic result to lowest order: 21 2 222 2 e )½()½( 1 − −+++− += α α Zjjn Z cmE jn 80 2017 MRT + − + −−= L 4 3 ½ )( 2 1 2 1 1 4 222 2 22 2 e j n n Z n Z cmE jn αα So far, we have only considered the Dirac electron interacting with a classical Coulomb potential. Therefore, it is not surprising that this formula neglects smaller quantum corrections in the hydrogen energy levels. In particular, the 2S½ and 2P½ levels have the same n and j values, so they should be degenerate according to the Dirac formula. However, experimentally these two levels are found to be split by a smaller amount, called the Lamb shift. It was not until 1949, with the correct formulation of the quantum theory of electrodynamics (QED), that one could successfully calculate these small quantum corrections to the Dirac energy levels. The calculation of these quantum corrections in QED was one of the finest achievements of quantum field theory.
- Because of gauge invariance, there are also complications when we quantize the theory. A naïve quantization of the Maxwell theory fails for a simple reason: the propagator does not exist! To see this, let us write down the action in the following form: Quantizing the Maxwell Field 81 2017 MRT where: ν νµ µ APA 2 2 1 ∂=L 2 ∂ ∂∂ −= νµ νµνµ ηP The problem with this propagator is that it is not invertible, and hence we cannot construct a propagator for the theory. In fact, this is typical of any gauge theory, not just Maxwell’s theory. The origin of the noninvertibility of this operator is because Pµν is a projection operator (i.e., its square is equal to itself, Pµν Pνλ =P λ µ, and it projects out longitudinal states, ∂µPµν =0). The fact that Pµν is a projection operator goes to the heart of why Maxwell’s theory is a gauge theory. This projection operator projects out any states with the form ∂µλ, which is just the statement of gauge invariance.
- The solution to this problem is that we must break this invariance by choosing a gauge. In our case, choosing a gauge means that we are making a particular choice of the scalar and vector potentials. Because we have the freedom to add ∂µχ to Aµ, we will choose a specific value of χ which will break gauge invariance. We are guaranteed this degree of freedom if the variation δ Aµ =∂µχ can be inverted. There are several ways in which we can fix the gauge and remove this infinite redundancy. We can, for example, place constraints directly on the gauge field Aµ, or we may add the following term to the action: where ξ is arbitrary. We list some common gauges: 2 )( 2 1 µ µ ξ A∂− 82 2017 MRT ∞= = = =∂ = = =∇ ξ ξ ξ µ µ :gaugeUnitary :gaugeFeynman :gaugeLandau :gaugeLandau :gaugeTemporal :gaugeAxial :gaugeCoulomb 1 0 0 0 0 0 0 3 A A A Aii Notice that there are two equivalent ways to represent the Landau gauge.
- In the Coulomb gauge, one extracts out the longitudinal modes of the field from the very start. To show that the gauge degrees of freedom allows us to make this choice, we write down: Solving for χ, we find: 0)( =•=′• χ∇∇∇∇++++∇∇∇∇∇∇∇∇ AA 83 2017 MRT ∫ ′• ′ ′−= • ∇ −= )( π4 1 1 3 2 xA xx x A ∇∇∇∇ −−−− ∇∇∇∇ d χ Likewise, the Landau gauge choice means that we can find a χ such that: 2 ∂ ∂ −= µ µ χ A
- To begin the process of canonical quantization, we will take the Coulomb gauge in which only the physical states are allowed to propagate. Let us first calculate the canonical conjugate to the various fields. Since A0 does not occur in the Lagrangian, this means that A0 does not appear to propagate, which is a sign that there are redundant modes in the action. The other modes, however, have canonical conjugates: We write the Lagrangian as: 2 0 2 EM 4 1 4 1 jiiii FFEE −−−=L 84 2017 MRT i i i i i EAA AA =∂−−==== 0 0 0 0 & && δ δ π δ δ π LL and 22 0 2 EM 4 1 2 1 4 1 jii FFF −=−= νµL where Fµν 2=Fµν Fµν, &c. We introduce the independent Ei field via a trick. We rewrite the action as: By eliminating Ei by its equation of motion, we find Ei =−F0i . By inserting this value back into the Lagrangian, we find the original Lagrangian back again. In vector form, with Aµ= [A0,A] and Fµν = ∂µ Aν ‒∂ν Aµ, this equation reads: 22 0EM 4 1 2 1 jiFA −−•−•−= EEAE ∇∇∇∇&L
- Then we write the Lagrangian for QED as: where A0 is a Lagrange multiplier. If we solve for the equation of motion of this field, we find that there is the additional constraint: 85 2017 MRT ψγψ ψγψ µ µ µ µ νµ )( 4 1 2 1 )( 4 1 22 0 2 QED mDiFA mDiF ji −+−−•−•−= −+−= EEAE ∇∇∇∇& L ψγψρ 0e==•E∇∇∇∇:LawGauss’ Thus, Gauss’ Law emerges only after solving for the equation of motion of A0. If we now count the independent degrees of freedom, we find we have only two degrees left, which correspond to the two independent transverse helicity states. Of the original four components of Aµ, we see that A0 can be eliminated by its equation of motion, and that we can gauge away the longitudinal mode, therefore leaving us with 4−2=2 degrees of freedom which means that massless states have two helicity states (N.B., these correspond to a state where the Pauli-Lubanski tensor Wλ =½ε λµνσ Mµν Pσ ‒ c.f., PART VI – GROUP THEORY ‒ is aligned parallel to the momentum vector and also aligned antiparallel to the momentum vector). )(),()( 2 1 0 0 22 ψγψψ eA −•−++−•−= EABEAE ∇∇∇∇LL & or:
- Intuitively, this means that a photon moving in the z-direction can vibrate in the x- and y-direction, but not in the z-direction or the timelike direction. This corresponds to the intuitive understanding of transverse photons. In the Coulomb gauge, we can reduce all fields to their transverse components by eliminating their longitudinal components. Let us separate out the transverse (⊥) and longitudinal (||) parts as follows: where: ||EEE ++++⊥= 86 2017 MRT ρ2 1 ∇ = ∇∇∇∇||E We see also that ∇×∇×∇×∇×E|| =0. If we insert this back into the Lagrangian, then we find that all longitudinal contributions cancel, leaving only the transverse parts, except for the piece: ∫= ∇ = yx yyxx yxE || −−−− ),(),(),(),( π8 1 2 1 2 1 †† 33 2 2 2 tttt dd e ψψψψ ρρ The last term is called the instantaneous four-fermion Coulomb term, which seems to violate special relativity since this interaction travels instantly across space. How, we shall show at the end of the chapter that this term precisely cancels against another term in the propagator, so Lorentz symmetry is restored. Let us now solve for E|| in terms of ρ. Then we have: ρ=•=• ⊥ ||0 EE ∇∇∇∇∇∇∇∇ and
- If we impose canonical commutation relations, we find a further complication. Naïvely, we might want to impose [Ai(x,t),πj (y,t)]=−iδijδ 3(x−−−−y). However, this cannot be correct because we can take the divergence of both sides of the equation. The divergence of Ai is zero, so the left-hand side is zero, but the right-hand side is not. As a result, we must modify the canonical commutation relations as follows: where the right-hand side must be transverse; that is: 87 2017 MRT )( ~ )],(,),([ yxyx −−−−jiji ittA δπ −= ∫ −= ′• 2 )( 3 3 e )π2( ~ k k xxk ji ji i ji kkd δδ −−−− As before, our next job is to decompose the Maxwell field in terms of its Fourier modes, and then show that they satisfy the commutation relations. However, we must be careful to maintain the transversality condition, which imposes a constraint on the polarization vector εεεε α(k). The decomposition is given by: ∫ ∑= ⋅⋅− += 2 1 † 0 3 3 ]e)(e)([)( 2)π2( )( α ααα xkixki kakak k d x εεεε k A In order to reserve the condition that A is transverse, we take the divergence of this equation and set it to zero. This means that we must impose: βαβαα δ=•=• εεεεεεεεεεεε and0k
- By inverting these last relations, we can solve for the Fourier moments in terms of the fields: In order to satisfy the canonical commutation relation among the fields, we must impose the following commutation relations among the Fourier moments: 88 2017 MRT and ∫ •∂= ⋅ )()(e 2)π2( )( 0 0 3 3 xk k d ika xki A k αα εεεε t ∫ •∂−= ⋅− )()(e 2)π2( )( 0 0 3 3 † xk k d ika xki A k αα εεεε t )(])(,)([ 3† kk ′=′ −−−−δδ βαβα kaka Recall the operator: BABABA )(∂−∂≡∂ t
- Let us now insert this Fourier decomposition into the expression for the energy and momentum: After normal ordering, once again the energy is positive definite. ∫ ∑= ′−⋅− ⊥ + =′=′− 2 1 2 )( 4 4 )()( e )π2( 0)]()([0)]([ α α ν α µνµνµ εε kk εik kd ixAxATxxDi xxki F 89 2017 MRT ∫ ∑∫= =+= 2 1 †3223 ])()([ω)( 2 1 α αα kakaddH kBEx and ∫ ∑∫ = == 2 1 †33 )()(:)(: α αα kakadd kkBExP ×××× This expression is not Lorentz invariant since we are dealing with transverse states. This is a bit troubling, until we realize that Green functions are off-shell objects and are not measurable. However, these Lorentz violating terms should vanish in the full scattering matrix. Finally, we wish to calculate the propagator for the theory. Again, there is a complication because the field is transverse. The simplest way to construct the propagator is to write down the time-ordered vacuum expectation value of two fields. The calculation is almost identical to the one for scalar and spinor fields, except we have the polarization tensor to insert:
- To see this explicitly, let us choose a new orthogonal basis of four vectors, given by εµ 1(k), εµ 2(k), kµ, and a new vector eµ =[1,0,0,0]. Any tensor can be power expanded in terms of this new basis. Therefore, the sum over polarization vectors appearing in the propagator can always be expanded in terms of the tensors gµν, eµ eν, kµ kν , and kµ eν. We can calculate the coefficient of this expansion by demanding that both sides of the equation be transverse. Then we get: Fortunately, the noninvariant terms involving e can all be dropped. The terms proportional to kµ vanish when inserted into a scattering amplitude. This is because the propagator couples to two currents, which in turn are conserved by gauge invariance. 22 2 2222 2 1 )()( ))(( )( )()( kek eek kek kkkkek kek kk kk −⋅ − −⋅ +⋅ + −⋅ −−=∑= νµµννµνµ νµ α α ν α µ ηεε 90 2017 MRT If we drop terms proportional to kµ in the propagator, we are left with: xx ′− ′− −=′−∆−=′−⊥ π4 )( )0;()]([ tt eemxxxxD FF δ η νµνµνµ The first term is what we want since it is covariant. The second term proportional to eµ eν is called the instantaneous Coulomb term. In any Feynman diagram, it occurs between two currents, creating (ψ †ψ )∇−2(ψ †ψ ). This term precisely cancels the Coulomb term found in the Hamiltonian when we solved for E|| above.
- The purpose of this Appendix is to derive the basic Hamiltonian of electrodynamics sub- ject to the requirement that the three-vector potential A satisfies the transversality con- dition. We shall first show that it is always possible to consider a gauge transformation of type Aµ → A′µ + ∂µ χ such that in the new gauge the transversality condition (e.g., Fourier transforming this to ik•A(k,t)=0,the transversalitycondition for a wave with wavevectork): 91 2017 MRT ),(),(),( ),(1 ),(),( oldnewold 0 new 0 ttAtA t t c tAtA xxx x xx χ χ ∇∇∇∇++++= ∂ ∂ −= and is satisfied (i.e., the vector potential is a transverse vector function). But for now, we set: Appendix I: Radiation Gauge 0),( =• txA∇∇∇∇ where: ∫ ′ ′•′ ′= xx xA xx −−−− ∇∇∇∇ π4 ),( ),( old 3 t dtχ We then have: 0)( oldnew =•=• χ∇∇∇∇++++∇∇∇∇∇∇∇∇ AA which shows that the transversality condition is satisfied for the new vector potential. On the other hand, from Aµ −∂µ (∂ν Aν )= J µ (c.f., Maxwell’s Equation chapter), where J µ is the charge-current density, which holds quite generally, we get for the A0 component: ρ−= ∂ ∂ +• ∂ ∂ + ∂ ∂ −∇ t A ctct A c A 0 2 0 2 20 2 111 A∇∇∇∇ ~
- When the three-vector potential satisfies the transversality condition, it is seen that A0 is determined by Poisson’s equation: 92 2017 MRT ),(),(0 2 ttA xx ρ−=∇ which does not contain any time derivative. The solution to this differential equation satisfying the boundary condition A0 →0 at infinity is: ∫ ′ ′ ′= xx x xx −−−− ),( π4 1 ),( 3 0 t dtA ρ We wish to emphasize that A0 in this last relation is an instantaneous Coulomb potential in the sense that the value of A0 at t is determined by the charge distribution at the same instant of time; in particular, it should not be confused with the retarded Coulomb potential: ∫ ′ −=′ ′ ′ ′= ctt t dtA xxxx x xx −−−−−−−− ),( π4 1 ),( 3 0 ρ which we obtain by solving the d’Alembertian equation A0 =0 for A0. To summarize what we have done so far: 1) we can always choose the vector potential so that the transversality condition ∇∇∇∇•A=0 is satisfied; 2) if A satisfies the transversality condition, the A0 is given by the instantaneous Coulomb potential integral above. The gauge in which A satisfies ∇∇∇∇•A=0 and A0 is given by the integral above is called the radiation (or Coulomb) gauge.
- The total Hamiltonian for the charged particles and the electromagnetic field is given by the space integral of: 93 2017 MRT nInteractioMatterEM HHHH ++= From the Hamiltonian density of the electromagnetic field, we have, from (c.f., Maxwell’s Equation chapter): )()( 2 1 )( 2 1 )( )( 00 22 0 22 EM0 0 EM EM EBE EBE AA AA A •+−−= •−−=−∂ ∂ = ∇∇∇∇ ∇∇∇∇ ρ δ δ µ µ L L H where the last term of the last line is of no significance since is vanishes when we evaluate H=∫d3x HEM. As for the interaction part of the Hamiltonian density, we write is as: µ µ JA c 1 nInteractio −=H Combining these last two equations yields: ∫∫∫ •−−=+=+ JAxBExx 3223 nInteractioEM 3 nInteractioEM 1 )( 2 1 )( d c ddHH HH At first sight this appears to be a very strange result because of the ρA0 term has completely disappeared! We shall come back to this puzzle in a moment…
- Our next step is to show that this last Hamiltonian (i.e., HEM +HInteraction) can be written in a somewhat different form when we work in the radiation gauge. First, we recall that, quite generally, the electric field E is derivable from: 94 2017 MRT tc A ∂ ∂ −= A E 1 0 −−−−∇∇∇∇ We claim that in the radiation gauge the irrotational and solenoidal part of E, defined by: ⊥= EEE ++++|| 0EE ==• ⊥ ||0 ××××∇∇∇∇∇∇∇∇ and where: are uniquely given by: yc A ∂ ∂ −=−= ⊥ ⊥ A EE 1 0|| and∇∇∇∇ where the subscript ⊥ is to remind us of the transversality condition ∇∇∇∇•A=0. This can be proved by noting: 1) the vector identity ∇∇∇∇××××(∇∇∇∇A0) vanishes; 2) the divergence of ∂A⊥/∂t vanishes because of the transversality condition ∇∇∇∇•A=0; and 3) the decomposition of E into E|| and E⊥ is unique.
- Meanwhile, we can show that the space integral of |E|2 is actually given by: 95 2017 MRT ∫∫∫ += ⊥ 0 32323 Addd ρxExEx in the radiation gauge. ∫ ∫∫∫∫∫ += =•+=+•−= ⊥ ⊥⊥⊥⊥ )( )(2])(2[ 2 || 23 2 || 3 0 3232 ||0 2323 EEx ExExExEEExEx d dAddAdd ∇∇∇∇∇∇∇∇ ∫ ∫∫∫∫ = ∇−•=•= 0 3 0 2 0 3 00 3 00 32 || 3 )()()( Ad AAdAAdAAdd ρx xxxEx ∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇ Furthermore: Combining these last two results, we see that the space integral of |E|2 is indeed given by ∫d3x|E|2 =∫d3x|E⊥ |2 +∫d3xρA0 . To prove this, we first note:
- Coming back to the expression for the Hamiltonian given by HEM +HInteraction above, we note that the ρA0-term which, we thought, had disappeared is actually buried in the space integral of |E|2: 96 2017 MRT ∫ ∫∫ ′ ′ ′= xx xx xxx −−−−π4 ),(),( 2 1 2 1 33 0 3 tt ddAd ρρ ρ Note, however, that the ρA0-term now appears with a factor ½. ∫∫ •−−−=+ ⊥⊥ AJxBEx c AddHH 1 2 1 )( 2 1 0 3223 nInteractioEM ρ We are finally in a position to eliminate A0 altogether (or equivalently E||) from the Hamiltonian. Using the instantaneous Coulomb potential integral for A0(x,t) above, we can write the ρA0-term of HEM +HInteraction above as: Hence: ∫ ∫∫∫ ′ ′ ′+•− ∂ ∂ =+ ⊥⊥ ⊥ xx xx xxAJxA A x −−−− ××××∇∇∇∇−−−− π4 ),(),( 2 111 2 1 3332 2 3 nInteractioEM tt ddd ctc dHH ρρ or, in terms of A⊥, J and ρ only: ∫∫ ∫∫ ⊥⊥ •− ′ ′ ′−−=+ AJx xx xx xxBEx 333223 nInteractioEM 1 π4 ),(),( 2 1 )( 2 1 d c tt dddHH −−−− ρρ In this form the electrostatic potential A0 (or irrotational part of E) has been completely eliminated in favor of the instantaneous Coulomb interaction.

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