Download pdf - Relativistic Quantum Mechanics - TCM Groupbds10/aqp/handout_relqu.pdfAdvanced Quantum Physics. 168 ... urally from a relativistic treatment of quantum mechanics. ... In non-relativistic

Chapter 15

Relativistic QuantumMechanics

The aim of this chapter is to introduce and explore some of the simplest aspectsof relativistic quantum mechanics. Out of this analysis will emerge the Klein-Gordon and Dirac equations, and the concept of quantum mechanical spin.This introduction prepares the way for the construction of relativistic quantumfield theories, aspects touched upon in our study of the quantum mechanicsof the EM field. To prepare our discussion, we begin first with a survey ofthe motivations to seek a relativistic formulation of quantum mechanics, andsome revision of the special theory of relativity.

Why study relativistic quantum mechanics? Firstly, there are many ex-perimental phenomena which cannot be explained or understood within thepurely non-relativistic domain. Secondly, aesthetically and intellectually itwould be profoundly unsatisfactory if relativity and quantum mechanics couldnot be united. Finally there are theoretical reasons why one would expect newphenomena to appear at relativistic velocities.

When is a particle relativistic? Relativity impacts when the velocity ap-proaches the speed of light, c or, more intrinsically, when its energy is largecompared to its rest mass energy, mc2. For instance, protons in the accelera-tor at CERN are accelerated to energies of 300GeV (1GeV= 109eV) which isconsiderably larger than their rest mass energy, 0.94 GeV. Electrons at LEPare accelerated to even larger multiples of their energy (30GeV compared to5! 10!4GeV for their rest mass energy). In fact we do not have to appeal tosuch exotic machines to see relativistic e!ects – high resolution electron mi-croscopes use relativistic electrons. More mundanely, photons have zero restmass and always travel at the speed of light – they are never non-relativistic.

What new phenomena occur? To mention a few:

! Particle production: One of the most striking new phenomena toemerge is that of particle production – for example, the production ofelectron-positron pairs by energetic "-rays in matter. Obviously oneneeds collisions involving energies of order twice the rest mass energy ofthe electron to observe production.

Astrophysics presents us with several examples of pair production. Neu-trinos have provided some of the most interesting data on the 1987 su-pernova. They are believed to be massless, and hence inherently rela-tivistic; moreover the method of their production is the annihilation ofelectron-positron pairs in the hot plasma at the core of the supernova.High temperatures, of the order of 1012K are also inferred to exist in thenuclei of some galaxies (i.e. kBT " 2mc2). Thus electrons and positrons

Advanced Quantum Physics

168

Figure 15.1: Anderson’s cloud chamber picture of cosmic radiation from 1932 show-ing for the first time the existence of the positron. A cloud chamber contains a gassupersaturated with water vapour (left). In the presence of a charged particle (suchas the positron), the water vapour condenses into droplets – these droplets mark outthe path of the particle. In the picture a charged particle is seen entering from thebottom at high energy. It then looses some of the energy in passing through the 6mmthick lead plate in the middle. The cloud chamber is placed in a magnetic field andfrom the curvature of the track one can deduce that it is a positively charged particle.From the energy loss in the lead and the length of the tracks after passing though thelead, an upper limit of the mass of the particle can be made. In this case Andersondeduces that the mass is less that two times the mass of the electron. Carl Anderson(right) won the 1936 Nobel Prize for Physics for this discovery. (The cloud chambertrack is taken from C. D. Anderson, The positive electron, Phys. Rev. 43, 491 (1933).

are produced in thermal equilibrium like photons in a black-body cavity.Again a relativistic analysis is required.

! Vacuum instability: Neglecting relativistic e!ects, we have shown thatthe binding energy of the innermost electronic state of a nucleus of chargeZ is given by,

E = #!

Ze2

4#$0

"2m

2!2.

If such a nucleus is created without electrons around it, a peculiar phe-nomenon occurs if |E| > 2mc2. In that case, the total change in energyof producing an electron-positron pair, subsequently binding the elec-tron in the lowest state and letting the positron escape to infinity (itis repelled by the nucleus), is negative. There is an instability! Theattractive electrostatic energy of binding the electron pays the price ofproducing the pair. Nuclei with very high atomic mass spontaneously“screen” themselves by polarising the vacuum via electron-positron pro-duction until the they lower their charge below a critical value Zc. Thisimplies that objects with a charge greater than Zc are unobservable dueto screening.

! Info. An estimate based on the non-relativistic formula above gives Zc $270. Taking into account relativistic e!ects, the result is renormalised down-wards to 137, while taking into account the finite size of the nucleus one finallyobtains Zc % 165. Of course, no such nuclei exist in nature, but they can bemanufactured, fleetingly, in uranium ion collisions where Z = 2 ! 92 = 184.Indeed, the production rate of positrons escaping from the nucleus is seen toincrease dramatically as the total Z of the pair of ions passes 160.

! Spin: Finally, while the phenomenon of electron spin has to be grafted


169

artificially onto the non-relativistic Schrodinger equation, it emerges nat-urally from a relativistic treatment of quantum mechanics.

When do we expect relativity to intrude into quantum mechanics? Accord-ing to the uncertainty relation, "x"p & !/2, the length scale at which thekinetic energy is comparable to the rest mass energy is set by the Comptonwavelength

"x & h

mc' %c.

We may expect relativistic e!ects to be important if we examine the motionof particles on length scales which are less than %c. Note that for particles ofzero mass, %c = (! Thus for photons, and neutrinos, relativity intrudes atany length scale.

What is the relativistic analogue of the Schrodinger equation? Non-relativisticquantum mechanics is based on the time-dependent Schrodinger equationH& = i!'t&, where the wavefunction & contains all information about a givensystem. In particular, |&(x, t)|2 represents the probability density to observe aparticle at position x and time t. Our aim will be to seek a relativistic versionof this equation which has an analogous form. The first goal, therefore, is tofind the relativistic Hamiltonian. To do so, we first need to revise results fromEinstein’s theory of special relativity:

! Info. Lorentz Transformations and the Lorentz Group: In the specialtheory of relativity, a coordinate in space-time is specified by a 4-vector. A con-travariant 4-vector x = (xµ) ' (x0, x1, x2, x3) ' (ct,x) is transformed into thecovariant 4-vector xµ = gµ!x! by the Minkowskii metric

(gµ!) =

#

$%

1#1

#1#1

&

'( , g !µ g!" = (µ" ,

Here, by convention, summation is assumed over repeated indicies. Indeed, sum-mation covention will be assumed throughout this chapter. The scalar product of4-vectors is defined by

x · y = xµyµ = xµy!gµ! = xµyµ .

The Lorentz group consists of linear Lorentz transformations, #, preserving x·y,i.e. for xµ )* x!µ = #µ

!x! , we have the condition

gµ!#µ##!

$ = g#$ . (15.1)

Specifically, a Lorentz transformation along the x1 direction can be expressed in theform

#µ! =

#

$%

" #"v/c#"v/c "

1 00 1

&

'(

where " = (1# v2/c2)"1/2.1 With this definition, the Lorentz group splits up intofour components. Every Lorentz transformation maps time-like vectors (x2 > 0) into

1Equivalently the Lorentz transformation can be represented in the form

! = exp[!K1], [K1]µ! =

0

BB@

0 !1!1 0

0 00 0

1

CCA ,

where ! = tanh!1(v/c) is known as the rapidity, and K1 is the generator of velocitytransformations along the x1-axis.


15.1. KLEIN-GORDON EQUATION 170

time-like vectors. Time-like vectors can be divided into those pointing forwards intime (x0 > 0) and those pointing backwards (x0 < 0). Lorentz transformations donot always map forward time-like vectors into forward time-like vectors; indeed #does so if and only if #0

0 > 0. Such transformations are called orthochronous.(Since #µ

0#µ0 = 1, (#00)2 # (#j

0)2 = 1, and so #00 += 0.) Thus the group splits

into two according to whether #00 > 0 or #0

0 < 0. Each of these two componentsmay be subdivided into two by considering those # for which det # = ±1. Thosetransformations # for which det # = 1 are called proper.

Thus the subgroup of the Lorentz group for which det # = 1 and #00 > 0 is

called the proper orthochronous Lorentz group, sometimes denoted by L#+. Itcontains neither the time-reversal nor parity transformation,

T =

#

$%

#11

11

&

'( , P =

#

$%

1#1

#1#1

&

'( . (15.2)

We shall call it the Lorentz group for short and specify when we are including T or P .In particular, L#+, L# = L#+ , L

#" (the orthochronous Lorentz group), L+ = L#+ , L

$+

(the proper Lorentz group), and L0 = L#+ , L$" are subgroups, while L$" = PL#+,L#" = TL#+ and L$+ = TPL#+ are not.

Special relativity requires that theories should be invariant under Lorentz trans-formations xµ )#* #µ

!x! , and, more generally, Poincare transformations xµ *#µ

!x! + aµ. The proper orthochronous Lorentz transformations can be reached con-tinuously from identity.2 Loosely speaking, we can form them by putting togetherinfinitesimal Lorentz transformations #µ

! = (µ! +)µ

! , where the elements of )µ! - 1.

Applying the identity g#$ = #µ##µ$ = g#$ + )#$ + )$# + O()2), we obtain the

relation )#$ = #)$#. )#$ has six independent components: L#+ is a six-dimensional(Lie) group, i.e. it has six independent generators: three rotations and three boosts.

Finally, according to the definition of the 4-vectors, the covariant and contravari-ant derivative are respectively defined by 'µ = %

%xµ = (1c

%%t ,.), 'µ = %

%xµ=

( 1c

%%t ,#.). Applying the scalar product to the derivative we obtain the d’Alembertian

operator (sometimes denoted as !), '2 = 'µ'µ = 1c2

%2

%t2 #.2.

15.1 Klein-Gordon equation

Historically, the first attempt to construct a relativistic version of the Schrodingerequation began by applying the familiar quantization rules to the relativisticenergy-momentum invariant. In non-relativistic quantum mechanics the cor-respondence principle dictates that the momentum operator is associated withthe spatial gradient, p = #i!., and the energy operator with the time deriva-tive, E = i!'t. Since (pµ ' (E/c,p) transforms like a 4-vector under Lorentztransformations, the operator pµ = i!'µ is relativistically covariant.

Oskar Benjamin Klein 1894-1977A Swedish theo-retical physicist,Klein is creditedfor inventingthe idea, partof Kaluza-Kleintheory, that extradimensions maybe physically realbut curled upand very small, an idea essential tostring theory/M-theory.

Non-relativistically, the Schrodinger equation is obtained by quantizingthe classical Hamiltonian. To obtain a relativistic version of this equation,one might apply the quantization relation to the dispersion relation obtainedfrom the energy-momentum invariant p2 = (E/c)2 # p2 = (mc)2, i.e.

E(p) = +)m2c4 + p2c2

*1/2 / i!'t& =+m2c4 # !2c2.2

,1/2&

where m denotes the rest mass of the particle. However, this proposal posesa dilemma: how can one make sense of the square root of an operator? Inter-preting the square root as the Taylor expansion,

i!'t = mc2& # !2.2

2m& # !4(.2)2

8m3c2& + · · ·

2They are said to form the path component of the identity.


15.1. KLEIN-GORDON EQUATION 171

we find that an infinite number of boundary conditions are required to specifythe time evolution of &.3 It is this e!ective “non-locality” together with theasymmetry (with respect to space and time) that suggests this equation maybe a poor starting point.

A second approach, and one which circumvents these di$culties, is to applythe quantization procedure directly to the energy-momentum invariant:

E2 = p2c2 + m2c4, #!2'2t & =

)#!2c2.2 + m2c4

*&.

Recast in the Lorentz invariant form of the d’Alembertian operator, we obtainthe Klein-Gordon equation

)'2 + k2

c

*& = 0 , (15.3)

where kc = 2#/%c = mc/!. Thus, at the expense of keeping terms of secondorder in the time derivative, we have obtained a local and manifestly covariantequation. However, invariance of & under global spatial rotations implies that,if applicable at all, the Klein-Gordon equation is limited to the consideration ofspin-zero particles. Moreover, if & is the wavefunction, can |&|2 be interpretedas a probability density?

To associate |&|2 with the probability density, we can draw intuition fromthe consideration of the non-relativistic Schrodinger equation. Applying theidentity &"(i!'t& + !2#2

2m &) = 0, together with the complex conjugate of thisequation, we obtain

't|&|2 # i!

2m. · (&".& # &.&") = 0 .

Conservation of probability means that density * and current j must satisfythe continuity relation, 't* + . · j = 0, which states simply that the rate ofdecrease of density in any volume element is equal to the net current flowingout of that element. Thus, for the Schrodinger equation, we can consistentlydefine * = |&|2, and j = #i !

2m(&".& # &.&").Applied to the Klein-Gordon equation (15.3), the same consideration im-

plies

!2't (&"'t& # &'t&")# !2c2. · (&".& # &.&") = 0 ,

from which we deduce the correspondence,

* = i!

2mc2(&"'t& # &'t&

") , j = #i!

2m(&".& # &.&") .

The continuity equation associated with the conservation of probability canbe expressed covariantly in the form

'µjµ = 0 , (15.4)

where jµ = (*c, j) is the 4-current. Thus, the Klein-Gordon density is thetime-like component of a 4-vector.

From this association it is possible to identify three aspects which (at leastinitially) eliminate the Klein-Gordon equation as a wholey suitable candidatefor the relativistic version of the wave equation:

3You may recognize that the leading correction to the free particle Schrodinger equationis precisely the relativistic correction to the kinetic energy that we considered in chapter 9.


15.2. DIRAC EQUATION 172

! The first disturbing feature of the Klein-Gordon equation is that thedensity * is not a positive definite quantity, so it can not represent aprobability. Indeed, this led to the rejection of the equation in the earlyyears of relativistic quantum mechanics, 1926 to 1934.

! Secondly, the Klein-Gordon equation is not first order in time; it isnecessary to specify & and 't& everywhere at t = 0 to solve for latertimes. Thus, there is an extra constraint absent in the Schrodingerformulation.

! Finally, the equation on which the Klein-Gordon equation is based,E2 = m2c4 + p2c2, has both positive and negative solutions. In factthe apparently unphysical negative energy solutions are the origin of thepreceding two problems.

To circumvent these di$culties one might consider dropping the negativeenergy solutions altogether. For a free particle, whose energy is thereby con-stant, we can simply supplement the Klein-Gordon equation with the conditionp0 > 0. However, such a definition becomes inconsistent in the presence oflocal interactions, e.g.

)'2 + k2

c

*& = F (&) self # interaction

-(' + iqA/!c)2 + k2

c

.& = 0 interaction with EM field.

The latter generate transitions between positive and negative energy states.Thus, merely excluding the negative energy states does not solve the problem.Later we will see that the interpretation of & as a quantum field leads to aresolution of the problems raised above. Historically, the intrinsic problemsconfronting the Klein-Gordon equation led Dirac to introduce another equa-tion.4 However, as we will see, although the new formulation implied a positivenorm, it did not circumvent the need to interpret negative energy solutions.

Paul A. M. Dirac 1902-1984Dirac was bornon 8th August,1902, at Bristol,England, his fa-ther being Swissand his motherEnglish. He waseducated at the Merchant Venturer’sSecondary School, Bristol, then wenton to Bristol University. Here, hestudied electrical engineering, obtain-ing the B.Sc. (Engineering) degreein 1921. He then studied mathemat-ics for two years at Bristol University,later going on to St. John’s Col-lege, Cambridge, as a research stu-dent in mathematics. He received hisPh.D. degree in 1926. The followingyear he became a Fellow of St.John’sCollege and, in 1932, Lucasian Pro-fessor of Mathematics at Cambridge.Dirac’s work was concerned with themathematical and theoretical aspectsof quantum mechanics. He beganwork on the new quantum mechan-ics as soon as it was introduced byHeisenberg in 1928 – independentlyproducing a mathematical equivalentwhich consisted essentially of a non-commutative algebra for calculatingatomic properties – and wrote a seriesof papers on the subject, leading upto his relativistic theory of the elec-tron (1928) and the theory of holes(1930). This latter theory requiredthe existence of a positive particlehaving the same mass and charge asthe known (negative) electron. This,the positron was discovered experi-mentally at a later date (1932) byC. D. Anderson, while its existencewas likewise proved by Blackett andOcchialini (1933) in the phenomenaof “pair production” and “annihila-tion”. Dirac was made the 1933 No-bel Laureate in Physics (with ErwinSchrodinger) for the discovery of newproductive forms of atomic theory.

15.2 Dirac Equation

Dirac attached great significance to the fact that Schrodinger’s equation ofmotion was first order in the time derivative. If this holds true in relativisticquantum mechanics, it must also be linear in '. On the other hand, forfree particles, the equation must imply p2 = (mc)2, i.e. the wave equationmust be consistent with the Klein-Gordon equation (15.3). At the expense ofintroducing vector wavefunctions, Dirac’s approach was to try to factorise thisequation:

("µpµ #m)& = 0 . (15.5)

(Following the usual convention we have, and will henceforth, adopt the short-hand convention and set ! = c = 1.) For this equation to be admissible, thefollowing conditions must be enforced:

! The components of & must satisfy the Klein-Gordon equation.4The original references are P. A. M. Dirac, The Quantum theory of the electron, Proc.

R. Soc. A117, 610 (1928); Quantum theory of the electron, Part II, Proc. R. Soc. A118,351 (1928). Further historical insights can be obtained from Dirac’s book on Principles ofQuantum mechanics, 4th edition, Oxford University Press, 1982.



! There must exist a 4-vector current density which is conserved and whosetime-like component is a positive density.

! The components of & do not have to satisfy any auxiliary condition. Atany given time they are independent functions of x.

Beginning with the first of these requirements, by imposing the condition["µ, p! ] = "µp! # p!"µ = 0, (and symmetrizing)

("! p! + m) ("µpµ #m) & =!

12{"! , "µ} p! pµ #m2

"& = 0 ,

the latter recovers the Klein-Gordon equation if we define the elements "µ suchthat they obey the anticommutation relation,5 {"! , "µ} ' "!"µ +"µ"! = 2gµ!

– thus "µ, and therefore &, can not be scalar. Then, from the expansion ofEq. (15.5), "0("0p0 # " · p #m)& = i't& # "0" · p& #m"0& = 0, the Diracequation can be brought to the form

i't& = H&, H = ! · p + +m, (15.6)

where the elements of the vector ! = "0" and + = "0 obey the commutationrelations,

{,i, ,j} = 2(ij , +2 = 1, {,i, +} = 0 . (15.7)

H is Hermitian if, and only if, !† = !, and +† = +. Expressed in terms of", this requirement translates to the condition ("0")† ' "†"0† = "0", and"0† = "0. Altogether, we thus obtain the defining properties of Dirac’s "matrices,

"µ† = "0"µ"0, {"µ, "!} = 2gµ! . (15.8)

Given that space-time is four-dimensional, the matrices " must have dimen-sion of at least 4! 4, which means that & has at least four components. It isnot, however, a 4-vector; it does not transform like xµ under Lorentz trans-formations. It is called a spinor, or more correctly, a bispinor with specialLorentz transformations which we will shall discuss presently.

! Info. An explicit representation of the " matrices which most easily capturesthe non-relativistic limit is the following,

"0 =!

I2 00 #I2

", " =

!0 ### 0

", (15.9)

where # denote the familiar 2 ! 2 Pauli spin matrices which satisfy the relations,-i-j = (ij + i$ijk-k, #† = #. The latter is known in the literature as the Dirac-Pauli representation. We will adopt the particular representation,

-1 =!

0 11 0

", -2 =

!0 #ii 0

", -3 =

!1 00 #1

".

Note that with this definition, the matrices ! and + take the form,

! =!

0 ## 0

", + =

!I2 00 #I2

".

5Note that, in some of the literature, you will see the convention [ , ]+ for the anticom-mutator.



15.2.1 Density and Current

Turning to the second of the requirements placed on the Dirac equation, wenow seek the probability density * = j0. Since & is a complex spinor, * hasto be of the form &†M& in order to be real and positive. Applying hermitianconjugation to the Dirac equation, we obtain

[("µpµ #m)&]† = &†(#i"†µ0#' µ #m) = 0 ,

where &†0#' µ ' ('µ&)†. Making use of (15.8), and defining & ' &†"0, theDirac equation takes the form &(i0#+' + m) = 0, where we have introducedthe Feynman ‘slash’ notation +a ' aµ"µ. Combined with Eq. (15.5) (i.e.(i#*+' #m)& = 0), we obtain

&/0#+' +#*+'

0& = 'µ

)&"µ&

*= 0 .

From this result and the continuity relation (15.4) we can identify

jµ = &"µ& , (15.10)

(or, equivalently, (*, j) = (&†&,&†!&)) as the 4-current. In particular, thedensity * = j0 = &†& is, as required, positive definite.

15.2.2 Relativistic Covariance

To complete our derivation, we must verify that the Dirac equation remainsinvariant under Lorentz transformations. More precisely, if a wavefunction&(x) obeys the Dirac equation in one frame, its counterpart &$(x$) in a Lorentztransformed frame x$ = #x, must obey the Dirac equation,

)i"µ'$µ #m

*&$(x$) = 0 . (15.11)

In order that an observer in the second frame can reconstruct &$ from & theremust exist a local transformation between the wavefunctions. Taking thisrelation to be linear, we therefore must have,

&$(x$) = S(#)&(x) ,

where S(#) represents a non-singular 4 ! 4 matrix. Now, using the identity,'$µ ' "

"x"µ = "x!

"x"µ"

"x! = (#!1)!µ

""x! = (#!1)!

µ'! , the Dirac equation (15.11)in the transformed frame takes the form,

)i"µ(#!1)!

µ'! #m*S(#)&(x) = 0 .

The latter is compatible with the Dirac equation in the original frame if

S(#)"!S!1(#) = "µ(#!1)!µ . (15.12)

To define an explicit form for S(#) we must now draw upon some of thedefining properties of the Lorentz group discussed earlier. For an infinitesi-mal proper Lorentz transformation we have #!

µ = g!µ + )!

µ and (#!1)!µ =

g!µ # )!

µ + · · ·, where the matrix )µ! is antisymmetric and g!µ ' (!

µ. Corre-spondingly, by Taylor expansion in ), we can define

S(#) = I# i

4%µ!)

µ! + · · · , S!1(#) = I +i

4%µ!)

µ! + · · · ,



where the matrices %µ! are also antisymmetric in µ.. To first order in ),Eq. (15.12) yields (a somewhat unrewarding exercise!)

["! ,%#$ ] = 2i)g!

#"$ # g!$"#

*. (15.13)

The latter is satisfied by the set of matrices (another exercise!)6

%#$ =i

2["#, "$] . (15.14)

In summary, if &(x) obeys the Dirac equation in one frame, the wavefunctioncan be obtained in the Lorentz transformed frame by applying the transforma-tion &$(x$) = S(#)&(#!1x$). Let us now consider the physical consequencesof this Lorentz covariance.

15.2.3 Angular momentum and spin

To explore the physical manifestations of Lorentz covariance, it is instructiveto consider the class of spatial rotations. For an anticlockwise spatial rotationby an infinitesimal angle / about a fixed axis n, x )* x$ = x # /x ! n. Interms of the “Lorentz transformation”, #, one has

x$i = [#x]i ' xi # )ijxj

where )ij = $ijknk/, and the remaining elements #µ0 = #0

µ = 0. Applied tothe argument of the wavefunction we obtain a familiar result,7

&(x) = &(#!1x$) = &(x$0,x$ + x$ ! n/) = (1# /n · x$ !.+ · · ·)&(x$)

= (1# i/n · L + · · ·)&(x$),

where L = x! p represents the non-relativistic angular momentum operator.Formally, the angular momentum operators represent the generators of spatialrotations.8

However, we have seen above that Lorentz covariance demands that thetransformed wavefunction be multiplied by S(#). Using the definition of )ij

above, one finds that

S(#) ' S(I + )) = I# i

4$ijknk%ij/ + · · ·

Then drawing on the Dirac/Pauli representation,

%ij =i

2["i, "j ] =

i

2

1!0 -i

#-i 0

",

!0 -j

#-j 0

"2= # i

2[-i, -j ]1 I2 = $ijk-k 1 I2,

one obtains

S(#) = I# in · S/ + · · · , S =12

!# 00 #

".

Combining both contributions, we thus obtain

&$(x$) = S(#)&(#!1x$) = (1# i/n · J + · · ·)&(x$) ,

6Since finite transformations are of the form S(!) = exp[!(i/4)""#!"# ], one may showthat S(!) is unitary for spatial rotations, while it is Hermitian for Lorentz boosts.

7Recall that spatial rotataions are generated by the unitary operator, U(") = exp(!i"n ·L).

8For finite transformations, the generator takes the form exp[!i"n · L].


15.3. FREE PARTICLE SOLUTION OF THE DIRAC EQUATION 176

where J = L + S can be identified as a total e!ective angular momentum ofthe particle being made up of the orbital component, together with an intrin-sic contribution known as spin. The latter is characterised by the definingcondition:

[Si, Sj ] = i$ijkSk, (Si)2 =14

for each i . (15.15)

Therefore, in contrast to non-relativistic quantum mechanics, the concept ofspin does not need to be grafted onto the Schrodinger equation, but emergesnaturally from the fundamental invariance of the Dirac equation under Lorentztransformations. As a corollary, we can say that the Dirac equation is arelativistic wave equation for particles of spin 1/2.

15.2.4 Parity

So far, our discussion of the covariance properties of the Dirac equation haveonly dealt with the subgroup of proper orthochronous Lorentz transformations,L%+ – i.e. those that can be reached from # = I by a sequence of infinites-imal transformations. Taking the parity operation into account, relativisticcovariance demands

S!1(P )"0S(P ) = "0, S!1(P )"iS(P ) = #"i.

This is achieved if S(P ) = "0ei%, where 0 denotes some arbitrary phase.Taking into account the fact that P 2 = I, 0 = 0 or #, and we find

&$(x$) = S(P )&(x) = 1"0&(P!1x$) = 1"0&(ct$,#x$) , (15.16)

where 1 = ±1 represents the intrinsic parity of the particle.

15.3 Free Particle Solution of the Dirac Equation

Having laid the foundation we will now apply the Dirac equation to the prob-lem of a free relativistic quantum particle. For a free particle, the plane wave

&(x) = exp[#ip · x]u(p) ,

with energy E ' p0 = ±3

p2 + m2 will be a solution of the Dirac equationif the components of the spinor u(p) are chosen to satisfy the equation (+p #m)u(p) = 0. Evidently, as with the Klein-Gordon equation, we see that theDirac equation therefore admits negative as well as positive energy solutions!Soon, having attached a physical significance to the former, we will see thatit is convenient to reverse the sign of p for the negative energy solutions.However, for now, let us continue without worrying about the dilemma posedby the negative energy states.

In the Dirac-Pauli block representation,

"µpµ #m =!

p0 #m ## · p# · p #p0 #m

".

Thus, defining the spin elements u(p) = (2, 1), where 2 and 1 represent two-component spinors, we find the conditions, (p0 #m)2 = # · p 1 and # · p 2 =



(p0 +m)1. With (p0)2 = p2 +m2, these equations are consistent if 1 = #·pp0+m2.

We therefore obtain the bispinor solution

u(r)(p) = N(p)

#

%3(r)

# · pp0 + m

3(r)

&

( ,

where 3(r) represents any pair of orthogonal two-component vectors, and N(p)is the normalisation.

Concerning the choice of 3(r), in many situations, the most convenient basisis the eigenbasis of helicity – eigenstates of the component of spin resolvedin the direction of motion,

S · p|p|3

(±) ' #

2· p|p|3

(±) = ±123(±) ,

e.g., for p = p3e3, 3(+) = (1, 0) and 3(!) = (0, 1). Then, for the positiveenergy states, the two spinor plane wave solutions can be written in the form

&(±)p (x) = N(p)e!ip·x

#

%3(±)

± |p|p0 + m

3(±)

&

(

Thus, according to the discussion above, the Dirac equation for a free particleadmits four solutions, two states with positive energy, and two with negative.

15.3.1 Klein paradox: anti-particles

While the Dirac equation has been shown to have positive definite density,as with the Klein-Gordon equation, it still exhibits negative energy states!To make sense of these states it is illuminating to consider the scatteringof a plane wave from a potential step. To be precise, consider a beam ofrelativistic particles with unit amplitude, energy E, momentum pe3, and spin2 (i.e. 3 = (1, 0)), incident upon a potential V (x) = V /(x3) (see figure).At the potential barrier, spin is conserved, while a component of the beamwith amplitude r is reflected (with energy E and momentum #pe3), and acomponent t is transmitted with energy E$ = E # V and momentum p$e3.According to the energy-momentum invariant, conservation of energy acrossthe interface dictates that E2 = p2 + m2 and E$2 = p$2 + m2.

Being first order, the boundary conditions on the Dirac equation requireonly continuity of & (cf. the Schrodinger equation). Therefore, matching & atthe step, we obtain the relations

#

$$%

10p

E+m0

&

''( + r

#

$$%

10

# pE+m0

&

''( = t

#

$$%

10p"

E"+m0

&

''( ,

from which we find 1+r = t, and pp0+m(1#r) = p"

p"0+m t. Setting 4 = p"

p(E+m)(E"+m) ,

these equations lead to

t =2

1 + 4,

1 + r

1# r=

14, r =

1# 4

1 + 4.

To interpret these solutions, let us consider the current associated withthe reflected and transmitted components. Making use of the equation for thecurrent density, j = &†!&, and using the Dirac/Pauli representation wherein

,3 = "0"3 =!

I2#I2

" !-3

#-3

"=

!-3

-3

",



the current along e3-direction is given by

j3 = &†!

-3

-3

"&, j1 = j2 = 0 .

Therefore, up to an overall constant of normalisation, the current densities aregiven by

j(i)3 =

2p

p0 + m, j(t)

3 =2(p$ + p$")p$0 + m

|t|2, j(r)3 = # 2p

p0 + m|r|2.

From these relations we obtain

j(t)3

j(i)3

= |t|2 (p$ + p$")2p

p0 + m

p$0 + m=

4|1 + 4|2

12(4 + 4")

j(r)3

j(i)3

= #|r|2 = #44441# 4

1 + 4

44442

from which current conservation can be confirmed:

1 +j(r)3

j(i)3

=|1 + 4|2 # |1# 4|2

|1 + 4|2 =2(4 + 4")|1 + 4|2 =

j(t)3

j(i)3

.

Interpreting these results, it is convenient to separate our considerationinto three distinct regimes of energy:

! E$ ' (E # V ) > m: In this case, from the Klein-Gordon condition (theenergy-momentum invariant) p$2 ' E$2 # m2 > 0, and (taking p$ > 0– i.e. beam propagates to the right) 4 > 0 and real. From this resultwe find |j(r)

3 | < |j(i)3 | – as expected, within this interval of energy, a

component of the beam is transmitted and the remainder is reflected(cf. non-relativistic quantum mechanics).

! #m < E$ < m: In this case p$2 ' E$2 # m2 < 0 and p$ is purelyimaginary. From this result it follows that 4 is also pure imaginary and|j(r)

3 | = |j(i)3 |. In this regime the under barrier solutions are evanescent

and quickly decay to the right of the barrier. All of the beam is reflected(cf. non-relativistic quantum mechanics).

! E$ < #m: Finally, in this case p$2 ' E$2#m2 > 0 and, depending on thesign of p$, j(r)

3 can be greater or less than j(i)3 . But the solution has the

form e!i(p"x!E"t). Since we presume the beam to be propagating to theright, we require E$ < 0 and p$ > 0. From this result it follows that 4 < 0and we are drawn to the surprising conclusion that |j(r)

3 | > |j(i)3 | – more

current is reflected that is incident! Since we have already confirmedcurrent conservation, we can deduce that j(t)

3 < 0. It is as if a beam ofparticles were incident from the right.

The resolution of this last seeming unphysical result, known as the Kleinparadox,9 in fact gives a natural interpretation of the negative energy solu-tions that plague both the Dirac and Klein-Gordon equations: Dirac particlesare fermionic in nature. If we regard the vacuum as comprised of a filled Fermisea of negative energy states or antiparticles (of negative charge), the KleinParadox can be resolved as the stimulated emission of particle/antiparticle

9Indeed one would reach the same conclusion were one to examine the Klein-Gordonequation.



Figure 15.2: The photograph shows asmall part of a complicated high energyneutrino event produced in the Fermi-lab bubble chamber filled with a neonhydrogen mixture. A positron (red)emerging from an electron-positronpair, produced by a gamma ray, curvesround through about 180 degrees.Then it seems to change charge: it be-gins to curve in the opposite direction(blue). What has happened is that thepositron has run head-on into a (more-or-less from the point of view of particlephysics) stationary electron – transfer-ring all its momentum. This tells usthat the mass of the positron equals themass of the electron.

pairs, the particles moving o! towards x3 = #( and the antiparticles towardsx3 =(. What about energy conservation? One might worry that the energyfor these pairs is coming from nowhere. However, the electrostatic energy re-covered by the antiparticle when its created is su$cient to outweigh the restmass energy of the particle and antiparticle pair (remember that a repulsivepotential for particles is attractive for antiparticles). Taking into account thefact that the minimum energy to create a particle/antiparticle pair is twicethe rest mass energy 2 !m, the regime where stimulated emission is seen tooccur can be understood.

Negative energy states: With this conclusion, it is appropriate to revisitthe definition of the free particle plane wave state. In particular, for energiesE < 0, it is more sensible to set p0 = +

3(p2 + m2), and redefine the plane

wave solution as &(x) = v(p)eip·x, where the spinor satisfies the condition(+p + m)v(p) = 0. Accordingly we find,

v(r)(p) = N(p)

5 # · pp0 + m

3(r)

3(r)

6.

So, to conclude, two relativistic wave equations have been proposed. Thefirst of these, the Klein-Gordon equation was dismissed on the grounds thatit exhibited negative probability densities and negative energy states. Bycontrast, the states of the Dirac equation were found to exhibit a positivedefinite probability density, and the negative energy states were argued tohave a natural interpretation in terms of antiparticles: the vacuum state doesnot correspond to all states unoccupied but to a state in which all the negativeenergy states are occupied – the negative energy states are filled up by a Fermisea of negative energy Fermi particles. For electron degrees of freedom, if apositive energy state is occupied we observe it as a (positive energy) electronof charge q = #e. If a negative energy state is unoccupied we observe it as a(positive energy) antiparticle of charge q = +e, a positron, the antiparticleof the electron. If a very energetic electron interacts with the sea causing atransition from a negative energy state to positive one (by communicating anenergy of at least 2m) this is observed as the production of a pair of particles,an electron and a positron from the vacuum (pair production) (see Fig. 15.2).

However, the interpretation attached to the negative energy states providesgrounds to reconsider the status of the Klein-Gordon equation. Evidently, the


15.4. QUANTIZATION OF RELATIVISTIC FIELDS 180

Dirac equation is not a relativistic wave equation for a single particle. If itwere, pair production would not appear. Instead, the interpretation aboveforces us to consider the wavefunction of the Dirac equation as a quantumfield able to host any number of particles – cf. the continuum theory of thequantum harmonic chain. In the next section, we will find that the consid-eration of the wavefunction as a field revives the Klein-Gordon equation as atheory of scalar (interger spin) particles.

15.4 Quantization of relativistic fields

15.4.1 Info: Scalar field: Klein-Gordon equation revisited

Previously, the Klein-Gordon equation was abandoned as a candidate for a relativis-tic theory on the basis that (i) it admitted negative energy solutions, and (ii) thatthe probability density associated with the wavefunction was not positive definite.Yet, having associated the negative energy solutions of the Dirac equation with an-tiparticles, and identified & as a quantum field, it is appropriate that we revisit theKlein-Gordon equation and attempt to revive it as a theory of relativistic particles ofspin zero.

If 0 were a classical field, the Klein-Gordon equation would represent the equationof motion associated with the Lagrangian density (exercise)

L =12'µ0 'µ0# 1

2m202 ,

(cf. our discussion of the low energy modes of the classical harmonic chain and theMaxwell field of the waveguide in chapter 11). Defining the canonical momentum#(x) = '&L(x) = 0(x) ' '00(x), the corresponding Hamiltonian density takes theform

H = #0# L =12

+#2 + (.0)2 + m202

,.

Evidently, the Hamiltonian density is explicitly positive definite. Thus, the scalarfield is not plagued by the negative energy problem which beset the single-particletheory. Similarly, the quantization of the classical field will lead to a theory in whichthe states have positive energy.

Following on from our discussion of the harmonic chain in chapter 11, we arealready equipped to quantise the classical field theory. However, there we worked ex-plicitly in the Schrodinger representation, in which the dynamics was contained withinthe time-dependent wavefunction &(t), and the operators were time-independent. Al-ternatively, one may implement quantum mechanics in a representation where thetime dependence is transferred to the operators instead of the wavefunction — theHeisenberg representation. In this representation, the Schrodinger state vector &S(t)is related to the Heisenberg state vector &H by the relation,

&S(t) = e"iHt&H , &H = &S(0) .

Similarly, Schrodinger operators OS are related to the Heisenberg operators OH(t) by

OH(t) = eiHtOSe"iHt .

One can easily check that the matrix elements 3&!S |OS |&S4 and 3&!

H |OH |&H4 areequivalent in the two representations, and which to use in non-relativistic quantummechanics is largely a matter of taste and convenience. However, in relativistic quan-tum field theory, the Heisenberg representation is often preferable to the Schrodingerrepresentation. The main reason for this is that by using the former, the Lorentzcovariance of the field operators is made manifest.



In the Heisenberg representation, the quantisation of the fields is still enforced bypromoting the classical fields to operators, # )* # and 0 )* 0, but in this case, weimpose the equal time commutation relations,

-0(x, t), #(x!, t)

.= i(3(x# x!),

-0(x, t), 0(x!, t)

.= [#(x, t), #(x!, t)] = 0 ,

with # = '00. In doing so, the Hamiltonian density takes the form

H =12

-#2 + (.0)2 + m202

..

To see the connection between the quantized field and particles we need to Fouriertransform the field operators to obtain the normal modes of the Hamiltonian,

0(x) =7

d4k

(2#)40(k)e"ik·x .

However the form of the Fourier elements 0(k) is constrained by the following con-ditions. Firstly to maintain Hermiticity of the field operator 0(x) we must chooseFourier coe$cients such that 0†(k) = 0(#k). Secondly, to ensure that the field op-erator 0(x) obeys the Klein-Gordon equation,10 we require 0(k) % 2#((k2 # m2).Taking these conditions together, we require

0(k) = 2#((k2 #m2))/(k0)a(k) + /(#k0)a†(#k)

*,

where k0 ' )k ' +5

k2 + m2, and a(k) represent the operator valued Fourier co-e$cients. Rearranging the momentum integration, we obtain the Lorentz covariantexpansion

0(x) =7

d4k

(2#)42#((k2 #m2)/(k0)

+a(k)e"ik·x + a†(k)eik·x,

.

Integrating over k0, and making use of the identity7

d4k

(2#)42#((k2 #m2)/(k0) =

7d4k

(2#)3((k2

0 # )2k)/(k0)

=7

d4k

(2#)3( [(k0 # )k)(k0 + )k)] /(k0) =

7d4k

(2#)31

2k0[((k0 # )k) + ((k0 + )k)] /(k0)

=7

d3k(2#)3

7dk0

2k0((k0 # )k)/(k0) =

7d3k

(2#)32)k,

one obtains

0(x) =7

d3k(2#)32)k

)a(k)e"ik·x + a†(k)eik·x*

.

More compactly, making use of the orthonormality of the basis

fk =13

(2#)32)k

e"ik·x,

7f%k (x)i

&' 0 fk!(x)d3x = (3(k# k!),

where A&'0 B ' A'tB # ('tA)B, we obtain

0(x) =7

d3k3(2#)32)k

+a(k)fk(x) + a†(k)f%k (x)

,.

10Note that the field operators obey the equation of motion,

#(x, t) = ! $H$%(x, t)

= "2%!m2% .

Together with the relation # = %, one finds ($2 + m2)% = 0.



Similarly,

#(x) ' '00(x) =7

d3k3(2#)32)k

i)k

+#a(k)fk(x) + a†(k)f%k (x)

,.

Making use of the orthogonality relations, the latter can be inverted to give

a(k) =3

(2#)32)k

7d3xf%k (x)i

&'0 0(x), a†(k) =

3(2#)32)k

7d3x0(x)i

&'0 fk(x) ,

or, equivalently,

a(k) =7

d3x/)k0(x)# i#(x)

0e"ik·x, a†(k) =

7d3x

/)k0(x) + i#(x)

0eik·x .

With these definitions, it is left as an exercise to show

+a(k), a†(k!)

,= (2#)32)k(3(k# k!), [a(k), a(k!)] =

+a†(k), a†(k!)

,= 0 .

The field operators a† and a can therefore be identified as operators that create andannihilate bosonic particles. Although it would be tempting to adopt a di!erentnormalisation wherein [a, a†] = 1 (as is done in many texts), we chose to adopt theconvention above where the covariance of the normalisation is manifest. Using thisrepresentation, the Hamiltonian is brought to the diagonal form

H =7

d3k(2#)32)k

)k

2+a†(k)a(k) + a(k)a†(k)

,,

a result which can be confirmed by direct substitution.Defining the vacuum state |&4 as the state which is annhiliated by a(k), a single

particle state is obtained by operating the creation operator on the vacuum,

|k4 = a†(k)|&4 .

Then 3k!|k4 = 3&|a(k!)a†(k)|&4 = 3&|[a(k!), a†(k)]|&4 = (2#)32)k(3(k! # k). Many-particle states are defined by |k1 · · ·kn4 = a†(k1) · · · a†(kn)|&4 where the bosonicstatistics of the particles is assured by the commutation relations.

Associated with these field operators, one can define the total particle numberoperator

N =7

d3k3(2#)32)k

a†(k)a(k) .

Similarly, the total energy-momentum operator for the system is given by

Pµ =7

d3k3(2#)32)k

kµa†(k)a(k) .

The time component P 0 of this result can be compared with the Hamiltonian above.In fact, commuting the field operators, the latter is seen to di!er from P 0 by an infiniteconstant,

8d3k)k/2. Yet, had we simply normal ordered11 the operators from the

outset, this problem would not have arisen. We therefore discard this infinite constant.

15.4.2 Info: Charged Scalar Field

A generalization of the analysis above to the complex scalar field leads to the La-grangian,

L =12'µ0'µ0# 1

2m2|0|2 .

11Recall that normal ordering entails the construction of an operator with all the annihi-lation operators moved to the right and creation operators moved to the left.



The latter can be interpreted as the superposition of two independent scalar fields0 = (01+i02)/

52, where, for each (real) component 0†

r(x) = 0r(x). (In fact, we couldas easily consider a field with n components.) In this case, the canonical quantisationof the classical fields is achieved by defining (exercise)

0(x) =7

d3k3(2#)32)k

+a(k)fk(x) + b†(k)f%k (x)

,.

(similarly 0†(x)) where both a and b obey bosonic commutation relations,+a(k), a†(k!)

,=

+b(k), b†(k!)

,= (2#)32)k(3(k# k!),

[a(k), a(k!)] = [b(k), b(k!)] =+a(k), b†(k!)

,= [a(k), b(k!)] = 0 .

With this definition, the total number operator is given by

N =7

d3k3(2#)32)k

+a†(k)a(k) + b†(k)b(k)

,' Na + Nb ,

while the energy-momentum operator is defined by

Pµ =7

d3k3(2#)32)k

kµ+a†(k)a(k) + b†(k)b(k)

,.

Thus the complex scalar field has the interpretation of creating di!erent sorts ofparticles, corresponding to operators a† and b†. To understand the physical interpre-tation of this di!erence, let us consider the corresponding charge density operator,j0 = 0†(x)i

&' 0 0(x). Once normal ordered, the total charge Q =

8d3xj0(x) is given

by

Q =7

d3k3(2#)32)k

+a†(k)a(k)# b†(k)b(k)

,= Na # Nb .

From this result we can interpret the particles as carrying an electric charge, equalin magnitude, and opposite in sign. The complex scalar field is a theory of chargedparticles. The negative density that plagued the Klein-Gordon field is simply a man-ifestation of particles with negative charge.

15.4.3 Info: Dirac Field

The quantisation of the Klein-Gordon field leads to a theory of relativistic spin zeroparticles which obey boson statistics. From the quantisation of the Dirac field, weexpect a theory of Fermionic spin 1/2 particles. Following on from our considerationof the Klein-Gordon theory, we introduce the Lagrangian density associated with theDirac equation (exercise)

L = & (i"µ'µ #m) & ,

(or, equivalently, L = &( 12 i"µ

&' µ #m)&). With this definition, the corresponding

canonical momentum is given by ''L = i&"0 = i&†. From the Lagrangian density,we thus obtain the Hamiltonian density,

H = & (#i" ·.+ m) & ,

which, making use of the Dirac equation, is equivalent to H = &i"0'0& = &†i't&.For the Dirac theory, we postulate the equal time anticommutation relations

9&#(x, t), i&†

$(x!, t):

= i(3(x# x!)(#$ ,

(or, equivalently {&#(x, t), i&$(x!, t)} = "0#$(3(x# x!)), together with

{&#(x, t), &$(x!, t)} =;&#(x, t), &$(x!, t)

<= 0 .


15.5. THE LOW ENERGY LIMIT OF THE DIRAC EQUATION 184

Using the general solution of the Dirac equation for a free particle as a basis set,together with the intuition drawn from the study of the complex scalar field, we maywith no more ado, introduce the field operators which diagonalise the Hamiltoniandensity

&(x) =2=

r=1

7d3k

(2#)32)k

-ar(k)u(r)(k)e"ik·x + b†r(k)v(r)(k)eik·x

.

&(x) =2=

r=1

7d3k

(2#)32)k

-a†

r(k)u(r)(k)eik·x + br(k)v(r)(k)e"ik·x.

,

where the annihilation and creation operators also obey the anticommutation rela-tions,

;ar(k), a†

s(k!)

<=

;br(k), b†s(k

!)<

= (2#)32)k(rs(3(k# k!)

{ar(k), as(k!)} =;a†

r(k), a†s(k

!)<

= {br(k), bs(k!)} =;b†r(k), b†s(k

!)<

= 0 .

The latter condition implies the Pauli exclusion principle a†(k)2 = 0. With thisdefinition, a(k)u(k)e"ik·x annilihates a postive energy electron, and b†(k)v(k)eik·x

creates a positive energy positron.From these results, making use of the expression for the Hamiltonian density

operator, one obtains

H =2=

r=1

7d3k

(2#)32)k)k

+a†

r(k)ar(k)# br(k)b†r(k),

.

Were the commutation relations chosen as bosonic, one would conclude the existenceof negative energy solutions. However, making use of the anticommutation relations,and dropping the infinite constant (or, rather, normal ordering) one obtains a positivedefinite result. Expressed as one element of the total energy-momentum operator, onefinds

Pµ =2=

r=1

7d3k

(2#)32)kkµ

+a†

r(k)ar(k) + b†r(k)br(k),

.

Finally, the total charge is given by

Q =7

j0d3x =7

d3x&†& = Na # Nb .

where N represents the total number operator. Na =8

d3k a†(k)a(k) is the numberof the particles and Nb =

8d3k b†(k)b(k) is the number of antiparticles with opposite

charge.

15.5 The low energy limit of the Dirac equation

To conclude our abridged exploration of the foundations of relativistic quan-tum mechanics, we turn to the interaction of a relativistic spin 1/2 particlewith an electromagnetic field. Suppose that & represents a particle of chargeq (q = #e for the electron). From non-relativistic quantum mechanics, we ex-pect to obtain the equation describing its interaction with an EM field givenby the potential Aµ by the minimal substitution

pµ )#* pµ # qAµ ,

where A0 ' 5. Applied to the Dirac equation, we obtain for the interaction ofa particle with a given (non-quantized) EM field, ["µ (pµ # qAµ) #m]& = 0,or compactly

(+p# q +A#m)& = 0 .



Previously, in chapter 9, we explored the relativistic (fine-structure) cor-rections to the hydrogen atom. At the time, we alluded to these as the leadingrelativistic contributions to the Dirac theory. In the following section, we willexplore how these corrections are derived.

In the Dirac-Pauli representation,

! =!

0 ## 0

", + =

!I2 00 #I2

".

we have seen that the plane-wave solution to the Dirac equation for particlescan be written in the form

&p(x) = N

!3

c#·pmc2+E 3

"ei(px!Et)/! ,

where we have restored the parameters ! and c. From this expression, we cansee that, at low energies, where |E #mc2| - mc2, the second component ofthe bispinor is smaller than the first by a factor of order v/c. To obtain thenon-relavistic limit, we can exploit this asymmetry to develop a perturbativeexpansion of the coe$cients in v/c.

Consider then the Dirac equation for a particle moving in a potential(0,A). Expressed in matrix form, the Dirac equation H = c! · (#i!. #ecA) + mc2+ + e0 is expressed as

H =!

mc2 + e0 c# · (#i!.# ecA)

c# · (p# qcA) #mc2 + q0

".

Defining the bispinor &T (x) = (&a(x), &b(x)), the Dirac equation translates tothe coupled equations,

(mc2 + e0)&a + c# · (p# q

cA)&b = E&a

c# · (p# q

cA)&a # (mc2 # q0)&b = E&b .

Then, if we define W = E # mc2, a rearrangement of the second equationobtains

&b =1

2mc2 + W # q0c# · (p# q

cA)&a .

Then, at zeroth order in v/c, we have &b $ 12mc2 c# ·(p# q

cA)&a. Substi-tuted into the first equation, we thus obtain the Pauli equation Hnon!rel&a =W&a, where

Hnonrel =1

2m

-# · (p# q

cA)

.2+ q0 .

Making use of the Pauli matrix identity -i-j = (ij + i$ijk-k, we thus obtainthe familiar non-relativistic Schrodinger Hamiltonian,

Hnon rel =1

2m(p# q

cA)2 # q!

2mc# · (.!A) + q0 .

As a result, we can identify the spin magnetic moment

µS =q!

2mc# =

q!mc

S ,



with the gyromagnetic ratio, g = 2. This compares to the measured valueof g = 2 ! (1.0011596567 ± 0.0000000035), the descrepency form 2 being at-tributed to small radiative corrections.

Let us now consider the expansion to first order in v/c. Here, for sim-plicity, let us suppose that A = 0. In this case, taking into account the nextorder term, we obtain

&b $1

2mc2

!1 +

V #W

2mc2

"c# · p&a

where V = q0. Then substituted into the second equation, we obtain1

12m

(# · p)2 +1

4m2c2(# · p)(V #W )(# · p) + V

2&a = W&a .

At this stage, we must be cautious in interpreting &a as a complete non-relavistic wavefunction with leading relativistic corrections. To find the truewavefunction, we have to consider the normalization. If we suppose that theoriginal wavefunction is normalized, we can conclude that,

7d3x&†(x, t)&(x, t) =

7d3x

/&†

a(x, t)&a(x, t) + &†b(x, t)&b(x, t)

0

$7

d3x&†a(x, t)&a(x, t) +

1(2mc)2

7d3x&†

a(x, t)p2&a(x, t) .

Therefore, at this order, the normalized wavefunction is set by, &s = (1 +1

8m2c2 p2)&a or, inverted,

&a =!

1# 18m2c2

p2

"&s .

Substituting, then rearranging the equation for &s, and retaining terms oforder (v/c)2, one ontains (exercise) Hnon!rel&s = W&s, where

Hnon!rel =p2

2m# p4

8m3c2+

14m2c2

(# · p)V (# · p) + V # 18m2c2

(V p2 + p2V ) .

Then, making use of the identities,

[V, p2] = !2(.2V ) + 2i!(.V ) · p(# · p)V = V (# · p) + # · [p, V ](# · p)V (# · p) = V p2 # i!(.V ) · p + !# · (.V )! p ,

we obtain the final expression (exercise),

Hnon!rel =p2

2m# p4

8m3c2+

!4m2c2

# · (.V )! p> ?@ A

spin!orbit coupling

+!2

8m2c2(.2V )

> ?@ ADarwin term

.

The second term on the right hand side represents the relativistic correctionto the kinetic energy, the third term denotes the spin-orbit interaction andthe final term is the Darwin term. For atoms, with a central potential, thespin-orbit term can be recast as

HS.O. =!2

4m2c2# · 1

r('rV )r! p =

!2

4m2c2

1r('rV )# · L .

To address the e!ects of these relativistic contributions, we refer back to chap-ter 9.
