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Chapter 22 The Hamiltonian and Lagrangian …

Chapter 22 The Hamiltonian and Lagrangian densities − from my book: Understanding Relativistic Quantum Field Theory Hans de Vries January 2, 2009

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Text of Chapter 22 The Hamiltonian and Lagrangian …

Chapter 22The Hamiltonian and Lagrangian densities from my book:Understanding Relativistic Quantum Field TheoryHans de VriesJanuary 2, 20092ChapterContents22 The Hamiltonian and Lagrangian The relativistic Hamiltonian and Lagrangian . . . . . . . . Principle of least action / least proper time . . . . . . . . The Hamiltonian and Lagrangian density . . . . . . . . . The Euler-Lagrange equation for Fields . . . . . . . . . . . Lagrangian of the scalar Klein Gordon field . . . . . . . . Hamiltonian of the scalar Klein Gordon field . . . . . . . . The complex Klein Gordon field . . . . . . . . . . . . . . . Expressing the Lagrangian in a 4d environment . . . . . . Electromagnetic and Proca Langrangian . . . . . . . . . . The electromagnetic equation of motion . . . . . . . . . . The electromagnetic interaction Lagrangian . . . . . . . . Electromagnetic and Proca Hamiltonian . . . . . . . . . . 20Chapter 22The Hamiltonian andLagrangian densities2Chapter 22. The Hamiltonian and Lagrangian The relativistic Hamiltonian and LagrangianThe Hamiltonian and Lagrangian which are rather abstract constructionsin classical mechanics get a very simple interpretation in relativistic quan-tum mechanics. Both are proportional to the number of phase changes perunit of time. The Hamiltonian runs over the time axis while the Lagrangianruns over the trajectory of the moving particle, the t : The Hamiltonian and LagrangianFigure shows the relativistic de Broglie wave in a Minkowski dia-gram. The triangle represents the relation between the Lagrangian an theHamiltonian, which holds in both relativistic and non-relativistic H( )The Hamiltonian counts the phase-changes per unit of time on the verticalaxis while the termpvcounts the phase-changes per unit on the horizontalaxis:vis the distance traveled per unit of time whilepis proportionalwith the phase-changes per unit of distance, hence the termpv. We cannow understand the classical relation. (with q= x=v) L q=p( ) The relativistic Hamiltonian and Lagrangian3For the free classical relativistic particle we have for the Hamiltonian (En-ergy) and 1 v2c2, pv=mv2 1 v2c2( )Calculating the Lagrangian we see that the Hamiltonian is proportional to while the Lagrangian is proportional to 1/ .L= (H pv) = c2 v2 1 v2c2m= 1 v2c2mc2( )This is what we expect from time dilation. The moving particle has lessclock-ticks by a factor due to the time dilation, We now check that. L q= v{ 1 v2c2mc2}=mv 1 v2c2=p( )For sofar we have not yet discussed the potential energy. To obtain theequation of motion of a relativistic particle in a potential field we have toadd the potential energy termV(q). In the non-relativistic case we ( q) +V(q), L=T( q) V(q)( )WhereT( q) =12mv2is the kinetic energy. The relativistic Hamiltonianand Lagrangian we have discussed however also include the restmass en-ergy. The restmass energy can be considered as being part of the potentialenergy. The kinetic partTin the relativistic case can be obtained +L= 2T=pv T=12pv=12mv2 1 v2c2 12mv2( forv c)( )Using the termL=12mv2in eq. ( ) gives usp=mvfor the non-relativistic momentum4Chapter 22. The Hamiltonian and Lagrangian Principle of least action / least proper timeTo obtain the relativistic equation of motion of a particle in a potential fieldwe use Lagrange equation of motion derived from Hamilton s variationalprinciple of Least Action. Relativistic quantum mechanics gives us thedeeper perspective of this principle which now becomes the principle ofleast phase-change and thus the least proper time: The relativistic particlefollows the path which will bring it there in the shortest proper will briefly recall the derivation here, it is one of the most fundamentalprinciples of physics. Somewhat abstract we can write. S= t2t1L(q, q)dt= 0( )Where Sis the variation of the action, the variation of the proper timein quantum mechanics, is zero, meaning that we have a minimum or max-imum just like the a zero first order derivative of a function indicates alocal minimum or maximum. The particle traverses a path between twofixed pointsx1andx2betweent1andt2which are also fixed. We assumethat the Lagrangian only depends on the positionqof the particle and itsvelocity q. Now we allow small arbitrary variations inqand qalong thepath. S= t2t1( L q q+ L q q)dt= 0( )As long as these variations are small we can determine the change inLwith the help of the first order derivatives. The last term above is alsothe last term in the expression below according to the product rule { L q q}=ddt{ L q} q+ L qddt{ q}( )We can instantly integrate the left hand term which gives us the following. S= t2t1( L q ddt{ L q}) q dt+ L q q t2t1= 0 ( ) Principle of least action / least proper time5The integrated term at the end vanishes since qis defined as zero at theend-points. Since qis arbitrary along the rest of the path we have to setthe term between brackets to zero. This then gives us the ( L q) L q= 0( )It defines a local minimum at each point along the path. We can usethe metaphor of a ball rolling at the bottom of a valley. Inserting theLagrangian of the classical relativistic 1 v2c2mc2 V(q)( )Gives us theequation of motionof the relativistic particle in a dVdq,withp=mv 1 v2/c2( )A gradient of the potential field causes a change in the relativistic momen-tum: If we use the non-relativistic Lagrangian with,L 12mv2 V(q)( )then we get the non-relativistic equation of dVdq( )As long as the derivation is correct for the relativistic particle, then we canbe assured that it is valid for the non-relativistic limit as 22. The Hamiltonian and Lagrangian The Hamiltonian and Lagrangian densityWe can define the Hamiltonian and Lagrangiandensityfor any extendedobject, being either classical or a quantum field, Hdx3, L= Ldx3( )Let us see how these quantities transform under Lorentz did see in ( ) and ( ) that the integrated quantities, the Hamil-tonian H, the Lagrangian L and the term pv transform as pvtransforms as 2 Ltransforms as 1/ ( )The volume of a wave-functions transforms like 1/ due to Lorentz con-traction. So, the densities become higher by a factor , hence the densityH, the Lagrangian densityLand the density of thepvterm transform as 2PVtransforms as 2 2Ltransforms as 1 ( )We see that the Lagrangian density is the same in all reference is a Lorentz scalar. This makes the Lagrangian density a fundamentalquantity in quantum field theory. The Standard Model of physics is basedon the Lagrangian density which in quantum physics is generally calledjust the Lagrangian, without the equations of motion are based on the derivatives of the Lagrangiandensity which is a Lorentz scalar. Done in the right way assures that thewhole Standard Model of physics transforms in the right way, that is, thelaws of physics are the same in every reference triangular equationH pv= Lbasically counts phase change clock-pulses on thet,xandt -axis. The corresponding relation of thedensities transforms like the basic energy/momentum PV= Ltransforms asE2 p2=m2( ) The Euler-Lagrange equation for The Euler-Lagrange equation for FieldsThe Euler-Lagrange equation for fields operates on a lagrangian whichdepends only on the (generalized) coordinateqand velocity qof the par-ticle. It is valid for relativistic particles even though it was developed byLeonhard Euler and Joseph-Louis Lagrange in the 1750 L(q, q)( )In quantum field theory we do not have a quantity likeqexplicitly work with a field (t,r) instead. We assume that the Lagrangiandensity only depends on (t,r) and its first order L( , t , x , y , z )( )We will avoid in this book the widespread custom to present the correctrelativistic Lagrangian density for the (scalar) Klein Gordon field and thenjustify it by making substitutions ,12mv2 12 2( )The term12mv2should worry the reader. Indeed the Hamiltonian densitysubsequently derived does transform in the wrong way and its integral overspace does not correspond with the Hamiltonian of the classical origin of these substitutions can be understood by looking at ourinitial mechanical spring-mass model of the Klein Gordon equation shownin figure??. In this model is a displacement which could be associatedwithq. These substitutions however are to naive, worse, they lead toviolations of special then trying to make associations between classical and field termswe want to stress the fact that the Euler-Lagrange mechanism to derive theequations of motion is an entirelymathematicalmechanism to find a min-imum/maximum. It does not matter by what quantities the Lagrangiandensity is expressed as long as they express the right Euler-Lagrange equation for quantum fields goes well beyond thescalar Klein Gordon field. It holds for all quantum fields fields. The8Chapter 22. The Hamiltonian and Lagrangian densitiesderivation is the same as the derivation for the classical one. We willfollow the same four steps as in equations ( ) through ( ) for theclassical wave equation. The variation of the action is symbolized by. S= t2t1dt + L( , t , x , y , z )dx3= 0 ( )More concretely we can write it expressed in variations of the field andits (four) derivatives. S= t2t1( L + L ( ) ))dx4= 0( )The last term above is equal to the last term in the equation below whichjust expresses the product rule for differentiation. In fact there are fourterms in total, one for each derivative. { L ( ) }= { L ( )} + L ( ) { }( )The left hand term above (representing four terms) can be directly inte-grated over one axis, each of the four along its own axis. The result ofthese integrations is zero since the variations at the end points are definedas zero. So, we can omit the integration over the other three axis and con-tinue with the remaining terms which are all proportional to the variationof itself. S= t2t1( L { L ( )}) dx4= 0( )Since is totally arbitrary we conclude that the equation between brack-ets has to hold at each point (This means we have nothing to do anymorewith where the borders are in which reference frame). We have obtainedthe Euler Lagrange equation for the relativistic Lagrangian density: ( L ( )) L = 0( ) Lagrangian of the scalar Klein Gordon Lagrangian of the scalar Klein Gordon fieldThe form of the Hamiltonian and Lagrangian densities of the Klein Gordonfield are determined by the fact the Klein Gordon field is a scalar field. Thismeans that the values of the field are Lorentz invariant. They are thesame in any reference we perform a Lorentz transform on the field like (x ) = (x ) thenis suffices to transform the coordinatesx to obtain (x ) (x ) = (x ) = ( 1x )( )We did see that the Hamiltonian density transforms asE2, while the kinetic(density) termTtransforms likep2, see ( ). Since the scalar field itself doesn t transform we might expect differential operators which obtainE2andp2from the quantum field instead. We will see that this is indeedthe the classical non relativistic theory the Lagrangian is given byL=T case of a relativistic free particle there is no potential energy, howeverthere is the energy corresponding with the mass of the particle. This self-energy term, which plays a somewhat similar role as the potential energy,is absent in the non-relativistic did see that the kinetic termTbecomes12pvin the relativistic want to write the relativistic case in a form ofL=T W, whereWrelates to the mass-energy. Using the generally validL=pv Hwe canrewriteLlike:L= 12H+12pv+12L( )For a classical relativistic particle we should substitute the right handterms as follows (see section )L= 12 mc2+12 mv2 12 1mc2( )Going from the Lagrangian to the Lagrangian density we have to multiplythe right hand terms with and extra factor to compensate for Lorentz10Chapter 22. The Hamiltonian and Lagrangian densitiescontraction which confines the wave-function into a smaller volume, hencethe density goes up by a factor . At this point we will also multiply allright hand terms with the constantmc2for convenance. We will see , all the right hand terms are multiplied by mc2when going from thelagrangianLto the Lagrangian 12[ mc2]2+12[ mv]2c2 12[mc2]2( )The terms between square brackets we recognize as E, p and the rest-massenergy. We can now make the step from the classical relativistic particleto the Klein Gordon field theory:L=12[~ t]2 12[~ xi]2c2 12[mc2 ]2( )Setting (c=~= 1) gives us the familiar form of the Lagrangian density ofthe Klein Gordon 2 12 12m2 2( )Note that the middle term at the right hand side corresponds with theclassical kinetic term T which becomes the kinetic energy in the non rel-ativistic theory of the classical particle. We have derived the Lagrangiandensity for the scalar field. By assuming that is a Lorentz scalar weobtained the above Lagrangian density. We now use the Euler Lagrangeequation. ( L ( )) L = 0( )In order to obtain the equation of motion. What we get is the Klein Gordonequation. 2 t2 2 x2i+m2 = 0( )The scalar quantum field representation derived directly from the classicalrelativistic particle Hamiltonian of the scalar Klein Gordon Hamiltonian of the scalar Klein Gordon fieldFor the Hamiltonian density we go back to the expression for the classicalrelativistic particle since we want both Hamiltonians to correspond. Usingequation ( ) andH= 2T Lwe get for the Hamiltonian density ofthe classical [ mc2]2+12[ mv]2c2+12[mc2]2( )This corresponds with the following Hamiltonian density for the KleinGordon 12 2 12 +12m2 2( )With (c=~= 1). We can see how this Hamiltonian density transformsby applying it to a plane-wave of the form exp( iEt+ipx). We get:H E2+p2+m2 2+ 2 2+ 1 2( )Thus: the Hamiltoniandensitytransforms like 2where one factor stemsfrom the Hamiltonian being the 0 th component of the 4-momentum andthe second factor comes from the Lorentz contraction of the volumewhich confines the Hamiltonian density corresponds with the classical particle but differsin signs with the second quantization related Hamiltonian density11The Hamiltonian related with the second quantization of the Klein Gordon fieldis given byH=12 2+12( )2+12m2 2. The reason of the difference in signsis that the term is considered as the momentum in an internal or unspecified space. The term12 2is then considered to be theTinH= 2T L. The generalproblem in applying second order quantization in the relativistic theory is that the mixof internal or unspecified space and the usual space-time coordinates doesn t leadto the correct Lorentz transformation. A second problem is that12 2corresponds to thenon-relativistic expression12mv2instead of the relativistic version12pv12Chapter 22. The Hamiltonian and Lagrangian The complex Klein Gordon fieldThe complex Klein Gordon equation comes in when we need to describeboth particles and anti-particles. becomes a field with two components 1and 2. = 1+i 2( )How can we interpret these components. One possibility is to take onecomponent as a position (in some internal or unspecified space andthe other component as the momentum. The two would be 90oout ofphase in an oscillatory motion which we could associate with the particle sfrequency exp( iEt/~)Another way is to interpret both as coordinates on a plane of rotation. Theexpression exp( iEt/~) would then correspond with a circular Lagrangian for the complex field must contain both components andthe Euler-Lagrange equation must lead to the usual Klein Gordon we us the Lagrangian of the real Klein Gordon equation,L=12 2 12 12m2 2( )and simply use as a complex variable. Evaluating this with a plane wavesolution however doesn t produce a real the local phaseof the wave-function, its not a real ( 12E2+12p2 12m2) 2= m2 2( )The result we want is m2| |2or m2 which more explicitly expressedin its individual components is m2( 2+ 2)without the imaginary val-ued cross-term 2im2 1 2included inm2 2. We get the Lagrangiandensity we want by simply adding the two Lagrangians of the individualcomponents together. The same can be done for the interpretations lead to a specific direction in space, either the direction ofoscillation or the spin pointer. This is an issue for a scalar theory such as the onerepresented by the scalar Klein Gordon equation, which is not supposed to have anyspecial direction in The complex Klein Gordon field13Lagrangian density for the complex scalar field 1+i 2L= +(12 21 12 1 1 12m2 21)+(12 22 12 2 2 12m2 22)( )L=12( + m2 )( )Hamiltonian density for the complex scalar field 1+i 2H= +( 12 21 12 1 1+12m2 21)+( 12 22 12 2 2+12m2 22)( )H(x) =12( + +m2 )( )Some care is required with the signs here since = ( 21+ 22). Equation( ) is often found with reversed signs. One can easily check the requiredsigns by inserting a plane-wave eigenfunction into the Lagrangian density: = 1+i 2= cos( Et+px) +isin( Et+px),One should ( E2+p2 m2) = m2 ( )The equation of motion can be derived in a mathematically proper way3byapplying the Euler-Lagrange equation on ( ), taking the derivatives inthe fields 1, 2and their derivatives, and then defining the combined fieldagain as = 1+i 2. The result is the familiar Klein Gordon equation. 2 t2 2 x2i=m2 ( )3Despite what is often seen, the Euler Lagrange equation can not be applied directlyon expressions containing terms like . A derivative like / is not zero but un-determined since it violates the Cauchy-Riemann equations for complex differentiability:The derivative is not independent of the direction of in the complex 22. The Hamiltonian and Lagrangian Expressing the Lagrangian in a 4d environmentThe reader may have noticed that there seems to be an ambiguity in how wedefine the Lagrangian density. The expression for the Lagrangian densitywe obtained ( 2+( )2 m2 2)( )While we in fact could also have m2 2orL= 2+( )2( )Both expressions lead to the amount of phase change per unit of time perunit of volume over the trajectory of the particle, at least for we end up with a linear combination of the latter two For the Hamiltonian it seems that we could equally well 2orH=( )2+m2 2( )To get the amount of phase change per unit of time per unit of volumeover the time-axis. Instead we ended up (x) =12( 2+( )2+m2 2)( )One important argument is that we do not only want to know the La-grangian or Hamiltonian density in one particular reference frame, but wewant to know these quantities in all reference frames. Knowing a phasechange rate in one particular space-time direction doesn t say anythingabout the other directions. We need sufficient information to be ableto transform the Lagrangian and Hamiltonian density into any might suspect that Nature itself also needs such a definition in 4dspace-time, and that therefor, even though the expressions seem to beambiguous from a single reference point of view, they actually do representthe physics as required in a 4d-dimensional world obeying the rules ofspecial Electromagnetic and Proca Electromagnetic and Proca LangrangianIn correspondence with the Lagrangian densities discussed sofar we mightexpect the Lagrangian density for the electromagnetic four-vector to beexpressed by the +(12 A20 12 A0 A0 12m2A20) (12 A2x 12 Ax Ax 12m2A2x) (12 A2y 12 Ay Ay 12m2A2y) (12 A2z 12 Az Az 12m2A2z)( )Where all four components ofA have independently the form of the (clas-sical) Lagrangian field density. The requirement that the total Lagrangiandensity transforms like a Lorentz scalar imposes the (+,-,-,-) metric on thetime/space components, the signs in the first column of ( ). Theseexpressions can be written more compact as:Lp=12 A A 12m2c2A A ( )If the massmis not zero then we call the field aProca field. This expressionis however not complete. We have to replace the derivatives ofA in thefollowing sense. (We ll discuss the reason for this in a minute) A = A A , A = A A ( )So the Lagrangian density becomes. (in the massless case)L=12( A A ) ( A A ) =F F ( )Which we can write (using the normalization in SI) 22. The Hamiltonian and Lagrangian densitiesL=14 F F =12(BH - DE)( )This Lagrangian density is zero in case of electromagnetic radiation wherethe relation|E|=|cB|holds always. This shows us that the invariantphoton mass is zero. (The partial Lagrangian density ( ) is zero aswell in this case)Now, why do we need to subtract these extra terms? Well typically theenergy-momentum of the electromagnetic field is derived by calculatinghow the field acts on charges on a capacitor and currents in an inductor,so charge is involved in one way or the other and charge is represented byan U(1) symmetry: exp(i ).The terms we need to subtract induce an equal and indistinguishable U(1)phase as the regular Lagrangian components and therefor need to be takencare of. The total induced phase on a charged scalar field by the four-vectorA is defined in the following way. = exp[ i~ (po+eAo)dxo+i~3 i=1 (pi+eAi)dxi]( )Wherepis the inertial momentum defined by the invariant mass. We cansplit of the factor dependent onA as follows = p A. Since is ascalar we can write = , the order of the differential operatorsdoes not combination (p +eA ) must be curl free in any of the six 2D planesof 4D space-time otherwise otherwise the expression within the squarebrackets can not be a single valued scalar function. (p +eA )ds= 0( )The individual termsp andeA can have curl, so any curl ineA mustbe canceled by an opposite curl inp . This relationship gives rise tothe electromagnetic Lorentz Force. The expressioni~ pyields Electromagnetic and Proca Langrangian17changes of the momentap in the directionsx . All of these terms end upin the expression which describes the change of momentum of a particlemoving at some dt= p t+ p xvx+ p yvy+ p zvz( )Where the right hand side is just the mathematical expansion of the lefthand side (vx= x/ t, et-cetera ) It is no surprise that the expression Agives rise to the terms which end up in the electromagnetic fieldtensorF which yields the changing dt=e( A A )v =eF v ( )Where the term between brackets represents the curl ofA . Since theorder of differentiation is irrelevant we can deduce. = = (p +eA ) = (p +eA ) ( )The righthand side can be reordered as. p p = e( A A )( )This expression simplifies for a particle with~v= 0, and thus~p= 0 (atevery point in space) to. pi xo= e Ai xo e Ao xi=eE( )Which is just the electric part of the Lorentz force. A non-zero velocity~v, constant over space, gives rise to the full Lorentz force including themagnetic see that the assumption that somehow charge is involved in the La-grangian leads to extra terms which are indistinguishable from the basic18Chapter 22. The Hamiltonian and Lagrangian densitiesterms because of the U(1) symmetry. Certainly a term like12DEsuggestslikewise. It is the energy needed for the vacuum displacement currentDto move through the electric fieldEwhich builds up in the process as The electromagnetic equation of motionWe can derive the equations of motion with the use of the Euler Lagrangeequation starting from the Lagrangian o( A A )( A A )=12 o(B2 1c2E2)( )Multiplying the terms and realizing that a nd are dummy variables toadd all terms together to a single scalar result, we can rewrite this in theform which is generally used to derive the equations of o( A A A A )( )The first of the two terms is identical to the initial Lagrangian density( ) while the second term corresponds to the extra terms. We recallthe Euler Lagrange equation. ( L ( A )) L A = 0( )Applying it gives us the equation of motion. A A = 0( )The term A is zero ifA is a conserved current. We see that theequation of motion is not changed by the extra terms if this is the case. Incase of a vacuum without net charge-current density we obtain. A = 0( ) The electromagnetic interaction The electromagnetic interaction LagrangianThe electromagnetic interaction Lagrangian density represents the extraphase change rates induced by the four-potentialA on a charged (KleinGordon) field . It is given A ( )Which transforms as it should like a Lorentz scalar. This gives us for thetotal equation of motion. A A = oj ( )Which becomes under the assumption thatA is a conserved current(Lorentz gauge condition). A = oj ( )Here we recover the classical the classical wave equation ofA with thecharge-current densityj as its source. We can write the equation ofmotion as. ( A A )= oj ( )This allows us to express it with the use ofF instead ofA . F = oj ( )Which is a compact way of writing of the inhomogeneous Maxwell equa-tions, which we have hereby derived from the Lagrangian density. E=c ojo=1 o ( ) B 1c2 E t= o~j( )20Chapter 22. The Hamiltonian and Lagrangian Electromagnetic and Proca HamiltonianWe ll follow the same steps here as we did for the electromagnetic La-grangian density. In correspondence with the Hamiltonian densities dis-cussed sofar we expect the Hamiltonian density for the electromagneticfour-vector to be expressed by the +( 12 A20 12 A0 A0+12m2A20)+( 12 A2x 12 Ax Ax+12m2A2x)+( 12 A2y 12 Ay Ay+12m2A2y)+( 12 A2z 12 Az Az+12m2A2z)( )Where all four components ofA have independently the form of the (clas-sical) Hamiltonian field density. These expressions can be written morecompact as:Hp= 12 A A +12mc2A A ( )We used instead of here for later convenience. Both give the sameend result due to the square. This expression is again not complete. Wehave to replace the derivatives ofA in the following sense. A = A A ,( )For the same reason as explained in the section on the Lagrangian a zero mass field normalized for SI units this leads us to the wellknow expression for the energy density of the electromagnetic o( A A )( A A )( )H=14 oF F =12(BH+DE)( )

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