dirac_RQM_2006

Relativistic Quantum Mechanics SUPA Graduate School October/November 2006 David J. Miller University of Glasgow Recommended Text: Quarks & Leptons by F. Halzen and A. Martin (though this is not really necessary) http://www.physics.gla.ac.uk/~dmiller/lectures/RQM_ 2006.ppt 2Rough outline of topics /xrhombusNon-relativistic Quantum Mechanics /square4The Schrodingerequation /xrhombusRelativistic Quantum Mechanics /square4The Klein-Gordon Equation, The DiracEquation, Angul ar Momentum and Spin, Symmetries of the DiracEquation /xrhombusQuantum Electrodynamics /square4Classical Electromagnetism, The DiracEquation in an electromagnetic field /xrhombusScattering and Perturbation Theory /square4The QED Feynman rules, cross-sections, crossing sym metry, identical particles, the fermion propagator, decay rates /xrhombusQuantum Chromodynamics /square4Quarks, gluons and color, renormalisation, running couplings 3Non-relativistic Quantum Mechanics Consider a plane wave: This plane wave has energy E = /planckover2pi1ωand momentum p = /planckover2pi1k These values can be extracted using the energy and momentum operators defined by: Classically we know the relation between total ener gy and momentum is Writing this in terms of operators provides us with the Schrödinger equation : Erwin Schrödinger 4In the pre-school problems, you should have shown t hat the quantity satisfies the continuity equation But how do we interpret the Schrödinger equation an d the associated wavefunction? with Now, integrating over a volume V: Volume V enclosed by Area A J and using Gauss’Law Any change in the total ρin the volume must come about through a current Jthrough the surface of the volume. is a conserved density and we interpret it as the probability density for finding a particle at a particular position. Notice that ρis positive definite, as required. 5Relativistic Quantum Mechanics The Schrödinger Equation only describes particles i n the non-relativistic limit. To describe the particles at particle colliders we nee d to incorporate special relativity. Let’s have a quick review of special relativity We construct a position four-vector as An observer in a frame S /primewill instead observe a four-vector where denotes a Lorentztransformation. e.g. under a Lorentzboost by vin the positive x direction: 6The quantity xµxµis invariant under a Lorentztransformation where is the metric tensor of Minkowski space-time. A particle’s four-momentum is defined by where is proper time , the time in the particle’s own rest frame. Proper t ime is related to an observer’s time via Its four-momentum’s time component is the particle’ s energy, while the space components are its three-momentum and its length is an invariant, its mass : note the definition of a covector ⇔ 7Finally, I define the derivative Note that you will sometimes use the vector express ion Watch the minus sign! For simplicity, from now on I will use natural units . Instead of writing quantities in terms of kg, m and s, we could write them in terms of c, /planckover2pi1and eV: c = 299 792 458 ms -1 /planckover2pi1= 6.582 118 89(26) ×10 -16 eVs 1 eV = 1.782 661 731(70) c 2kg So any quantity with dimensions kg ambsccan be written in units of eV α/planckover2pi1βcγwith, Then we omit /planckover2pi1and c in our quantites (you can work them out from t he dimensions).transforms as so 8The Klein-Gordon Equation The invariance of the four-momentum’s length provid es us with a relation between energy, momentum and mass: (We have set V=0 for simplicity.) ( ∂2≡∂µ∂µis sometimes written as
) Oskar Klein This has plane-wave solutions normalization Replacing energy and momentum with , gives the Klein-Gordon equation: Alternatively, in covariant notation: with gives 9This is the relativistic wave equation for a spin zero particle, which conventionally is denoted φ. Under a Lorentztransformation the Klein-Gordon op erator is invariant: with So (Lorentztrans. preserve the norm) real (since is real) Under continuous Lorentztransformations, S must be the same as for the identity, ie. S = 1 If S = 1, then φis a scalar If S = -1, then φis a pseudoscalar But for a parity inversion it can take either sign 10 Since φis invariant, then | φ|2does not change with a Lorentztransformation. This is at odds with our previous interpretation of |φ|2as a probability density, since densities do change with Lorentztransformations (the probabilit y P = ρV = constant). One possible choice is: (this is the same current as before, just with a different normalisation) As a four-vector, We need new definitions for the density ρand current Jwhich satisfy the continuity equation or, convariantly, with 11 Consider our plane-wave solution: We have solutions with negative energy , and even worse, so these negative energy states have negative probability distributions ! We can’t just ignore these solutions since they wil l crop up in any Fourier decomposition. This is why Schrödinger abandoned this equation and developed the non- relativistic Schrödinger equation instead –he (impl icitly) took the positive sign of the square root so that he could ignore the nega tive energy solutions. 12 Feynman-StuckelbergInterpretation You will see in your QFT course that positive energ y states must propagate forwards in time in order to preserve causality. Feynman and Stuckelberg suggested that negative ener gy states propagate backwards in time . If the field is charged, we may reinterpret as a charge density, instead of a probability density: If E < 0, just move the sign into the time: Particles flowing backwards in time are then reinte rpreted as anti-particles flowing forwards in time. Now ρ= j0, so for a particle of energy E: while for an anti-particle of energy E: which is the same as the charge density for an elec tron of energy -E 13 In reality, we only ever see the final state partic les, so we must include these anti-particles anyway. Quantum mechanics does not adequately handle the cr eation of particle—anti-particle pairs out of the vacuum. For that you will need Quantum Field Theory .positive energy state flowing forwards in time time space negative energy state flowing backwards in time ≡ positive energy anti-particle state flowing forwards in time 14 The particle (or charge) density allows us to norma lize the KG solutions in a box.Normalization of KG solutions So if we normalize to 2Eparticles per unit volume , then N= 1 Notice that this is a covariant choice. Since the n umber of particles in a box should be independent of reference frame, but the volume of t he box changes with a Lorentzboost, the density must also change with a boost. In fact, the density is the time component of a four-vector j0. so in a box of volume V the number of particles is : 15 The Dirac Equation The problems with the Klein-Gordon equation all cam e about because of the square root required to get the energy: Dirac tried to get round this by finding a field equ ation which was linear in the operators. All we need to do is work out and Paul Dirac 16 So, comparing with we must have:Now, we have where iand jare summed over  ,, and are anti-commuting objects – not just numbe rs! 17 These commutation relations define αand β. Anything which obeys these relations will do. One possibility, called the Dirac representation , is the 4 ×4 matrices: 2×2 matrices where σiare the usual Paulimatrices: Since these act on the field ψ, ψitself must now be a 4 component vector, known as a spinor . 18 We can write this equation in a four-vector form by defining a new quantity γµ: And the DiracEquation is: (with ) The anti-commutation relations become: Often is written as 19 Does the DiracEquation have the right properties? Is the probability density positive definite? A appropriate conserved quantity is now with In four-vector notation, ( Note ) with Clearly always! /checkbld 20 Does the DiracEquation only have positive energy so lutions? For a particle at rest , Solutions: with with OR Since we want the energy, it is easier to work with out four-vector notation: Look for plane wave solutions: 4 component spinor 2 component spinors 21 Dirac got round this by using the PauliExclusion principle . He reasoned that his equation described particles w ith spin (e.g. electrons) so only two particles can occupy any particular energy level (o ne spin-up, the other spin-down). E=0 DiracSea …Energy If all the energy states with E<0 are already filled, the electron can’t fall into a negative energy state. … Moving an electron from a negative energy state to a positive one leaves a hole which we interpret as an anti-particle. Note that we couldn’t have used this argument for b osons (no exclusion principle) so the Feynman-Stuckleberg interpretation is more useful.Oops! We still have negative energy solutions! /xmarkbld 22 A General Solution We need to choose a basis for our solutions. Choose , Check these are compatible: since 23 Negative Energy Solutions, E < 0, are Antiparticle spinor, normalised to make Conventions differ here: sometimes the order is inverted Positive Energy Solutions, E > 0, are [Normalization choice (see next slide)] Typically, we write this in terms of the antipartic le’s energy and momentum: 24 Normalization of solutions Notice the normalization choice made for the spinors. So a covariant normalization is unit volume Just like for the KG equation, we can choose to hav e 2E particles per unit volume Occasionally, I will instead normalize spinorsto 2 Eparticles per volume V Then and only set V =1 at the end. But 25 Orthagonalityand completeness With the normalization of 2 Eparticles per unit volume, it is rather obvious tha t: This is a statement of orthogonality. Less obvious, but easy to show, are the completenes s relations: 26 Angular Momentum and Spin The angular momentum of a particle is given by . If this commutes with the Hamiltonian then angular momentum is conserved. So the quantity is conserved! This is not zero, so is not conserved! But, if we define then 27 is the orbital angular momentum, whereas is an intrinsic angular momentum Notice that our basis spinors are eigenvectors of with eigenvalues Note that an E < 0 electron with spin ≡an E > 0 positron with spin This is why we switched the labelling of the anti-p articles earlier. 28 Helicityof massless fermions If the mass is zero, our wave equation becomes These two component spinors, called Weyl spinors , are completely independent, and can even be considered as separate particles! Notice that each is an eigenstate of the operator with eigenvalues for masslessstate Writing then we find the equations decouple and 29 For the full Diracspinor, we define the Helicity operator as This is the component of spin in the direction of motion. A particle with a helicityeigenvalue is right handed p A particle with a helicityeigenvalue is left handed p Since an antiparticle has opposite momentum it will have opposite helicity. left handed particle right handed antiparticle 30 We can project out a particular helicityfrom a Dira cspinorusing γmatrices. Then a spinor PLuwill be left handed, while PRuwill be right handed.Define and projection operators This is the Dirac representation. e.g. but so PRuis right handed. 31 We can make this more explicit by using a different representation of the γmatrices. Now The left-handed Weylspinorsits in the upper part of the Diracspinor, while the right handed Weylspinorsits in the lower part. e.g. The chiral representation (sometimes called the Weyl representation ) is: 32 Since parity transforms left handed particles onto right handed ones (and vice versa), i.e. the weak interactions is parity violating. Also, helicityis only a good quantum number for massless particles. If a particle has a mass, I can always move to a re ference framewhere I am going faster than it, causing the momentum to reverse dir ection. This causes the helicityto change sign. For a massless particle there is no such frame and h elicityis a good quantum number.The weak interaction acts only on left handed particles. You will explicitly see in your QFT course that a m ass term in the Lagrangian looks like so mass terms mix left and right handed states. (chiral rep.) 33 Symmetries of the DiracEquation The LorentzTransformation How does the field behave under a Lorentz transfromation? ( γµand mare just numbers and don’t transform) Premultiplyby : This notation differs in different texts. e.g. Peskin and Schroeder would write 34 [I jumped a few steps here] write (just a parameterisation) [ignoring terms ] We can find S for an infinitesimal proper transformation [antisymmetric] This tells us how a fermion field transforms under a Lorentzboost. 35 scalar pseudoscalar vector axial vector tensor Common fermion bilinears:The adjoint transforms as [since for the explicit form of Sderived above] So is invariant. And so our current is a four-vector. 36 Can you derive the parity transformations of the bi linears given on the last slide? You should see that ηdrops out, so there is no loss of generality settin g η= 1 Parity A parity transformation is an improper Lorentztransformation described by Again , so and Since commutes with itself (trivially) and a nticommutes with , a suitable choice is 37 Charge Conjugation Another discrete symmetry of the Dirac equation is t he interchange of particle and anti- particle. Therefore we need Csuch that Premultiplyby and the Dirac equation be comes: Take the complex conjugate of the Dirac equation: used and 38 The form of C changes with the representation of th e γ-matrices. For the Dirac representation a suitable choice is How does this transformation affect the stationary solutions? We have mapped particle states onto antiparticle st ates, as desired.etc 39 Time Reversal A naive transformation of the wavefunction is not su fficient for time reversal. Since the momentum of a particle, is a rate of change , it too must change sign. Changing the momentum direction for a plane wave gi ves: We must (again!) make a complex conjugation: Take complex conjugation of DiracEquation, switch and pre-multiply by T: 40 Need: A suitable choice is: 41 CPT For the discrete symmetries, we have shown: So if is an electron, is a positron travelling backwards in space-time multiplied by a factor . This justifies the Feynman-Stuckleberginterpretatio n! Doing all of these transformations gives us 42 Quantum Electro-Dynamics (QED) Maxwell’s equations: Maxwell wrote these down in 1864 , but amazingly they are covariant! James Clerk Maxwell Classical Electromagnetism Writing and they are Note: the ability to write Maxwell’s Equations in this form is not a proof of covariance!] [ 43 Maxwell’s equations can also be written in terms of a potential Aµ Choose λsuch that Now, notice that I can change Aµby a derivative of a scalar and leave Fµ νunchanged Writing we have This is a gauge transformation , and the choice is know as the Lorentzgauge . In this gauge: 44 The wave equation with no source, has solutions with polarisation vector So has only 3 degrees of freedom (two transve rse d.o.f. and one longitudinal d.o.f.) The Lorentzcondition We still have some freedom to change Aµ, even after our Lorentzgauge choice: is OK, as long as Usually we choose such that . This is known as the Coulomb gauge . So only two polarisation states remain (both transv erse). 45 The Dirac Equation in an Electromagnetic Field So far, this has been entirely classical. So how do we incorporate electromagnetism into the quantum Dirac equation? Often we write Dµis called the Covariant Derivative , and the DiracEquation in an electromagnetic field becomes We do the ‘obvious’ thing and replace the momentum o perator Then charge of the electron = -e Beware: conventions differ, e.g. Halzen and Martin have while Peskin& Schroeder have as above 46 The Magnetic Moment of the Electron We saw that the interaction of an electron with an electromagnetic field is given by Writing as before, Coulomb gauge ⇒A0=0 and So Also, 47 So we have, The magnetic moment is composed of a contribu tion from theorbital angular momentum, and the intrinsic spin angular momentum gyromagnetic ratio In the non-relativistic limit, a nd , so we can write the Dirac equation as approximately: This is an magnetic moment interaction with 48 The Diracequation predicts a gyromagnetic ratio g = 2 We can compare this with experiment: gexp = 2.0023193043738 ±0.0000000000082 The discrepancy of g-2 from zero is due to radiativecorrections The electron can emit a photon, interact, and reabs orb the photon. The muon’smagnetic moment is more interesting becau se it is more sensitive to new physics.Theory: Experiment: excellent agreement!If one does a more careful calculation, including t hese effects, QED predicts: 49 Now we have the Dirac equation in an Electromagnetic field we can calculate the scattering of electrons (via electromagnetism) . We will assume that the coupling eis small, and that far away from the interaction, i.e. outside the shaded area, the electrons are fre e particles. γa bc dScattering and Perturbation theory The DiracEquation (in a field) can be written: with c.f. the Schrödinger equation in a potential V [Remember γ0γ0= 1 ] 50 Let’s assume that the state at time is an momentum eigenstate of the free Dirac equation (V=0) with energy i.e. with Our Dirac equation in an external field is We need to solve this equation for . Now, since form a complete set, any solution must be of the form with Let’s expand this in powers of e: 51 Let’s stick this in and see what we get: cancel To order e: To order e: We can now extract using the orthogonal ityof : 52 But at time the initial state is , Integrate over t: zero By time the interaction has stopped. The probability of finding the system in a state is given by to order ewith: 53 Explicitly putting in our gives OK, so now we know the effect of the field Aµon the electron, but what Aµdoes the other electron produce to cause this effect? b da c 54 Putting this all together: a bc dqforces momentum conservation 55 Feynman Diagrams: The QED Feynman Rules We can construct transition amplitudes simply by as sociating a mathematical expression with the diagram describing the interaction. • for each incoming electron • for each outgoing electron • for each incoming positron • for each outgoing positron • for each incoming photon • for each outgoing photon • for each internal photon • for each internal electron • for each vertex p Remember that γ-matrices and spinors do not commute, so be careful with the order in spin lines. Write left to right, against the fermion flow. Richard Feynman For each diagram, write: p 56 2 details: •Closed loops : Integrate over loop momentum and include an extra factor of -1 if it is a fermion loop. •Fermi Statistics : If diagrams are identical except for an exchange of electrons, include a relative –sign. k p pp pp pp p– These rules provide , and the transition amplitude is The probability of transition from initial to final state is 57 An example calculation: k p/primek/prime pe-e- µ- µ-This is what we had before. To get the total probability we must square this, average over initial spins , and sum over final spins . But 58 (summation over α, β, γ, δ= 1,2) But don’t forget that the uare 4-component spinorsand the γare 4 ×4 matrices: So We need some trace identities! We can simplify this using the completeness relatio n for spinors: beware normalization here – this is only true for 2E particles per unit volume Then 59 Trace Identities [This is true for any odd number of γ-matrices] Using these identities: So be careful with this one! 60 If we are working at sufficiently high energies, th en and we may ignore the masses. Then Often this is written in terms of MandlestamVariables , which are defined: [Note that ] 61 Cross-sections So we have but we are not quite there yet –we need to turn this into a cross-section. But we need the transition probability per unit tim e and per unit volume is:since Recall ⇒ 62 The cross-section is the probability of transition per unit volume, p er unit time ×the number of final states / initial flux. Initial Flux In the lab frame, particle A, moving with velocity , hits particle B, which is stationary. A B The number of particles like A in the beam, passing through volume V per unit time is The number of particles like B per volume V in the target is So the initial flux in a volume V is For a collider , where A and B are both moving, this becomes:×# final states initial flux 63 # final states How many states of momentum can we fit in a vol ume V? In order to not have any particle flow through the boundaries ofthe box, we must impose periodic boundary conditions . LSo in a volume V we have Note that so this is covariant!But there are 2 EV particles per volume V, so # final states per particle = so the number of states between pxand px+dp xis 64 Putting all this together, the differential cross-section is: where the Flux Fis given by, and the Lorentz invariant phase space is, momentum conservation on-shell conditions integration measure 65 In the centre-of-mass , this becomes much simpler Then the Flux becomes This frame is defined by a nd Remember So and with 66 Also and with relations analogous to those for paand pb Putting this together: CM since The phase space measure becomes: 67 Returning to our process With me = mµ= 0 Τhe fine structure constant θa bc dIn terms of the angle between aand c The differential cross-section is: Notice that this is divergent for small angles: as This is exactly the same divergence as is in the Ru therford scattering formula. 68 Crossing symmetry crossing Generally, in a Feynman diagram, any incoming parti cle with momentum pis equivalent to an outgoing antiparticle with momentum –p. This lets us use our result for e-µ-→e-µ-to easily calculate the differential cross-section for e+e-→µ+µ-. crossing i.e. s ↔t 69 e+e- µ+µ- ⇒ Be careful not to change the s from flux and phase space! Writing θas the angle between the e-and µ-, as before ⇒Notice the singularity is gone! The total cross-section is 70 Identical particles in initial or final state So far, in the reactions we have looked at, the fin al state particles have all been distinguishable . form one another. If the final state particles ar e identical, we have additional Feynman diagrams. pcand pdinterchanged interchange of identical fermions ⇒minus sign e.g. e-e-→e-e- p pp pe- e-e- e-p pp pe-e- e- e-[See Feynman rules] 71 Since the final state particles are identical, thes e diagrams are indistinguishable and must be summed coherently . We have interference between the two contributions. 72 Compton Scattering and the fermion propagator Compton scattering is the scattering of a photon with an electron. e- e-e-e-γ γ γ γ + I just quoted the Feynman rule for the fermion propagator , but where did it come from? Let’s go back to the photon propagator first. Recall the photon propagator is The is the inverse of the photon’s wave equ ation: 73 The gµν is coming from summing the photon polarization vect ors over spins: So, the photon propagator is then For a massless fermion propagator we follow the same procedure The masslessfermion spin sum is so the massless fermion propagator is sometimes written this is for virtual photons 74 But what about massive propagators? Lets think about a massive scalar propagator since it is easier (we can forget the sp in-sum). We can consider the mass term as a perturbation on the ‘free’(i.e. massless) theory. with = + + +…. So the massive scalar propagator is 75 The Klein-Gordon equation leads to a propagator The same procedure on the Diracequation gives a propagator More precisely, the propagator is the momentum spac e Fourier transform of the wave equation’s Greens function Green’s function Sobeys: Writing , and pre-multiplying by gives ⇒ More details in your QFT course! 76 So now we are armed with enough information to calc ulate Compton Scattering + Putting in the Feynman rules, and following through , with me= 0 Can you reproduce this?You will need to use 77 Decay Rates So far we have only looked at 2 →2 processes, but what about decays? ×# final states # of decaying particles per unit volume A decay width is given by: This replaces the Flux. # of decay particles per unit volume For a decay we have ⇒# final states 78 In the rest frame of particle a: ab cwith The decay is back-to-back But 79 Remember, to get the total decay rate, you need to sum over all possible decay processes. The inverse of the total width will gi ve the lifetime of the particle: If the number of particles = Na then, 80 Quantum Chromo Dynamics (QCD) QCD describes the interaction of quarks and gluons . It is very similar to QED, except we have 3 types o f ‘charge’instead of just one. Conventionally we call these charges red , green and blue , and each quark can be written as a vector in “colorspace”: The force between the quarks is mediated by gluons which can also change the colorof the quarks. RGB However, QCD is symmetric under rotations in this c olor-space, so we can always rotate the quarks to pure colorstates and say they are either red, green or blue. This symmetry is known as SU(3) color , and parallels the U(1) QED symmetry of QED.Quarks, Gluons and Color 81 Since we have 3different sorts of quark (red, green and blue), to connect them all together we na ϊvelyneed 3 ×3 = 9 different gluons. Since we are connecting together quarks of different color, the gluons must be colored too.B B particle flow colorflow B RR RRB_ ≡ So, for example, we could have gluons: three orthogonal combinations of Conventionally these last 3 are Since QCD is symmetric to rotations in color-space, the first 8 of these must have related couplings. However, the last one is a color singlet, so in principle can have an arbitrary coupling. In QCD, its coupling is zero. ⇒We have 8 gluons 82 In order to transform one quark color-vector onto a nother, we need eight 3 ×3 matrices. These matrices are generators of the SU(3) group an d obey the SU(3) algebra , are conventionally normalised by and are traceless . The above matrix is not a very convenient choice (i t is actuallya ladder operator). Instead we normally write TAin terms of the Gell-Mann λmatrices .For example to turn a red quark into a blue quark w e need a gluon represented by i.e. Hence the removal of SU(3) structure constants 83 The Gell-Mann matrices are: 84 The full QCD Feynman rules will be given to you in the Standard Model course.At a vertex between quark and gluons we need to inc lude a factor b cA α i j The gluons also carry color , so we must also include a gluon-gluon interaction . This is given by AB Cα γβ p p p 85 Renormalisation When we calculate beyond leading order in our pertu rbative expansion, we will find that we have diagrams with loops in them. For example, the corrections to our e+e-→µ+µ-would include the diagram k p p k+ p This integral is infinite! But momentum conservation at all vertices leaves th e momentum flowing around the loop unconstrained! We need to integrate over this loop momentum, and find a result proportional to 86 To see that it is infinite, lets look at this integ ral in the limit as k→ ∞. Then we can neglect the momentum pand the mass m. The integral becomes this is a fake, because our approximation doesn’t work for k→0Ultra-Violet (UV) singularity This is not really that surprising. Even in classic al electromagnetism we have singularities when we go to small distances/high en ergies. For example, in classical electromagnetism, the ene rgy associated with a charged sphere of radius Ris: So classically, a point charge should have infinite energy! 87 Our theories such as QED and QCD make predictions of physical quantities . While infinities may make the theory difficult to work with, there is no real problem as long as o ur predictions of physical quantities are finite and match experiment . Are infinities really a problem? + + = finite In order for the physically measured mass to be finite , the bare mass must be infinite and cancel the divergence from the loop. We absorb infinities into unmeasurable bare quantiti es. To understand this, lets think about the one-loop c alculation of the electron mass We find, that in both QED and QCD, that our physica l observables are finite: they are renormalizable theories. Freeman Dyson 88 In reality, what we are doing is measuring differen ces between quantities. Since the loop contains a dependence on the momentu m scale, Q, the mass changes with probed energy. The difference between two masses at different scales is:+ + =Q2 infinities are the same in both m1’s ⇒finite The difference between the masses is finite . 89 Both philosophies, absorption or subtraction of sin gularities, are doing the same thing. We replace the infinite bare quantities in the Lagrangianwith finite physical ones . This is called renormalization . The beauty of QED (and QCD) is that we don’t need t o do this for every observable (which would be rather useless). Once we have done it for certain observables, everything is finite! This is a very non-trivial st atement. We say that QED and QCD are renormalizable . In QED we choose to absorb the divergences into: electron charge electron mass electron wave-function photon wave-function Instead of writing our observables in terms of the infinite bare quantities , we write them in terms of the physical measurable ‘ renormalized’ quantities In order to do this, we must first regularize the divergences in our integrals. 90 Regularization by a Momentum cut-off The most obvious regularization is to simply forbid any momenta above an scale Λ. Then, the integral becomes The UV divergence has been regularized . This isn’t very satisfactory though, since this breaks gauge invariance. Dimensional Regularization The most usual way to regulate the integrals is to work in dimensions rather than 4 dimensions. we have increased the power of kin the denominator, making the integral finite 91 More precisely, out original integral (ignoring mas ses for simplicity) gives: finite divergent as Notice that it is rather arbitrary which bit one wa nts to absorb or subtract off. One could subtract off only the pole in epsilon, i. e. for the above integral. This is known as the Minimal Subtraction , denoted MS . Also notice the renormalization scale Q .[Euler-Mascheroni Constant] This choice is known as MS Alternatively we could have removed some of the fin ite terms too, e.g. 92 Running couplings How does the QED coupling echange with quantum corrections? = + + + + … these cancel, due to a Ward Identity I can include some extra loops by…. = + + + … 93 In terms of , we find but since this was general, I could have chosen to evaluate my coupling at a different scale e.g. cut-off I can use this second equation to eliminate α0(which is infinite) from my first equation. The QED coupling changes with energy. 94 We can do the same thing for QCD, except we have so me extra diagrams e.g. We find, where Nc= # of colors= 3 Nf= # of active flavors At higher orders in perturbation theory we will hav e more contributions. The complete evolution of the coupling is described by the beta function 95 For . the QCD and QED couplings run in the opposite direction. Asymptotic freedom QCD confinement QED absurdly high energy At low energies QCD becomes strong enough to confine quarks inside hadrons. (The βfunction is not proof of this!) At high energies QCD is asymptotically free , so we can use perturbation theory.