Lecture notes Quantum field Theory

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Quantum Field Theory Mark Srednicki University of California, Santa Barbara markphysics.ucsb.edu c 2006 by M. Srednicki All rights reserved. Please DO NOT DISTRIBUTE this document. Instead, link to http://www.physics.ucsb.edu/∼mark/qft.html 11: Attempts at relativistic quantum mechanics 19 1 Attempts at relativistic quantum mechanics Prerequisite: none In order to combine quantum mechanics and relativity, we must first un- derstand what we mean by “quantum mechanics” and “relativity”. Let us begin with quantum mechanics. Somewhere in most textbooks on the subject, one can find a list of the “axioms of quantum mechanics”. These include statements along the lines of The state of the system is represented by a vector in Hilbert space. Observables are represented by hermitian operators. The measurement of an observable yields one of its eigenvalues as the result. And so on. We do not need to review these closely here. The axiom we need to focus on is the one that says that the time evolution of the state of the system is governed by the Schr¨odinger equation, ∂ ψ,ti=Hψ,ti, (1.1) i¯h ∂t where H is the hamiltonian operator, representing the total energy. Let usconsider avery simplesystem: aspinless, nonrelativistic particle with no forces acting on it. In this case, the hamiltonian is 1 2 H = P , (1.2) 2m where m is the particle’s mass, and P is the momentum operator. In the position basis, eq.(1.1) becomes 2 ∂ ¯h 2 ih ¯ ψ(x,t) =− ∇ ψ(x,t), (1.3) ∂t 2m whereψ(x,t)=hxψ,ti is the position-space wave function. We would like to generalize this to relativistic motion. The obvious way to proceed is to take q 2 2 2 4 H =+ P c +m c , (1.4)1: Attempts at relativistic quantum mechanics 20 which yields the correct relativistic energy-momentum relation. If we for- mally expand this hamiltonian in inverse powers of the speed of lightc, we get 1 2 2 H =mc + P +... . (1.5) 2m This is simply a constant (the rest energy), plus the usual nonrelativistic hamiltonian, eq.(1.2), plus higher-order corrections. With the hamiltonian given by eq.(1.4), the Schr¨odinger equation becomes q ∂ 2 2 2 2 4 ih ¯ ψ(x,t) =+ −¯h c ∇ +m c ψ(x,t). (1.6) ∂t Unfortunately, this equation presents us with a number of difficulties. One is that it apparently treats space and time on a different footing: the time derivative appears only on the left, outside the square root, and the space derivatives appear only on the right, under the square root. This asymme- try between space and time is not what we would expect of a relativistic 2 theory. Furthermore, if we expand the square root in powers of∇ , we get an infinite number of spatial derivatives acting onψ(x,t); this implies that eq.(1.6) is not local in space. We can alleviate these problems by squaring the differential operators on each side of eq.(1.6) before applying them to the wave function. Then we get 2   ∂ 2 2 2 2 2 4 −h ¯ ψ(x,t)= −h ¯ c ∇ +m c ψ(x,t). (1.7) 2 ∂t This is the Klein-Gordon equation, and it looks a lot nicer than eq.(1.6). It is second-order in both space and time derivatives, and they appear in a symmetric fashion. To better understand the Klein-Gordon equation, let us consider in more detail what we mean by “relativity”. Special relativity tells us that physicslooksthesame inallinertialframes. Toexplainwhatthismeans,we first suppose that a certain spacetime coordinate system (ct,x) represents 0 μ (by fiat) an inertial frame. Let us define x = ct, and write x , where μ=0,1,2,3, in place of (ct,x). It is also convenient (for reasons not at all 0 i obvious at this point) to define x = −x and x = x , where i = 1,2,3. 0 i This can be expressed more elegantly if we first introduce the Minkowski metric,   −1   +1   g = , (1.8) μν   +1 +1 ν where blank entries are zero. We then have x =g x , where a repeated μ μν index is summed.1: Attempts at relativistic quantum mechanics 21 Toinvertthisformula,weintroducetheinverseofg,whichisconfusingly also called g, except with both indices up:   −1   +1 μν   g = . (1.9)   +1 +1 μν μ μ We then haveg g =δ , whereδ is the Kronecker delta (equal to one νρ ρ ρ if its two indices take on the same value, zero otherwise). Now we can also μ μν write x =g x . ν It is a general rule that any pair of repeated (and therefore summed) indices must consist of one superscript and one subscript; these indices are said to be contracted. Also, any unrepeated (and therefore unsummed) indices must match (in both name and height) on the left- and right-hand sides of any valid equation. Now we are ready to specify what we mean by an inertial frame. If the μ coordinatesx represent an inertial frame (which they do, by assumption), μ then so do any other coordinates x¯ that are related by μ μ ν μ x¯ =Λ x +a , (1.10) ν μ μ whereΛ isaLorentz transformation matrix anda isatranslation vector. ν μ μ μ Both Λ and a are constant (that is, independent of x ). Furthermore, ν μ Λ must obey ν μ ν g Λ Λ =g . (1.11) μν ρ σ ρσ Eq.(1.11) ensures that the interval between two different spacetime points μ ′μ μ ′μ that are labeled by x and x in one inertial frame, and by x¯ and x¯ in another, is the same. This interval is defined to be ′ 2 ′ μ ′ ν (x−x) ≡ g (x−x) (x−x) μν ′ 2 2 ′ 2 = (x−x) −c (t−t) . (1.12) In the other frame, we have ′ 2 ′ μ ′ ν (x¯−x¯) = g (x¯−x¯) (x¯−x¯) μν μ ν ′ ρ ′ σ = g Λ Λ (x−x) (x−x) μν ρ σ ′ ρ ′ σ = g (x−x) (x−x) ρσ ′ 2 = (x−x) , (1.13) as desired. When we say that physics looks the same, we mean that two observers (Alice and Bob, say) using two different sets of coordinates (representing1: Attempts at relativistic quantum mechanics 22 two different inertial frames) should agree on the predicted results of all possibleexperiments. Inthecaseofquantummechanics,thisrequiresAlice andBobtoagreeonthevalueofthewavefunctionataparticularspacetime point, a point that is called x by Alice and x¯ by Bob. Thus if Alice’s ¯ predicted wave function is ψ(x), and Bob’s is ψ(x¯), then we should have ¯ ¯ ψ(x) =ψ(x¯). Furthermore, in order to maintain ψ(x) =ψ(x¯) throughout ¯ spacetime, ψ(x) andψ(x¯) should obey identical equations of motion. Thus acandidatewave equation shouldtake thesameformin any inertial frame. LetusseeifthisistrueoftheKlein-Gordonequation. Wefirstintroduce some useful notation for spacetime derivatives:   ∂ 1 ∂ ∂ ≡ = + ,∇ , (1.14) μ μ ∂x c∂t   ∂ 1 ∂ μ ∂ ≡ = − ,∇ . (1.15) ∂x c∂t μ Note that μ ν μν ∂ x =g , (1.16) so that our matching-index-height rule is satisfied. ¯ If x¯ and x are related by eq.(1.10), then ∂ and ∂ are related by μ μ ν ¯ ∂ =Λ ∂ . (1.17) ν To check this, we note that ρ σ ρ μ σ ν μ ρ σ μ ν ρ σ μν ρσ ¯ ∂ x¯ =(Λ ∂ )(Λ x +a )=Λ Λ (∂ x )=Λ Λ g =g , μ ν μ ν μ ν (1.18) as expected. Thelastequality in eq.(1.18) is another formof eq.(1.11); see section 2. We can now write eq.(1.7) as 2 2 2 2 2 2 2 4 −h ¯ c ∂ ψ(x) =(−¯h c ∇ +m c )ψ(x). (1.19) 0 2 μ 2 2 After rearranging and identifying ∂ ≡∂ ∂ =−∂ +∇ , we have μ 0 2 2 2 2 (−∂ +m c/h ¯ )ψ(x)=0. (1.20) This is Alice’s form of the equation. Bob would write 2 2 2 2 ¯ ¯ (−∂ +m c/¯h )ψ(x¯)=0. (1.21) Is Bob’s equation equivalent to Alice’s equation? To see that it is, we set ¯ ψ(x¯)=ψ(x), and note that 2 μ ν μ μ ρ σ 2 ¯ ¯ ¯ ∂ =g ∂ ∂ =g Λ Λ ∂ ∂ =∂ . (1.22) μν μν ρ σ1: Attempts at relativistic quantum mechanics 23 Thus, eq.(1.21) is indeed equivalent to eq.(1.20). The Klein-Gordon equa- tion is therefore manifestly consistent with relativity: it takes the same form in every inertial frame. Thisisthegoodnews. ThebadnewsisthattheKlein-Gordonequation violates oneoftheaxiomsofquantummechanics: eq.(1.1), theSchr¨odinger equation in its abstract form. The abstract Schr¨odinger equation has the fundamentalpropertyofbeingfirstorderinthetimederivative,whereasthe Klein-Gordon equation is second order. This may not seem too important, but in fact it has drastic consequences. One of these is that the norm of a state, Z Z 3 3 ∗ hψ,tψ,ti = dxhψ,txihxψ,ti = dxψ (x)ψ(x), (1.23) is not in general time independent. Thus probability is not conserved. The Klein-Gordon equation obeys relativity, but not quantum mechanics. Dirac attempted to solve this problem (for spin-one-half particles) by introducing an extra discrete label on the wave function, to account for spin: ψ (x), a=1,2. He then tried a Schr¨odinger equation of the form a   ∂ j 2 ih ¯ ψ (x)= −ih ¯c(α ) ∂ +mc (β) ψ (x), (1.24) a ab j ab b ∂t j where all repeated indices are summed, andα andβ are matrices in spin- space. This equation, the Dirac equation, is consistent with the abstract Schr¨odinger equation. The state ψ,a,ti carries a spin label a, and the hamiltonian is j 2 H =cP (α ) +mc (β) , (1.25) ab j ab ab where P is a component of the momentum operator. j Since the Dirac equation is linear in both time and space derivatives, it has a chance to be consistent with relativity. Note that squaring the hamiltonian yields 2 2 j k 3 j j 2 2 2 (H ) =c P P (α α ) +mc P (α β+βα ) +(mc ) (β ) . (1.26) j j ab k ab ab ab j k Since P P is symmetric on exchange of j and k, we can replace α α by j k 1 j k its symmetric part, α ,α , where A,B = AB +BA is the anticom- 2 mutator. Then, if we choose matrices such that j k jk j 2 α ,α =2δ δ , α ,β =0, (β ) =δ , (1.27) ab ab ab ab ab we will get 2 2 2 2 4 (H ) =(P c +m c )δ . (1.28) ab ab 2 2 Thus,theeigenstates ofH aremomentumeigenstates, withH eigenvalue 2 2 2 4 p c +m c . This is, of course, the correct relativistic energy-momentum1: Attempts at relativistic quantum mechanics 24 relation. While it is outside the scope of this section to demonstrate it, it turnsoutthattheDiracequationisfullyconsistentwithrelativityprovided the Dirac matrices obey eq.(1.27). So we have apparently succeeded in constructing a quantum mechanical, relativistic theory There are, however, some problems. We would like the Dirac matrices to be 2× 2, in order to account for electron spin. However, they must in fact be larger. To see this, note that the 2× 2 Pauli matrices obey i j ij i σ ,σ =2δ ,andarethuscandidatesfortheDiracα matrices. However, thereisnofourthmatrixthatanticommutes withthesethree(easilyproven by writing down the most general 2×2 matrix and working out the three anticommutators explicitly). Also, we can show that the Dirac matrices must be even dimensional; see problem 1.1. Thus their minimum size is 4×4,anditremainsforustointerpretthetwoextrapossible“spin”states. However, these extra states cause a more severe problem than a mere overcounting. Acting on a momentum eigenstate, H becomes the matrix 2 cα·p+mc β. In problem 1.1, we find that the trace of this matrix is zero. Thus the four eigenvalues must be +E(p), +E(p),−E(p),−E(p), where 2 2 2 4 1/2 E(p) = +(p c +m c ) . The negative eigenvalues are the problem: they indicate that there is no ground state. In a more elaborate theory that included interactions with photons, there seems to be no reason why a positive energy electron could not emit a photon and drop down into a negative energy state. This downward cascade could continue forever. (The same problem also arises in attempts to interpret the Klein-Gordon equation as a modified form of quantum mechanics.) Dirac made a wildly brilliant attempt to fix this problem of negative energy states. His solution is based on an empirical fact about electrons: they obey the Pauli exclusion principle. It is impossible to put more than one of them in the same quantum state. What if, Dirac speculated, all the negative energy states were already occupied? In this case, a positive energy electron could not drop into one of these states, by Pauli exclusion. Many questions immediately arise. Why don’t we see thenegative elec- tric charge of this Dirac sea of electrons? Dirac’s answer: because we’re used to it. (More precisely, the physical effects of a uniform charge density dependontheboundaryconditionsatinfinitythatweimposeonMaxwell’s equations, and there is a choice that renders such a uniform charge density invisible.) However, Dirac noted, if one of these negative energy electrons were excited into a positive energy state (by, say, a sufficiently energetic photon), it would leave behind a hole in the sea of negative energy elec- trons. This hole would appear to have positive charge, andpositive energy. Dirac therefore predicted (in 1927) the existence of the positron, a particle with the same mass as the electron, butopposite charge. The positron was found experimentally five years later.1: Attempts at relativistic quantum mechanics 25 However, we have now jumped from an attempt at a quantum descrip- tion of a single relativistic particle to a theory that apparently requires an infinite number of particles. Even if we accept this, we still have not solved the problem of how to describe particles like photons or pions or alpha nuclei that do not obey Pauli exclusion. At this point, it is worthwhile to stop and reflect on why it has proven to be so hard to find an acceptable relativistic wave equation for a sin- gle quantum particle. Perhaps there is something wrong with our basic approach. And there is. Recall the axiom of quantum mechanics that says that “Observables are represented by hermitian operators.” This is not entirely true. Thereisoneobservableinquantummechanicsthatisnot represented by a hermitian operator: time. Time enters into quantum mechanics only when we announce that the “state of the system” depends on an extra parameter t. This parameter is not the eigenvalue of any operator. This is in sharp contrast to the particle’s positionx, which is the eigenvalue of an operator. Thus, space and time are treated very differently, a fact that is obscuredbywritingtheSchr¨odingerequationintermsoftheposition-space wave function ψ(x,t). Since space and time are treated asymmetrically, it is not surprisingthat we are having trouble incorporating asymmetry that mixes them up. So, what are we to do? In principle, the problem could be an intractable one: it might be im- possible to combine quantum mechanics and relativity. In this case, there would have to be some meta-theory, one that reduces in the nonrelativistic limittoquantummechanics,andintheclassicallimittorelativisticparticle dynamics, but is actually neither. This, however, turns out not to be the case. We can solve our problem, but we must put space and time on an equal footing at the outset. There are two ways to do this. One is to demote position from its status as an operator,andrenderitasanextralabel,liketime. Theotheristopromote time to an operator. Letusdiscussthesecondoptionfirst. Iftimebecomesanoperator,what do we use as the time parameter in the Schr¨odinger equation? Happily, in relativistic theories, there is more than one notion of time. We can use the proper time τ of the particle (the time measured by a clock that moves with it) as the time parameter. Thecoordinate timeT (the time measured by a stationary clock inan inertial frame)is then promoted to an operator. In the Heisenberg picture (where the state of the system is fixed, but the operatorsarefunctionsoftimethatobeytheclassicalequationsofmotion), μ 0 we would have operators X (τ), where X = T. Relativistic quantum mechanics can indeed be developed along these lines, but it is surprisingly1: Attempts at relativistic quantum mechanics 26 complicated to do so. (The many times are the problem; any monotonic function ofτ is justas good a candidate asτ itself for the propertime, and this infinite redundancy of descriptions must beunderstoodand accounted for.) One of the advantages of considering different formalisms is that they may suggest different directions for generalizations. For example, once μ we have X (τ), why not consider adding some more parameters? Then μ we would have, for example, X (σ,τ). Classically, this would give us a continuous family of worldlines, what we might call a worldsheet, and so μ X (σ,τ) would describe a propagating string. This is indeed the starting point for string theory. Thus, promoting time to an operator is a viable option, but is compli- cated in practice. Let us then turn to the other option, demoting position to a label. The first question is, label on what? The answer is, on oper- ators. Thus, consider assigning an operator to each point x in space; call these operators ϕ(x). A set of operators like this is called a quantum field. In the Heisenberg picture, the operators are also time dependent: iHt/¯h −iHt/¯h ϕ(x,t) =e ϕ(x,0)e . (1.29) Thus, bothposition and (in theHeisenberg picture) time arenow labels on operators; neither is itself the eigenvalue of an operator. So,nowwehavetwodifferentapproachestorelativisticquantumtheory, approaches that might, in principle, yield different results. This, however, is not the case: it turns out that any relativistic quantum physics that can be treated in one formalism can also be treated in the other. Which we use is a matter of convenience and taste. And, quantum field theory, the formalism in which position andtime are bothlabels on operators, is much more convenient and efficient for most problems. There is another useful equivalence: ordinary nonrelativistic quantum mechanics, for a fixed number of particles, can be rewritten as a quantum fieldtheory. Thisisaninformativeexercise,sincethecorrespondingphysics is already familiar. Let us carry it out. Begin with the position-basis Schr¨odinger equation for n particles, all with the same mass m, moving in an external potential U(x), and inter- acting with each other via an interparticle potential V(x −x ): 1 2 " n n j−1 2 X XX ∂ h ¯ 2 ih ¯ ψ = − ∇ +U(x ) + V(x −x ) ψ, (1.30) j j k j ∂t 2m j=1 j=1k=1 where ψ =ψ(x ,...,x ;t) is the position-space wave function. The quan- 1 n tum mechanics of this system can be rewritten in the abstract form of1: Attempts at relativistic quantum mechanics 27 eq.(1.1) by first introducing (in, for now, the Schr¨odinger picture) a quan- † tum fielda(x) and its hermitian conjugate a (x). We take these operators to have the commutation relations ′ a(x),a(x) = 0, † † ′ a (x),a (x) = 0, † ′ 3 ′ a(x),a (x) = δ (x−x), (1.31) 3 † whereδ (x) is the three-dimensional Dirac delta function. Thus,a (x) and a(x) behave like harmonic-oscillator creation and annihilation operators that are labeled by a continuous index. In terms of them, we introduce the hamiltonian operator of our quantum field theory, Z   2 3 † ¯h 2 H = dxa (x) − ∇ +U(x) a(x) 2m Z 1 3 3 † † + dxd yV(x−y)a (x)a (y)a(y)a(x). (1.32) 2 Now consider a time-dependent state of the form Z 3 3 † † ψ,ti= dx ...dx ψ(x ,...,x ;t)a (x )...a (x )0i, (1.33) 1 n 1 n 1 n whereψ(x ,...,x ;t)is somefunctionof thenparticle positionsandtime, 1 n and0i is the vacuum state, the state that is annihilated by all thea’s, a(x)0i =0. (1.34) It is now straightforward (though tedious) to verify that eq.(1.1), the ab- stract Schr¨odinger equation, is obeyed if andonly if the functionψ satisfies eq.(1.30). Thuswecaninterpretthestate0iasastateof“noparticles”, thestate † † † a (x )0iasastatewithoneparticleatpositionx ,thestatea (x )a (x )0i 1 1 1 2 as a state with one particle at position x and another at position x , and 1 2 so on. The operator Z 3 † N = dxa (x)a(x) (1.35) counts the total number of particles. It commutes with the hamiltonian, as is easily checked; thus, if we start with a state ofn particles, we remain with a state of n particles at all times. However, we can imagine generalizations of this version of the theory (generalizations that would not be possible without the field formalism) in which the number of particles is not conserved. For example, we could try adding to H a term like Z h i 3 † 2 ΔH∝ dx a (x)a (x)+h.c. . (1.36)1: Attempts at relativistic quantum mechanics 28 Thisterm doesnot commute withN, andsothenumberof particles would not be conserved with this addition to H. Theoriesinwhichthenumberofparticlescanchangeastimeevolvesare a good thing: they are needed for correct phenomenology. We are already familiar with the notion that atoms can emit and absorb photons, and so wehadbetterhave aformalism thatcanincorporatethis phenomenon. We are less familiar with emission and absorption (that is to say, creation and annihilation) of electrons, but this process also occurs in nature; it is less common because it must be accompanied by the emission or absorption of a positron, antiparticle to the electron. There are not a lot of positrons + − aroundto facilitate electron annihilation, whilee e pair creation requires 2 us to have on hand at least 2mc of energy available for the rest-mass energy of these two particles. The photon, on the other hand, is its own antiparticle, and has zero rest mass; thus photons are easily and copiously produced and destroyed. There is another important aspect of the quantum theory specified by eqs.(1.32) and (1.33). Because the creation operators commute with each other, only the completely symmetric part of ψ survives the integration in eq.(1.33). Therefore, without loss of generality, we can restrict our attention to ψ’s of this type: ψ(...x ...x ...;t)=+ψ(...x ...x ...;t). (1.37) i j j i This means that we have a theory of bosons, particles that (like photons or pionsoralphanuclei)obeyBose-Einsteinstatistics. IfwewantFermi-Dirac statistics instead, we must replace eq.(1.31) with ′ a(x),a(x) = 0, † † ′ a (x),a (x) = 0, † ′ 3 ′ a(x),a (x) = δ (x−x), (1.38) whereagainA,B =AB+BAis theanticommutator. Now onlythefully antisymmetric part of ψ survives the integration in eq.(1.33), and so we can restrict our attention to ψ(...x ...x ...;t)=−ψ(...x ...x ...;t). (1.39) i j j i Thus we have a theory of fermions. It is straightforward to check that the abstract Schr¨odinger equation, eq.(1.1), still implies that ψ obeys the 1 differential equation (1.30). Interestingly, there is no simple way to write 1 † Now, however, the ordering of the a and a operators in the last term of eq.(1.32) becomes significant, and must be as written.1: Attempts at relativistic quantum mechanics 29 downaquantumfieldtheorywithparticles thatobeyBoltzmann statistics, corresponding to a wave function with no particular symmetry. This is a hint of the spin-statistics theorem, which applies to relativistic quantum field theory. It says that interacting particles with integer spin must be bosons, and interacting particles with half-integer spin must be fermions. In our nonrelativistic example, the interacting particles clearly have spin zero (because their creation operators carry no labels that could be inter- preted as corresponding to different spin states), but can be either bosons or fermions, as we have seen. Now that we have seen how to rewrite the nonrelativistic quantum me- chanics of multiple bosons or fermions as a quantum field theory, it is time to try to construct a relativistic version. Reference Notes The history of the physics of elementary particles is recounted in Pais. A brief overview can be found in Weinberg I. More details on quantum field theory for nonrelativistic particles can be found in Brown. Problems 1.1) Show that the Dirac matrices must be even dimensional. Hint: show that the eigenvalues ofβ are all±1, and that Trβ =0. To show that 2 Trβ =0, consider, e.g., Trα β. Similarly, show that Trα =0. i 1 1.2) With the hamiltonian of eq.(1.32), show that the state defined in eq.(1.33) obeys the abstract Schr¨odinger equation, eq.(1.1), if and onlyifthewavefunctionobeyseq.(1.30). Yourdemonstrationshould applybothtothecaseofbosons,wheretheparticlecreationandanni- hilation operators obey the commutation relations of eq.(1.31), and to fermions, where the particle creation and annihilation operators obey the anticommutation relations of eq.(1.38). 1.3) Show explicitly that N,H = 0, where H is given by eq.(1.32) and N by eq.(1.35).2 Lorentz Invariance Prerequisite: 1 A Lorentz transformation is a linear, homogeneous change of coordinates μ μ from x to x¯ , μ μ ν x¯ =Λ x , (2.1) ν 2 μ that preserves the interval x between x and the origin, where 2 μ μ ν 2 2 2 x ≡x x =g x x =x −c t . (2.2) μ μν μ This means that the matrix Λ must obey ν μ ν g Λ Λ =g , (2.3) μν ρ σ ρσ where   −1   +1   g = . (2.4) μν   +1 +1 is the Minkowski metric. Notethatthissetoftransformationsincludesordinaryspatialrotations: 0 0 i i take Λ = 1, Λ = Λ = 0, and Λ = R , where R is an orthogonal 0 i 0 j ij rotation matrix. The set of all Lorentz transformations forms a group: the product of any two Lorentz transformations is another Lorentz transformation; the μ μ product is associative; there is an identity transformation, Λ = δ ; ν ν and every Lorentz transformation has an inverse. It is easy to demonstrate thesestatements explicitly. For example,tofindtheinversetransformation −1 μ ν (Λ ) , note that the left-hand side of eq.(2.3) can be written as Λ Λ , ν νρ σ ρ ν ρ and that we can raise the ρ index on both sides to get Λ Λ =δ . On ν σ σ −1 ρ ν ρ the other hand, by definition, (Λ ) Λ =δ . Therefore ν σ σ −1 ρ ρ (Λ ) =Λ . (2.5) ν ν Another useful version of eq.(2.3) is μν ρ σ ρσ g Λ Λ =g . (2.6) μ ν To get eq.(2.6), start with eq.(2.3), but with the inverse transformations −1 μ −1 ν (Λ ) and (Λ ) . Then use eq.(2.5), raise all down indices, and lower ρ σ all up indices. The result is eq.(2.6). For an infinitesimal Lorentz transformation, we can write μ μ μ Λ =δ +δω . (2.7) ν ν ν2: Lorentz Invariance 31 Eq.(2.3) can be used to show that δω with both indices down (or up) is antisymmetric: δω =−δω . (2.8) ρσ σρ Thus there are six independent infinitesimal Lorentz transformations (in four spacetime dimensions). These can be divided into three rotations (δω =−ε nˆ δθ for a rotation by angle δθ about the unit vector nˆ) and ij ijk k three boosts (δω =nˆ δη for a boost in the direction nˆ by rapidity δη). i0 i Not all Lorentz transformations can be reached by compounding in- −1 finitesimal ones. If we take thedeterminant of eq.(2.5), we get (detΛ) = detΛ, which implies detΛ = ±1. Transformations with detΛ = +1 are proper, and transformations with detΛ =−1 are improper. Note that the product of any two proper Lorentz transformations is proper, and that infinitesimal transformations of the form Λ = 1+δω are proper. There- fore, any transformation thatcan bereached by compoundinginfinitesimal ones is proper. The proper transformations form a subgroup of the Lorentz group. Anothersubgroupisthatoftheorthochronous Lorentztransformations: 0 0 2 i i those for which Λ ≥+1. Note that eq.(2.3) implies (Λ ) −Λ Λ =1; 0 0 0 0 0 0 thus, either Λ ≥ +1 or Λ ≤ −1. An infinitesimal transformation is 0 0 clearly orthochronous, and it is straightforward to show that the product of two orthochronous transformations is also orthochronous. Thus,theLorentztransformationsthatcanbereachedbycompounding infinitesimal ones are both proper and orthochronous, and they form a subgroup. We can introduce two discrete transformations that take us out of this subgroup: parity and time reversal. The parity transformation is   +1   −1 μ −1 μ   P =(P ) = . (2.9) ν ν   −1 −1 It is orthochronous, but improper. The time-reversal transformation is   −1   +1 μ −1 μ   T =(T ) = . (2.10) ν ν   +1 +1 It is nonorthochronous and improper. Generally, when a theory is said to be Lorentz invariant, this means under the proper orthochronous subgroup only. Parity and time reversal are treated separately. It is possible for a quantum field theory to be invariant under the proper orthochronous subgroup, but not under parity and/or time-reversal.2: Lorentz Invariance 32 From here on, in this section, we will treat the proper orthochronous subgroup only. Parity and time reversal will be treated in section 23. In quantum theory, symmetries are represented by unitary (or antiu- nitary) operators. This means that we associate a unitary operator U(Λ) to each proper, orthochronous Lorentz transformation Λ. These operators must obey the composition rule ′ ′ U(ΛΛ)=U(Λ)U(Λ). (2.11) For an infinitesimal transformation, we can write i μν δω M , (2.12) U(1+δω)=I + μν 2¯h μν νμ where M =−M is a set of hermitian operators called the generators −1 ′ −1 ′ of the Lorentz group. If we start withU(Λ) U(Λ)U(Λ)=U(Λ ΛΛ), let ′ ′ Λ =1+δω , and expand both sides to linear order in δω, we get −1 μν μ ν ρσ δω U(Λ) M U(Λ)=δω Λ Λ M . (2.13) μν μν ρ σ Then, since δω is arbitrary (except for being antisymmetric), the anti- μν symmetric part of its coefficient on each side must be the same. In this μν case, because M is already antisymmetric (by definition), we have −1 μν μ ν ρσ U(Λ) M U(Λ)=Λ Λ M . (2.14) ρ σ μν We see that each vector index on M undergoes its own Lorentz trans- formation. This is a general result: any operator carrying one or more vector indices should behave similarly. For example, consider the energy- μ 0 i momentum four-vector P , whereP is the hamiltonian H andP are the components of the total three-momentum operator. We expect −1 μ μ ν U(Λ) P U(Λ)=Λ P . (2.15) ν If we now let Λ = 1+δω in eq.(2.14), expand to linear order in δω, and equate the antisymmetric part of the coefficients of δω , we get the μν commutation relations   μν ρσ μρ νσ M ,M =ih ¯ g M −(μ↔ν) −(ρ↔σ). (2.16) These commutation relations specify the Lie algebra of the Lorentz group. We can identify the components of the angular momentum operator J as 1 jk i0 J ≡ ε M , and the components of the boost operatorK asK ≡M . i ijk i 2 We then find from eq.(2.16) that J,J = i¯hε J , i j ijk k J,K = i¯hε K , i j ijk k K,K = −i¯hε J . (2.17) i j ijk k2: Lorentz Invariance 33 The first of these is the usual set of commutators for angular momentum, and the second says that K transforms as a three-vector under rotations. The third implies that a series of boosts can be equivalent to a rotation. Similarly, we can let Λ=1+δω in eq.(2.15) to get   μ ρσ μσ ρ P ,M =i¯h g P −(ρ↔σ) , (2.18) which becomes J,H = 0, i J,P = ih ¯ε P , i j ijk k K,H = i¯hP , i i K,P = i¯hδ H , (2.19) i j ij μ Also, the components of P should commute with each other: P,P = 0, i j P,H = 0. (2.20) i Together,eqs.(2.17), (2.19),and(2.20)formtheLiealgebraofthePoincar´e group. Letusnow consider whatshouldhappentoa quantumscalar fieldϕ(x) under a Lorentz transformation. We begin by recalling how time evolution works in the Heisenberg picture: +iHt/h ¯ −iHt/¯h e ϕ(x,0)e =ϕ(x,t). (2.21) Obviously, this should have a relativistic generalization, −iPx/¯h +iPx/h ¯ e ϕ(0)e =ϕ(x), (2.22) μ where Px = P x = P·x−Hct. We can make this a little fancier by μ defining the unitary spacetime translation operator μ T(a)≡exp(−iP a /¯h). (2.23) μ Then we have −1 T(a) ϕ(x)T(a) =ϕ(x−a). (2.24) For an infinitesimal translation, i μ T(δa) =I− δa P . (2.25) μ h ¯ Comparing eqs.(2.12) and (2.25), we see that eq.(2.24) leads us to expect −1 −1 U(Λ) ϕ(x)U(Λ) =ϕ(Λ x). (2.26)2: Lorentz Invariance 34 Derivatives ofϕthencarryvector indicesthattransformintheappropriate way, e.g., −1 μ μ ρ −1 ¯ U(Λ) ∂ ϕ(x)U(Λ) =Λ ∂ ϕ(Λ x), (2.27) ρ wherethebaronaderivative means thatitiswith respectto theargument −1 x¯=Λ x. Eq.(2.27) also implies −1 2 2 −1 ¯ U(Λ) ∂ ϕ(x)U(Λ) =∂ ϕ(Λ x), (2.28) 2 2 2 2 so that the Klein-Gordon equation, (−∂ +m /h ¯ c )ϕ = 0, is Lorentz invariant, as we saw in section 1. Reference Notes A detailed discussion of quantum Lorentz transformations can be found in Weinberg I. Problems 2.1) Verify that eq.(2.8) follows from eq.(2.3). −1 ′ −1 ′ 2.2) Verify that eq.(2.14) follows from U(Λ) U(Λ)U(Λ)=U(Λ ΛΛ). 2.3) Verify that eq.(2.16) follows from eq.(2.14). 2.4) Verify that eq.(2.17) follows from eq.(2.16). 2.5) Verify that eq.(2.18) follows from eq.(2.15). 2.6) Verify that eq.(2.19) follows from eq.(2.18). 2.7) What property should be attributed to the translation operatorT(a) that could be used to prove eq.(2.20)? 2.8) a) Let Λ=1+δω in eq.(2.26), and show that μν μν ϕ(x),M =L ϕ(x), (2.29) where μν h ¯ μ ν ν μ (x ∂ −x ∂ ). (2.30) L ≡ i μν ρσ μν ρσ b) Show that ϕ(x),M ,M =L L ϕ(x). c) Prove the Jacobi identity, A,B,C+B,C,A+C,A,B =0. Hint: write out all the commutators. d) Use your results from parts (b) and (c) to show that μν ρσ μν ρσ ρσ μν ϕ(x),M ,M =(L L −L L )ϕ(x). (2.31)2: Lorentz Invariance 35 e) Simplify the right-hand side of eq.(2.31) as much as possible. f) Use your results from part (e) to verify eq.(2.16), up to the possi- bility of a term on the right-hand side that commutes withϕ(x) and its derivatives. (Such a term, called a central charge, in fact does not arise for the Lorentz algebra.) 2.9) Let us write μν ρ ρ i ρ Λ =δ + δω (S ) , (2.32) τ τ μν V τ 2¯h where μν ρ μρ ν νρ μ h ¯ (S ) ≡ (g δ −g δ ) (2.33) V τ τ τ i arematriceswhichconstitutethevector representation oftheLorentz generators. a) Let Λ=1+δω in eq.(2.27), and show that ρ μν μν ρ μν ρ τ ∂ ϕ(x),M =L ∂ ϕ(x)+(S ) ∂ ϕ(x). (2.34) V τ μν b) Show that the matrices S must have the same commutation V μν relations as the operators M . Hint: see the previous problem. c) For a rotation by an angle θ about the z axis, we have   1 0 0 0   0 cosθ −sinθ 0 μ   Λ = . (2.35) ν   0 sinθ cosθ 0 0 0 0 1 Show that 12 Λ=exp(−iθS /h ¯). (2.36) V d) For a boost by rapidity η in thez direction, we have   coshη 0 0 sinhη   0 1 0 0 μ   Λ = . (2.37) ν   0 0 1 0 sinhη 0 0 coshη Show that 30 Λ=exp(+iηS /h ¯). (2.38) V3: Canonical Quantization of Scalar Fields 36 3 Canonical Quantization of Scalar Fields Prerequisite: 2 Let us go back and drastically simplify the hamiltonian we constructed in section 1, reducing it to the hamiltonian for free particles: Z   3 † 1 2 H = dxa (x) − ∇ a(x) 2m Z 3 1 2 † e e = dp p a (p)a(p), (3.1) 2m where Z 3 dx −ip·x e a(p)= e a(x). (3.2) 3/2 (2π) Here we have simplified our notation by setting h ¯ =1. (3.3) The appropriate factors ofh ¯ can always be restored in any of our formulas viadimensionalanalysis. Thecommutation (oranticommutation) relations † e e of thea(p) and a (p) operators are ′ ae(p),ae(p) = 0, ∓ † † ′ ae (p),ae (p) = 0, ∓ † ′ 3 ′ ea(p),ae (p) = δ (p−p), (3.4) ∓ where A,B is either the commutator (if we want a theory of bosons) ∓ † e or the anticommutator (if we want a theory of fermions). Thus a (p) can be interpreted as creating a state of definite momentum p, and eq.(3.1) describes a theory of free particles. The ground state is the vacuum 0i; it is annihilated by ae(p), ae(p)0i =0, (3.5) and so its energy eigenvalue is zero. The other eigenstates of H are all of † † the form ae (p )...ae (p )0i, and the corresponding energy eigenvalue is 1 n 1 2 E(p )+...+E(p ), whereE(p) = p . 1 n 2m Itis easy toseehow togeneralize thistheorytoarelativistic one; allwe 2 2 2 4 1/2 need todois usetherelativistic energy formulaE(p) =+(p c +m c ) : Z 3 2 2 2 4 1/2 † H = dp(p c +m c ) ae (p)ae(p). (3.6) Now we have a theory of free relativistic spin-zero particles, and they can be either bosons or fermions.3: Canonical Quantization of Scalar Fields 37 Is thistheory really Lorentz invariant? We willanswer thisquestion (in the affirmative) in a very roundabout way: by constructing it again, from a rather different point of view, a point of view that emphasizes Lorentz invariance from the beginning. We will start with the classical physics of a real scalar field ϕ(x). Real means thatϕ(x) assigns a real number to every point in spacetime. Scalar μ meansthatAlicewhousescoordinatesx andcalls thefieldϕ(x)andBob μ μ μ ν ν whousescoordinatesx¯ ,relatedtoAlice’scoordinatesbyx¯ =Λ x +a , ν and calls the field ϕ¯(x¯), agree on the numerical value of the field: ϕ(x) = ϕ¯(x¯). This then implies that the equation of motion for ϕ(x) must be the same as that for ϕ¯(x¯). We have already met an equation of this type: the Klein-Gordon equation, 2 2 (−∂ +m )ϕ(x)=0. (3.7) Here we have simplified our notation by setting c=1 (3.8) in addition to h ¯ = 1. As with h ¯, factors of c can restored, if desired, by dimensional analysis. We will adopt eq.(3.7) as the equation of motion that we would like ϕ(x) to obey. It should be emphasized at this point that we are doing classical physics of a real scalar field. We are not to think of ϕ(x) as a quantum wave function. Thus, there should not be any factors ofh ¯ in this version of the Klein-Gordon equation. This means that the parameter m must have dimensions of inverse length; m is not (yet) to be thought of as a mass. The equation of motion can be derived from variation of an action R S = dtL, where L is the lagrangian. Since the Klein-Gordon equation is local, we expect that the lagrangian can be written as the space integral of R R 3 4 a lagrangian density L: L = dxL. Thus, S = dxL. The integration 4 μ μ ν measure dx is Lorentz invariant: if we change to coordinates x¯ =Λ x , ν 4 4 4 we have dx¯ = detΛdx = dx. Thus, for the action to be Lorentz in- ¯ variant, the lagrangian density must be a Lorentz scalar: L(x) = L(x¯). R R 4 4 ¯ ¯ Then we have S = dx¯L(x¯) = dxL(x) = S. Any simple function of ϕ is a Lorentz scalar, and so are products of derivatives with all indices μ contracted, such as ∂ ϕ∂ ϕ. We will take forL μ μ 2 2 1 1 L=− ∂ ϕ∂ ϕ− m ϕ +Ω , (3.9) μ 0 2 2 whereΩ isanarbitraryconstant. Wefindtheequationmotion(alsoknown 0 astheEuler-Lagrange equation)bymakinganinfinitesimalvariationδϕ(x)

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