The Dirac equation. L= i ψ(x)[ 1 2 2

Transcription

1 The Dirac equation Infobox 0.1 Chapter Summary The Dirac theory of spinor fields ψ(x) has Lagrangian density L= i ψ(x)[ m]ψ(x) / 2 where/ γ µ µ. Applying the Euler-Lagrange equation yields the Dirac equation [/ +m]ψ(x)=0 TheDiracmatricesobey{γ µ,γ ν }=2η µν withη µν themetricofminkowski space-time. The equal time anti-commutation relations are {ψ a (x),ψ b (y)}δ(x 0 y 0 )=δ b aδ 4 (x y) {ψ a (x),ψ b (y)}δ(x 0 y 0 )=0, {ψ a (x),ψ b (y)}δ(x 0 y 0 )=0 The Dirac theory has a phase symmetry which results in the conserved Noether current corresponding to particle number, J µ V (x)= ψ(x)γ µ ψ(x) The Dirac theory is Poincare invariant. The Noether currents associated with this symmetry,j µ (x)= T µν (x)f ν (x), can be formed from the appropriate Killing vectors f µ (x) and the symmetric, conserved energymomentum tensor T µν (x)= i 4 ψ(x)[γ µ ν +γ ν µ γ µ ν γ ν µ ]ψ(x) With the explicit representation γ i =[ 0 σi σ i 0 ], γ0 =[ 0 I I 0 ] (1) 1

2 and the ortho-normal helicity eigenvectors, the Dirac field is ψ(x)= d 3 k ei k x iet 2(2π) d 3 k e i k x+iet 2(2π) 3 2 σ ku + = k u +, σ ku = k u i 1 k /Eu + 1+ k /Eu a + ( k)+ + i 1+ k /Eu 1 k /Eu a ( k) i 1 k /Eu + 1+ k /Eu b +( i 1+ k)+ k /Eu + 1 k /Eu b ( k) (2) where E( k)= k2 +m 2. and the non-vanishing anti-commutation relations are {a + ( k),a +( k )}=δ( k k ), {a ( k),a ( k )}=δ( k k ) (3) {b + ( k),b +( k )}=δ( k k ), {b + ( k),b +( k )}=δ( k k ) (4) The vacuum is defined by a ± (k) O>=0, b ± (k) O>=0 k and particle and anti-particle states are created by a ±(k) and b ±(k), respectively. The Noether charges for phase and translation symmetries are given by N= d 3 xjv(x)= 0 d 3 k (a s(k)a s (k) b s(k)b s (k)) s=± H= d 3 xt 00 (x)= d 3 k k2 +m 2 (a s(k)a s (k)+b s(k)b s (k)) s=± P i = d 3 xt 0i (x)= d 3 k s=± k i (a s(k)a s (k)+b s(k)b s (k)) 2

3 1 The Dirac Equation So far, we have formulated an approach to the quantum mechanics of a many-particle system which led us to the non-relativistic field equation, in the absence of interactions, i h t ψ( x,t)=[ h 2 2m 2 +ǫ F ]ψ σ ( x,t)=0 (5) In this Lecture,we shall discuss the equation which replaces this one in a relativistic theory. From our point of view, the main difference between the two is symmetry. The field equation above has Galilean symmetry. We want to trade it for an equation that has Lorentz symmetry. We will continue to discuss a many-fermion system and to give the discussion a physical context, we will call the fermions electrons with the idea that they will eventually become the electrons of quantum electrodynamics. We have ignored the interaction terms in the above equation. We will continue to do this, to assume that the fermions are noninteracting. Later on, once we have a relativistic field equation, we will let the fields interact. To seek the appropriate relativistic wave equation, we could recall our discussion of the previous chapters. If we simply postulate a Hamiltonian with a relativistic dispersion relation, so that the wave equation (5) is replaced by i h t ψ σ( x,t)= m 2 c 4 c 2 h 2 2 ψ σ ( x,t) (6) the resulting theory has difficulties with causality. There is a finite probability of the particle propagating faster than the speed of light. The difficulty lies in the fact that the hamiltonian operator on the right-hand-side of this equation is not a polynomial in derivatives. Having all orders in derivatives, it is not a local differential operator. Dirac found a way to replace this equation by one where the hamiltonian has the same spectrum, but the operator is a polynomial in derivatives. We also have this goal. Let us begin by postulating the existence of an anti-particle which would satisfy and equation similar to (8) but with negative energy, i h t ψ σ ( x,t)= m 2 c 4 c 2 h 2 2 ψ σ ( x,t) (7) Assuming that both the particle and the anti-particle occur in the same theory, we could We could combine the two into the same multi-component 3

4 field to find i h t [ψ σ( x,t) ψ σ ( x,t) ]=[ m2 c 4 c 2 h m 2 c 4 c 2 h 2 2[[ψ σ( x,t) ψ σ ( x,t) ] (8) One might wonder whether there is a matrix hamiltonian which has eigenvalues ± m 2 c 2 +c 2 h 2 k 2 and which is polynomial in derivatives. There is no such 2 2 matrix. We thus need to involve the spin indices and consider [ ψ σ( x,t) ] as a four-component object. It is Dirac s great insight that our ψ σ ( x,t) problem can be solved with a 4 4 matrix. (We caution the reader that equation (8) is still not quite correct, as it has that form only in a particular basis for spins.) Consider the four Hermitian 4 4 matrices which have the algebraic properties β, α 1, α 2, α 3 ββ=i, βα i +α i β=0, α i α j +α j α i =2δ ij I wherei is the 4 4 unit matrix. (Alternatively, if we consider matrices with the above properties, there is a way to show that the minimal size of such matrices is 4 4. Then, we consider the wave equation i h ψ 1 ( x,t) ψ 1 ( x,t) ψ 2 ( x,t) =[βmc t ψ 3 ( x,t) 2 ψ +i hc α ] 2 ( x,t) ψ 3 ( x,t) ψ 4 ( x,t) ψ 4 ( x,t) The Dirac hamiltonian h D =[βmc 2 +i hc α ] (9) is a hermitian operator. It must have real eigenvalues. What is more h 2 D =m 2 c 4 h 2 c 2 2 (10) so, since the eigenvalues of 2 are k 2, h D has eigenvalues± m 2 c 4 + h 2 c 2 k 2, which is the desired property. Moreover, h D is a polynomial, in fact it is at most linear in derivatives, and it is therefore a local operator. 4

5 To make the Dirac equation look more covariant, we can define the matrices β=iγ 0, α=γ 0 γ (11) γ 0 = γ 0, γ i =γ i (12) {γ µ,γ ν }=2η µν (13) γ µ are called the Dirac gamma-matrices. Using them, the Dirac equation is the matrix differential equation 4 b=1 or, with implicit summations over indices, [γ µ ab µ+ mc h δ ab]ψ b ( x,t)=0 (14) [/ µ +m]ψ(x)=0 (15) where we shall use the slash notation / γ µ µ. We will hereafter assume that we are using a system of units where h=1 and where c=1. We shall find it useful to define the following quantity ψ(x) ψ (x)γ 0 Using this definition and taking a hermitian conjugate of the equation in (16), we obtain ψ(x)[ / m]=0 (16) The Dirac equation has a structure similar to the Schrödinger wave equation with the difference that, what we would call the single-particle Hamiltonian, h D, is a matrix and it is linear, rather than quadratic, in derivatives. We might expect that the Hamiltonian of the quantum field theory is given by H 0 = d 3 xψ (x)h D ψ(x)= d 3 xh(x) (17) H(x)=i ψ(x)[ γ +m]ψ(x) (18) and that the time derivative of ψ( x,t) is generated by this Hamiltonian by taking a commutator, i t ψ(x)=[ψ(x),h 0] 5

6 In fact this will indeed be the case if the field operator obeys the equal-time anti-commutation relations {ψ a ( x,t),ψ b ( y,t)}=δ b aδ 3 ( k y) {ψ a ( x,t),ψ b ( y,t)}=0, {ψ a ( x,t),ψ b ( y,t)}=0 (19) Our task in the following will be to assume that ψ( x,t) indeed obeys the Dirac equation and this anti-commutation relation and to find a solution of them. We have not discussed why the Dirac equation is relativistic. We will put off the details until later. Here, we observe that, given the anti-commutation algebra of the Dirac matrices, (γ µ µ ) 2 =γ µ γ ν µ ν = 1 2 {γµ,γ ν } µ ν = 2 Using this identity, we can operate the matrix valued differential operator ( γ ν ν +m) on the Dirac equation from the left to obtain ( γ ν ν +m)(γ µ µ +m)ψ=0 ( 2 +m 2 )ψ=0 We see that, if ψ(x) obeys the Dirac equation, it also obeys the relativistic wave equation( 2 +m 2 )ψ(x)=0. This implies that the solutions of the Dirac equation also obey this relativistic wave equation, and must therefore propagate like relativistic matter waves. 2 Solving the Dirac equation To see how the Dirac equation is solved, it is useful to choose a specific form for the Dirac matrices γ i =[ 0 σi σ i 0 ], γ0 =[ 0 I I 0 ] (20) wherei is the 2 2 unit matrix and σ i are the 2 2 Pauli matrices σ 1 =[ ], σ2 =[ 0 i i 0 ], σ3 =[ ] (21) The Pauli matrices have the properties σ i σ j +σ j σ i =2δ ij I, σ i σ j σ j σ i =2iǫ ijk σ k 6

7 where ǫ ijk is the totally antisymmetric tensor with ǫ 123 = 1. It is easy to confirm that the explicit form (20) indeed have the correct anti-commutation algebra for Dirac matrices. With the matrices in (20), the Dirac equation is m I [ 0 + σ ][ u(x) ]=0 (22) I 0 + σ m v(x) where we have split the four-component Dirac spinor into u(x) an v(x) which are two 2-component objects. To solve the differential equation, we use the ansatz [ u(x) v(x) ]=eipµ η µνx ν [ u v ]=e iωt+i k x [ u v ] (23) mu i[iω σ k]v=0 (24) mv+i[iω+ σ k]u=0 (25) We have now reduced the Dirac equation to two matrix equations. Equation (25) determines the 2-component object v in terms of u, that is, if u were known, we could determine v as Plugging this into (24) yields the condition v= i m [Iω+ σ k]u (26) ω 2 = k 2 +m 2 (27) which has two solutions for ω, ω= k2 +m 2 and ω= k2 +m 2, the positive and negative energy solutions, respectively. In the following, we will use the notation where ω is the frequency which can be either positive or negative, and E( k)= k2 +m 2 is positive, and sometimes abbreviated by E. Next, we note that σ k is a Hermitian matrix which can be diagonalized. Once diagonal, it has real eigenvalues. The eigenstates of this matrix are said to be eigenstates of helicity. It is left as an exercise to the reader to show that there exist two eigenvectors, σ ku + = k u +, σ ku = k u u +u + =1=u u, u +u =0=u u + 7

8 We shall find the following identities very useful u + u += k + σ k 2 k, u u = k σ k 2 k Then, putting it all together, we have four linearly independent solutions which we can superimpose to form the Dirac field, ψ(x)= d 3 k ei k x iet i 1 3 k /Eu + 2(2π) 2 1+ k /Eu a + ( k)+ + i 1+ k /Eu 1 k /Eu a ( k) + d 3 k e i k x+iet i 1 3 k /Eu + 2(2π) 2 1+ k /Eu b +( i 1+ k)+ k /Eu + 1 k /Eu b ( k) (28) where E( k)= k2 +m 2. This solution will obey the anti-commutation relation for the Dirac field(19) if the Fourier coefficients satisfy the non-vanishing anti-commutation relations are {a + ( k),a +( k )}=δ( k k ), {a ( k),a ( k )}=δ( k k ) (29) {b + ( k),b +( k )}=δ( k k ), {b + ( k),b +( k )}=δ( k k ) (30) All other combinations have vanishing anti-commutators. We can easily check that, with these anti-commutation relations for creation and annihilation operators, the solution in equation (28) obeys equations (19). With this solution, the Hamiltonian is H 0 = d 3 k k2 +m 2 (a +( k)a + ( k)+a ( k)a ( k)+b +( k)b + ( k)+b ( k)b ( k)) and the number operator is N= d 3 xψ (x,t)ψ( x,t) (31) = d 3 k(a +( k)a + ( k)+a ( k)a ( k) b +( k)b + ( k) b ( k)b ( k)) (32) In both of the above expressions, we have dropped infinite constants. Unlike in the non relativistic theory, the vacuum energy density and the vacuum 8

9 charge density both contain infinite constants which we have to simply drop in order to have a sensible Hamiltonian and number operator. We see from the expression (32), that, in direct analogy to the non relativistic system that we have studied, electrons, which are associated with a ± ( k) and a ±( k) contribute positively to the particle number whereas holes, or positrons, which are associated with b ± ( k) and b ±( k), have negative particle number. What differs from the non-relativistic theory is the fact that electrons and positrons have the same energy spectrum. Since the energy of a single electron, k2 +m 2, can be arbitrarily large, for large values of k, it is also so for holes (or positrons). This means that, as we shall seen, unlike the Fermi sea, the relativistic analog, which we can call the Dirac sea, is infinitely deep. This is what leads to the infinite values of the energy and number densities (which we have dropped). Another difference with the nonrelativistic theory, where the electron had a state of well-defined spin is that in the relativistic theory, it is the helicity (the states labeled by subscripts (+) and (-)), which are important. The helicity can be thought of as the projection of the spin in the direction of motion of the fermion. We construct the basis of the Hilbert space beginning with the vacuum O> which we assume is normalized, <O O>=1 and has the property that it is annihilated by all of the annihilation operators, a + ( k) O>=0, a ( k) O>=0, b + ( k) O>=0, b ( k) O>=0 for all values of k. Then, multi particle and anti-particle states are created by operating creation operators a +( k 1 )...a +( k m )a ( k 1)...a ( k m )b +( l 1 )...b +( l n )b ( l 1)...b ( l n ) O> These states are eigenstates of particle number, N, with eigenvalue N=m+m n n and they are eigenstates of the Hamiltonian, H 0, with eigenvalue the total energy, E= m 1 k 2 i +m2 + m 1 ( k i )2 +m n 1 l 2 i +m2 + n 1 ( l i )2 +m 2

10 3 Lorentz Invariance The Dirac equation is clearly invariant under translations of the space-time coordinates. If ψ( x,t) is a solution then ψ( x+ a,t+τ), with constants a and τ, is also a solution. What about Lorentz transformations. Let us begin by recalling how a Lorentz transformation of a scalar field is implemented. In our discussion of Lorentz invariance, we have derived the transformation property of the scalar field, with infinitesimal transformation φ(x)=φ(λ 1 x) δφ x =ω µν x µ ν φ(x) We would expect this transformation to be a symmetry of any relativistic waveequationthatascalarfieldwouldobey. Weexpectthat,underaLorentz transformation, the argument of the Dirac field also changes. However, the Dirac field has four components and the Lorentz transformation could also mix the components. ψ(x)=s(λ)ψ(λ 1 x) Here, S(Λ) is a 4 4 matrix which depends on the Lorentz transformation. Since a Lorentz transformation should be invertible, we expect that S is an invertible matrix, that is, that dets 0. This transformation is a symmetry of the Dirac equation if the transformed field also satisfies the equation, that is, if 0=[/ +m] ψ(x)=[/ +m]s(λ)ψ(λ 1 x)=s[s 1 γ µ SΛ 1ν µ x) (Λ 1 x) ν+m]ψ(λ 1 The equation is unchanged if S 1 γ µ S=Λ µ νγ ν If the transformation is infinitesimal, so that Λ ν µ=δ ν µ+ω ν µ and S=1+s, the Dirac equation will be invariant if A matrix which does the job is γ µ s sγ µ =ω µν γ ν s= 1 4 [γρ,γ σ ]ω ρσ 10

11 This can be seen by using the algebra for Dirac matrices. The upshot is that the infinitesimal Lorentz transformation of the Dirac spinor is δψ(x)=ω µν [x µ ν [γµ,γ ν ]]ψ(x) (33) For a rotation about the i-axis, where ω kj =ǫ jki θ i, this gives what we would expect, θ ( x δψ(x)=i[ 1 i σ) 0 0 θ ( x 1 i σ)]ψ(x) that is, a rotation is generated by a combination of an angular momentum operator L i =( x p) i =( x 1 i ) i and a spin operator 1 2 σi. Then, in particular, under infinitesimal rotations, the electron density transforms like a scalar field, δ(ψ (x)ψ(x))=( θ x )(ψ (x)ψ(x)) However, for an Lorentz transformation, that is, a boost with small velocity v δψ(x)=v i [x 0 i x i γ0 γ i ]ψ(x) (34) and the density is not a scalar, but has the transformation law δ(ψ (x)ψ(x))=v i (x 0 i x i 0 )(ψ (x)ψ(x))+ v ψ (x)γ 0 γψ(x) (35) Its infinitesimal transformation is like the time-component of a vector field, j µ (x)=(ψ (x)ψ(x),ψ (x)γ 0 γ i ψ(x)) (36) Indeed, we could examine the full Lorentz transformation law and see that j µ (x) transforms like a vector field. 4 Electric current In the above, we have examined the transformation law for the electron density and we found that it transforms like the time-component of a vector field (36). We will see shortly that this vector field obeys the continuity equation, µ j µ (x)=0, which we would expect for a conserved current and, 11

12 when our Dirac fermions are identified with electrons, it will be identified with the electric current. In non-relativistic terminology, j 0 (x) is the charge density and j(x) is the current density. In relativistic physics, we simply call j µ (x) a current or a conserved current. We consider the Dirac equation and its conjugate [/ +m]ψ=0, ψ [γ µ µ +m]=0 Then,wenotethat,γ µ γ 0 = γ 0 γ µ,sothat,ifwemultiplythesecondequation above from the right by γ 0 we obtain [/ µ +m]ψ=0, ψ γ 0 [ / +m]=0 We shall shorten the notation by defining Then, we can write the current as ψ(x) ψ (x)γ 0 (37) j µ (x)= ψ(x)γ µ ψ(x) (38) Now, we are ready to use derive the continuity equation. We form the current and we assume that the Dirac spinor satisfies the Dirac equation. Then, µ j µ (x)= µ ( ψ(x)γ µ ψ(x))= ψ[/ + / ]ψ(x)= ψ(x)[ m+m]ψ(x)=0 This continuity equation for the current implies the conservation of charge, which we have so far called the number operator, d dt N= d dt d 3 xψ (x)ψ(x)= d 3 x ψ(x) γψ(x)= d 2 σˆn ψ(x) γψ(x) where we have used Gauss theorem to write the last term as a surface integral at the infinite boundary of three dimensional space. If our boundary conditions are such that the quantum expectation value of the current density goes to zero sufficiently rapidly there, the number operator is conserved. 12

13 5 The Energy-Momentum Tensor The Dirac equation can be derived from an action S= d 4 xl(x) where L(x) = i ψ(x)[ 1 1 +m]ψ(x) / 2 2 Here we have defined the action as in integral of a Lagrangian density. The i in front and the symmetrization of the derivative operators are present to make the Lagrangian density real. The Dirac equation is easily recovered from this action using a variational principle. The equal time anticommutation relation that we have used is also compatible with this action. Here, although it has little effect what we shall do in the rest of this chapter, we note that the classical theory of the Dirac field requires that we use anticommuting classical fields for Fermions. In this sense, Fermions are never truly classical in that even the classical fields should anti-commute with each other. Now, consider an infinitesimal space-time translation x µ x µ +ǫ µ, Under this transformation δψ(x)= ǫ µ µ ψ(x), δ ψ(x)= ǫ µ µ ψ(x) δl(x)= µ [ǫ µ L(x)] The fact that the Lagrangian density varies by a total derivative term means that the infinitesimal translation is a symmetry of the theory. Of course, we already knew that this should be the case, since we have already seen that the equation of motion has this symmetry. To find the Noether current, we assume that the transformation parameter now depends on the coordinates, ǫ µ (x). The variation of the Lagrangian now has the form δl(x)= µ [ǫ µ (x)l(x)] λ ǫ ρ (x) i 2 ψ(x)(γ λ ρ ρ γ λ )ψ(x) Now, if ǫ µ (x) vanishes sufficiently rapidly at the boundaries of the integral so that boundary terms can be ignored, and if we assume that the equations 13

14 of motion are obeyed, the action must vanish for any variation, in particular, with any ǫ µ (x) with any profile. Then, we must have λ [ i 2 ψ(x)(γ λ ρ γ λ ρ )ψ(x)]=0 It is easy to confirm that this is indeed the case by using the equation of motion. We have identified a conserved quantity which is related to translation symmetry. It is called the energy-momentum tensor T λρ 0 (x)= i 2 ψ(x)[γ λ ρ γ λ ρ ]ψ(x), λ T λρ 0 (x)=0 We note for future reference, that the four-divergence of the energy-momentum tensor on its second index also vanishes, ρ T λρ 0 (x)=0 It is easy to see this using the equation of motion. The space integral of the time-component of this energy-momentum tensor is the generator of space-time translations, P ρ = d 3 xt 0ρ 0 (x)= i 2 d 3 xψ (x)[ ρ ρ ]ψ(x) We should eliminate the time derivative from the integrand for P 0 by using the equation of motion. This means replacing 0 ψ by ih D ψ where h D is defined in equation (9). The result is that P 0 is the second quantized Dirac hamiltonian. Now, consider an infinitesimal Lorentz transformation, δψ(x)=ω µν [x µ ν [γµ,γ ν ]]ψ(x) δ ψ(x)=ω µν x µ ν ψ(x) ψ(x) 1 4 [γµ,γ ν ] Under this transformation, δl(x)=ω µν x µ ν L(x)= ν [ω µν x µ L(x)] where, in the last step, we used the fact that ω µν = ω νµ. The fact that the Lagrangian density varies by a total derivative term means that the Lorentz 14

15 transformation is a symmetry of the theory. Of course, we already knew that this should be the case, since we have already demonstrated that the equation of motion has this symmetry. To find the Noether current, we assume that the transformation parameter depends on the coordinates, ω µν (x). The variation of the Lagrangian now has the form δl(x)= ν [ω µν (x)x µ L(x)] λ ω ρσ i 2 ψ(x)(γ λ [x ρ σ [γρ,γ σ ]] [x ρ σ 1 4 [γρ,γ σ ]]γ λ )ψ(x) Now, if ω µν (x) vanishes sufficiently rapidly at the boundaries of the integral so that boundary terms can be ignored, and if we assume that the equations of motion are obeyed, the action must vanish for any variation, in particular, with any profile of ω ρσ (x). Then, we must have where λ M λρσ (x)=0 M λρσ (x)= T λρ 0 xσ +T λσ 0 x ρ + ψ(x){γ λ, i 4 [γρ,γ σ ]}ψ(x) (39) where T µν 0 (x) is the energy-momentum tensor which was associated with translations. It is easy to confirm explicitly that λ M λρσ (x)=0 by using the equation of motion and the algebra of gamma-matrices. Again, the space integral of the time component M ρσ = d 3 xm 0ρσ (x) has vanishing time derivative and it generates the Lorentz transformation of the Dirac field. There are good reasons why it is convenient to have a symmetric energymomentum tensor. By modifying T ρσ 0 (x) to make it symmetric, we will be able to unify the generator of translations and Lorentz transformations. Let us first begin by using the above development in order to get a symmetric energy-momentum tensor. We will do this by adding a term to it which is where T ρσ (x)=t ρσ 0 (x)+ λs λρσ (x) 15

16 1. λ S λρσ (x)= 1 2 (Tρσ 0 (x) Tσρ (x)) 2. ρ λ S λρσ (x)=0 and σ λ S λρσ (x)=0 (which are actually implied by the above). 3. d 3 x λ S λ0σ (x)=0 so that the new, symmetric energy-momentum tensor is the same generator of translations as the old one. We begin with the observation that the last term on the right-handside in M λρσ (x) in equation (39) is due to the spin of the electron. It is sometimes called the spin tensor. Note that, µ M λρσ (x)=0, λ T λρ 0 (x)=0 and ρ T λρ 0 (x)=0 imply the divergence of the last term is proportional to the anti-symmetric part of T λρ 0 (x), λ ( ψ(x){γ λ, i 4 [γρ,γ σ ]}ψ(x))=t ρσ 0 (x) Tσρ 0 (x) This accomplishes the first step, it demonstrates that that the anti-symmetric part of T ρσ 0 (x) is equal to the four-divergence. What is more, since, as we have observed above, T ρσ (x) has vanishing four-divergence on both of its indices, we also know that ρ [ λ ( ψ(x){γ λ, i 4 [γρ,γ σ ]}ψ(x))]=0, σ [ λ ( ψ(x){γ λ, i 4 [γρ,γ σ ]}ψ(x))]=0 Finally d 3 x λ ( ψ(x){γ λ, i 4 [γ0,γ σ ]}ψ(x))= d 3 x t ( ψ(x){γ 0, i 4 [γ0,γ σ ]}ψ(x))=0 where, in the middle term, we have dropped the spatial integral of the spatial divergence since we assume that the fields fall off sufficiently at spatial infinity that the surface integral that one would obtain by using Gauss theorem would vanish. The last term vanishes as a result of the gamma-matrix algebra. We have shown that the antisymmetric part of the energy momentum tensor satisfies all of the criteria for λ S λρσ. This allows us to simply drop the anti-symmetric part of the energy-momentum tensor without altering the fact that it is conserved and that the spatial integral of its time component is a generator of translations. We then keeps its symmetric part, which it is easiest to obtain by simply symmetrizing T µν 0 (x), T µν (x)= 1 2 (Tµν 0 (x)+tνµ (x)) 16

17 or T µν (x)= i 4 ψ(x)[γ µ ν +γ ν µ γ µ ν γ ν µ ]ψ(x), λ T νµ (x)=0 (40) Now, finally, we note that the trace of the energy-momentum tensor is T µ µ (x) η µν T µν (x)= im ψ(x)ψ(x) (41) is proportional to the mass of the Dirac field. When the mass vanishes, the theory has an extended space-time of symmetry. We will comment on this below. If we recall that a space-time symmetry is a coordinate transformation which is generated by a Killing vector ˆf µ (x), we might make a candidate for a conserved current by contracting the energy-momentum tensor with the vector field which generates the co-ordinate transformation, T µ f (x) Tµ ν(x)f ν (x) Then, to have a conservation law, we need µ T µ f (x)=0 With a vector field f µ (x), we will have such a conservation law if: 1. T µν (x) is conserved, i.e. µ T µν (x)=0. Then µ T µˆf(x)=0 if ˆf µ = a µ, a constant vector. A translation invariant field theory should have a conserved energy-momentum tensor. A field theory can be translation invariant without being Lorentz invariant. It would still have a conserved energy-momentum tensor. However, if the theory is not Lorentz invariant, it should not be possible to symmetrize the energymomentum tensor. 2. T µν (x) is conserved and T µν (x)=t νµ (x) is symmetric. Then T µˆf(x) is conserved when ˆf µ (x) obeys the Killing equation µ ˆfν (x)+ ν ˆfµ (x)=0 A conserved, symmetric energy-momentum tensor can thus be used to generate all of the symmetries of Minkowski space. A translation and Lorentz invariant field theory should have a conserved and symmetric energy-momentum tensor. 17

18 3. Finally,T µν (x)isconserved,symmetricandhasvanishingtrace,t µ µ(x)= 0. Then T µ f (x) is conserved when fµ (x) satisfies the conformal Killing equation µ f ν (x)+ ν f µ (x) η µν 2 λf λ (x)=0 and it generates a conformal transformation. Notice that all solutions of the Killing equation are also solutions of the conformal Killing equation. However, the conformal Killing equation is less restrictive. It has more solutions than the Killing equation. The extra solutions correspond to conformal transformations. A conformal field theory should have a conserved, symmetric and traceless energy-momentum tensor. Our example of the non-interacting Dirac field is a conformal field theory when m=0. 18