3 Quantization of the Dirac equation

Transcription

1 3 Quantization of the Dirac equation 3.1 Identical particles As is well known, quantum mechanics implies that no measurement can be performed to distinguish particles in the same quantum state. Elementary particles are thus naturally indistinguishable and the permutation operator P commutes with the hamiltonian of any many-particle quantum system. Since P is hermitian and P 2 = 1, the corresponding eigenvalues are ±1. Those particles corresponding to the plus eigenvalue are called bosons and those corresponding to the negative eigenvalue are fermions. As we have seen in the previous lecture, the canonical cuantization rules when imposed on a scalar field require the creation and annihilation operators to satisfy the commutation relations (in a simplified normalization and notation) [a n, a n] = 0, [a n, a n ] = 0, [a n, a n ] = δ n,n. (104) This necessarily implies that the particles (or states) created by a n acting on 0 are of a bosonic nature as a na n 0 = a n a n 0. (105) Fermionic particles thus require creation and annihilation operators that anticommute. In this way a n a n 0 = a n a n 0. (106) Thus {a n, a n } = 0, {a n, a n } = 0, {a n, a n } = δ n,n. (107) By proceeding backwards, this immediately implies that for fermionic fields the canonical commutation relations have to be implemented by the following ETC {ψ i (x), ψ j (y)} = {π i (x), π j (y)} = 0, x 0 = y 0 (108) {ψ j (x), π k (y)} = iδ j,k δ (3) ( x y), x 0 = y 0. (109) Let us now go back to the bosonic case and let us take a look again to formula (101). The hamiltonian we can rewrite as H = 1 2 (2π) 3 2E p E p (2a pa p + [a p, a p]), (110) 21

2 or using the commutation relations H = (2π) 3 E p a 2E pa p + 1 p 2 δ(3) (0) E p. (111) The last term corresponds obviously to the sum of zero point energies (the well known hω/2 of the harmonic oscillator). This is an (admittedly infinite) constant term that can be subtracted by just redefining the origin of energies. This is acomplished by introdcing normal ordered operators, where all annihilation operators are, by construction, moved to the right, regardless of the commutation relations : H : (2π) 3 2E p E p a pa p. (112) This is equivalent to defining : H := H 0 H 0. (113) Now imagine that the a n had satisfied anticommutation relations. First of all, you should trust me that the total energy would still be given by eq. (101); i.e. in arriving to this equation, no commutation o anticommutation relation has been assumed yet. clear that, assuming anticommutation relations H = 1 2 Then it is (2π) 3 {a p, a p }, (114) that is, the normal ordering subtraction piece is the only term to survive and, if we normal order H 0!! A scalar field can be quantized only consistently using canonical commutation relations. It necessarily describes bosons. 3.2 Quantization of a fermion field Let us now write the mode expansion of the general solution of the Dirac equation ψ(x) = (2π) 3 [u( p, )c( p, )e ipx + v( p, )d ( p, )e ipx ]. (115) 2E p The index describes the possible states of angular momentum, to which we shall return later. u and v are positive and negative energy solutions of the Dirac equation, respectively ( p m)u( p, ) = 0, ( p + m)v( p, ) = 0. (116) 22

3 According to the arguments in the previous chapter, c and d must satisfy anticommutation relations. They will be {c( p, ), c ( p, )} = {d( p, ), d ( p, )} = (2π) 3 2E p δ, δ (3) ( p p ), (117) {c( p, ), c( p, )} = {d( p, ), d( p, )} = 0, (118) {c ( p, ), c ( p, )} = {d ( p, ), d ( p, )} = 0. (119) c accompanies positive energy solutions, while d goes with negative energy solutions. They are independent degrees of freedom and they therefore commute in any case. The physical action of the operators c and d is as follows (also by comparison to the bosonic case): c( p, ) annihilates a positive energy state with momentum p and polarization. d ( p, ) annihilates a negative energy state with momentum p and polarization. Therefore d ( p, ) creates an antiparticle (of positive energy) and momentum p. The (Dirac) vacuum will be defined by the relations c( p) 0 = 0, d( p) 0 = 0. (120) The first one defines the vacuum as the state where there are no positive energy solutions (and hence none can be destroyed). The second relation defines the vacuum as the state where negative energy solutions are all filled up (and hence none can be created, due to the Pauli exclusion principle). This makes clear that we are indeed dealing with fermions. Likewise the adjoint field is expanded as ψ(x) = (2π) 3 [ v( p, )d( p, )e ipx + ū( p, )c ( p, )e ipx ].. (121) 2E p This field has a component that annihilates antiparticles and another one that creates particles, when acting on the vacuum. It is immediate to construct, in close analogy with the bosonic case, conserved currents and charges associated to the Dirac lagrangian L(x) = i ψ ψ m ψψ. (122) For instance T µν = i 2 ( ψγ µ ν ψ ν ψγ µ ψ) (123) 23

4 P µ = (after normal-ordering). (2π) 3 2E p p µ (c ( p, )c( p, ) + d ( p, )d( p, )) (124) Let us investigate the effect of applying a charge conjugation transformation on the secondquantized Dirac field (126) ψ ψ c = iγ 2 ψ. (125) By comparing the expansion of ψ c ψ c (x) = (2π) 3 [u c ( p, )c c ( p, )e ipx + v c ( p, )d 2E c( p, )e ipx ]. (126) p and iγ 2 ψ we shall immediately conclude that c c = d, d c = c, (127) which is exactly as it should be, and u c = iγ 2 v u c = iγ 0 γ 2 v T, v c = iγ 2 u. (128) Of course, formally, u c = u and v c = v since nowhere the charges appear explicitly and the mas of any particle and its antiparticle are the same. 3.3 More on angular momentum and helicity We introduce three matrices Σ It is quite clear that they satisfy ( ) σ 0 Σ =. (129) 0 σ [ 1 2 Σ 1, 1 2 Σ 2] = i 1 2 Σ 3, (130) and ( 1 2Σ) 2 = 3 4I. They do constitute a representation of the familiat quantum-mechanical momentum for j = 1 2. They are in fact spin 1 2 operators appropriate to the rest frame and u r, v r are eigenstates. Now recall that the Hamiltonian of the Dirac field, i.e. the energy operator is just α p + βm that does not commute with Σ unless p = 0. However since there is clearly some degeneracy 24

5 (two linearly independent solutions), some hermitian operator must commute with H. The choice is not unique, but one that does commute is the helicity operator ) h( p) = ( σ p p 0 σ p 0 p (131) Unless explicitly stated otherwise, we shall understand that the label in the spinors will refer to helicity. 3.4 Symmetries A two particle state will be of the form p 1, 1 ; p 2, 2 c ( p 1, 1 )c ( p 2, 2 ) 0. (132) However, due to the fermion nature, and contrary to the boson case p 2, 2 ; p 1, 1 = p 1, 1 ; p 2, 2. (133) It is interesting to consider a particle-antiparticle state of the form c ( p 1 )d ( p 2 ) 0 (134) and investigate the effects of charge conjugation on such a state. We already know from the previous subsection that this should simply be d ( p 1 )c ( p 2 ) 0. (135) If we exchange the space degrees of freedom p 1 and p 2 we still get a minus sign relative to the original state! The Dirac lagrangian has also a global U(1) symmetry, namely ψ(x) ψ (x) = e iα ψ(x). (136) The corresponding Noether current is immediately obtained J µ = ψγ µ ψ. (137) And the corresponding charge is Q = (2π) 3 (c ( p, )c( p, ) d ( p, )d( p, )). (138) 2E p 25

6 The physical interpretation of this symmetry depends on the particle assigment. It may correspond to lepton number, baryon number,... even electric charge, eventually (after coupling to electromagnetism and making the symmetry local). As we see, particles and antiparticles have oposite values of this conserved charge. It will turn out to be extremely important the answer to the following question: what happens if α = α(x); namely if the phase rotation is local rather than global? Obviously the Dirac lagrangian is not invariant under such a transformation, due to the derivative term acting on α, but we can compensate for the additional piece if we include a gauge field L(x) = i ψ( + ie A)ψ m ψψ, (139) and assume that A µ A µ 1 e µα. (140) Note that the variation of L(x) (before introducing the gauge field transformation) is precisely the derivative of the Noether current associated to global U(1) invariance. The new (and clearly greater) symmetry is called local U(1) or simply gauge U(1). Indeed the transformation on A µ corresponds to the well-known gauge transformations of Maxwell electrodynamics. What it amounts to is to make unobservable the phase of the field ψ(x). 3.5 Majorana fermions When discussing the scalar field we have assumed that it is real, φ (x) = φ(x) and this is manifest in the expansion in creation and annihilation operators. φ(x) = (2π) 3 2p 0 [a pe ipx + a p eipx ]. (141) This field describes a neutral particle. Therefore a particle that coincides with its own antiparticle. Probably the easiest way to see that is to try to couple it to an electromagnetic field. Let us attempt to do using minimal coupling (we shall see in the next lecture that this is in fact the proper way). Thus we write L(x) = 1 2 (D µφ) (D µ φ) 1 2 m2 φ 2. (142) Where D µ is the covariant derivative introduced in Lecture 1. The complex conjugate is needed to ensure the reality of the lagrangian. Expanding these derivatives we get 1 2 ( µφ)( µ φ) e2 A µ A µ φ m2 φ 2, (143) 26

7 but no term that is linear in e. This would imply that the electromagnetic interactions cannot distinguish a (scalar) particle from its antiparticle. This is absurd and proves that one needs a complex field to describe charged particles. It is thus tempting to require a similar reality condition on fermionic fields. In fact we already know the condition for a fermion to coincide with its antifermion ψ c = Cψ. (144) At this point we can ask if we can take C = 1. This is indeed the case if we use Majorana representation, that obeys γ µ = γµ and thus C = 1. In this representation the solution to Dirac equation can be taken real. Consequently the c and d creation and annihilation operators are no longer independent. Notice that ψγ µ ψ is no longer a conserved current (as phase redefinitions are no longer possible since ψ is real. Then the corresponding charge is not conserved. Since neutrinos carry no charge, they could perfectly be Majorana particles. It should not be confused the fact that a fermion is a Majorana particle with the fact that we may use the Majorana representation to describe usual Dirac fermions. 27