4 Version of February 4, 005 CHAPTER. DIRAC EQUATION (0, 0) is a scalar. (/, 0) is a left-handed spinor. (0, /) is a right-handed spinor. (/, /) is a vector. Before discussing spinors in detail, let us mention the discrete transformations that preserve the form x t. These are Parity P : x x, t t, (.6a) Time reversal T : x x, t t. (.6b) (We know experimentally that these are not exact symmetries of nature.) The Lorentz group generators transform as follows: so P : J J (pseudoscalar or axial vector), (.7a) N N (vector or polar vector), (.7b) T : J J, (.7c) N N, (.7d) P : J (+) J ( ), T : J (±) J (±). (.8) In particular, under P, the representations change accoring to (s +, s ) (s, s + ), (.9) that is, the left-handed spinor changes into the right-handed spinor, (/, 0) (0, /). So if we want a description of spin-/ particles which respects parity, we must use ( ) (, 0 0, ), (.0) which is what is called a Dirac spinor, Incidentally, note that we can construct everything from spinors by using (s +, s ) (s +, s ) s ±+s ± r ± s ± s ± (r +, r ). (.) The transformation properties of a left-handed spinor (/, 0) are given by the explicit realization of the generators, J (+) σ +, J ( ) 0, (.) which means that under finite rotations and boosts ( ), 0 : U (+) R [i exp ] θ e σ +, U (+) B [ exp ] φe σ +. (.3)
.4. LORENTZ COVARIANCE 5 Version of February 4, 005 Similarly, the transformations of a right-handed spinor (0, /) are given by ( 0, ) : U ( ) R [i exp ] [ ] θ e σ, U ( ) B exp φe σ. (.4) Therefore, a Dirac spinor (/, 0) (0, /) transforms by U R U (+) R U( ) R + iδθe (σ + + σ ), (.5a) U B U (+) B U( ) B δφe (σ + σ ), (.5b) that is, the rotation and boost generators are J (σ + + σ ), N i (σ + σ ). (.6).4. Direct Product versus Direct Sum We are familiar with adding two angular momenta in quantum mechanics, for example, m.m, m +m, m +m m,, m + 0, 0 0, 0 m,, m, (.7) where the coefficients of the two vectors of the right-hand-side of this equation are the Clebsch-Gordan coefficients. This is a direct product, the product of two two-component objects. Thus, the generator of rotations is the third component of which is J 3 J (σ + τ) (σ + τ), (.8) ( ) 0 + ( ) τ3 0 0 0 τ 3 0 0 0 0 0 0 0 0 0 0 0. (.9) 0 0 0 The possible eigenvalues of J 3 are (+, 0, ) and 0, corresponding to the eigenvalue of J being j(j + ) with j or 0. Symbolically, we write 0. (.30) This is the situation we have for the vector representation of the Lorentz group (/, /). A Dirac spinor is a direct sum, a four-component object, so the angular momentum operator is J (σ + + σ ) ( ) σ+ 0, (.3) 0 σ
6 Version of February 4, 005 CHAPTER. DIRAC EQUATION so the third component if 0 0 0 J 3 0 0 0 0 0 0, (.3) 0 0 0 which has two eigenvalues of + and two eigenvalues of..5 Lorentz transformations of the Dirac equation Now let us return to the Dirac equation (.4), i t ψ (α p + βm)ψ, p. (.33) i The αs are related to the angular momentum (spin) by (.3), or [α i, α j ] iǫ ijk Σ k, S k Σ k. (.34) Comparing this to the boost equation (.0) we must have α i ±in i, (.35) which may be easily checked to satisfy (.94) with J k S k. Thus, choosing the lower sign, we find that an infinitesimal Lorentz transformation on a Dirac spinor is given by U + iδω Σ δv α. (.36) Two relatively moving observers will write the Dirac equation as O : i ( ) t ψ(x, t) i α x + βm ψ(x, t), (.37a) where O : For a boost, because we have i t ψ(x, t) ( i α x + βm ) ψ(x, t), (.37b) ψ(x, t) Uψ(x, t). (.38) t t + δv x, x x + δv t, (.39) t xµ t x µ t + δv x, x xµ x x µ x + δv t. (.40a) (.40b)
.5. LORENTZ TRANSFORMATIONS OF THE DIRAC EQUATION7 Version of February 4, 005 So the Dirac equation as seen by the barred observer is ( ) ( i t + δv δv ) x α ψ(x, t) ( i α x iα δv ) ( t + βm δv ) α ψ(x, t), (.4) Premultiply this equation by ( δv α): ( i t + δv x δv α ) ψ(x, t) t ( i α x iα δv { }) t + βm i α + βm, δv x α ψ(x, t). (.4) In view of the commutation relations (.6a) and (.6b) we see that all the terms proportional to δv cancel, so the barred equation agrees with the unbarred one to order δv. In homework, you will verify rotational invariance. We can do all of this more compactly if we adopt covariant notation. Let us take new Dirac matrices defined by γ 0 is Hermitian while γ i is anti-hermitian, Then the anticommutation relations become γ 0 β, γ i βα i. (.43) γ 0 γ 0, γ i γ i. (.44) {γ µ, γ ν } g µν, (.45) and the Dirac equation reads ( γ µ ) i µ + m ψ 0. (.46) Now write a general infinitesimal Lorentz transformation as x µ x µ δx µ, δx µ δω νµ x ν, δω µν δω νµ. (.47) In terms of the parameters corresponding to rotations and boosts, δω ij ǫ ijk δω k, δω i0 δv i. (.48) This induces a unitary transformation U + ig, where the generator is G Jµν δω µν, (.49)
8 Version of February 4, 005 CHAPTER. DIRAC EQUATION where [J µν, J κλ ] i (g µκ J νλ g µλ J νκ g νκ J µλ + g νλ J µκ ), (.50) which you will prove in homework. For spin /, you will also prove in homework that Now, under a general Lorentz transformation J µν σµν, σ µν i [γµ, γ ν ]. (.5) µ xν x µ x ν µ + δω µν ν. (.5) Therefore, the transformed Dirac equation is (ψ(x) Uψ(x)) ( γ µ ) [ i µ + m ψ(x) γ µ ] i ( µ + δω µν ν ) + m ( + i4 ) σαβ δω αβ ψ(x) ( + i4 ) ( σαβ δω αβ γ µ ) i µ + m ψ(x) [ + γ µ, i ] 4 σαβ δω αβ i µψ(x) + γ µ i δω µν ν ψ(x). The commutator occuring here is [γ µ, σ αβ ] i [γµ, [γ α, γ β ]] (.53) i ( {γ µ, γ α }γ β γ α {γ µ, γ β } (α β) ) i(g µβ γ α g µα γ β ), (.54) so then we see that the second term in (.53) cancels the third term exactly. Thus the Dirac equation transforms covariantly: If it holds in one inertial frame, it holds in all..5. Discrete transformation How does a Dirac field behave under a parity transformation, x x? Since then 0 0, i i, we require ψ Uψ, where U γ i U γ i, U γ 0 U γ 0. (.55) A solution to this is so we take U U γ 0, (.56) ψ(x, t) γ 0 ψ( x, t). (.57)
.6. POINCARÉ GROUP 9 Version of February 4, 005 Under time reversal, 0 0, i i, so now with ψ Uψ, we must have U γ 0 U γ 0, U γ i U γ i (.58) A solution is where U U iγ 5 iγ 0, (.59) γ 5 γ 0 γ γ γ 3. (.60) It may be easily shown that the matrices iγ 5 are Hermitian, and that (iγ 5 ), {iγ 5, γ µ } 0. (.6) We also should recall that the implementation of time reversal on quantum states and operators is represented by an antiunitary operator, T UK, (.6) where K represents the operation of complex conjugation..6 Poincaré Group In addition to the Lorentz transformations, which are described by six generators J µν J νµ or J,N, we have, as discussed earlier in Section.3, translations in time and space. The four generators of translation are the energy and momentum, G P µ δǫ µ, P µ (H,P). (.63) The full set of commutators define the Poincaré group algebra: [P µ, P ν ] 0, (translations commute), (.64a) i [P µ, J νλ ] g µλ P ν g µν P λ, (P µ is a four-vector), (.64b) i [J µν, J κλ ] g µκ J νλ g µλ J νκ g νκ J µλ + g νλ J µκ, (J µν is a second-rank tensor). (.64c) The irreducible representations of this group, which correspond to particles, are characterized by the values of where m is the mass, and of P P (P 0 ) m, (.65) W W µ W µ, W µ P ν ǫµνλσ J λσ, P µ W µ 0. (.66)
0 Version of February 4, 005 CHAPTER. DIRAC EQUATION It is evident that W µ is translationally invariant, [P µ, W ν ] 0. W is a Lorentz scalar, [J µν, W ], as you will explicitly show in homework. Here ǫ µνλσ is the totally antisymmetric invariant tensor, ǫ µνλσ (g µµ δω µµ )(g νν δω νν )(g λλ δω λλ )(g σσ δω σσ )ǫ µ ν λ σ ǫ µνλσ, (.67) (because ǫ µνλσ vanishes if any two indices are the same), where ǫ 03 +. In the rest frame of a particle, and so P 0, P 0 E m, (.68) W 0 ǫ0ijk J ij P k 0, W i ǫijk0 J ij m m ǫijk J jk mj i, (.69a) (.69b) where the latter is the spin. Thus, the eigenvalues of W are W m s(s + ). (.70) This means for a particle with nonzero rest mass, m > 0, the irreducible representations belong to the values s 0, /,,.... For a given s, the possible value of J 3 are s 3 s, s +, s +,...,s, s. The massless limit has to be taken carefully (see homework): m 0 : W µ λp µ, λ P S P 0. (.7) λ is called the helicity, which is the spin projected along the direction of motion. There are other representations of the Poincaré group, such as tachyons, where m < 0, but they seem not to be realized in nature.