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1 Fermionic representation Our goal is to write a Lorentz invariant action describing a spin half massive charged field (such as the electron). First a brief reminder about Lorentz representations. If we denote the rotation generators by Ji and the boost generators by Ki, then (very similar to the Hydrogen atom exercise) we can define 1 1 A = (J + iK ) ;B = (J − iK ) (1) i 2 i i i 2 i i and they satisfy the algebra [Ai;Aj] = iijkAk ; [Bi;Bj] = iijkBk ; [Ai;Bj] = 0 : (2) We get two commuting su(2)0s generated by the A's and the B's. Therefore, we can classify the Lorentz representations by the reps of the two su(2)'s. The irreps of su(2) are classified using one integer or half integer number 1 j = 0; 2 ; 1:::. The Lorentz irreps are then classified by two numbers (jL; jR). For example, • (0,0): Lorentz scalar. • (1/2,0): Left Weyl spinor L. • (0,1/2): Right Weyl spinor R. • (1/2,1/2): Lorentz Vector Aµ ;@µ::: . Weyl spinors transform as σi σi ! Λ = exp (iθ + φ ) ; ! Λ = exp (iθ − φ ) R R R i i 2 R L L L i i 2 L (3) i where θi and φi parametrize rotations and boosts respectively and σ are y the Pauli matrices. what about L=R? σi y σi y ! exp (iθ + φ ) = y exp (−iθ + φ ) = y Λ−1 : R i i 2 R R i i 2 R L (4) and similarly σi y ! y exp (−iθ − φ ) = y Λ−1 : (5) L L i i 2 L R An immediate result is that σi σi y ! y exp (−iθ − φ ) exp (iθ − φ ) (6) L L L i i 2 i i 2 L 1 y is not Lorentz invariant (and also R R). But σi σi y ! y exp (−iθ − φ ) exp (iθ + φ ) = y (7) L R L i i 2 i i 2 R L R y is invariant, and the same for R L. If we want to write a mass term for that will be Lorentz invariant and quadratic in the fields it seems like we cannot do it for L( R) by itself and we need to couple them together. This observation motivates us to define the Dirac fermion which can be written as L = (8) R and its Lorentz representation is (1=2; 0) ⊕ (0; 1=2). The transformation of in this basis is simply ΛL ! Λ1=2 = : (9) ΛR We wrote in the chiral basis where the decomposition to the left and right spinors is manifest. Before we continue and write the action, we want to write the Dirac fermion and its properties in a basis independent form. 2 Clifford algebra and the γ matrices Define a set of four 4 × 4 matrices γµ; µ = 0; 1; 2; 3 which obeys µ ν µν fγ ; γ g = 2η I4×4 : (10) From the γ matrices we can construct a representation of the Lorentz gen- erators i Sµν = [γµ; γν] : (11) 4 One can check that the Sµν's satisfy the correct Lorentz algebra. The Dirac spinor has the following transformation rule i µν − !µν S ! Λ1=2 = e 2 : (12) In the chiral representation the γ matrices are 0 σµ γµ = ; σµ = ( ; σi) ; σ¯µ = ( ; −σi) (13) σ¯µ 0 I2×2 I2×2 and the generators are i σi 0 1 σk 0 S0i = − ;Sij = ijk : (14) 2 0 −σi 2 0 σk 2 Lets see what happens to under rotation around the z axis. To do this we choose !xy = −!yx = θ and all the rest are zero. 0e−iθ=2 1 3 iθ=2 iθ σ 0 B e C ! exp − = B C : (15) 2 0 σ3 @ e−iθ=2 A eiθ=2 If we take θ = 2π we find ! − as expected from a fermion. 2.1 Chirality We can also define the fifth γ matrix i γ5 = γµγνγργσ = iγ0γ1γ2γ3 (16) 4! µνρσ which satisfies fγ5; γµg = 0 ; [γ5;Sµν] = 0 : (17) In the chiral basis − 0 γ5 = I2×2 : (18) 0 I2×2 The left/right Weyl spinors are eigenspinors of γ5 with eigen values 5 5 γ L = − L ; γ R = R : (19) The eigenvalue of some spinor under γ5 is called its chirality and the Weyl spinors are chirality eigenstates. Since γ5 commutes with all the Lorentz generators, the chirality is a Lorentz invariant property. We can also define 1−γ5 1+γ5 L/R projection operators PL = 2 ;PR = 2 . They satisfy 2 2 L 0 PL = PL ;PR = PR ;PLPR = PRPL = 0 ;PR = ;PL = : 0 R (20) 3 Dirac action We want to write an action for the Dirac fermion that will be hermitian, Lorentz invariant and quadratic in the fields. We already saw that under y boosts Λ1=2 is not unitary and therefore is not Lorentz invariant. We want to define the Dirac conjugate of denoted by ¯ that will have the ¯ ¯ −1 ¯ transformation rule ! Λ1=2 and therefore will be Lorentz invariant y y (and hermitian). In terms of Weyl spinors we already saw that L R + R L is Lorentz invariant and it is hermitian. Therefore in the chiral basis ¯ y y y 0 = R L = γ (21) 3 will do the work. We will show that this definition ¯ = yγ0 satisfies what we want in every basis. We will use the following identities 0 µ 0 µ y 0 2 γ γ γ = (γ ) ; (γ ) = I (22) which lead to i y i (Sµν)y = (γµγν − γνγµ) = − ((γν)y(γµ)y − (γµ)y(γν)y) 4 4 i i = − (γ0γνγ0γ0γµγ0 − γ0γµγ0γ0γνγ0) = γ0(γµγν − γνγµ)γ0 = γ0Sµνγ0 : 4 4 (23) The transformation of ¯ is i µν y i µν y i 0 µν 0 i µν ¯ − 2 !µν S 0 y 2 !µν (S ) 0 y 2 !µν γ S γ 0 y 0 2 !µν S ¯ −1 ! e γ = e γ = e γ = γ e = Λ1=2 : (24) We found the way to write a good mass term. Now lets try to write a kinetic term. Using representation arguments, we have ; ¯ that live in the (1=2; 0)⊕(0; 1=2) representation, and the derivative @µ that live in the vector representation (1=2; 1=2). We can combine two fermions and one derivative to create a Lorentz scalar ((1=2; 0) ⊕ (0; 1=2)) ⊗ (1=2; 1=2) ⊗ ((1=2; 0) ⊕ (0; 1=2)) = ((0 ⊕ 1; 1=2) ⊕ (1=2; 0 ⊕ 1)) ⊗ ((1=2; 0) ⊕ (0; 1=2)) = ((0; 1=2) ⊕ (1; 1=2) ⊕ (1=2; 0) ⊕ (1=2; 1)) ⊗ ((1=2; 0) ⊕ (0; 1=2)) (25) = (1=2; 1=2) ⊕ (1 ⊗ 1=2; 1=2) ⊕ (0 ⊕ 1; 0) ⊕ (0 ⊕ 1; 1) ⊕ (0; 0 ⊕ 1) ⊕ (1; 0 ⊕ 1) ⊕ (1=2; 1=2) ⊕ (0 ⊕ 1; 1) which contains the (0; 0) representation, but also many other representa- b tions. We want to write a term of the form ¯a@µ and we need to contract the a; b; µ indices in such a way that we will get the (0; 0) rep. It happens to be that the way we need to contract the indices is with the γ matrices µ and ¯γ @µ is Lorentz invariant. It can be shown that −1 ν µ µ Λ1=2γ Λ1=2Λν = γ (26) and therefore under Lorentz transformation ¯ µ ¯ −1 µ ν ¯ µ γ @µ ! Λ1=2γ Λµ @νΛ1=2 = γ @µ : (27) Now we are ready to write the Dirac action Z 4 ¯ µ ¯ SDirac = d x i γ @µ − m : (28) We just need to show that the kinetic term is hermitian: Z Z Z Z 4 µ y 4 y µ y 0 y 4 µ 4 µ d x i ¯γ @µ = d x −i@µ (γ ) (γ ) = −i d x@µ ¯γ = i d x ¯γ @µ : (29) 4 4 Dirac equation The Dirac equation can be derived from the Dirac action using the EL equations for and ¯. @L @L @ − = −iγµ@ + m = 0 µ @(@ ¯) @ ¯ µ µ (30) @L @L µ @µ − = i@µ ¯γ + m ¯ = 0 : @(@µ ) @ We will use the Feynman slash notation µ A= = γ Aµ : (31) The Dirac equation can be written as (i@= − m) = 0 : (32) First we will show that each component of the Dirac spinor satisfies the KG equation (and therefore the relativistic dispersion relation): 0 = (i@= + m)(i@= − m) = −(@=@= + m2) = −(@2 + m2) = 0 : (33) Now we will solve the equation. Consider one Fourier mode with positive energy (x) = u(p)e−ip·x ; p0 > 0 : (34) The equation simplifies to µ (p= − m)u(p) = (γ pµ − m)u(p) = 0 : (35) In the chiral basis it takes the form −m p σµ u (p) µ L = 0 p σ¯µ −m u (p) µ R (36) p σµ p σ¯µ u (p) = µ u (p) ; u (p) = µ u (p) : L m R R m L This is solved by p p · σξ u(p) = p ; p0 > 0 : (37) p · σξ¯ ξ has two arbitrary components, and we can expand the solutions in the following basis p p · σξ 1 0 us(p) = p ; ξ1 = ; ξ2 = : (38) p · σξ¯ 0 1 5 On the same way we can solve for negative energy modes (x) = v(p)eip·x ; p0 > 0 : (39) The equation is µ µ m pµσ (γ pµ + m)v(p) = µ v(p) = 0 (40) pµσ¯ m which is solved by p p · σηs 1 0 vs(p) = p ; η1 = ; η2 = : (41) − p · ση¯ s 0 1 The most general solution to the Dirac equation can be written as Z d3p 1 X (x) = asus(p)e−ip·x + bsvs(p)eip·x ; p0 > 0 (2π)3 p p p (42) 2Ep s=1;2 s s where ap; bp are arbitrary coefficients.
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  • Time Reversal Symmetry

    Time Reversal Symmetry

    Contents Contents i List of Figures ii List of Tables iii 7 Time Reversal 1 7.1 The Poincar´eGroup ........................................ 1 7.1.1 Space inversion and time-reversal ............................ 1 7.1.2 Representations of the Poincar´eLie algebra ...................... 2 7.1.3 Whither time-reversal? .................................. 3 7.2 Antilinearity : The Solution to All Our Problems ........................ 4 7.2.1 Properties of antilinear operators ............................ 4 7.2.2 Position and momentum eigenstates .......................... 6 7.2.3 Change of basis for time-reversal ............................ 7 7.2.4 Time reversal with spin .................................. 8 7.2.5 Kramers degeneracy ................................... 9 7.2.6 Externalfields ....................................... 9 7.3 Time Reversal and Point Group Symmetries .......................... 10 7.3.1 Complex conjugate representations ........................... 11 7.3.2 Generalized time-reversal ................................ 12 7.4 Consequences of Time-Reversal ................................. 13 i ii CONTENTS 7.4.1 Selection rules and time-reversal ............................ 13 7.4.2 Onsager reciprocity .................................... 14 7.5 Colorgroups ............................................ 16 7.5.1 Magnetic Bravais lattices and magnetic space groups ................. 19 7.5.2 Corepresentations of color groups ............................ 23 7.6 Appendix : The Foldy-Wouthuysen Transformation .....................