Quantum Mechanics 2 Tutorial 10: The Dirac Fermion Yehonatan Viernik January 3, 2020 1 Recap 1.1 The Dirac Representation A Weyl fermion transforms under Lorentz boosts and rotations as 1 ¯ ¯ 0 2 (iθ±φ)·σ¯ L=R = e L=R: (1.1) Stacking a left spinor and a right spinor in a column gives us the Dirac fermion: 1 iθ¯+φ¯ ·σ¯ ! χ e 2 ( ) 0 χ = L ! 0 = L : (1.2) D D 1 iθ¯−φ¯ ·σ¯ ξR 0 e 2 ( ) ξR Recall that it is organized as a list of four complex components, but it is not a Lorentz four-vector! 1.2 Lorentz Transformations from (m; n) Representations Question: Why work with su(2)C ⊕su(2)C and not directly with so(1; 3)? What's \wrong" with working directly with Lorentz generators? The answer is nothing's wrong! But since the complexified algebra is isomorphic to su(2)C ⊕ su(2)C, we can just take what we know about su(2) and directly apply it. So the real question is why complexify in the first place? The answer here is very simple: we don't know how to do it otherwise. Let's go from another direction to the same outcome, and see if we can gather insights along the way: 1. SO+(1; 3) is not simply connected, so we need its covering group1. This turns out to be SL(2; C), i.e. the set of 2 × 2 invertible complex matrices with det = 1. We can denote the algebra by sl(2), which is a real Lie algebra, with the natural basis2 0 1 0 0 1 0 A = ;B = ;C = : (1.3) 0 0 1 0 0 −1 2. Unfortunately, we don't know how to classify irreps of sl(2) systematically. We do know, however, that if we allow complex linear combinations of the generators, we get sl(2)C ' sl(2) ⊕ sl(2) ' su(2)C ⊕ su(2)C: (1.4) 3. These guys we know how to deal with. The key insight here is that a representation for a group (or algebra) is also a representation for any subgroup (or subalgebra)! A representation is just a bunch of matrices, so we can simply ignore matrices coming from complex linear combinations of elements of the algebra, and just focus on the bunch of matrices that come from real linear combinations, which represent what we were originally interested in (Lorentz transformations in this case). 4. Note that this is not the same as reducing back from the covering group to the original group. The covering group is still important, as explained in the previous TA. It's the complexification that we ignore once we are done. We complexify, construct representations for the enlarged algebra, and then choose to ignore a bunch of matrices that don't correspond to things that have physical meaning. 1Recall that every Lie algebra is uniquely identified with exactly one simply connected Lie group 2 We wrote sl(2) and not sl(2; C) because it's also the algebra for SL(2; R). In both cases, we only take real linear combinations. 1 This is basically the same reason why we use complex numbers in physics in general. They're just easier to work with. It's easier to allow complex numbers and demand physical results to be real than to stick with the reals all the way through. No one in their right mind would ever give that up... ± ± Now let's put this in the context of the connection between Ai and Ji;Ki. Since Ai rotate among themselves, + − + giving us su(2)C ⊕su(2)C, we need to assign each copy of su(2)C with either A or A . In TA 9, we chose A to − 1 be the left copy of su(2)C and A the right copy. Therefore, when we looked at the ( 2 ; 0) representation, we got that A+ was represented by Pauli matrices, while A− we represented trivially by 0. Since linear combinations of generators induce linear combinations of their representations, if we have representations for A±, we also have representations for J; K! This is how we extract representations for the Lorentz algebra from this somewhat awkward procedure. 2 Parity As an explicit matrix acting in Minkowski space, parity is given by 01 1 B −1 C P = B C ; (2.1) @ −1 A −1 i.e. it flips all spatial indices (hence the name)3. One can show by explicit calculation that T T PJiP = Ji ;PKiP = −Ki; (2.2) and therefore 1 1 PA±P T = P (J ± iK ) P T = (J ∓ iK ) = A∓: (2.3) i 2 i i 2 i i i 1 1 This means that under parity, a ( 2 ; 0) representation becomes a 0; 2 and vice versa. This is why we call Weyl spinors left and right. One can verify that this holds also for finite transformations. In particular, a finite boost transforms under parity as φ¯·K¯ P −φ¯·K¯ B 1 (φ) = e −! e = B 1 (φ) (2.4) ( 2 ;0) (0; 2 ) and vice versa, while finite rotations remain invariant. Implication: If parity is a symmetry of our system, we cannot have only one type of spinor! We must have both left and right spinors to comply with the symmetry. This motivates using Dirac/Majorana spinors. Let's see how D transforms under parity. To do that, first 1 P 1 note that we found that under parity we have 2 ; 0 ! 0; 2 . This implies that 1 1 1 1 ; 0 ⊕ 0; P! 0; ⊕ ; 0 : (2.5) 2 2 2 2 We now inspect the transformation properties of a Lorentz transformation (boost and/or rotation) on the parity transformed object: P P 0 ΛL P P P P ΛR P D = D = Λ( 1 ;0)⊕(0; 1 ) D = Λ(0; 1 )⊕( 1 ;0) D = D (2.6) ΛR 2 2 2 2 ΛL We found that a Dirac fermion under parity transforms like an object with the first spinor being of right chirality and the second of left chirality. We conclude: χL P P ξR D = −! D = : (2.7) ξR χL Note that parity does not apply the transformation χL ! χR or anything like that! Because the Dirac fermion is so incredibly important, we develop an array of tools and notations just to make working with Dirac fermions more convenient. 3Really what you want is to flip an odd number of indices, so that the determinant will change sign. In 3 spatial dimensions, flipping one is enough, and flipping all three is equivalent to a reflection followed by a rotation, which also does the job. However, the above definition of parity is common in Minkowski space. Therefore, it is important to keep this comment in mind, because in 2d for example, flipping all spatial dimensions is just a rotation by π. 2 3 The Clifford Algebra and Lorentz Transformations 3.1 Reorganizing the Generators We saw in class we can organize the generators in an anti-symmetric object, by defining Jij = "ijkJk ;J0i = −Ji0 = Ki ) Jµν = −Jνµ: (3.1) One can show by explicit calculation that4 [Jµν ;Jρσ] = i (ηµσJνρ − ηµρJνσ − ηνσJµρ + ηνρJµσ) : (3.2) Note that Jµν are real linear combinations of the generators, so Eq. (3.2) is strictly equivalent to the Lorentz algebra we wrote in terms of [Ji;Jj]; [Ki;Kj] and [Ji;Kj]: no covering groups, no complexification, just a different basis for the algebra (antisymmetry (3.1) implies we only have 6 linearly independent Jµν 's, so it spans the same vector space). In the fundamental representation, our Ji;Ki generators are 4 × 4 matrices, which we quote here for conve- nience: 00 0 0 0 1 00 0 0 01 00 0 0 01 B0 0 0 0 C B0 0 0 1C B0 −1 0 0C J1 = i B C ;J2 = i B C ;J3 = i B C ; @0 0 0 −1A @0 0 0 0A @1 0 0 0A 0 0 1 0 0 −1 0 0 0 0 0 0 (3.3) 00 1 0 01 00 0 1 01 00 0 0 11 B1 0 0 0C B0 0 0 0C B0 0 0 0C K1 = −i B C ;K2 = i B C ;K3 = i B C @0 0 0 0A @1 0 0 0A @0 0 0 0A 0 0 0 0 0 0 0 0 1 0 0 0 This means that each Jµν is also a 4 × 4 matrix, with indices which we can label with another set (ρ, σ). Interestingly, any Jµν matrix can be written compactly as ρ ρ ρ (Jµν ) σ = i η µηνσ − η ν ηµσ : (3.4) A finite Lorentz transformation is given by exponentiation of this antisymmetric tensor of the generators i Λ = exp !µν J : (3.5) 2 µν µν The antisymmetry of Jµν imposes antisymmetry also on the parameter ! (can you see why?). Example: Boost in the x direction 01 10 Consider the choice ! = −! = φ with the rest being zero. Only the J01 and J10 survive in this case 0 0 −i 0 01 ρ ρ ρ B−i 0 0 0C ρ (J01) = i η η1σ − η η0σ = B C = (K1) : (3.6) σ 0 1 @ 0 0 0 0A σ 0 0 0 0 Before we exponentiate it, notice that this matrix is not Hermitian, meaning that the representation under consideration is non-unitary. The exponentiation parameter, φ, is accordingly non-compact. It is easy to show that 0−1 0 0 01 2 B 0 −1 0 0C 2k k 2 2k+1 k K = B C =) K = − (−1) K ;K = (−1) K1 : (3.7) 1 @ 0 0 0 0A 1 1 1 0 0 0 0 4This section follows Tong's QFT notes (see Eqs.