Quantum Mechanics 2 Tutorial 10: the Dirac Fermion

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¯ ψL/R = e ψL/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 ψ = L → ψ0 = 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

1   −1  P =   , (2.1)  −1  −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 ψD 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 ψD = ψD = Λ( 1 ,0)⊕(0, 1 ) ψD = Λ(0, 1 )⊕( 1 ,0) ψD = ψD (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 ψD = −→ ψD = . (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 = εijkJk ,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: 0 0 0 0  0 0 0 0 0 0 0 0 0 0 0 0  0 0 0 1 0 −1 0 0 J1 = i   ,J2 = i   ,J3 = i   , 0 0 0 −1 0 0 0 0 1 0 0 0 0 0 1 0 0 −1 0 0 0 0 0 0 (3.3) 0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 K1 = −i   ,K2 = i   ,K3 = i   0 0 0 0 1 0 0 0 0 0 0 0 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 ωµν J . (3.5) 2 µν

µν The antisymmetry of Jµν imposes antisymmetry also on the parameter ω (can you see why?).

Example: Boost in the x direction

01 10 Consider the choice ω = −ω = φ with the rest being zero. Only the J01 and J10 survive in this case

 0 −i 0 0 ρ ρ ρ −i 0 0 0 ρ (J01) = i η η1σ − η η0σ =   = (K1) . (3.6) σ 0 1  0 0 0 0 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

−1 0 0 0 2  0 −1 0 0 2k k 2 2k+1 k K =   =⇒ K = − (−1) K ,K = (−1) K1 . (3.7) 1  0 0 0 0 1 1 1 0 0 0 0

4This section follows Tong’s QFT notes (see Eqs. (4.11)–(4.12)), but with the notation from class. To see the equivalence µν µν between the two notation schemes, simply redeﬁne Jµν to have upper indices instead, and then J = iM and ωµν = −Ωµν . Our notation is also the same as Hugh Osborne’s Group Theory notes, see Eq. (4.32).

3 We thus have

∞ k X (iφK1) eiφK1 = k! k=0 ∞ 2k ∞ 2k+1 X (iφK1) X (iφK1) = + + I (2k)! (2k + 1) ! k=1 k=0 ∞ k 2k ∞ k 2k+1 X (−1) (iφ) X (−1) (iφ) = + K2 − K2 + K I 1 1 (2k)! 1 (2k + 1) ! k=0 k=0 (3.8) 0 0 0 0 1 0 0 0  0 −i 0 0 0 0 0 0 0 1 0 0 −i 0 0 0 = + cos(iφ) + sin(iφ) 0 0 1 0 0 0 0 0  0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 cosh (φ) sinh (φ) 0 0 sinh (φ) cosh (φ) 0 0 =   .  0 0 1 0 0 0 0 1

In the last step we used sin(iφ) = sinh(φ) and cos(iφ) = cosh(φ). We thus derived the boost transformation in thex ˆ direction. The boost parameter φ is called the rapidity, and it is related to the velocity β by 1 cosh (φ) = γ = , sinh (φ) = βγ . (3.9) p1 − β2

3.2 The Clifford Algebra The Clifford algebra is defined as a set of objects with the following property

{γµ, γν } ≡ γµγν + γν γµ = 2ηµν . (3.10)

Explicitly, this reads  (γ0)2 = 1  (γi)2 = −1 i = 1, 2, 3 (3.11) γµγν = −γν γµ µ 6= ν.

The simplest representation of the Cliﬀord algebra is in terms of 4 × 4 matrices, where γµ are called gamma matrices. We can use the Pauli matrices as a basis, and take for example

i 0 0 I2 i 0 σ γ = , γ = i . (3.12) I2 0 −σ 0 One can show that such a choice satisﬁes (3.10). This is called the Weyl basis. Alternatively, we can use the Dirac basis, in which γ0 is given by

0 γ0 = I2 , (3.13) 0 −I2 which also satisﬁes (3.10). Consider now the following set of matrices: ( 1 1 1 0 µ = ν Sµν ≡ [γµ, γν ] = γµγν − ηµν = (3.14) 4 2 2 1 µ ν 2 γ γ µ = ν, where again µν labels which matrix we are looking at, i.e. every choice of µν identiﬁes a full 4 × 4 matrix. We will now show that the Sµν ’s satisfy the Lorentz algebra (3.2), where [·, ·] is a matrix commutator. This means that Sµν form a representation of the Lorentz algebra, that is constructed from gamma matrices. The fact that Sµν are 4 × 4 and are made of blocks of Pauli’s hints that they are useful for Dirac fermions. We will see this explicitly. First, let’s show that they satisfy (3.2). Claim:

[Sµν , γρ] = γµηνρ − γν ηρµ. (3.15)

4 Proof: When µ = ν, Sµµ = 0 and the LHS vanishes, while for the RHS we have γµηµρ − γµηµρ = 0. When µ 6= ν, 1 1 1 [Sµν , γρ] = [γµγν , γρ] = γµγν γρ − γργµγν 2 2 2 1 1 1 1 = γµ{γν , γρ} − γµγργν − {γρ, γµ}γν + γµγργν (3.16) 2 2 2 2 = γµηνρ − ηρµγν = γµηνρ − γν ηρµ.

Now, for µ 6= ν and ρ 6= σ, we have 1 [Sµν ,Sρσ] = [Sµν , γργσ] 2 1 1 = [Sµν , γρ]γσ + γρ[Sµν , γσ] (3.17) 2 2 1 1 1 1 (Eq. (3.15)) = γµγσηνρ − γν γσηρµ + γργµηνσ − γργν ησµ. 2 2 2 2 We now use (3.14) to write γµγν = 2Sµν + ηµν , such that

[Sµν ,Sρσ] = Sµσηνρ − Sνσηρµ + Sρµηνσ − Sρν ησµ. (3.18)

The leftover η’s cancel, and we get our required result. If either µ = ν or ρ = σ, then the LHS in (3.18) vanishes by deﬁnition (3.14), while the RHS vanishes due to cancellations between the various terms. Therefore, (3.18) holds in all cases, as desired.

3.3 Lorentz Transformations µν We now need to see how S look like in terms of rotations and boosts. Recall that Jij were rotation generators, ij µν and J0i were boost generators. Therefore, we expect S to be representations for rotations. The fact that S are 4 × 4 doesn’t mean that they will be the 4 × 4 matrices given in Eq. (3.3)! Indeed, in the Weyl basis, we have the following representations:

Rotations

1 0 σi 0 σj i σk 0 Sij = = − εijk , (3.19) 2 −σi 0 −σj 0 2 0 σk

i j ij ijk k where we used the general identity for Pauli matrices: σ σ = δ I2 + iε σ . Since we need Ωµν = −Ωνµ, we k ijb b 5 can write Ωij = −εijkθ . Using the identity εijaε = 2δa, we ﬁnd

i θ¯·σ¯ 1 ij e 2 0 S[Λ] = exp ΩijS = i θ¯·σ¯ . (3.20) 2 0 e 2

Therefore, an object that transforms under such a representation will behave as

i θ¯·σ¯ i 0 α β e 2 0 ψ ψ → ψ = S[Λ] βψ = i θ¯·σ¯ 2 (3.21) 0 e 2 ψ

i i i θ¯·σ¯ i where ψ are two component objects that transform as ψ → e 2 ψ . In particular, a 2π rotation around the z axis, parameterized by θ¯ = (0, 0, 2π) is given by

eiπ  iπσ3 ! i −iπ i e 0 ψ  e  ψ ψ → 3 =   = −ψ. (3.22) 0 eiπσ ψ2  eiπ  ψ2 e−iπ

This is indicative that ψ is not a Lorentz vector, as the latter rotates around z under the exponentiation of J3 given in (3.3).

5 This is now using Tong’s notation, because this is what is shown in the lecture notes in this section. But similarly to Jµν vs µν i µν ˜ M , we can just plug in i’s in various places to produce an equivalent exp 2 ω Sµν version of the same thing.

5 Boosts For boosts, we need the S0i matrices.

i i 0i 1 0 I2 0 σ 1 −σ 0 S = i = i . (3.23) 2 I2 0 −σ 0 2 0 σ

Taking Ω0i = −Ωi0 = φi, we have

1 φ¯·σ¯ e 2 0 S[Λ] = − 1 φ¯·σ¯ . (3.24) 0 e 2

Together with S[Λ] for rotations, we just showed that an object ψ that transforms under such a representation 1 2 S[Λ] is a Dirac fermion, namely we can identify ψ = χL and ψ = ξR!

3.4 Summary In this section we did three things:

1. We started by rewriting the Lorentz algebra in terms of a rank 2 tensor Jµν with the speciﬁc Lie brackets of (3.2). We said that everything that follows these Lie brackets is equivalent to the Lorentz algebra, and showed in (3.5) how Lorentz transformations are given in this notation. We then concluded that if we have a bunch of matrices that follow the commutator version of (3.2), then they constitute a representation of the Lorentz algebra. 2. Next, we introduced the Cliﬀord algebra and the gamma matrices, and used them to construct a set of matrices Sµν that satisfy the commutation relations for representations of the Lorentz algebra.

3. Finally, we saw how rotations and boosts act in this representation, and identiﬁed the object that transforms under them as none other than a Dirac fermion. This may have felt redundant, as we already had everything we needed when we classiﬁed irreps of su(2). But as explained, Dirac fermions are extremely important objects in physics (electrons anyone?), and gamma matrices will be the main workhorse for the rest of the course and possibly your careers for fermion calculations.

4 Symmetries in The Dirac Action

Recall the Lagrangian of a Dirac fermion:

L = iψ¯∂ψ/ − mψψ,¯ (4.1)

¯ † 0 µ where ψ ≡ ψ γ and ∂/ ≡ γ ∂µ. Taking the Euler Lagrange equations gives us the Dirac equation

∂/ − m ψ = 0 . (4.2)

4.1 U(1) Symmetry The internal symmetry of (4.1) is a global U(1), where

ψ → eiωψ , ψ¯ → e−iωψ¯ . (4.3)

Recall Noether’s theorem from TA 6: If we have a continuous symmetry of the action with

µ δL = ∂µF , (4.4) we get a current

X ∂L J µ = δφ − F µ (4.5) ∂(∂ φ ) a a µ a

µ which is conserved, i.e. ∂µj = 0. In our case, the variation of the ﬁelds are

ψ0 ≈ ψ + iωψ ⇒ δψ = iψ , (4.6) ψ¯0 ≈ ψ¯ − iωψ¯ ⇒ δψ¯ = −iψ¯ .

6 Since this is a global symmetry, we don’t worry about d4x, and the variation of the Lagrangian is

0 2 L = (ψ¯ − iωψ¯) i∂/ − m (ψ + iωψ) + O(ω ) 2 = L + ψ¯ i∂/ − m iωψ − iωψ¯ i∂/ − m ψ + O(ω ) (4.7) = L + O(ω2), thus F µ = 0. The conserved current and the charge associated with this symmetry are therefore 0 ∂L ∂L > J µ = iψ + −iψ¯ = −ψγ¯ µψ ∂ (∂ ψ) ¯ µ ∂ ∂µψ (4.8) Z Z Z Q = − d3x ψγ¯ 0ψ = − d3x ψ†γ0γ0ψ = − d3x ψ†ψ .

4.2 N Dirac fermions Consider a theory of N Dirac fermions (j, k = 1,...,N)

¯ µ ¯ L = iψjγ ∂µψj − ψjMjkψk , (4.9) where M is, without loss of generality, a diagonal positive-definite mass matrix. The symmetry in the case of no degeneracy is U(1)N , as the fields are distinct. If we assume all masses are degenerate, it means that we can no longer distinct between the different copies of the fields, so the symmetry is now U(N). Let us assume that the matrix M vanishes identically. Naively, this is just a special case of a diagonal mass matrix. But in fact, we will now show that we cannot even distinguish between left and right. We rewrite the Lagrangian in terms ψ of the Weyl spinors ψ = L ψR

j† µ j j† µ j L = iψL σ¯ ∂µψL + iψR σ ∂µψR . (4.10)

From this expression we can clearly see a U(N)2 symmetry that mixes the left-handed and right-handed ﬁelds ∗ ∗ separately. However, we saw that using σ2σj = −σjσ2, the combination σ2ψL transforms as

σ∗ h σ i σ ψ∗ −→ σ Λ∗ ψ∗ = σ exp (−iθ − φ ) i ψ∗ = exp (iθ + φ ) j σ ψ∗ = Λ σ ψ∗ . (4.11) 2 L 2 L L 2 j j 2 L j j 2 2 L R 2 L

We can now rename

∗ T j j N+j j∗ j N+j j† N+j χR ≡ ψR , χR ≡ σ2ψL ⇒ ψL = −σ2 χR , ψL = − χR σ2 . (4.12)

The Lagrangian is now

T ∗ N+j µ N+j j† µ j L = i χR σ2σ¯ σ2∂µ χR + iχR σ ∂µχR T ∗ N+j µ N+j j† µ j (integration by parts) = −i ∂µχR σ2σ¯ σ2 χR + iχR σ ∂µχR † N+j µ T N+j j† µ j (4.13) (transpose + anticommutation) = i χR σ2 (¯σ ) σ2∂µχR + iχR σ ∂µχR † µ T µ ∗ N+j µ N+j j† µ j (¯σ ) = (¯σ ) = i χR σ ∂µχR + iχR σ ∂µχR n† µ n (n = 1,..., 2N) = iχR σ ∂µχR . The symmetry is enhanced to a U(2N) symmetry. In the transpose step, we used anticommutation of the fermion ﬁelds. The reason for that is beyond the scope of this course, and we tried to avoid it, but it turns out this chiral symmetry only makes sense if the ﬁelds are anti-commuting also prior to quantization. Because we only care about the quantum theory anyway, we might as well adopt this anti-commutativity here as well. Below is a more elaborate explanation. Please do not waste time on it before the exam! This is strcitly for casual discussions over beer after the exam!

7 Appendix - Fermions and anti-commutation

Note: This section is strictly outside the course material!!! This is for your own amusement only!

We started our journey in classical field theory by considering a bunch of (generally complex) functions of spacetime, which satisfy certain wave equations and follow well defined transformation rules under Lorentz. We constructed Lagrangians from various combinations of these fields, and noticed we have symmetries, so that if we stack a bunch fields into vectors, matrices, complex linear combinations etc., we can construct objects that transform under known representations of those symmetries of mixing the various fields. At this point, we did not yet make any contact with quantization. In particular, the field ψ that transforms under the Dirac representation of the Lorentz group, and that satisfies the Dirac equation, has no a priori relation to fermions in their quantum sense of statistics (Pauli exclusion etc.). Of course, once we quantized the theory, namely promoted the fields to operators in a Hilbert space, we noticed we must impose anti-commutation relations to ensure the energy is bounded from below, and by doing so we automatically got the statistics of fermions! Looking back at the classical theory, one can show that if we define fermion fields to anti-commute at the classical level as well, then their promotion to operators naturally gives rise to canonical anti-commutation relations. Due to this hindsight perspective, you will be hard-pressed to find literature that talks about Dirac fields as anything other than “anti-commuting numbers”. One can show that when ψ, ψ† are taken to be either commuting or anti-commuting numbers, the Dirac action is real in both cases! Just play a similar game as Eq. (4.13), but without trying to change chiralities. Simply take the complex conjugate, integrate by parts (clumsily ignoring the surface term), and then claim the Lagrangian is a number so you are allowed to transpose for free. The Dirac equation is also agnostic about the “parity” of ψ because it is linear in it, and more generally, essentially everything we did in the course at the classical level works just the same whether you take ψ to be commuting or anti-commuting! Since we did not want to bother you with these issues (you will encounter them in QFT), we simply avoided this concept altogether. It seems to be the case, though, that the chiral symmetry really does hinge on fermion fields being anti- commuting at the classical level. It just doesn’t work otherwise. What does that mean? It kind of means that this is not a classical symmetry. Or to be more precise, it is a classical symmetry only when ψ’s are defined as anticommuting, and when you avoid thinking about the ”classical theory” as being related to ”classical physics”. This is perfectly acceptable for the following two reasons: 1. Fields themselves don’t represent anything physical, they are not observable quantities and they are not to be interpreted as “probability amplitudes” or anything like that. So what do you care how they are mathematically defined? As long as they satisfy the things you expect them to satisfy, you should be 100% happy. 2. There is no classical analogue or notion of being a fermion. There are no classical objects that display fermionic properties. An electron is a strictly quantum creature. Then, since there is no natural classical limit for fermions, the classical symmetry is to be interpreted in the sense of a set of transformations under which the action is invariant, without reference to a prejudicial “physical” interpretation. This is in contrast to the quantum symmetry, which is the symmetry not of the action, but rather of all the correlation functions of the theory, i.e. including the path integral, its measure and the vacuum state, and also of the Hamiltonian side of the theory. Since we care about the quantum theory, we might as well choose the fields to anti-commute, such that the symmetry of the action prior to quantization agrees with that of the quantized theory. To conclude, fermion fields are required to anti-commute in order to naturally produce the canonical quantum anti-commutation relations and the fermionic nature required from spin 1/2 fields due to spin-statistics. This identification made no different for none of the classical analysis we did, except this statement of the chiral symmetry, which forced our hands into imposing anticommutativity of fermions. Finally, take what’s written here with a grain of salt. These are the conclusions we’ve come to thinking about this minus sign, for the past day or so, this is far from set in stone!