Notes on Dirac Field

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 ψL.

• (0,1/2): Right Weyl spinor ψR.

• (1/2,1/2): Lorentz Vector Aµ , ∂µ... . Weyl spinors transform as

σi σi ψ → Λ ψ = exp (iθ + φ ) ψ , ψ → Λ ψ = exp (iθ − φ ) ψ 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 † the Pauli matrices. what about ψL/R?

σi † σi ψ† → exp (iθ + φ ) ψ = ψ† exp (−iθ + φ ) = ψ† Λ−1 . R i i 2 R R i i 2 R L (4) and similarly

σi ψ† → ψ† exp (−iθ − φ ) = ψ† Λ−1 . (5) L L i i 2 L R

An immediate result is that σi σi ψ† ψ → ψ† exp (−iθ − φ ) exp (iθ − φ ) ψ (6) L L L i i 2 i i 2 L

1 † is not Lorentz invariant (and also ψRψR). But σi σi ψ† ψ → ψ† exp (−iθ − φ ) exp (iθ + φ ) ψ = ψ† ψ (7) L R L i i 2 i i 2 R L R

† is invariant, and the same for ψRψL. If we want to write a mass term for ψ that will be Lorentz invariant and quadratic in the ﬁelds it seems like we cannot do it for ψL(ψR) by itself and we need to couple them together. This observation motivates us to deﬁne the Dirac fermion which can be written as ψ ψ = L (8) ψR and its Lorentz representation is (1/2, 0) ⊕ (0, 1/2). The transformation of ψ in this basis is simply ΛL ψ → Λ1/2ψ = ψ . (9) ΛR

We wrote ψ 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 {γ , γ } = 2η I4×4 . (10)

From the γ matrices we can construct a representation of the Lorentz generators 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 − ωµν S ψ → Λ1/2ψ = e 2 ψ . (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 ψ under rotation around the z axis. To do this we choose ωxy = −ωyx = θ and all the rest are zero.

e−iθ/2  3 iθ/2 iθ σ 0  e  ψ → exp − ψ =   ψ . (15) 2 0 σ3  e−iθ/2  eiθ/2 If we take θ = 2π we ﬁnd ψ → −ψ 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 {γ5, γµ} = 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 ψL = −ψL , γ ψR = ψR . (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 ψL 0 PL = PL ,PR = PR ,PLPR = PRPL = 0 ,PRψ = ,PLψ = . 0 ψR (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 † boosts Λ1/2 is not unitary and therefore ψ ψ is not Lorentz invariant. We want to deﬁne the Dirac conjugate of ψ denoted by ψ¯ that will have the ¯ ¯ −1 ¯ transformation rule ψ → ψΛ1/2 and therefore ψψ will be Lorentz invariant † † (and hermitian). In terms of Weyl spinors we already saw that ψLψR +ψRψL is Lorentz invariant and it is hermitian. Therefore in the chiral basis ¯ † † † 0 ψ = ψR ψL = ψ γ (21)

3 will do the work. We will show that this deﬁnition ψ¯ = ψ†γ0 satisﬁes what we want in every basis. We will use the following identities 0 µ 0 µ † 0 2 γ γ γ = (γ ) , (γ ) = I (22) which lead to i † i (Sµν)† = (γµγν − γνγµ) = − ((γν)†(γµ)† − (γµ)†(γν)†) 4 4 i i = − (γ0γνγ0γ0γµγ0 − γ0γµγ0γ0γνγ0) = γ0(γµγν − γνγµ)γ0 = γ0Sµνγ0 . 4 4 (23) The transformation of ψ¯ is

i µν † i µν † i 0 µν 0 i µν ¯ − 2 ωµν S 0 † 2 ωµν (S ) 0 † 2 ωµν γ S γ 0 † 0 2 ωµν S ¯ −1 ψ → e ψ γ = ψ e γ = ψ e γ = ψ γ e = ψΛ1/2 . (24) We found the way to write a good mass term. Now lets try to write a kinetic term. Using representation arguments, we have ψ, ψ¯ 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 ψ¯a∂µψ 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 ψγ¯ ∂µψ 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 ψγ ∂µψ → ψΛ1/2γ Λµ ∂νΛ1/2ψ = ψγ ∂µψ . (27) Now we are ready to write the Dirac action Z 4 ¯ µ ¯ SDirac = d x iψγ ∂µψ − mψψ . (28)

We just need to show that the kinetic term is hermitian: Z Z Z Z 4 µ † 4 † µ † 0 † 4 µ 4 µ d x iψγ¯ ∂µψ = d x −i∂µψ (γ ) (γ ) ψ = −i d x∂µψγ¯ ψ = i d xψγ¯ ∂µψ . (29)

4 4 Dirac equation

The Dirac equation can be derived from the Dirac action using the EL equations for ψ and ψ¯. ∂L ∂L ∂ − = −iγµ∂ ψ + mψ = 0 µ ∂(∂ ψ¯) ∂ψ¯ µ µ (30) ∂L ∂L µ ∂µ − = i∂µψγ¯ + mψ¯ = 0 . ∂(∂µψ) ∂ψ We will use the Feynman slash notation

µ A/ = γ Aµ . (31)

The Dirac equation can be written as

(i∂/ − m)ψ = 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)ψ = −(∂/∂/ + m2)ψ = −(∂2 + m2)ψ = 0 . (33)

Now we will solve the equation. Consider one Fourier mode with positive energy ψ(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 · σξ u(p) = √ , p0 > 0 . (37) p · σξ¯ ξ has two arbitrary components, and we can expand the solutions in the following basis √ p · σξ 1 0 us(p) = √ , ξ1 = , ξ2 = . (38) p · σξ¯ 0 1

5 On the same way we can solve for negative energy modes

ψ(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 · σηs 1 0 vs(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 ψ(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.

5 Quantization

The Lagrangian is L = ψ¯(i∂/ − m)ψ (43) from which we derive the conjugate momentum ∂L Π = = iψ† (44) ∂(∂tψ) and the Hamiltonian Z Z Z 3 3 i 3 0 H = d x[Πψ˙ − L] = d xψ¯(−iγ ∂i + m)ψ = i d xψγ¯ ∂0ψ . (45)

ψ and ψ† satisfy the canonical anti-commutation relations

† (3) † † {ψa(~x), ψb(~y)}ET = δ (~x − ~y)δab , {ψ, ψ}ET = {ψ , ψ }ET = 0 . (46)

We will expand ψ and ψ† using

Z d3p 1 X ψ(x) = asus(p)e−ip·x + bsvs(p)eip·x , p0 > 0 (2π)3 p p p 2Ep s=1,2 Z d3p 1 X h i ψ†(x) = as†us†(p)eip·x + bs†vs†(p)e−ip·x , p0 > 0 . (2π)3 p p p 2Ep s=1,2 (47)

6 One can show that the following identities hold

† † † † ur(p)us(p) = 2Epδrs , vr(p)vs(p) = 2Epδrs , ur(p)vs(−p) = vr(−p)us(p) = 0 . (48) For example, √ † √ T √ T p · σξs ur(p)us(p) = p · σξr p · σξ¯ r √ = p · σδrs + p · σδ¯ rs = 2Epδrs . p · σξ¯ s (49) Write the Hamiltonian in terms of these modes Z X 3 0 H = i d xψγ¯ ∂0ψ s,r=1,2 R 3 3 X Z d pd k 1 h i = i d3x ar†ur†(k)eik·x + br†vr†(k)e−ik·x ∂ asus(p)e−ip·x + bsvs(p)eip·x (2π)6 p k k 0 p p s,r=1,2 2 EpEk R 3 3 R 3 X d pd k i d x h i = ar†ur†(k)eik·x + br†vr†(k)e−ik·x −iE asus(p)e−ip·x + iE bsvs(p)eip·x (2π)6 p k k p p p p s,r=1,2 2 EpEk R 3 3 p X d pd k Ep Z h = √ d3x ar†ur†(k)ei(k−p)·xasus(p) − ar†ur†(k)ei(k+p)·xbsvs(p) 2(2π)6 E k p k p s,r=1,2 k r† r† −i(k+p)·x s s r† r† −i(k−p)·x s s i +bk v (k)e apu (p) − bk v (k)e bpv (p) R 3 X d p h i = ar†ur†(p)asus(p) − ar† ur†(−p)bsvs(p) + br† vr†(−p)asus(p) − br†vr†(p)bsvs(p) 2(2π)3 p p −p p −p p p p s,r=1,2 R 3 R 3 X d p h i X d p h i = 2E as†as − 2E bs†bs = E as†as − bs†bs . 2(2π)3 p p p p p p (2π)3 p p p p p s=1,2 s=1,2 (50) In order to study the spectrum, we need to ﬁnd the (anti-) commutation relations of the a, b, a†, b† operators. One can isolate Z Z s 1 s† 3 iky s 1 s† 3 −iky ak = p u (k) d yψ(y)e , bk = p v (k) d yψ(y)e 2Ep 2Ep Z Z s† 1 3 † s −iky s† 1 3 † iky s ak = p d yψ (y)u (k)e , bk = p d yψ (y)e v (k) . 2Ep 2Ep (51) From these we can show that Z s r† 1 s† 3 3 † r ipx−iky {ap, ak } = p ua (p) d xd y{ψa(x), ψb(y)}ub(k)e 2 EpEk Z 3 1 s† 3 r ipx−ikx (2π) s† r (3) ~ 3 (3) ~ = p ua (p) d xua(k)e = ua (p)ua(p)δ (~p − k) = (2π) δ (~p − k)δrs 2 EpEk 2Ep (52)

7 and on the same way

r s† 3 (3) ~ {bp, bk } = (2π) δ (~p − k)δrs (53) and the rest are zero. Compute the commutation relations with the Hamil- tonian: R 3 X d k [H, as†] = E ar†[ar , as†] = E as† . p (2π)3 k k k p p p (54) r=1,2 For the b’s we can see that because of the minus sign,

s† s† [H, bp ] = −Epbp (55) so b† lower the energy. Fortunately, since the fermionic operators are quan- tized with anti-commutators and b2 = (b†)2 = 0, there is no real difference between a creation and an annihilation operator. The way to decide which one is a creation operator and which one is an annihilation operator is by checking which of the operators raises the energy and which lowers it. There- fore we can rename the operators b → b† , b† → b and write the Hamiltonian as R 3 R 3 X d p h i X d p h i H = E as†as − bsbs† = E as†as + bs†bs + const (2π)3 p p p p p (2π)3 p p p p p s=1,2 s=1,2 (56) s† s† At this form of H it is clear that ap , bp can be interpreted as creation s s operators while ap , bp are annihilation operators. To construct the Fock space of the theory, we first define the vacuum state which is annihilated by all of the annihilation operators

s r ap|0 >= bk|0 >= 0 . (57) We can act with the creation operators on |0 > to get multiparticle states. s† 2 s† 2 Since (ap ) = (bp ) = 0, each state can be occupied once as expected from fermions. We have two kinds of particles, a† and b† which are the particle and its anti-particle. Each one of them also carry the index s = 1, 2 which are the two spin states.

6 Symmetries of the Dirac action

Recall the Dirac action Z 4 ¯ µ ¯ SDirac = d x iψγ ∂µψ − mψψ . (58)

The action is invariant under the global U(1) transformations

ψ → eiαψ , ψ¯ → ψe¯ −iα . (59)

8 The conserved current and the charge associated with this symmetry are ∂L Z J µ = iψ = −ψγ¯ µψ , Q = − d3xψ†ψ . (60) ∂(∂µψ) In terms of the creation and annihilation operators R 3 X d p Q = [as†as − bs†bs] ∼ Nˆ − Nˆ . (2π)3 p p p p a b (61) s=1,2 The conservation of Q implies that the number of particles minus the number of anti particles is a constant, and therefore pairs of p-a.p can annihilate each other or appear together such that Q is conserved. In the massless limit of the Dirac action there is another U(1) under which

5 ψ → eiαγ ψ (62) which rotates the left and right spinors with an opposite phase. You will prove it (in some version) in your exercise.

7 Complex scalar Vs Dirac ﬁeld

The free Lagrangian for the complex scalar field is ∗ µ 2 2 L = ∂µφ ∂ φ − m |φ| . (63) The EL equations are (∂2 + m2)φ = (∂2 + m2)φ∗ = 0 . (64) The quantization procedure is very similar to the real scalar with one im- portant difference. The expansion of the complex field is Z 3 d p 1 −ipx † ipx φc(x) = 3 p bpe + cpe . (65) (2π) 2Ep

For the real scalar, the operators b, c in the expansion must satisfy bp = cp from reality and there is only one set of creation and annihilation operators. For the complex scalar we have two sets, similar to the Dirac field. The Hamiltonian of the complex field is Z d3p H = E b†b + c†c + const . (66) (2π)3 p p p p p The interpretation is similar to that of the Dirac field, in the sense that the theory consists particles and anti particles. The U(1) charge in terms of the operators is Z d3p Q = b†b − c†c = Nˆ − Nˆ . (67) u(1) (2π)3 p p p p b c