Celso C. Nishi∗
Universidade Federal do ABC - UFABC, Santo Andr´e, SP, Brasil
I. REPRESENTATIONS OF THE LORENTZ GROUP
Based on Schwartz, Chap.10; Srednicki, Chap.2; Peskin, Chap.3. Beware the metric of Srednicki is opposite to those of Peskin and Schwartz.
For a quantized scalar field φ(x), a Lorentz transformation Λ acts as
φ(x) → ¯φ(x) = U (Λ)−1φ(x)U (Λ) = φ(Λ−1x) . (1) This transformation law implies that
U (Λ)−1∂µφ(x)U (Λ) = Λµν∂¯νφ(¯x) , (2) where ¯x = Λ−1x and ¯∂ν = ∂ ¯∂x
ν. By analogy we then expect that a vector field V
µ(x) transforms similarly as
U (Λ)−1Vµ(x)U (Λ) = ΛµνVν(Λ−1x) . (3) This is what we expect, for example, for a 3-vector field vi(x) under a rotation R:
hψ|U†(R)vi(Rx)U (R)|ψi = Rijhψ|vj(x)|ψi , (4) i.e., the direction of a classical vector field at a rotated position Rx is also rotated by the same transformation.
Analogously, we expect a rank 2 tensor field to transform as
U (Λ)−1Bµν(x)U (Λ) = ΛµαΛνβBαβ(Λ−1x) . (5) For a generic set of fields ϕA(x) which mix under Lorentz according to a representation L, we expect
U (Λ)−1ϕA(x)U (Λ) = LAB(Λ)ϕB(Λ−1x) . (6) The matrices L(Λ) should represent the Lorentz group as
LAB(Λ0)LBC(Λ) = LAC(Λ0Λ) . (7)
There are additional representations L for the Lorentz group besides the tensorial representations.
∗
We now seek the so called spinorial representations.
II. LORENTZ GROUP AND ALGEBRA
We have seen that the Lorentz group, also denoted as O(3, 1), is defined as a set of 4 × 4 real matrices Λ that obey
ΛTgΛ = g . (8)
One can check that this set satisfies the properties of a group G: (a) (xy)z = x(yz) , ∀x, y, z, (b) ∃ e | ex = xe ∀x ,
(c) ∀x, ∃ x−1| x−1x = e = xx−1.
The Lorentz group is composed of four branches and the branch connected to the identity is the proper and orthochronous Lorentz group denoted as SO+(3, 1). We can pick an element arbitrarily close to the identity such that it can be approximated as
Λµν ≈ δµ
ν + ωµν, (9)
or, more compactly,
Λ ≈14+ ω . (10)
Plugging in this approximation into the defining property (8) tell us that
gωT= −ωg , (11)
or, with all indices lowered,
ωνµ= −ωµν. (12)
The possible matrices ω live in a vector space called the Lie algebra of SO+(3, 1), denoted as so(3, 1). With all the indices lowered, ωµν is a real 4 × 4 antisymmetric matrix which can be expanded using the canonical basis of 6 elements. Raising the first index, this basis is composed of
iK1 = 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 , iK2= 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 , iK3= 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 , iJ1 = 0 0 0 0 0 0 0 0 0 0 0 1 0 0 −1 0 , iJ2= 0 0 0 0 0 0 0 −1 0 0 0 0 0 1 0 0 , iJ3= 0 0 0 0 0 0 1 0 0 −1 0 0 0 0 0 0 . (13)
We put a factor i so that Ji, which generate spatial rotations, are hermitean. In contrast, the generators Ki which generate boosts are antihermitean. Then we can expand any ω as
ω = i~θ · ~J + i~β · ~K . (14)
We call the matrices Ji, Ki generators because any group element in SO+(3, 1) can be written as
Λ = exp(i~θ · ~J + i~β · ~K) . (15)
The boosts, given by ~θ = 0, are not unitary because Ki are antihermitean. Since βi is unbounded, we say the group is noncompact and this defining representation of the Lorentz group is not unitary. In fact, there is no nontrivial finite dimensional unitary representation for the Lorentz group.
Another way of writing the expansion (14) is to use the entries of ω itself as expansion parameters as
ωαβ= 12ωµν(iJµν)αβ. (16)
For example, ω01= β1 and ω23= −θ1. Our generators Ki and Jiare collectively contained in Jµν which is
antisymmetric in µ ↔ ν. Consistency requires that (iJµν)αβ= δαµδ
ν
β− (α ↔ β) . (17)
You can check that
(Ki)αβ= (J 0i)α β and (Ji)αβ= − 1 2εijk(J jk)α β. (18)
The quantum operators U (Λ) should also correspond to a representation of the Lorentz group such that
U (Λ0Λ) = U (Λ0)U (Λ) . (19)
Within SO+(3, 1), we can study the representations of the quantum generators in the algebra so(3, 1) instead of the group. For infinitesimal transformations,
U (14+ ω) ≈ I + 2iωµνJµν. (20)
Note the similarity with (16).
To confirm the transformation law for Jµν we expand Λ0 =14+ ω0 in
U (Λ)−1U (Λ0)U (Λ) = U (Λ−1Λ0Λ) . (21)
See Srednicki, p.17. We obtain the transformation law
U (Λ)−1JµνU (Λ) = ΛµαΛνβJαβ, (22) which is the same for all rank 2 tensors; see (5).
Taking infinitesimal Λ and U (Λ) in eq. (22), we obtain the commutation relation
[Jµν, Jαβ] = igµαJνβ− (µ ↔ ν)− (α ↔ β) . (23) These commutation relations define the Lie algebra of the Lorentz group. We can identify the generators of rotations Ji ≡ −12εijkJjk and the generators of boosts Ki ≡ J0i. In terms of these generators, the commutation relations (23) can be cast into the form
[Ji, Jj] = iεijkJk, [Ji, Kj] = iεijkKk, [Ki, Kj] = −iεijkJk.
(24)
The first relation is the familiar commutation relation for angular momentum. The second relation tell us that Ki behaves as a 3-vector under rotations.
Since any representation of the Lorentz group obey its Lie algebra, the relation (23) is obeyed by Jµν in the defining representation and equally (24) is satisfied for Ji, Ki.
If we also include the momentum operator Pµ= (H, P) we obtain the Lie algebra of the Poincar´e group; see eqs.(2.18)–(2.20) of Srednicki for the algebra.
A simple representation for Jµν is the generalization of the angular momentum differential operator in quantum mechanics:
Lµν = −i(xµ∂ν− xν∂µ) . (25)
One can check (23) is satisfied. The usual angular momentum operator in QM is Li = −12εijkLjk where, e.g., L3 = −L12.
III. REPRESENTATIONS OF so(3, 1)
To study the representations of so(3, 1) it is useful to define
Ji+≡ 12(Ji+ iKi) , Ji−≡ 12(Ji− iKi) . (26) In terms of these generators, the algebra (24) becomes
[Ji+, Jj+] = iεijkJk+, [Ji−, Jj−] = iεijkJk−, [Ji+, Jj−] = 0 .
(27)
We can identify two su(2) algebras that commute and we can write so(3, 1) ' su(2) ⊕ su(2). The representations of su(2) are well known from quantum mechanics: a spin j representation has dimension 2j + 1. In our case, we have two su(2) which should be characterized by two “spins”
su(2) ⊕ su(2) (0, 0) (1/2, 0) (0, 1/2) (1/2, 1/2) (1, 0) (1, 1)
so(3) 0 1/2 1/2 1 ⊕ 0 1 2 ⊕ 1 ⊕ 0
TABLE I: Representations of the Lorentz algebra and its so(3) subalgebra.
A, B as
su(2) with {Ji−} : A = 0, 1/2, 1, 3/2, 2, . . . ,
su(2) with {Ji+} : B = 0, 1/2, 1, 3/2, 2, . . . . (28) A generic representation of so(3, 1) is characterized by the pair (A, B) and its dimension is (2A + 1)(2B + 1).
On the other hand, the spin for quantum fields is characterized by the SO(3) subgroup of SO+(3, 1) which is related to the little group in Wigner’s construction (for the massive case). The spin j of this subgroup for a representation (A, B) spans
j = A + B, A + B − 1, . . . , |A − B| . (29)
We list the smallest representations in table I. .
For tensorial representations we have integer j and thus integer A + B. In contrast, for spinorial representations we have half-integer j and A + B.
IV. SPIN 1/2
Def. Spinors are vectors on which the spin 1/2 Lorentz representations act.
There are only two representations with j = 1/2: (0, 1/2) and (1/2, 0). They are based on the spin 1/2 representation of su(2) which are represented by the Pauli matrices σi/2.
The two representations are
(1/2, 0) : J~−= 12~σ , J~+= ~0 ,
(0, 1/2) : J~−= ~0 , J~+= 12~σ . (30) Or, if we invert the relations (26),
~
J = ~J−+ ~J+, K = i( ~~ J−− ~J+) , (31) we can write
(1/2, 0) : J =~ 12~σ , K =~ 2i~σ ,
(0, 1/2) : J =~ 12~σ , K = −~ 2i~σ , (32) For this representation too ~K are anti-hermitean and then the group elements they generate are not unitary.
These representations act on their respective spinors as (1/2, 0) : ψL→ e
i
2(~θ+i~β)·~σψL, left-handed, (33a)
(0, 1/2) : ψR→ e
i
2(~θ−i~β)·~σψR, right-handed. (33b)
We call the spinor transforming as (1/2, 0) as a left-handed Weyl spinor while the one transforming as (0, 1/2) as a right-handed Weyl spinor. Note that these spinors have two components.
V. FIELD EQUATIONS AND LAGRANGIANS
Let us seek field equations which are first order in ∂µ instead of the Klein-Gordon equation which is second order. The part that depends on the derivative comes from the kinetic term in the Lagrangian. Then we need a kinetic term which contains ∂µ only in first order.
We seek an structure such as
∂µ( )µ= scalar. (34)
So we need to construct a vector out of one field, say ψL. Let us define
¯
σµ≡ (12, −~σ) . (35)
We are going to show that ψ†Lσ¯µψL indeed transforms as a 4-vector. To show that, it is enough to study the infinitesimal version of (33):
δψL= 2i(~θ + i~β) · ~σψL, δψL† = ψL†(−i)2 (~θ − i~β) · ~σ . (36) This implies δ(ψ†Lσ¯µψL) = 12ψL† n (−i)(~θ − i~β) · ~σ ¯σµ+ ¯σµi(~θ + i~β) · ~σoψL (37) The braces in the middle can be simplified to
µ = 0 : { } = −2~β · ~σ , µ = i : { } = +2εijkθjσk+ 2βi. (38) If we denote Vµ= ψL†¯σµψL, we obtain δV0= βiVi, δVi= −εijkθjVk+ βiV0. (39)
This matches the behavior implied by (15) for which
We can now write the kinetic term as
kinetic = ψ†Li∂µσ¯µψL, (41)
where we have put an i to make the term hermitean (remember QM where ∇ is not an hermitean operator but −i∇ is.)
The equation of motion implied by this Lagrangian is
i∂µσ¯µψL= 0 . (42)
This is the Weyl equation for a massless left-handed fermion.
One can analogously check that ψ†RσµψR is a 4-vector when we define
σµ≡ (12, ~σ). (43)
So the kinetic term for a right-handed Weyl fermion is
kinetic = ψ†Ri∂µσµψR. (44)
One can also check that
ψL†ψL, ψ†RψR, (45)
are not Lorentz invariant as they correspond to the µ = 0 component of 4-vectors; take e.g. ψ†Lσ¯µψL for µ = 0. But the term
ψ†RψL (46)
is Lorentz invariant. So the Lorentz invariant and hermitean Lagrangian that we can write is L = ψ† Li∂µσ¯µψL+ ψ † Ri∂µσµψR+ m(ψ † RψL+ ψ † LψR) . (47)
The previous Lagrangian can be written in a more compact form if we define an object with 4 components as ψ ≡ ψL ψR ! , ψ ≡¯ ψR† ψL† . (48)
Notice the exchange of fields in ¯ψ. Then we can rewrite (47) as
L = ¯ψ(i /∂ − m)ψ . (49)
This is the Dirac Lagrangian. The slashed notation denotes /
where γµare 4 × 4 Dirac matrices which reads γ0 = 0 12 12 0 ! , γi = 0 σ i −σi 0 ! . (51)
This form for the Dirac matrices are in the Weyl or chiral representation. We can also redefine ¯
ψ = ψ†γ0, (52)
which is valid in any representation.
Considering ψα and ¯ψα, α = 1, 2, 3, 4, as independent fields, the Lagrangian in (49) leads to the following equations of motion:
(i /∂ − m)ψ(x) = 0 , (53a)
¯
ψ(x)(−i←−∂/µ− m) = 0 . (53b)
The first equation is the Dirac equation. The arrow to the left means the derivative acts on the fields on the left.
VI. DIRAC ALGEBRA AND LORENTZ COVARIANCE
The Dirac matrices satisfy the Dirac algebra
{γµ, γν} = 2gµν14, (54)
where braces indicate the anticommutator
{A, B} ≡ AB + BA . (55)
This is a special case of a Clifford algebra.
The first thing we can show with the Dirac algebra is that any 4-component ψ that satisfies the Dirac equation (53) also satisfies the Klein-Gordon equation. For the proof simply multiply
0 = (i /∂ + m)(i /∂ − m)ψ = −12{γµ, γν}∂µ∂ν − m2ψ = −( + m2)ψ . (56) Another remarkable feature of the Dirac algebra is that it automatically encodes the Lorentz algebra (23). We can see this by defining
σµν≡ i 2[γ µ, γν] (57) and Sµν ≡ −12σµν = −i 4[γ µ, γν] . (58)
means Sµν are the generator of the Lorentz group and we can write any group element in SO+(3, 1) as
Λ1/2= exp(2iωµνSµν) . (59)
This corresponds to the representation (1/2, 0) ⊕ (0, 1/2). Compare this representation with the one acting on 4-vectors in (15) using (16), i.e.,
Λ = exp(2iωµνJµν) . (60)
The relation with the rotation and boost generators are the same as for Jµν:
Ki = J0i → S0i= − i 4[γ 0, γi] , Ji = −12εijkJjk → −12εijkSjk = i 8εijk[γ j, γk] . (61)
In the Weyl representation,
Ki → S0i= i 2 σi −σi ! , Ji → −12εijkSjk = 12 σi σi ! . (62)
So the Lorentz transformation (59) is represented by
Λ1/2= exp h i~θ · ~J + ~β · ~K i = e i 2(~θ+i~β)·~σ e2i(~θ−i~β)·~σ ! , (63)
which corresponds exactly to the two blocks transforming as (1/2, 0) for ψLand as (0, 1/2) for ψR. It also shows that this representation is reducible and the reducibility is manifest when we use the Weyl representation for γµ.
To show that the Dirac algebra automatically leads to the Lorentz algebra, we can first show that
[γµ, Sαβ] = (Jαβ)µνγν. (64)
See, e.g., Peskin p.42. We can use this relation to write
(1 −2iωαβSαβ)γµ(1 +2iωαβSαβ) = (1 +2iωαβJαβ)µνγν. (65) This is just the infinitesimal version of
Λ−11/2γµΛ1/2= Λµνγν. (66)
This relation allows us to show that the Dirac equation is covariant, i.e., is form invariant when we change frames. See Peskin, p.42.
We end this part by reminding ourselves of the quantum operators representing Lorentz in (6). For the Dirac field, that equations specializes to
U (Λ)−1ψ(x)U (Λ) = Λ1/2ψ(x) . (67)
For the barred field we have
U (Λ)−1ψ(x)U (Λ) = ¯¯ ψ(x)Λ−11/2, (68) because
γ0Λ†1/2γ0 = Λ−11/2. (69)
Therefore, we obtain the nice result that ¯ψψ is a scalar.
[1] Peskin and Schroder. [2] Srednicki
