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Home , Casimir element, Lie group, Lorentz group

2 Tutorial 9: The Lorentz

Yehonatan Viernik December 27, 2020

1 Remarks on and Representations 1.1 Casimir elements and the Let us first give loose definitions, and then discuss their implications. A Cartan subalgebra for semi-simple1 Lie algebras is an abelian subalgebra with the nice property that op- erators in the of the Cartan can be diagonalized simultaneously. Often it is claimed to be the largest commuting subalgebra, but this is actually slightly wrong. A correct version of this statement is that the Cartan is the maximal subalgebra that is both commuting and consisting of simultaneously diagonalizable operators in the adjoint. Next, a is something that is constructed from the algebra, but is not actually part of the algebra. Its main property is that it commutes with all the generators of the algebra. The most commonly used one is the quadratic Casimir, which is typically just the sum of squares of the generators of the algebra 2 2 2 2 (e.g. L = Lx + Ly + Lz is a quadratic Casimir of so(3)). We now need to say what we mean by “square” 2 and “commutes”, because the itself doesn’t have these notions. For example, Lx is meaningless in terms of elements of so(3). As a , once a representation is specified, it of course gains meaning, because matrices are endowed with . But the Lie algebra only has the Lie bracket and linear combinations to work with. So instead, the Casimir is living inside a “larger” algebra called the universal enveloping algebra, where the Lie bracket is realized as an explicit . This is just a bunch of technicalities, with the sole intention to slightly de-mystify these concepts, and you are not expected to go and study . You only need to know that it is well defined within something very similar to the Lie algebra, and is actually even slightly less abstract (Lie bracket is explicitly the commutator), and its definition can be naturally extended to any actual representation of the algebra, e.g. when taking L2 as a 3 × 3 matrix. Now the implications. There is a famous lemma called Schur’s lemma, which says the following. Suppose you have an irrep R, where for every element g in your algebra, R(g) is some operator (matrix) acting in your V . Then any other operator T : V → V that commutes with all R(g) must be a scalar operator. That is, if ∀g ∈ g :[T,R(g)] = 0, then T = λ · In×n. The implication is that if we have a Casimir C, then since it commutes with everything in the algebra, for every irrep it will act as a scalar operator. Therefore, we can label each irrep by the value of the Casimir operator. We already know this for : L2 |l, mi = l(l + 1) |l, mi. We now just gave this idea a formal basis. As for the Cartan, since operators in the Cartan can be diagonalized simultaneously, they give us more labels for states within an irrep. For angular momentum, this is the m quantum number. For particles, it will be the conserved Noether charges: every particle is an irrep of the of our theory. Conserved charges commute with the Hamiltonian and can be diagonalized, thus labeling the different states of our particles (irreps). This applies both for internal as well as for . In the PS, you will use these concepts to construct all irreps of su(2).

1.2 Relationship between a and a Lie algebra Consider the following puzzle. You have three objects: A, B, C which act as generators for a real Lie algebra, satisfying

[A, B] = iC , [C,A] = iB , [B,C] = iA . (1.1)

What is the name for this Lie algebra? so(3)? su(2)? Something else?

1For more general Lie algebras, the definition is slightly more broad: instead of commuting, i.e. Lie bracket is zero [a, b] = 0, the Cartan is only nilpotent, which basically means that any chain of brackets will terminate after n steps for some fixed n: [a, [b, [...[c, [d, e]]]]] = 0.

1 Answer: There’s no concrete answer, because it’s a silly question to ask in the first place. All of them are isomorphic, and such labels are only there as mental “bookkeeping” devices to remind us from which group we started. But the Lie algebra is a well defined structure irrespective of any Lie group it may be related to.

Claim 1: A Lie group may contain more information than is available by exponentiating its corresponding algebra.

For example, O(N) and SO(N) have the same Lie algebra, but O(N) also has reflections, which are discrete transformations and cannot be obtained from repeated infinitesimal transformations. SO(N), on the other hand, is entirely characterized by the algebra. It is then tempting to claim that exponentiating so(3) uniquely con- structs SO(N). But then, what do we make of our puzzle? The truth is, specifying an algebra doesn’t uniquely identify a specific Lie group, even when we restrict ourselves to groups that are fully connected to the iden- tity like SO(3). To get an identification of a Lie algebra with a specific Lie group, we need a stronger restriction.

Claim 2: Every Lie algebra is identified with exactly one simply-connected Lie group.

This is a very important statement, which we’re not going to prove, but we will get back to its implications in a bit.

1.3 SO(3) vs SU (2) We would like to ask the following question: which group is more “fundamentally spherical”? Is there a sensible notion through which we might prefer one over the other?

Claim 3: SU(2) is equivalent, as a geometric object2, to a 3-sphere S3.

 a −b Proof: SU(2) can be defined as the set of 2×2 complex matrices of the form with |a|2 +|b|2 = 1. b∗ a∗ 2 2 2 2 Denoting a = a1 + ia2 and b = b1 + ib2, the last condition is written as a1 + a2 + b1 + b2 = 1, which is just the unit 3-sphere.

Claim 4: SO(3) ' SU(2)/Z2.

In words: SU(2) is a double of SO(3). In more words: there is a 2-to-1 surjective from SU(2) onto SO(3). In even more words: there is a mapping between the two groups, where every element in SO(3) is identified with two elements in SU(2), and this mapping preserves the group structure. The geometric implication of this statement, is that if SU(2) can be identified with the 3-sphere, then SO(3) is also identified with S3, but with every pair of antipodal points identified as the same point as well! In a sense, SO(3) is only “half a sphere”. Together, Claims 2-4 are telling us that SU(2) is a more complete, and perhaps more fundamental, descrip- tion of spherical symmetries: it is geometrically a sphere, it is uniquely identified by the Lie algebra, and it covers the group SO(3). This teaches us a very important lesson: We may think we have an understanding of the symmetry of a system, but when we attempt to systematically construct its blocks (the irreps of the algebra), it may lead us to a more fundamental symmetry that can cover the original one, and thereby reveal important physical objects that we’d otherwise miss! Therefore, when looking at the Lie algebra, we are encouraged to consider what the has to say, that is – the fundamental, simply-connected Lie group the algebra is “really” coming from. Let’s put this in the context of a concrete physical problem you are familiar with from undergrad.

1.4 Spherical symmetry in QM In non-relativistic QM, we sometimes want to solve systems with spherical symmetries, such as the . Studying symmetry groups proves to be a useful tool to systematically construct the eigenfunctions of such potentials. When solving Hψ = Eψ, if H has rotational symmetry, then the space of eigenfunctions VE will be under the action of SO(3). For each eigenenergy En, the space Vn can form various representations of SO(3), i.e. be constructed by objects that transform under SO(3) in different, well defined ways. As always with irreps: these are building blocks, in this case the building blocks for all wavefunctions that are solutions to the original Schrodinger equation. If we ignore the radial modes, these are the . For every value of the angular momentum eigenvalue ` = 0, 1, ..., n−1, we have an 2`+1-dimensional

2Recall Lie groups are smooth . Thus, by equivalent we mean diffeomorphic.

2 space spanned by polynomials of degree `. For example, for ` = 1, we have V1 = span(x, y, z), and for ` = 2 we 2 2 2 2 have V2 = span(xy, xz, yz, x − y , x − z ) with 2 × 2 + 1 = 5. However, we know this isn’t the whole story, since we also have fermions! Half- objects do not transform “natrually” under the rotation group. In particular, they don’t return to themselves under a 2π rotation. Indeed, formally fermions don’t form “ordinary” representations of SO(3), but instead they form something called projective representations, which is a slightly more general concept, which we will not define explicitly or worry about too much. We will, instead, see how it comes about in practice. Going back to the main machinery of constructing irreps, which is studying Lie algebras, we recall the following:

SO(3) → so(3) ' su(2). (1.2)

Thus, by using our knowledge of irreps of su(2), we get information on SO(3). Since SU(2) is a double cover of SO(3), such irreps should give what appears to be “redundant” objects. And indeed, in the previous TA you constructed irreps of SU(2) and found scalars and vectors, which appear also in the irreps of SO(3), but you also found spin 1/2 objects. Therefore, fermions are exactly these “superfluous” objects we got by looking at the “wrong” group. Was it a bad move on our part? Clearly not. A similar thing is going to happen when we talk about the next.

2 The Lorentz Group

Until now, we only encountered relatively simple groups, all were more-or-less spheres. The related to Lorentz transformations – the Lorentz group – is slightly more complicated, and it will take a bit more work to study its representations. Recall that the Lorentz group is defined as the set of all transformations Λ that preserve the Minkowski metric,

2 µ µ ν 0 2 X i 2 ds = x xµ = x ηµν x = (x ) − (x ) . (2.1) i

0µ µ ν That is, for x = Λ ν x , we have

0µ 0ν µ ρ ν σ x ηµν x = Λ ρx ηµν Λ σx , (2.2) and we require

µ ν ! Λ ρ ηµν Λ σ = ηρσ. (2.3) As an abstract group, we denote it as O(1, 3), in a similar spirit to O(4), labeling the group that preserves the Euclidean sphere3 (x1)2 + ... + (x4)2.

2.1 Properties of the Lorentz group 2.1.1 O(1, 3) is not connected In class, we found two general constraints arising from (2.3):

detΛT ηΛ =! det η ⇒ (det Λ)2 =! 1 ⇒ det Λ = ±1 , (2.4) and s µ ν ! 0 2 X i 2 ! 0 X i 2 Λ 0 ηµν Λ 0 = η00 ⇒ Λ 0 − Λ 0 = 1 ⇒ Λ 0 = ± 1 + Λ 0 . (2.5) i i

0 This splits the Lorentz group into 4 parts, where only the piece with positive and Λ 0 > 1 is con- nected to the identity. This part is called the restricted Lorentz group, or sometimes the proper orthochronous 0 Lorentz group, meaning orientation preserving (det = 1) and time-direction preserving (Λ 0 > 1), respectively. + ↑ The restricted Lorentz group is denoted by SO (1, 3) or L+, and is a normal of the full Lorentz group: SO+(1, 3) /O(1, 3). The other 3 pieces can be obtained from SO+(1, 3) by acting on it with ∀i : xi → −xi and/or time-reversal x0 → −x0.

3Do not confuse this with the identification of SU(2) with S3! There, the identification was with the underlying structure, which is captured by the group’s parameters a, b ∈ C with their determinant constraint. Here, it is the vector space on which the group acts that looks like S3. The manifold structure of O(4) is similarly determined by its parameters and their constraints, making it 6-dimensional. In fact, SO(4) ' (SU(2) × SU(2)) /Z2, i.e. SU(2) × SU(2) is a double cover of SO(4).

3 Implication: classifying the irreps of the algebra only gives information on SO+(1, 3). Parity and time reversal must be studied separately.

Luckily, these are very simple discrete symmetries (two copies of Z2), that aren’t going to complicate things too much.

2.1.2 SO+(1, 3) is not simply-connected This can be seen from the between SO+(1, 3) and SO(4) (preserving similar metrics) and by extending a similar geometric argument we had for SO(3). This immediately tells us that when we look at irreps of the algebra, we’re going to study irreps for the covering group instead, as mentioned earlier.

2.1.3 The Lorentz group is not compact Unlike SU(2),SO(17) etc, the Lorentz group is non-compact. Consider a boost in the x-direction:

cosh β sinh β  sinh β cosh β    . (2.6)  1  1

The parameter β can take any real value. Geometrically, a boost will take us along a hyperbole that asymptotes to the , but we can never actually get there (see diagram in Gilad’s notes). You can think about it in a similar way to the open line segment (a, b), where you can get as close as you like to the endpoints, but never touch them. The open segment is non compact, and so is the space strictly inside a light cone. The implication here comes from the following statement:

Simple, non-compact Lie groups do not have non-trivial, finite-dimensional, unitary representations.

We will therefore consider two options: 1. Finite dimensional, non-unitary representations. 2. Infinite dimensional, unitary representations.

When do we care about unitarity? The answer is Hilbert spaces. These are complex vector spaces, so states need to transform under unitary transformations in to preserve probabilities4:

φ0†φ0 = φ†U †Uφ = φ†φ (2.7)

However, Hilbert spaces are quite often infinite dimensional anyway, in particular in field theories (second in QM 1 language), so we don’t mind our representations to be infinite dimensional. In contrast, working with fields as we are currently doing, there’s no reason to require unitarity, but finite-dimensionality is quite convenient to construct objects that are easy to manipulate. Since we are currently interested in fields, we will consider the finite dimensional representations in what follows, and just note that for the Poincar´egroup, we look at the infinite-dimensional ones.

3 Representations of the Restricted Lorentz Group

To study SO+(1, 3), we use a similar trick to (1.2). After “modding out” P and T (parity and time reversal), we are left with spatial rotations Ji and boosts Ki as generators for the restricted Lorentz group. In class, you found the commutation relations

[Ji,Jj] = iεijkJk , [Ki,Kj] = −iεijkJk , [Ji,Kj] = iεijkKk. (3.1)

± 1 You then used them to construct two objects Ai = 2 (Ji ± iKi) that satisfy

± ± ± ± ∓ [Ai ,Aj ] = iεijkAk , [Ai ,Aj ] = 0. (3.2) Since so(1, 3) is a real Lie algebra (coming from a real Lie group), every element in the algebra should be a real ± linear combination of the elements in the algebra. Since Ai are complex linear combinations of the generators, 4see Wigner’s theorem if interested in further details.

4 they are not part of the algebra! Instead, they are part of the complexified algebra, which just means the same algebra, but as a vector space over C instead of R. Then, looking at the commutation relations, we find that

so(1, 3)C ' su(2)C ⊕ su(2)C . (3.3) Once again, we can use our knowledge of su(2) to talk about the Lorentz group. This time, the underlying covering group of SO+(1, 3) is SL(2, C), the group of 2 × 2 complex matrices with unit determinant. Since we are dealing with two copies of su(2), our representations are naturally labeled by two numbers (j1, j2) with 2ji ∈ N, and the reps are of dimension (2j1 + 1)(2j2 + 1). This leads to the following shortlist of common representations you saw in class: • (0, 0): scalar (trivial representation)

1  • 2 , 0 : left Weyl 1  • 0, 2 : right Weyl spinor 1  1  • 2 , 0 ⊕ 0, 2 : 1 1  • 2 , 2 : vector (fundamental representation) • (1, 0) ⊕ (0, 1): Antisymmetric 2-tensor

We note that the representations T(j1,j2) of the algebra are in general not irreducible. Even without careful mathematical reasoning, this is to be intuitively expected, because we have two copies of algebras. Because this algebra is semisimple, we have the nice feature that it is completely reducible, i.e. can be fully decomposed to irreps (can be block diagonalized). In general, this means the representation can be written as a direct sum: M T(n,m) = T(p) , p = |n − m|, ... , n + m (3.4) p

This decomposition is exactly the Clebsch-Gordan you may have come across in undergrad QM. If not, that’s okay, we’re not going to go deeper than that, and we will only look at stuff we have in the above shortlist.

Scalar

iT The representation for the algebra is given from Eq. (3.4) T(0,0) = T(0) = 0, hence e = 1, and so scalars transform trivially under the action of the Lorentz group φ → 1φ.

Weyl Spinor 1  Consider first the left spinor 2 , 0 . Looking again at (3.4) and recalling the irreps of su(2) we obtained in TA 8, we find

i i 1 i T( 1 ,0) = T( 1 ) ⊕ T(0) = σ ⊕ 0 . (3.5) 2 2 2

i ± Recall that T(1/2) and T(0) are representations for the generators Ai , because these are the generators for the ± two copies of su(2) we constructed. Since Ai are related to the generators Ji,Ki in so(1, 3)C, we get that ± a representation for Ai provides us with a representation for Ji,Ki! In this way, we find how rotations and boosts look like in a given representation. There is arbitrarity in choosing which A± is related to the left index and which to the right index. But once we make this choice, we need to stick to it. For example, once we make choice for T 1 , then T 1 is determined. Let’s see this in action. We choose the identification ( 2 ,0) (0, 2 ) + − − T(0) = R(A ) and T 1 = R(A ), where R(·) means ”the representation of”. That is, we take A to live in the ( 2 ) trivial representation of su(2), and A+ to live in its spin 1/2 representation. We then obtain (suppressing the R, but still implying the representation and not the abstract object):

− 1 0 = Ai = (Ji − iKi) ⇒ Ji = iKi 2 (3.6) σ 1 i = A+ = (J + iK ). 2 i 2 i i Solving for the (representation of the) generators, we find i 1 K = − σ ,J = σ , (3.7) i 2 i i 2 i

5 from which we can determine how rotations and boosts act in this representation:

iθ¯·J¯ i θ¯ ·σ¯ R(θ) = e = e 2 , ¯ (3.8) iφ¯·K¯ φ ·σ¯ B(φ) = e = e 2 . Finally, a generic acting on a left Weyl spinor is given by

 1 1 ¯ ¯ 1 ¯ ¯ χ 0 2 (iθ+φ)·σ¯ 2 (iθ+φ)·σ¯ L χL = Λ( 1 ,0)χL ≡ ΛL χL = e χL ≡ e 2 . (3.9) 2 χL In the last step, we introduced the conventional notation for Weyl as a two-component vector. 1  For the 0, 2 representation, we have

i i 1 i T(0, 1 ) = T(0) ⊕ T( 1 ) = 0 ⊕ σ . (3.10) 2 2 2 Now, recall we must make the inequivalent identification, where A+ is now the trivial one and A− is in the spin 1/2 representation. Repeating the same procedure as before, we have:

+ 1 0 = Ai = (Ji + iKi) ⇒ Ji = −iKi 2 (3.11) σ 1 i = A− = (J − iK ) 2 i 2 i i

¯ i 1 i θ¯ ·σ¯ − φ ·σ¯ ⇒ K = σ ,J = σ ⇒ R(θ) = e 2 ,B(φ) = e 2 . (3.12) i 2 i i 2 i We see that rotations are the same, but boosts operate in the opposite direction. For a right handed spinor under Lorentz, we thus find:

 1 1 ¯ ¯ 1 ¯ ¯ χ 0 2 (iθ−φ)·σ¯ 2 (iθ−φ)·σ¯ R χR = Λ(0, 1 )χR ≡ ΛR χR = e χR ≡ e 2 . (3.13) 2 χR Important: The way spinors transform under boosts and rotations is the defining property of their hand- edness!

This is of course true in general: to tell which representation an object is associated with, you need to see how it transforms under the symmetry action of the group. We note in passing that left and right spinors are currently just a name, but later we will show that they are in fact related by parity P in a way that justifies these names. They are also related by something called conjugation, which we define later on and denote by C, and we will discuss the difference between C and P as well.

Dirac fermion If we take two Weyl spinors and stick them together in a 4-component vector, we get what is called a Dirac 1  1  fermion. This corresponds to taking a direct sum of these two representations: 2 , 0 ⊕ 0, 2 (recall the discussion in TA 7 on combining φi with Ai into a 5-component vector). The decomposition is now:

T 1 ⊕ T 1 = T 1 ⊕ T 1 . (3.14) ( 2 ,0) (0, 2 ) ( 2 ) ( 2 ) In words, a Dirac spinor will have a component that transforms like a left handed spinor, and another that transforms like a right handed spinor. Explicitly, this looks like   χL ψD = (3.15) ξR where we use ξR and not χR to emphasize that these are two independent spinors, i.e. ξR is not related to χL (we will show how left and right spinors are related in the following section). Under a Lorentz transformation, we have     0 ΛL χL ψD = Λ( 1 , 1 )ψD = . (3.16) 2 2 ΛR ξR Dirac fermions transform under Lorentz in a block diagonal form, emphasizing it being a reducible representation. This is telling us that even though this is a 4-dimensional representation, it is not the fundamental representation, which we will mention next. Nevertheless, when parity is taken into consideration, the Dirac fermion becomes a useful concept, as discussed later on.

6 Vector representation 1 1 Next up is ( 2 , 2 ). Looking at (3.4), we get the decomposition

T 1 1 = T(1) ⊕ T(0) (3.17) ( 2 , 2 ) In the PS, you will show that this in fact transforms like a 4-vector, i.e. it is equivalent to the fundamental representation of the Lorentz group, so we are leaving the details for later.

4 Spinor Indices and Charge Conjugation

Spinor indices is a notation that is useful for distinguishing between left and right spinors. We define left(right) spinors as having a lower(upper), undotted(dotted) index:

a˙ χL = χa , χR = χ . (4.1)

The spinor metric is defined as

 0 1 εab = , (4.2) −1 0 and the charge conjugation operator is defined as

c ∗ χL ≡ χL, (4.3) where ∗ is the usual complex conjugation. Similarly to η, ε will be used to raise and lower spinor indices, and together with C, we will be able to transform between dotted and undotted indices, i.e. between left and right . c Let’s perform a boost on χL:

∗ 0 ∗  φ¯  φ¯ ∗ † φ¯ φ¯ c 0 2 ·σ¯ 2 ·σ¯ ∗ − 2 ·σ¯ ∗ − 2 ·σ¯ c (χL) = εχL = ε e χL = εe χL = e εχL = e χL. (4.4)

∗ † - Here we used εσi = −σiε, which can be easily verified explicitly, to move ε to the right. We found that a charge-conjugated left spinor is boosted like a right handed spinor! Under rotation, the extra i in the exponent gives an extra minus sign under complex conjugation, such that the result is unchanged:

0 θ¯ c i 2 ·σ¯ c (χL) = e χL. (4.5)

Since left and right chiralities are defined by their transformation properties under Lorentz, we conclude:

c ∗ χL = εχL = χR! (4.6)

c ∗ ! The same happens if we define the action of C on a right spinor as χR ≡ −εχR = χL. In a similar spirit to raising and lowering indices using η, we define

0 −1 εχ = εabχ ≡ χa, ε χb ≡ χ , ε ≡ , (4.7) L b ab a ab 1 0 which implies that in this notation, complex conjugation amounts to dotting and undotting indices:

∗ a∗ a˙ χa = χa˙ , χ = χ . (4.8)

To finish constructing this notational framework, we still need to analyze the transformation properties for the a remaining combinations χa˙ , χ .

 b ∗ b˙ ∗ h 1 ¯ ¯ i h 1 ¯ ¯ ∗ i 0 0 2 (iθ+φ)·σ¯ ∗ 2 (−iθ+φ)·σ¯ χa˙ = χL = e χb = e χb˙ . (4.9) a a˙

Similarly,

b a0 a˙ ∗ h 1 (−iθ¯−φ¯)·σ¯∗ i b χ = χ = e 2 χ . (4.10) a

7 ∗ T † Finally, using (4.9), (4.10) and that (σi ) = σi = σi, we can see that

a T a˙ T (χ ) χa and χ χa˙ (4.11) are both invariant to Lorentz transformations! In contrast,

a˙ T a T χ χa and (χ ) χa˙ (4.12) are not (the reader can verify both statements explicitly). We thus conclude that, similarly to the case of vector µ indices x yµ, Lorentz invariance is achieved by contracting (un)dotted spinor indices with (un)dotted spinor indices. To wrap things neatly, we write Lorentz transformations in spinor notation as

0 b 0 a˙ b˙ χL = (ΛL)a χb , χR = (ΛR) b˙ χ (4.13)

Finally, going back to Dirac fermions, we said that ξR is independent of χL. Now that we defined C, we can c also consider the special case where ξR ≡ χR = χL, such that   χL ψM = . (4.14) χR

What we got here is called a Majorana fermion.

5 Parity

As an explicit matrix acting in , parity is given by

1   −1  P =   , (5.1)  −1  −1 i.e. it flips all spatial indices (hence the name)5. One can show by explicit calculation that

T T PJiP = Ji ,PKiP = −Ki, (5.2) and therefore 1 1 PA±P T = P (J ±) P T = P (J ∓) P T = A∓. (5.3) i 2 i 2 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) (5.4) ( 2 ,0) (0, 2 ) and vice versa.

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 . (5.5) 2 2 2 2

5Really what you want is to flip an odd number of indices, so that the determinant will change sign. In 3 spatial , 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.

8 We now inspect the transformation properties of a Lorentz transformation (boost and/or rotation) on the parity transformed object:

0 Λ 1  P P P P (0, 2 ) P ψD = Λ( 1 ,0)⊕(0, 1 ) ψD = Λ(0, 1 )⊕( 1 ,0) ψD = ψD (5.6) 2 2 2 2 Λ 1 ( 2 ,0) We found that a Dirac fermion under parity transforms like an object with the first spinor being of right and the second of left chirality. We conclude:     χL P P ξR ψD = −→ ψD = . (5.7) ξR χL

Note that parity does not apply the transformation χL → χR or anything like that!

9