Note: This was the final paper my friend Hannah Mayerfield and I submitted for PHYS 24500 - Intro to Quantum Field Theory with Professor Jeff Harvey this Spring. Enjoy!

1. Introduction

In the Standard Model of particle physics, particles are either bosons (with integral spin) or fermions (with half-integral spin). The latter represent all the particles making up matter, from our familiar friends electrons and quarks to more exotic, elusive particles like muons and tau neutrinos.

There are three ways to describe these elementary fermions. The simplest is the Weyl representation, which utilizes two-component complex spinors to describe massless spin-\(\frac{1}{2}\) particles with some inherent chirality. This chiral particle has an antiparticle with opposite chirality (e.g. a left-handed Weyl particle would have a corresponding right-handed Weyl antiparticle). The Majorana representation describes a massive particle that encodes both particle and antiparticle states. Finally, the Dirac representation, which generates spinors with four components rather than two and describes all sixteen fermions present in the Standard Model, describes a massive particle and its distinct antiparticle [Ramond 1981].

Dirac fermions are represented using solutions to the Dirac equation:

\[(i\gamma^\mu\partial_\mu-m)\psi=0 \tag{1}\]

where \(\gamma^\mu\) are the four gamma matrices; each a \(4 \times 4\) matrix satisfying the relationships in 3+1D [Pal 2011]:

\[\{\gamma^\mu, \gamma^\nu\} = 2g^{\mu\nu} \tag{2}\] \[(\gamma^{0})^\dagger = \gamma^0, \qquad (\gamma^i)^\dagger=-\gamma^i \tag{3}\]

The solutions \(\psi\) are Dirac spinors: four-component basis elements of a complex vector space belonging to a reducible representation of the Lorentz group [Bonora et al. 2020]. In general, representations of fermions hinge on appropriate choices of gamma matrices to realize the action of the Lorentz group \(SO(3,1)\) (or its double cover, \(\mathrm{SL}(2,\mathbb{C})\)) on spinors.

We choose \(\gamma^\mu\) in the Dirac basis:

\[\gamma^0 = (\sigma^3\otimes\mathbb{I}), \qquad \gamma^i = (i\sigma^2\otimes\sigma^i)\]

for Pauli spin matrices \(\sigma^i\) where \(i\) runs from 1 to 3. When these \(\gamma^\mu\) act on column vectors through the Clifford algebra \(\mathrm{Cl}_{1,3}(\mathbb{R})\), those vectors form a space of Dirac spinors. The Dirac representation can be decomposed into different irreducible representations (irreps) when subject to different conditions: the Weyl representation and the Majorana representation.

These three types of elementary fermion — Dirac, Majorana, and Weyl — have different utilities in our efforts to describe our universe. In the rest of this paper, we will explore the differences between these fermions, their properties, and their use cases.

1.1 Derivation of the Majorana Spinor from Dirac and Weyl Building Blocks

The Dirac representation is reducible to a direct sum of two different 2-dimensional irreps of the Lorentz covering group \(\mathrm{SL}(2,\mathbb{C})\): \((\frac{1}{2}, 0) \oplus (0, \frac{1}{2})\). These are the left-handed and right-handed Weyl spinors respectively [Pal 2011].

To perform this decomposition, we can construct a Lorentz-invariant projection operator — i.e., an operator that commutes with all Lorentz generators \(\Sigma^{\mu\nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu]\). We can construct a fifth gamma matrix \(\gamma^5=i\gamma^0\gamma^1\gamma^2\gamma^3\) that is, by definition, Lorentz-invariant (this can be proven easily using the Clifford algebra outlined in Eq. (2) and the fact that \(\gamma^5\) anticommutes with each \(\gamma^\mu\)). \(\gamma^5\) has eigenvalues of \(\pm1\) — from this, we can find two Lorentz-invariant eigenspaces \(V_L\), \(V_R\) of \(\gamma^5\), both of which commute with the Lorentz generators.

From our \(\gamma^5\), we define projection operators \(P_L\) and \(P_R\) as follows:

\[P_L=\frac{1-\gamma^5}{2}, \qquad P_R=\frac{1+\gamma^5}{2} \tag{4}\]

where \(V_{L,R}=\mathrm{Im}(P_{L,R})\). Since \([\gamma^5, \Sigma^{\mu\nu}]=0\), the projectors satisfy \([P_{L,R}, \Sigma^{\mu\nu}] = 0\). Now, if \(\psi \in V_{R}\) (so that \(P_R \psi = \psi_R=\psi\)), then \(P_R(\Sigma^{\mu\nu}\psi) = \Sigma^{\mu\nu}\psi\). The two 2D irreducible representations of \(\mathrm{SL}(2,\mathbb{C})\) that arise when a Dirac spinor is acted upon by these projection operators are the left-handed Weyl spinor representation \((\frac{1}{2}, 0)\) and the right-handed Weyl spinor representation \((0, \frac{1}{2})\). This is an important result indicating that the most general fermion field can be reduced to a direct sum of two Weyl fields.

Weyl fermions as we treat them today exist primarily as mathematical tools — they were once thought to describe neutrinos, but it has since been shown that neutrinos are likely neither massless nor singly chiral (both noted attributes of the Weyl fermion). However, some collective excitations in certain semimetals have been observed to behave like Weyl fermions [Xu et al. 2015]. Henceforth, we use Weyl fermions as the building blocks for all other fermion fields.

1.1.1 The Majorana Condition

The motivation behind finding the Majorana representation was to ensure that the Dirac equation (Eq. (1)) is restricted such that it only has real solutions. Naively, one would assume that the easiest way to do this would be to make all the gamma matrices imaginary. Thus, we choose \(\tilde{\gamma}^\mu\) as followsFollowing the derivation in Pal [Pal 2011], we can transform from any one set of gamma matrices to any other using a set of unitary matrices. These unitary matrices can also be used to derive the charge conjugation operator.:

\[\tilde{\gamma}^0 = (\sigma^1\otimes\sigma^2), \quad \tilde{\gamma}^1 = ((i\sigma^1)\otimes\mathbb{I}), \quad \tilde{\gamma}^2 = (i\sigma^2\otimes \sigma^2), \quad \tilde{\gamma}^3 = ((i\sigma^3)\otimes \mathbb{I})\]

Additionally, we outline more properties of the gamma matrices:

\[\gamma_0\gamma_\mu\gamma_0 = \gamma_\mu^\dagger \ \ \therefore \ \ \gamma_0^\dagger = \gamma_0, \quad \gamma_i^\dagger = - \gamma_i\]

Following these conditions, the Dirac equation should only have real solutions. Physically, this means that all fermions found as solutions to the Dirac equation in this representation will be their own antiparticle.

Let us introduce a Dirac spinor

\[\psi = \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}, \qquad \psi\in\mathbb{C}^4 \ \text{ and } \ \psi_{L,R} \in\mathbb{C}^2\]

where \(\psi_{L,R}\) are the left- and right-handed Weyl spinors respectively, and a unitary \(4 \times 4\) charge conjugation matrix \(C\) obeying \(\psi=C\bar{\psi}^T\) for \(\bar{\psi}=\psi^{\dagger}\gamma^0\) for any \(\gamma^0\). For the Majorana representation, we will choose \(C=\gamma^0\).

Computing \(\psi=C\bar{\psi}^T\) yields two coupled equations describing the left- and right-handed Weyl spinor components of the Dirac spinor:

\[\psi = \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix} = \begin{pmatrix} \psi_L \\ i \sigma^2 \psi_L^* \end{pmatrix} \tag{5}\]

Here, \(\psi\) is our Majorana fermion. Note that Majorana fermions utilize the full \((\frac{1}{2}, 0) \oplus (0, \frac{1}{2})\) space, while Weyl fermions live in the smaller \((\frac{1}{2}, 0)\) and \((0, \frac{1}{2})\) spaces. This can be verified by trying to use the projection operators on a Majorana fermion — the final answer will not remain an element of the appropriate subspace. Thus, in 3+1D, Majorana and Weyl conditions cannot be met simultaneously. Also note that due to the reality constraint we imposed, the Majorana fermion has half the degrees of freedom that the Dirac fermion has, and the same number of degrees of freedom as the Weyl fermion. This will become important in our later discussion of mass terms.

The same chirality mixing that prevents the existence of Majorana-Weyl conditions has a beautiful consequence: we now have a tool to describe fermions with mass. The mass term in the Dirac Lagrangian (discussed in Section 4) requires two chiralities, so when analyzing a Weyl fermion, the mass term disappears. Now that we have fermions with two chiralities, the mass terms can survive!In the presentation, we will discuss what happens physically to cause mass to come from two massless fermions by reviewing Penrose’s electron “zigzags.”

2. Majorana-Weyl Fermions in \(n+1\)D Minkowski Space

Since we live in a 3+1D universe, we often take this dimensionality for granted in our physics. However, when we generalize Weyl and Majorana conditions to \(n+1\)D, we can understand the structure of these representations far more thoroughly.In the presentation, we will motivate this discussion further by talking briefly about CP/CPT conservation and why it’s of interest.

Recall the Clifford algebra, Eq. (2). For \(d\) dimensions, there are \(d\) gamma matrices. For even \(d=2m\), we can define fermionic creation and annihilation operators as follows:

\[a_k=\frac{1}{2}(\gamma^{2k-1}+i\gamma^{2k}), \qquad a_k^\dagger=\frac{1}{2}(\gamma^{2k-1}-i\gamma^{2k})\]

for \(k = 1, 2, \ldots, m\), satisfying \(\{ a_i, a_j^\dagger \} = \delta_{ij}\), \(\{ a_i, a_j \} = 0\).This is following Problem Set 3, Problem 1b. This means the spinor representation can be treated as a fermionic Fock space with \(m\) possible particles. Each particle can be in one of two states: it can either be occupied (\(a_k^\dagger \lvert 0\rangle\)), or it can be in the vacuum state (\(\lvert 0\rangle\)). This yields a total of \(2^m\) occupation configurations. Taking \(d = 2m\) and rounding to account for odd \(d\), the spinor space acted upon by the gamma matrices has a total dimension \(\dim(S)=2^{\lfloor d/2 \rfloor}\).

Continuing in even dimensions, we can derive a generalized Lorentz-invariant chirality operator \(\gamma_* = c\gamma^0\gamma^1\cdots\gamma^{d-1}\) where \(c\) is some phase ensuring \(\gamma_*^2=1\). Like the chirality operator \(\gamma^5\), \(\gamma_*\) anticommutes with all other gamma matrices and is Lorentz invariant. As shown in the Weyl derivation, this induces a \(\mathbb{Z}_2\) grading on the spinor representation, splitting it into even and odd pieces \(S = S_+ \oplus S_-\), where \(S_\pm=\ker(\gamma_*\mp1)\).

Weyl projection, as shown in Section 1.1, halves the dimension of the space by use of the projection operators in Eq. (4) (which can be generalized by substituting \(\gamma_*\) for \(\gamma^5\)). We can then decompose \(S\) into two chiral subspaces \(S_{L,R}=P_{L,R}S\). Because \(\gamma_*\) has eigenvalues \(\pm 1\), and they occur equally in the irrep, \(\dim(S_+) = \dim(S_-) = \frac{1}{2}\dim(S) = 2^{m-1}\). This projection is impossible in odd dimensions, because by our definition of \(\gamma_*\), it would now commute with the other gamma matrices instead of anticommuting and thus by Schur’s lemma gives no nontrivial grading. In summary: the general condition for the existence of a Weyl fermion is a spinor space with an even number of dimensions (ex. test cases 3+1D, 1+1D).

Now let’s generalize the Majorana condition. Charge conjugation is antilinear, so we can represent it as an antilinear map \(C: S \rightarrow S\). We can generalize the charge conjugation operator such that \(C\gamma^\mu C^{-1}= - (\gamma^{\mu *})^T\).

If charge conjugation preserves our chirality condition imposed by the Weyl projectors, that is, if \(C(S_\pm)\subseteq S_\pm\), then a fermion in that dimension can meet both Weyl and Majorana conditions. We have already shown that in 3+1D, \(C\) changes a particle’s chirality, therefore in 3+1D we cannot have a Majorana-Weyl fermion. But what about in lower dimensions?

Following De Vincenzo [De Vincenzo 2020], in 1+1D, we can define our charge conjugation operator as follows:

\[C = \exp(i\nu)\,\gamma^0 \gamma^1\]

for arbitrary phase \(\nu \in [0,2\pi)\). It is clearly evident that this choice of \(C\) commutes with our Weyl projectors (which now have \(\gamma^1\) in the place of \(\gamma_*\)) and thus, charge conjugation preserves chirality, meaning we can have a fermion which satisfies the Weyl and Majorana conditions simultaneously.

The reason why Majorana-Weyl spinors can exist in some \(n+1\)D spaces and not others lies in the structures inherent to Clifford algebras. The real Clifford algebras repeat with a periodicity of 8, and the complex Clifford algebras repeat with a periodicity of 2 [Henriquez 2016]. The specific nature of spinor representations possible given some Clifford algebra \(\mathrm{Cl}_{p,q}(\mathbb{R})\) depends on the signature of the vector space modulo 8, where if the Clifford algebra has \(d = p+q\) dimensions, it has signature \(s=q-p\) [nLab]. Because of Bott periodicity and the \(\mathbb{Z}_2\) grading of the chirality operator, spinors can only meet both Weyl and Majorana conditions in algebras with \(d = 2 \bmod 8\). For further reading, see [DeJonghe et al. 2012].

3. Majorana-Weyl Fermions in 4D Euclidean Space

Majorana-Weyl fermions cannot exist in 4D Euclidean space, either. An easy explanation would be to use our Bott periodicity argument — \(d\) is still 4, and in any case \(4 \neq 2 \bmod 8\). However, the reason why they cannot coexist in this space is different than that of the 3+1D case: Majorana spinors cannot exist at all in 4D Euclidean space [Wetterich 2011].

Take a Majorana spinor to be the real \(2^{\lfloor d/2 \rfloor}\)-component spinor representation of the generalized \(d\)-dimensional Lorentz group. In Euclidean space, the Lorentz group is \(SO(4)\) and has pseudorealWe call a representation \(\rho\) real or pseudoreal if \(\rho\cong\rho^*\). The distinction between real and pseudoreal representations refers to the product \(\pi^* \pi =\pm 1\) for their intertwiner \(\pi\), i.e. \(\pi\) symmetric vs. antisymmetric [UAlberta]. spinor representations [Wetterich 1983]. In Minkowski spacetime, where the Lorentz group is \(SO(1,3)\)Or \(SO(3,1)\) according to your metric convention. and has complex spinor representations \(\mathrm{SL}(2,\mathbb{C})\), the left- and right-handed Weyl spinors act the same way under conjugation. For Euclidean space, though, the intertwiner \(\pi=i\sigma^2\) between \(SU(2)_L\) and \(SU(2)_R\) is antisymmetric, leading to pseudoreal representations. Since our Majorana condition can be stated in terms of the intertwiner as \(\psi = \pi^{-1}\psi^*\), we see this implies

\[\psi = \pi^{-1}(\pi^{-1}\psi^*)^*=\pi^{-1}(\pi^*)^{-1} \psi = (\pi^* \pi)^{-1} \psi\]

For the pseudoreal case, this returns \(\psi=-\psi\), forcing \(\psi=0\).In the presentation, we will briefly discuss phenomena occurring in four spatial dimensions — namely, the 4D integer quantum Hall effect [Lohse et al. 2018].

4. Fermionic Mass Terms

So far, we have alluded to some questions of mass, but it has remained largely mysterious. Our building blocks, the Weyl fermions, are massless; Majorana and Dirac, both of which are built out of these massless things, somehow have mass. This alone merits a more in-depth exploration of how these mass terms behave.

4.1 Single Majorana Fermion

To explore the mass term of a single Majorana fermion, let us begin with the Dirac Lagrangian [Ramond 1981]

\[\mathcal{L} = \bar{\psi} (i\gamma^\mu \partial_\mu -m)\psi \qquad \bar{\psi} = \psi^{\dagger} \gamma^0\]

where the mass term for a single Dirac fermion field \(\psi\) in the 4-component notation is then \(\mathcal{L}_{D} = m\bar{\psi}\psi\).

What changes appear, if any, when we apply Majorana conditions? Let \(\psi\) be a single Majorana fermion field in 3+1D. Substituting in our Majorana condition we can write:

\[\mathcal{L}_M = m\left(\bar{\psi} C \bar{\psi}^T\right)\]

In the Majorana representation, the charge conjugation matrix \(C\) is simply \(\gamma^0\). So we find:

\[\mathcal{L}_M = m\,\bar{\psi}\,C\bar{\psi}^T = m\,\bar{\psi}\,\gamma^0\bar{\psi}^T = m\,(\psi^\dagger\gamma^0)\,\gamma^0\,(\psi^\dagger\gamma^0)^T = m\,\psi^\dagger\gamma^0\gamma^0\gamma^{0T}\psi^* = m\,\psi^\dagger\gamma^0\psi^* = m\,\bar{\psi}\,\psi^*\]

which, since for Majorana fields \(\psi^* = \psi\), is the same as our original mass term. However, as we’ve learned from the fact that mass terms can exist at all, sometimes unexpected emergent properties arise from applied constraints. Our result thus begs the question: are there other mass terms allowed in the Lagrangian of a singular Majorana field?

A fermion mass term must satisfy three conditions: (i) it is quadratic in the fieldsSince in natural units \([\mathcal{L}] = m^4\), \([\psi]=m^{3/2}\), and therefore \([m\psi^n]=m^{1+3n/2}=m^4 \Rightarrow n=2\).; (ii) it is a Lorentz-invariant scalar; and (iii) it yields a Hermitian Lagrangian, ensuring a real action and unitary time evolution. As explained in Peskin and Schroeder, the Dirac algebra is spanned by 16 independent matrices — bilinears built from these are candidates for the mass terms [Peskin & Schroeder 1995]. By examining how they act under Lorentz transformations, we can sort them into: \(\bar{\psi}\psi\) (scalar), \(\bar{\psi}\gamma^\mu\psi\) (vector), \(\bar{\psi}\sigma^{\mu\nu}\psi\) (antisymmetric tensor), \(\bar{\psi}\gamma^\mu \gamma^5\psi\) (pseudovector), and \(\bar{\psi}\gamma^5\psi\) (pseudoscalar). Property (ii) rules out all non-scalar options, confirming our assumptions about mass terms for Majorana fields — almost. The pseudoscalar is essentially a valid mass term.It is not invariant under parity transformation \(\psi \rightarrow \gamma^0 \psi\) since \(\{ \gamma^0, \gamma^5\} =0\), meaning our Lagrangian would violate parity, which isn’t disallowed, but is notable. We can absorb this pseudoscalar mass term \(m_{\text{pseudo}}\) by a redefinition of the field.See supplement.

Finally, applying the values for \(\psi\) derived in Eq. (5), we get that in the 2-component form, our mass term for a single Majorana fermion expands to

\[m\left(\psi_L^\dagger(i\sigma^2 \psi_L^*) + (i\sigma^2 \psi_L^*)^{\dagger} \psi_L\right) = -m\left(i \psi_L^T\sigma^2 \psi_L -i \psi_L^{\dagger} \sigma^2\psi_L^*\right) = -m \left(i \psi_L^T \sigma^2 \psi_L + \text{h.c.}\right)\]

up to a factor of \(1/2\) in the Majorana mass term. This factor intuitively appears because the Majorana field has half the degrees of freedom as the Dirac field, for reasons outlined in Section 1.1. We leave a full formal exploration to the supplement. After this evaluation, we find \(\mathcal{L}_M = -\frac{1}{2}m \bar{\psi} \psi\) and \(\mathcal{L}_M =-\frac{1}{2}m \left(i \psi_L^T \sigma^2 \psi_L + \text{h.c.}\right)\) — the Majorana mass term only differs from the Dirac mass term by a pseudoscalar artifact and a factor of \(\frac{1}{2}\).

4.2 Pair of Weyl Fermions

We have the same demands for a mass term as we did with the Majorana case. In order to compose a term meeting these conditions, we take the product of two two-component spinors \(\psi_L\beta_L\)For the two-component Weyl spinors \(\psi_L, \psi_R\), it can be shown that \(\sigma^2\psi_L^*\) transforms as a right-handed spinor [Peskin & Schroeder 1995]. Thus, we can express any right-handed spinor in terms of a left-handed one, and can reasonably restrict our work to the realm of the left-handed Weyl spinor. (which is inherently quadratic), and contract the free spinor indices using the spinor metric [Martin 1998] \(\epsilon_{ij}= i\sigma^2\) to enforce Lorentz-invariance, yielding the scalar \(\epsilon_{ij}\psi^i\beta^j\). Note that since the components are Grassmann numbers the spinor metric must be antisymmetric — this fact is the underpinning of fermion exchange statistics.

Returning to matrix notation, two left-handed spinor fields, \(\psi\) and \(\beta\), have the Lorentz scalar mass term \(\psi^T (i\sigma^2) \beta = \sum_{i,j} \epsilon_{ij} \psi^i \beta^j\). Note, due to Grassmann antisymmetry, \(i \psi^T \sigma^2\beta = i\beta^T \sigma^2 \psi\). Thus, for \(\psi, \beta\) left-handed spinor fields our candidate mass terms are: \(\psi^T (i\sigma^2) \beta\), \(\psi^T (i\sigma^2) \psi\), and \(\beta^T (i\sigma^2) \beta\). For right-handed fields, the metric must now have upper indices, flipping the sign. So for same-chirality pairs of Weyl spinors, our possible mass terms are

\[\textbf{Left-handed } (+) \textbf{ / Right-handed } (-) \ \psi, \beta: \qquad \pm i\psi^T\sigma^2 \beta, \quad \pm i\psi^T \sigma^2 \psi, \quad \pm i\beta^T \sigma^2 \beta \tag{6}\]

However, all these mass terms are complex, which would lead to a complex action, which is totally unphysical. To account for this, we simply add the Hermitian conjugates to correct each mass term, leaving the sum a Lorentz-invariant scalar.

As explained in Section 1.1, the mass term requires two chiralities, but our mass candidate term seems to be singly chiral. How can we have a mass term built from a field with a single fixed chirality? Well, we can’t — in taking the Hermitian conjugate, and recalling that \(\sigma^2\psi_L^*\) transforms as a right-handed field, we can see that this mass term actually does couple two chiralities. Thus, it cannot be interpreted as the mass term of a single Weyl field. In fact, we can see this is actually just the left-handed Majorana fermion mass term.The same goes for two right-handed Weyl fermions. The only difference here is the lack of the factor of \(1/2\), which results from the lack of a charge conjugacy constraint.

4.3 Mass Terms Summary

Fermion Type	4-Component	2-Component
Dirac	\(-m\bar{\psi}\psi\)	\(-m(\psi_L^\dagger\psi_R+\psi_R^\dagger\psi_L)\)
Majorana	\(-\frac{1}{2} m\bar{\psi}\psi\)	\(-\frac{1}{2} m\left(i\psi_L^T\sigma^2\psi_L+\text{h.c.}\right)\) or \(-\frac{1}{2} m\left(i\psi_R^T\sigma^2\psi_R+\text{h.c.}\right)\)
Two LH Weyl Fermions	—	\(m\left(i\psi_L^T\sigma^2\beta_L+\text{h.c.}\right)\)
Two RH Weyl Fermions	—	\(-m\left(i\psi_R^T\sigma^2\beta_R+\text{h.c.}\right)\)

To set us up well for our discussion of the fermion number symmetry, we try to unify these various mass terms and connections into one formulation. For a pair of distinct left-handed Weyl spinors \(\psi_1, \psi_2\), we know there are three possible mass terms. Thus, our most general Lagrangian for this pair is just the sum

\[\begin{aligned} \mathcal{L} &= m_{11}\left(i\psi_1^T\sigma^2\psi_1\right) +m_{12}\left(i\psi_1^T\sigma^2\psi_2\right) +m_{21}\left(i\psi_2^T\sigma^2\psi_1\right) +m_{22}\left(i\psi_2^T\sigma^2\psi_2\right) +\text{h.c.}\\[0.4em] &= 2m_{11}\psi_{12}\psi_{11} +2m_{12}\left(\psi_{12}\psi_{21}-\psi_{11}\psi_{22}\right) +2m_{22}\psi_{22}\psi_{21} +\text{h.c.} \end{aligned}\]

We can then write this compactly in matrix form by stacking \(\psi_1, \psi_2\) into a column, and define a general mass matrixIn the presentation, we will use this general mass matrix to discuss the seesaw mechanism in neutrino masses.:

\[\mathcal{M} = \begin{pmatrix} m_{11} & m_{12} \\ m_{12} & m_{22} \end{pmatrix} \qquad \mathcal{L} = -\frac{1}{2} \begin{pmatrix} \psi_1^T & \psi_2^T \end{pmatrix} (i\sigma^2)\mathcal{M} \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} + \text{h.c.}\]

We see that the \(m_{11}\) term is a left-handed Majorana mass term, the \(m_{22}\) is a right-handed Majorana mass term, and finally the \(m_{12}\) is essentially the same as the Dirac termYou might worry about the factor of \(1/2\), but the Dirac mass uses both \(m_{12}\) and \(m_{21}\), which eliminates the \(1/2\). since we mix the two independent Weyl fields. Thus, we denote \(m_{11} = m_L\), \(m_{22} = m_R\), and \(m_{12} = m_D\).

5. Fermion Number Symmetry in Massive Fermion Fields

At the heart of this exploration, and indeed at the heart of all physics, is the question of symmetry. In these fermion fields, what is conserved? When?

Even something as seemingly fixed as the number of particles in a system can be changed. We find that \(U(1)\) fermion number symmetry is only supported by the mass terms of some fermion fields. This symmetry is specifically the invariance of the Lagrangian (and thus the system dynamics) under the \(U(1)\) phase rotation \(\psi \rightarrow e^{i \theta} \psi\).\(\bar{\psi}\) transforms as \(\bar{\psi} \rightarrow e^{-i \theta} \bar{\psi}\). We observe that for our Majorana fermions our reality condition,

\[\psi = \psi^C \implies e^{i\theta} \psi = e^{-i\theta} \psi \implies e^{i\theta} = e^{-i\theta} \implies (e^{i\theta})^2= 1\]

which is a discrete \(\mathbb{Z}_2\) symmetry. This carries through to our Lagrangian mass term and we get \(-\frac{1}{2}m (i \psi_L^T \sigma^2 \psi_L) = e^{2i\theta}\left(-\frac{1}{2}m (i \psi_L^T \sigma^2 \psi_L)\right)\) implying \(e^{2i\theta} =1\), so our Majorana fermions violate the \(U(1)\) symmetry. The same happens for the pair of Weyl fermions, because they have the same form. However, the Dirac mass term transforms as

\[\psi^{\dagger}_L\psi_R + \psi^{\dagger}_R\psi_L \rightarrow e^{-i\theta} e^{i\theta} \psi^{\dagger}_L \psi_R + e^{i\theta} e^{-i\theta} \psi^{\dagger}_R \psi_L = \psi^{\dagger}_L\psi_R + \psi^{\dagger}_R\psi_L\]

and is thus invariant under the \(U(1)\) transformation. Thus, we find that \(U(1)\) symmetry can only be preserved when we require \(m_{11}=m_{22}=0\), leaving us only with the Dirac mass terms. But why does this rotation correspond to fermion number?

Given a Noether current \(j^\mu = -i\frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)} \psi = (\bar{\psi}\gamma^\mu)\psi\),See supplement for details fleshing out this section. we can compute our conserved charge \(Q =\int d^3x \, j^0 = \int d^3x \,(\bar{\psi} \gamma^0\psi) = \int d^3x \,(\psi^{\dagger} \psi)\). Rewriting in terms of creation and annihilation operators [Peskin & Schroeder 1995], we find

\[Q = \sum_s \int \frac{d^3p}{(2\pi)^3} \left[a_s^\dagger(p)a_s(p) -b_s^\dagger(p)b_s(p)\right]\]

We see that \(Q\) is just the difference between the number operators of the particle, \(a\), and antiparticle, \(b\), for a particle of momentum \(p\), spin \(s\). This is why we interpret the conserved charge to be the number of fermions minus the number of antifermions. This also makes it really easy to see why a single Majorana fermion fails to conserve fermion number symmetry, since for the Majorana case the particle is the same as its antiparticle, so the total fermion number must necessarily be 0.

6. Conclusions

In summary, we found that the Dirac representation, which describes all the fermions in the Standard Model and has four degrees of freedom, can be decomposed into two representations that are each home to a chiral Weyl spinor with 2 degrees of freedom. Weyl fermions are massless, but when they are paired, the overall system allows for three possible mass terms (Eq. (6)). Both the Dirac and Weyl fermions preserve \(U(1)\) fermion number symmetry. When we impose a reality condition, the Dirac representation yields a Majorana spinor with 2 d.o.f., which describes a fermion and its antiparticle. The Majorana mass term is \(1/2\) the Dirac mass term, due to the Majorana fermion having half the degrees of freedom. Majorana fermions do not preserve \(U(1)\) number symmetry, as they necessarily contain equal numbers of particles and antiparticles. Majorana and Weyl conditions can only be simultaneously met in dimensions \(d=2 \bmod 8\).

Supplement

S.1 Pseudoscalar

As noted in the main text, there is another valid mass term by our definition called the pseudoscalar mass term. In terms of the gamma matrices, a scalar is simply \(\bar{\psi}\psi\) and a pseudoscalar is of the form \(\bar{\psi} \gamma^5 \psi\). Peskin has sorted these bilinears in accordance with how they transform under the Lorentz group. A pseudoscalar is invariant under Lorentz transformations, which we can confirm with a quick computation following Peskin’s notation:

\[\psi(x)\;\longrightarrow\;\psi'(x') = S(\Lambda)\,\psi(x) \qquad S(\Lambda)=\exp\!\left(-\tfrac{i}{4}\,\omega_{\mu\nu}\,\sigma^{\mu\nu}\right)\]

where the \(\omega\) are antisymmetric parameters for the six Lorentz generators and \(\sigma^{\mu\nu}=\frac{i}{2}[\gamma^\mu,\gamma^\nu]\). Now since we have \(\{\gamma^5,\gamma^\mu\}=0 \ \forall \mu=0,1,2,3 \implies [\gamma^5, \gamma^\mu \gamma^\nu]=0 \implies [\gamma^5,\sigma^{\mu\nu}]=0\). Finally, this implies that \([\gamma^5,S(\Lambda)]=0\), which we compute quickly:

\[\gamma^5\,S(\Lambda) = \gamma^5\,\exp\!\left(-\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}\right) = \gamma^5\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}\right)^{\!n} = \sum_{n=0}^{\infty}\frac{1}{n!}\left(-\tfrac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}\right)^{\!n}\gamma^5 = S(\Lambda)\,\gamma^5\]

Evaluating the total bilinear \(\bar{\psi}\gamma^5\psi \rightarrow \bar{\psi} S^{-1}(\Lambda) \gamma^5S(\Lambda)\psi = \bar{\psi} S^{-1}(\Lambda) S(\Lambda)\gamma^5\psi = \bar{\psi}\gamma^5\psi\). Evidently, the pseudoscalar is exactly a Lorentz scalar. It differs from a regular scalar in its transformation under the parity operator: the regular scalar is parity invariant, where the pseudoscalar is not.

Now that we have a solid understanding of the pseudoscalar term, we show that it can be absorbed by redefining the field \(\psi\). We first note that the kinetic term in our Lagrangian is completely invariant under this transformation:

\[\bar{\psi}\, i\gamma^\mu \partial_\mu \psi \rightarrow \bar{\psi}\, e^{i\theta\gamma^5}\, i\gamma^\mu\, e^{i\theta\gamma^5}\, \partial_\mu \psi= \bar{\psi}\, i\gamma^\mu\, e^{-i\theta\gamma^5}\, e^{i\theta\gamma^5}\, \partial_\mu \psi = \bar{\psi}\, i\gamma^\mu \partial_\mu \psi\]

where we used \(e^{i\theta\gamma^5}\gamma^\mu = \gamma^\mu e^{-i\theta\gamma^5}\).

Our mass term, on the other hand, is not invariant under this transformation for a generic \(\theta\). Instead, it transforms as

\[\begin{aligned} \mathcal{L}_m &= - m\,\bar{\psi}\psi - m_{\mathrm{pseudo}}\,\bar{\psi} i\gamma^5 \psi \\ &= -(m\,\bar{\psi}\psi + m_{\mathrm{pseudo}}\,\bar{\psi} i\gamma^5 \psi)\\ &= -\bar{\psi}\left(m + i m_{\mathrm{pseudo}}\gamma^5\right)\psi \\ &= -\bar{\psi}\, e^{i\theta\gamma^5} \left(m + i m_{\mathrm{pseudo}}\gamma^5\right) e^{i\theta\gamma^5} \psi \\ &= -\bar{\psi}\left(m + i m_{\mathrm{pseudo}}\gamma^5\right) e^{2i\theta\gamma^5} \psi\\ &= -\bar{\psi} \Bigl[ m\cos(2\theta) - m_{\mathrm{pseudo}}\sin(2\theta) + i\gamma^5\bigl(m\sin(2\theta) + m_{\mathrm{pseudo}}\cos(2\theta)\bigr) \Bigr]\psi \end{aligned}\]

To absorb this pseudoscalar, we set \(m\sin(2\theta) + m_{\mathrm{pseudo}}\cos(2\theta)=0 \implies \tan(2\theta) = -m_{\mathrm{pseudo}}/m\). This leaves the Lagrangian mass term as \(\mathcal{L}_m = -\bar{\psi}\left[m \cos(2\theta) - m_{\mathrm{pseudo}}\sin(2\theta)\right]\psi\). Doing some quick trigonometric manipulations we find

\[\begin{aligned} \mathcal{L}_m &= -\bar{\psi}\left[m \cos(2\theta) - m_{\mathrm{pseudo}}\sin(2\theta)\right]\psi\\ &= -\bar{\psi}\left[\frac{m^2}{\sqrt{m^2 + m_{\mathrm{pseudo}}^2}} + \frac{m_{\mathrm{pseudo}}^2}{\sqrt{m^2 + m_{\mathrm{pseudo}}^2}}\right]\psi \\[0.5em] &= -\bar{\psi}\, \frac{m^2 + m_{\mathrm{pseudo}}^2}{\sqrt{m^2 + m_{\mathrm{pseudo}}^2}}\, \psi \\[0.5em] &= -\bar{\psi}\, \sqrt{m^2 + m_{\mathrm{pseudo}}^2}\, \psi \end{aligned}\]

Thus, this is the artifact of the pseudoscalar in the magnitude of our mass as mentioned in the main text. Whenever we write \(\mathcal{L}_m = -m\bar{\psi}\psi\) we’re understanding \(m := \sqrt{m^2 + m_{\mathrm{pseudo}}^2}\).

S.2 Factor of 1/2 in Majorana

Since we’re not focused on the kinetic term, we simply quote it for the Majorana Lagrangian from Pal [Pal 2011]: \(\mathcal{L}_{\text{kin}} = i\psi^{\dagger}\bar{\sigma}\partial_\mu \psi + \text{h.c.}\) where \(\bar{\sigma}=(1,\vec{\sigma})\). Thus, our full Majorana Lagrangian is

\[\mathcal{L}_M = i\psi^{\dagger}\bar{\sigma}\partial_\mu \psi - m i \psi_L^T \sigma^2 \psi_L + \text{h.c.} = i\psi^{\dagger}\bar{\sigma}\partial_\mu \psi - m i \psi_L^T \sigma^2 \psi_L + mi\psi_L^\dagger \sigma^2 \psi_L^*\]

Understanding \(\psi=\psi_L\) from now on, we apply the Euler-Lagrange equations to the mass term with respect to \(\psi^{\dagger}\):

\[\frac{\partial \mathcal{L}}{\partial \psi^{\dagger}} = \partial_\mu\left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi^{\dagger})}\right) \implies i\bar{\sigma}\partial_\mu\psi + \frac{\partial}{\partial \psi^\dagger}\left[im\sigma^2\psi^*\right]=0\]

Due to the fact that fermions are Grassmann fields, this second term is doubled. See that \(\frac{\partial}{\partial \psi^\dagger}[im\sigma^2\psi^*]\) expressed in terms of spinor indices is

\[\frac{\partial}{\partial \psi^\dagger_\mu} \left( i m \,\psi^\dagger_\nu \sigma^2_{\nu \xi} \psi^*_\xi \right) = i m \left( \delta_{\mu \nu} \sigma^2_{\nu \xi} \psi^*_\xi - \psi^\dagger_\nu \sigma^2_{\nu \xi} \delta_{\mu \xi} \right) = i m \left( \sigma^2_{\mu \xi} \psi^*_\xi - \psi^\dagger_\nu \sigma^2_{\nu\mu} \right)\]

where the minus sign in the second term appears because the Grassmann derivative \(\partial/\partial\psi^\dagger_\mu\) must anticommute past \(\psi^\dagger_\nu\) before acting on \(\psi^*_\xi\). But by the antisymmetry \(\sigma^2_{\nu\mu} = -\sigma^2_{\mu\nu}\) we see that this is actually equivalent to

\[i m \left( \sigma^2_{\mu \xi} \psi^*_\xi + \psi^\dagger_\nu \sigma^2_{\mu \nu} \right)\]

But \(\psi^\dagger_\nu = \psi^*_\nu\) are the same Grassmann field, so the second term can be relabeled as \(\sigma^2_{\mu\nu}\psi^*_\nu = (\sigma^2\psi^*)_\mu\). Hence both terms coincide and we obtain

\[i m \left( \sigma^2_{\mu \xi} \psi^*_\xi + \sigma^2_{\mu \nu} \psi^*_\nu \right) = 2im\,(\sigma^2 \psi^*)_\mu\]

And thus we confirm the doubling of this term and see our Euler-Lagrange equation altogether reads

\[i\bar{\sigma}\partial_\mu\psi + 2im\,(\sigma^2 \psi^*)_\mu = 0 \implies i\bar{\sigma}\partial_\mu\psi = -2im\,(\sigma^2 \psi^*)_\mu\]

Finally, to decouple \(\psi\) and \(\psi^*\), and thus yield the Klein-Gordon equation, we have to use the Dirac differential operator \(\sigma^\nu\partial_\nu\). This gives us \(\sigma^\nu\partial_\nu(\bar{\sigma}\partial_\mu\psi) = -2m\,\sigma^\nu\partial_\nu\left((\sigma^2 \psi^*)_\mu\right)\).

We present two quick lemmas which will be helpful in dealing with these \(\sigma\) matrices, from Peskin.

Lemma 1. Define \(\sigma^\mu \equiv (\mathbf{1}, \boldsymbol{\sigma})\) and \(\bar{\sigma}^\mu \equiv (\mathbf{1}, -\boldsymbol{\sigma})\). Then

\[\sigma^\mu \bar{\sigma}^\nu + \sigma^\nu \bar{\sigma}^\mu = 2g^{\mu\nu}\mathbf{1}, \qquad \bar{\sigma}^\mu \sigma^\nu + \bar{\sigma}^\nu \sigma^\mu = 2g^{\mu\nu}\mathbf{1}.\]

In the Weyl representation,

\[\gamma^\mu = \begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma}^\mu & 0 \end{pmatrix}.\]

Computing the off-diagonal blocks of \(\{\gamma^\mu, \gamma^\nu\} = 2g^{\mu\nu}\) [Peskin & Schroeder 1995] yields the result directly.

Lemma 2.

\[(\bar{\sigma}^\mu \partial_\mu \psi)^* = (-2m\sigma^2\psi^*)^* \implies (\bar{\sigma}^\mu)^*\partial_\mu \psi^* = 2m\sigma^2\psi\]

Multiply both sides on the left by \(\sigma^2\):

\[\sigma^2(\bar{\sigma}^\mu)^*\partial_\mu \psi^* = 2m\sigma^2\sigma^2\psi = 2m\psi\]

Using the Pauli matrix identity \(\sigma^2(\bar{\sigma}^\mu)^*\sigma^2 = \sigma^\mu\), we can write \(\sigma^2(\bar{\sigma}^\mu)^* = \sigma^\mu\sigma^2\). Substituting this matrix identity gives:

\[\sigma^\mu\sigma^2\partial_\mu \psi^* = 2m\psi\]

Now back to our EOM. On the LHS (using Lemma 1) we have

\[\begin{aligned} \sigma^\nu\partial_\nu(\bar{\sigma}\partial_\mu\psi) &= \sigma^\nu\partial_\mu(\bar{\sigma}\partial_\nu\psi)\\ &= \tfrac{1}{2}(\sigma^\nu\bar{\sigma}^\mu + \sigma^\mu \bar{\sigma}^\nu)\partial_\nu \partial_\mu \psi\\ &= \tfrac{1}{2}(2g^{\nu\mu}\mathbf{1})\partial_\nu \partial_\mu \psi\\ &= g^{\nu\mu}\mathbf{1}\partial_\nu \partial_\mu \psi \\ &= \square \psi \end{aligned}\]

And on the RHS using Lemma 2 we have

\[\begin{aligned} -2m\,\sigma^\nu\partial_\nu\left((\sigma^2 \psi^*)_\mu\right) &= -2m (\sigma^\nu \sigma^2 \partial_\nu \psi^*)\\ &= -2m (2m \psi)\\ &= -4m^2 \psi \end{aligned}\]

Finally, our EOM now reads as

\[\square \psi = -4m^2\psi \implies \square \psi + 4m^2\psi =0 \implies (\square + 4m^2)\psi=0\]

Since the Klein-Gordon relationship takes the form \((\square +m^2)\psi = 0\), we are led to believe the physical mass \(m_{\text{phys}} =2m\). To yield \(m_{\text{phys}} = m\), we normalize the Majorana Lagrangian by dividing by 2.Recall again that really the magnitude of \(m\) is directly dependent on the mass from the pseudoscalar term.

S.3 Conserved Charge

In the main text, in Section 5, we aim to derive a Noether charge from our Dirac Lagrangian. We’ve already noted the \(U(1)\) transformation, so its variation \(\delta \psi = i\theta\psi\) is clear. We use this to derive its Noether current:

\[\begin{aligned} \delta\mathcal{L} &= \frac{\partial \mathcal{L}}{\partial \psi} \delta \psi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)}\delta(\partial_\mu \psi) = \frac{\partial \mathcal{L}}{\partial \psi} \delta \psi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)}\left(\partial_\mu (\delta\psi)\right)\\ 0 &= \frac{\partial \mathcal{L}}{\partial \psi} \delta \psi - \left[ \partial_\mu\frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)}\right]\delta\psi + \partial_\mu \left[\frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)}\delta\psi\right] = \underbrace{\left[\frac{\partial \mathcal{L}}{\partial \psi} - \partial_\mu\frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)}\right]\delta\psi}_{\text{E.L.} \implies 0} + \partial_\mu \left[\frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)}\delta\psi\right] \\ 0 &= \partial_\mu \left[\frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)}\delta\psi\right] = \partial_\mu \left[i \frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)} \psi \right] = \partial_\mu \left[-i\frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)} \psi \right] \end{aligned}\]

Beginning with the quantized Dirac field as found in Peskin and Schroeder:

\[\psi(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_{\mathbf{p}}}} \sum_{s} \left( a_{\mathbf{p}}^s u^s(p)e^{-ip\cdot x} + b_{\mathbf{p}}^{s\dagger} v^s(p)e^{ip\cdot x} \right)\] \[\bar{\psi}(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_{\mathbf{p}}}} \sum_{s} \left( b_{\mathbf{p}}^s \bar{v}^s(p)e^{-ip\cdot x} + a_{\mathbf{p}}^{s\dagger} \bar{u}^s(p)e^{ip\cdot x} \right)\]

Expanding the Dirac fermion fields \(\psi, \psi^{\dagger}\) in \(p\)-modeWe make a slight change of notation from Peskin. yields:

\[\psi(x) = \sum_s \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_{\mathbf{p}}}} \left[ a_s(\mathbf{p})\, u_s(p)\, e^{-i p \cdot x} + b_s^\dagger(\mathbf{p})\, v_s(p)\, e^{+i p \cdot x} \right]\] \[\psi^{\dagger}(x) = \sum_r \int \frac{d^3q}{(2\pi)^3} \frac{1}{\sqrt{2E_{\mathbf{q}}}} \left[ a_r(\mathbf{q})\, u_r(p)\, e^{-i p \cdot x} + b_r^\dagger(\mathbf{p})\, v_r(p)\, e^{+i p \cdot x} \right]\]

Expanding the product in our \(Q\) integral gives \(\psi^\dagger \psi = \sum_{r,s} \int \frac{d^3p\, d^3q}{(2\pi)^6} \frac{1}{\sqrt{2E_{\mathbf{p}}\, 2E_{\mathbf{q}}}}\, ( c_1 + c_2 + c_3 + c_4 )\), where we have

\[\begin{aligned} c_1 &= a_r^\dagger(\mathbf{q})\, a_s(\mathbf{p})\, u_r^\dagger(q) u_s(p)\, e^{i(q - p)\cdot x} \\ c_2 &= a_r^\dagger(\mathbf{q})\, b_s^\dagger(\mathbf{p})\, u_r^\dagger(q) v_s(p)\, e^{i(q + p)\cdot x} \\ c_3 &= b_r(\mathbf{q})\, a_s(\mathbf{p})\, v_r^\dagger(q) u_s(p)\, e^{-i(q + p)\cdot x} \\ c_4 &= b_r(\mathbf{q})\, b_s^\dagger(\mathbf{p})\, v_r^\dagger(q) v_s(p)\, e^{-i(q - p)\cdot x} \end{aligned}\]

This leads to four integrals of the form \(\int d^3x \int \frac{d^3p \, d^3q}{(2\pi)^6}\frac{1}{\sqrt{4E_{\mathbf{p}}E_{\mathbf{q}}}}(c_i)\). For clarity, we explicitly compute the integral for \(c_1\) and then argue that the same mechanism follows for the other 3 terms.

\[\begin{aligned} \int d^3x \int \frac{d^3p \, d^3q}{(2\pi)^6}\frac{c_1}{\sqrt{4E_{\mathbf{p}}E_{\mathbf{q}}}} &= \int d^3x \int \frac{d^3p \, d^3q}{(2\pi)^6}\frac{1}{\sqrt{4E_{\mathbf{p}}E_{\mathbf{q}}}}\left(a_r^\dagger(\mathbf{q})\, a_s(\mathbf{p})\, u_r^\dagger(q) u_s(p)\, e^{i(q - p)\cdot x}\right)\\ &= \int \frac{d^3p \, d^3q}{(2\pi)^6}\frac{1}{\sqrt{4E_{\mathbf{p}}E_{\mathbf{q}}}}\left(a_r^\dagger(\mathbf{q})\, a_s(\mathbf{p})\, u_r^\dagger(q) u_s(p)\int d^3x \; e^{i(q - p)\cdot x}\right)\\ &= \int \frac{d^3p \, d^3q}{(2\pi)^6}\frac{1}{\sqrt{4E_{\mathbf{p}}E_{\mathbf{q}}}}\left(a_r^\dagger(\mathbf{q})\, a_s(\mathbf{p})\, u_r^\dagger(q) u_s(p)\, (2\pi)^3 \delta^{(3)}(q-p)\right)\\ &= \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\mathbf{p}}}\left(a_r^\dagger(\mathbf{p})\, a_s(\mathbf{p})\, u_r^\dagger(p) u_s(p)\right) \end{aligned}\]

From Peskin,Specifically, equations 3.60, 3.63, and 3.65. our \(u_r(p)\) behave as follows:

\[u_r^\dagger (p) u_s(p) = 2E_{\mathbf{p}}\delta^{rs} \qquad v_r^\dagger (p) v_s(p) = 2E_{\mathbf{p}}\delta^{rs} \qquad u_r^\dagger(p)v_s(-p) = v_r^\dagger(-p)u_s(p) = 0\]

Thus, our integral becomes

\[\begin{aligned} &= \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\mathbf{p}}}\left(a_r^\dagger(\mathbf{p})\, a_s(\mathbf{p})\, u_r^\dagger(p) u_s(p)\right) \\ &= \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\mathbf{p}}}\left(a_r^\dagger(\mathbf{p})\, a_s(\mathbf{p})\, 2E_{\mathbf{p}}\delta^{rs}\right) \\ &= \int \frac{d^3p}{(2\pi)^3}\left(a_r^\dagger(\mathbf{p})\, a_s(\mathbf{p}) \delta^{rs}\right) \end{aligned}\]

The spin sum also collapses, so this evaluates as \(\sum_s \int \frac{d^3p}{(2\pi)^3}\left(a_s^\dagger(\mathbf{p})\, a_s(\mathbf{p})\right)\). The same thing happens for \(c_4\), yielding \(\sum_s \int \frac{d^3p}{(2\pi)^3}\left(b_s(\mathbf{p})\, b_s^\dagger (\mathbf{p})\right)\). There is a slight difference for the \(c_2, c_3\) terms — we have different exponentials and mixed products \(u_r^\dagger v_s\) and \(v_r^\dagger u_s\). Evaluating \(c_2\):

\[\begin{aligned} \int d^3x \int \frac{d^3p \, d^3q}{(2\pi)^6}\frac{c_2}{\sqrt{4E_{\mathbf{p}}E_{\mathbf{q}}}} &= \int d^3x \int \frac{d^3p \, d^3q}{(2\pi)^6}\frac{1}{\sqrt{4E_{\mathbf{p}}E_{\mathbf{q}}}}\left(a_r^\dagger(\mathbf{q})\, b_s^\dagger(\mathbf{p})\, u_r^\dagger(q) v_s(p)\, e^{i(q + p)\cdot x}\right)\\ &= \int \frac{d^3p \, d^3q}{(2\pi)^6}\frac{1}{\sqrt{4E_{\mathbf{p}}E_{\mathbf{q}}}}\left(a_r^\dagger(\mathbf{q})\, b_s^\dagger(\mathbf{p})\, u_r^\dagger(q) v_s(p)\, (2\pi)^3\delta^{(3)}(q+p)\right)\\ &= \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{4E_{\mathbf{p}}E_{\mathbf{-p}}}}\left(a_r^\dagger(\mathbf{-p})\, b_s^\dagger(\mathbf{p})\, u_r^\dagger(-p) v_s(p)\right) \end{aligned}\]

Finally, by the identity \(u_r^\dagger(-p) v_s(p) = 0\) the whole term goes to zero. The same thing happens for \(c_3\). Our final Noether charge becomes:

\[Q = \sum_s \int \frac{d^3p}{(2\pi)^3}\left[a_s^\dagger(\mathbf{p})\, a_s(\mathbf{p}) + b_s(\mathbf{p})\, b_s^\dagger (\mathbf{p})\right]\]

We enforce normal ordering using the anticommutation relations of the creation/annihilation operators \(b^\dagger, b\):

\[\begin{aligned} Q &= \sum_s \int \frac{d^3p}{(2\pi)^3}\left[a_s^\dagger(\mathbf{p})\, a_s(\mathbf{p}) + b_s(\mathbf{p})\, b_s^\dagger (\mathbf{p})\right]\\ &= \sum_s \int \frac{d^3p}{(2\pi)^3}\left[a_s^\dagger(\mathbf{p})\, a_s(\mathbf{p}) - b_s^\dagger(\mathbf{p})\, b_s (\mathbf{p}) + \underbrace{\cancel{(2\pi)^3 \delta^{(3)}(0)}}_{\text{zero-point charge}}\right]\\ Q &= \sum_s \int \frac{d^3p}{(2\pi)^3}\left[a_s^\dagger(\mathbf{p})\, a_s(\mathbf{p}) - b_s^\dagger(\mathbf{p})\, b_s (\mathbf{p})\right] \end{aligned}\]

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