Biquaternionic Dirac Equation Predicts Zero Mass for Majorana Fermions

Abstract

A biquaternionic version of the Dirac Equation is introduced, with a procedure for converting four-component spinors to elements of the Pauli algebra. In this version, mass appears as a coefficient between the 4-gradient of a spinor and its image under an outer automorphism of the Pauli algebra. The charge conjugation operator takes a particulary simple form in this formulation and switches the sign of the mass coefficient, so that for a solution invariant under charge conjugation the mass has to equal zero. The multiple of the charge conjugation operator by the imaginary unit turns out to be a complex Lorentz transformation. It commutes with the outer automorphism, while the charge conjugation operator itself anticommutes with it, providing a second more algebraic proof of the main theorem. Considering the Majorana equation, it is shown that non-zero mass of its solution is imaginary.

1. Introduction

It is well known that Majorana fermions can exist as composite particles [1], but the question whether they can exist as single elementary particles remains open since the work of Majorana [2,3].

There is ongoing work attempting to detect Majorana neutrinos using the neutrinoless double beta decay [4,5].

The aim of this article is to show that if a solution of the Dirac Equation coincides with its own image under the charge conjugation operator, then its mass must necessarily be zero.

The very first step is to complete a four-component spinor with a second column, so that the four new scalar equations thus obtained will be exactly the four scalar equations of the conjugate Dirac Equation. Then the Dirac Equation would contain exactly the same eight scalar equations as its conjugate equation. This symmetry of the Dirac Equation exists only if mass is a real scalar.

In the four rows by two columns spinor the bottom square becomes a quaternion and the top square becomes a product of a quaternion with an imaginary unit (3). The sum of these two squares thus becomes a biquaternion, and the difference its biquaternion conjugate.

A biquaternionic form of the Dirac Equation is introduced, and shown to be equivalent to both the standard equation and its Hermitian conjugate (which both consist of the same eight scalar equations when completed with a second column).

The biquaternionic form of the Dirac Equation (16) involves the so-called bar-star outer automorphism of the Pauli algebra (1).

There is a simple expression for the conserved probability current in this formulation (29).

A transformation of two-column spinors (10) results in a symmetry of the Dirac Equation that reverses the signs of mass, leaving everything else unchanged. This symmetry takes a particularly simple form in the biquaternionic formulation (31), where it reverses the sign of mass. From there it is an easy conclusion that Majorana solutions of DE that are transformed by this charge conjugation symmetry to themselves must have zero mass.

When multiplied by the imaginary unit charge conjugation becomes a complex Lorentz transformation (2). This Lorentz transformation commutes with the bar-star automorphism while the charge conjugation operator itself anticommutes with the bar-star automorphism (12). This anticommutation is used in the second, more algebraic proof that the mass of a Majorana solution has to equal zero (13).

For the Majorana equation it is shown that if mass of its solution is non-zero then it is imaginary.

The main assumption used in this article is the validity of using the multiplicative structure of the biquaternionic spinor space; see also [6], which uses the multiplicative properties of the space-time algebra.

2. Notation and Terminology

The letter $\mathbb{P}$ is used for the Pauli algebra ${\mathit{Cl}}_{3,0}\left(\mathbb{R}\right)\simeq {\mathit{Cl}}_{1,2}\left(\mathbb{R}\right)\simeq M_2\left(\mathbb{C}\right)\simeq \mathbb{P}$. Since it is isomorphic to the algebra of biquaternions, these terms are used interchangeably. The upper case Greek letter $\mathrm{\Psi }$ is used for four-component Dirac spinors, the lower case $\psi$ for Pauli algebra, or equivalently, biquaternionic spinors.

The standard Dirac–Pauli representation is used throughout. The asterisk is used both for Hermitian conjugate and for complex conjugation. The space inversion is denoted by an overbar:

$$z=z_0+\underline{z}\overline{z}=z_0-\underline{z}\underline{z}=z_k{\sigma }^k$$

here $z\in \mathbb{P}$ is an arbitrary element of the Pauli algebra split into scalar and vector components.

Note. 

The use of overbar here is different from the Dirac adjoint!

In matrix form an element $z\in \mathbb{P}$ and its parity conjugate $\overline{z}$ look like:

$$z=\left[\begin{array}{cc}{z}_0+z_3& -i{z}_2+z_1\\ i{z}_2+z_1& z_0-z_3\end{array}\right]\overline{z}=\left[\begin{array}{cc}{z}_0-z_3& i{z}_2-z_1\\ -i{z}_2-z_1& z_0+z_3\end{array}\right]$$

The product of an element with its parity conjugate is a scalar: $z\overline{z}=\overline{z}z=det\left(z\right)I_2$.

This is formal and applies also to the product of the four-gradient with its parity conjugate:

$$\overline{\partial }\partial =\left({\partial }_0-\nabla \right)\left({\partial }_0+\nabla \right)=\left[\begin{array}{cc}{\partial }_0^2-{\nabla }^2& 0\\ 0& {\partial }_0^2-{\nabla }^2\end{array}\right]=\square I_2$$

where □ is the D’Alembertian, $\partial ={\partial }_{\mu },\overline{\partial }={\partial }^{\mu }$.

The algebra of quaternions is a subalgebra of Pauli algebra characterised by the condition:

$$\mathbb{H}=\{h\in \mathbb{P}|h^{*}=\overline{h}\}$$

Another way to characterise the subalgebra of quaternions is as those elements of Pauli algebra whose scalar component is real and whose vector components are purely imaginary:

$$h\in \mathbb{H}\iff h=h_0+i\underline{h}{h}_{\mu }\in \mathbb{R}$$

Still another way to characterise quaternions is that they are represented by complex matrices of the form:

$$u,v\in \mathbb{H}u=\left[\begin{array}{cc}c& -d^{*}\\ d& c^{*}\end{array}\right]iv=\left[\begin{array}{cc}a& b^{*}\\ b& -a^{*}\end{array}\right]$$

(1)

For any quaternion its product with its parity conjugate is the square of its norm:

$$q\in \mathbb{H}q\overline{q}={\left|\right|q\left|\right|}^2{I}_2{q}^{-1}=\frac{\overline{q}}{q\overline{q}}$$

The division is unambiguous because $q\overline{q}$ is a scalar.

Charge conjugation is denoted either as ${\psi }_c$ or as $\widehat{\mathbf{C}}\psi$ when using operator notation, (11) and (31).

2.1. The “Bar-Star" Outer Automorphism of the Pauli Algebra

Since Hermitian conjugation $z\to z^{*}$ and parity (space inversion) $z=z_0+\underline{z}\to \overline{z}=z_0-\underline{z}$ are both anti-automorphisms, their composition is an automorphism. It is straightforward to check that the composition of Hermitian conjugation and of parity does not depend on their order. This can be summed up as:

$$\overline{\left(xy\right)}{}^{*}=\overline{x}{}^{*}\overline{y}{}^{*}$$

Proposition 1.

“Bar-star" is an outer automorphism.

Proof. 

The proof is by checking the determinants:

$$\begin{array}{c}\hfill detz=\left|\begin{array}{cc}{z}_0+z_3& -i{z}_2+z_1\\ i{z}_2+z_1& z_0-z_3\end{array}\right|=z_0^2-z_1^2-z_2^2-z_3^2\ne det\overline{z}{}^{*}=\left|\begin{array}{cc}{z}_0{}^{*}-z_3{}^{*}& i{z}_2{}^{*}-z_1{}^{*}\\ -i{z}_2{}^{*}-z_1{}^{*}& z_0{}^{*}+z_3{}^{*}\end{array}\right|=\left(z_0^2-z_1^2-z_2^2-z_3^2\right){}^{*}\end{array}$$

 □

Proposition 2.

Quaternions are those elements of the Pauli algebra that are fixed by the bar-star automorphism.

Proof. 

Since both Hermitian conjugation and parity inversion are involutions it immediately follows that

$$h^{*}=\overline{h}⟺h=\overline{h}{}^{*}$$

 □

When writing the bar-star automorphism as an operator the letter iota is used: $\widehat{\iota }\psi =\overline{\psi }{}^{*}$.

2.2. The Algebra of Biquaternions

The algebra of biquaternions is the complexification of the quaternion algebra.

The Pauli algebra is the same as the algebra of biquaternions because any of its elements can be written in the form:

$$z\in \mathbb{P}z=u+iv=(u_0+i\underline{u})+i(v_0+i\underline{v})u,v\in \mathbb{H}{z}_0=u_0+i{v}_0\underline{z}=\underline{u}+i\underline{v}$$

Proposition 3.

The bar-star automorphism is the conjugation of the biquaternion algebra.

Proof. 

This follows from the action of the bar-star automorphism on both the real and imaginary quaternion components being the identity, so its only result is the switching the sign of the imaginary component. □

Note. 

“Real" and “Imaginary" parts of a biquaternion. Let

$$z=a\mathbf{u}+ib\mathbf{v},\mathbf{u},\mathbf{v}\in \mathbb{H},\left|\right|\mathbf{u}\left|\right|=\left|\right|\mathbf{v}\left|\right|=1a,b\in \mathbb{R}$$

be an arbitrary biquaternion written using two unit quaternions. Let us call it “real" if $b=0$ and “purely imaginary" if $a=0$. Of course in a “real" quaternion only the time coordinate is real, but the three spatial coordinates are purely imaginary. It is the opposite for the purely imaginary quaternion component of a biquaternion.

3. Symmetries of the Dirac Equation

For convenience, the four-component spinor is written with a standard choice of letters instead of numbered indices:

$$\left[\begin{array}{c}{\psi }_1\left(x\right)\\ {\psi }_2\left(x\right)\\ {\psi }_3\left(x\right)\\ {\psi }_4\left(x\right)\end{array}\right]=\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]$$

(2)

where the letters represent complex scalar functions. Then the spinor is completed with a second column as follows:

$$\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]⟶\mathrm{\Psi }=\left[\begin{array}{cc}a& b^{*}\\ b& -a^{*}\\ c& -d^{*}\\ d& c^{*}\end{array}\right]=\left[\begin{array}{c}{\psi }_{upp}\\ \cdots \\ {\psi }_{low}\end{array}\right]=\left[\begin{array}{c}iv\\ \cdots \\ u\end{array}\right]u,v\in \mathbb{H}$$

(3)

The spinor thus completed with the right column can be rewritten as a column of two Pauli algebra spinors, the upper and the lower. The lower spinor is a quaternion and the upper spinor is a product of a quaternion with the imaginary unit; see (1).

In what follows four-component spinor always refers to a two-column spinor completed in this manner. Later (14) the two resulting quaternions will be used to rewrite the Dirac Equation so that its unknown function is a biquaternion rather than a four-component spinor.

3.1. Symmetry between the Dirac Equation and Its Conjugate Equation

Next the Dirac Equation is written in a form so that both itself and its Hermitian conjugate equation contain the same eight scalar equations, four for the original Dirac Equation and four for its Hermitian conjugate (unfortunately this check requires the full writing out of matrices).

Using the spinors completed as above rewrite the Dirac equation

$$(i\cancel{\partial }-\frac{mc}{\hslash })\mathrm{\Psi }=0$$

(4)

first as:

$$\left({\gamma }^0{\partial }_0+{\gamma }^0{\alpha }^i{\partial }_i\right)\mathrm{\Psi }=-i\frac{mc}{\hslash }\mathrm{\Psi }$$

and then in block matrix form:

$$\left[\begin{array}{cc}{\partial }_0& \nabla \\ -\nabla & -{\partial }_0\end{array}\right]\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]=-i\frac{mc}{\hslash }\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]⟺\left[\begin{array}{cc}\frac{\widehat{E}}{c}& -\widehat{p}\\ \widehat{p}& -\frac{\widehat{E}}{c}\end{array}\right]\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]=mc\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]⟺\left[\begin{array}{cc}\frac{\widehat{E}}{c}& -\widehat{p}\\ \widehat{p}& -\frac{\widehat{E}}{c}\end{array}\right]\left[\begin{array}{c}iv\\ u\end{array}\right]=mc\left[\begin{array}{c}iv\\ u\end{array}\right]$$

Proposition 4.

Symmetry between DE and its Hermitian conjugate equation. The Dirac Equation in two-column form and its Hermitian conjugate equation:

$$\left[\begin{array}{cc}\frac{\widehat{E}}{c}& -\widehat{p}\\ \widehat{p}& -\frac{\widehat{E}}{c}\end{array}\right]\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]=mc\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]\left[\begin{array}{cc}{\psi }_{upp}{}^{*}& {\psi }_{low}{}^{*}\end{array}\right]\left[\begin{array}{cc}\frac{\widehat{E}}{c}& \widehat{p}\\ -\widehat{p}& -\frac{\widehat{E}}{c}\end{array}\right]=mc\left[\begin{array}{cc}{\psi }_{upp}{}^{*}& {\psi }_{low}{}^{*}\end{array}\right]$$

(5)

contain the same eight scalar equations, in different order.

Note. 

This can be rewritten in more compact form by denoting $\widehat{D}=\left[\begin{array}{cc}\frac{\widehat{E}}{c}& -\widehat{p}\\ \widehat{p}& -\frac{\widehat{E}}{c}\end{array}\right]$. Then the two equations become:

$$\widehat{D}\mathrm{\Psi }=mc\mathrm{\Psi }{\mathrm{\Psi }}^{*}{\widehat{D}}^T=mc{\mathrm{\Psi }}^{*}$$

(6)

since the Hermitian conjugate for $\widehat{D}$ is the same as transpose.

Proof. 

Write out both the D.E. and its Hermitian conjugate equation in full detail and check the resulting scalar equations:

Two-column Dirac Equation:

$$\left[\begin{array}{cccc}\frac{\widehat{E}}{c}& 0& -{\widehat{p}}_3& i{\widehat{p}}_2-{\widehat{p}}_1\\ 0& \frac{\widehat{E}}{c}& -i{\widehat{p}}_2-{\widehat{p}}_1& {\widehat{p}}_3\\ {\widehat{p}}_3& -i{\widehat{p}}_2+{\widehat{p}}_1& -\frac{\widehat{E}}{c}& 0\\ i{\widehat{p}}_2+{\widehat{p}}_1& -{\widehat{p}}_3& 0& -\frac{\widehat{E}}{c}\end{array}\right]\left[\begin{array}{cc}a& b^{*}\\ b& -a^{*}\\ c& -d^{*}\\ d& c^{*}\end{array}\right]=mc\left[\begin{array}{cc}a& b^{*}\\ b& -a^{*}\\ c& -d^{*}\\ d& c{\phantom{\int }}^{*}\end{array}\right]$$

(7)

Two-row conjugate Dirac Equation:

$$\left[\begin{array}{cccc}{a}^{*}& b^{*}& c^{*}& d^{*}\\ b& -a& -d& c\end{array}\right]\left[\begin{array}{cccc}\frac{\widehat{E}}{c}& 0& {\widehat{p}}_3& -i{\widehat{p}}_2+{\widehat{p}}_1\\ 0& \frac{\widehat{E}}{c}& i{\widehat{p}}_2+{\widehat{p}}_1& -{\widehat{p}}_3\\ -{\widehat{p}}_3& i{\widehat{p}}_2-{\widehat{p}}_1& -\frac{\widehat{E}}{c}& 0\\ -i{\widehat{p}}_2-{\widehat{p}}_1& +{\widehat{p}}_3& 0& -\frac{\widehat{E}}{c}\end{array}\right]=mc\left[\begin{array}{cccc}{a}^{*}& b^{*}& c^{*}& d^{*}\\ b& -a& -d& c\end{array}\right]$$

(8)

The proof consists of writing down the eight equations for Equation (7) and for Equation (8) and matching them with each other. For example, the equation in row 1, left column of Equation (7) is:

$$\frac{\widehat{E}}{c}a-{\widehat{p}}_{3}c-\left(-i{\widehat{p}}_2+{\widehat{p}}_1\right)d=mca$$

Furthermore, the equation in bottom row, column 2 of Equation (8) is:

$$-\frac{\widehat{E}}{c}a+\left(-i{\widehat{p}}_2+{\widehat{p}}_1\right)d+{\widehat{p}}_{3}c=-mca$$

(9)

The other checks are omitted. □

Note. 

The particle-antiparticle symmetry in this form of the Dirac Equation essentially depends on mass being a real scalar.

3.2. The Particle-Antiparticle Symmetry of the Two-Column Dirac Equation

The next proposition deals with a different symmetry of the free Dirac Equation, which reverses the signs of mass in all of its eight equations.

Definition 1.

For a two-column four-component spinor completed as in Equation (3) its charge conjugate spinor is defined as follows:

$$\mathrm{\Psi }=\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]⟶{\mathrm{\Psi }}_c=\left[\begin{array}{c}{\psi }_{low}{\sigma }^1\\ {\psi }_{upp}{\sigma }^1\end{array}\right]$$

(10)

In other words the symmetry between a spinor and its charge conjugate is as follows:

-

Switch the top and bottom Pauli algebra spinors;

-

For each of them, switch the left and right columns.

The result is shown below:

$$\mathrm{\Psi }=\left[\begin{array}{cc}a& b^{*}\\ b& -a^{*}\\ c& -d^{*}\\ d& c^{*}\end{array}\right]⟶{\mathrm{\Psi }}_c=\left[\begin{array}{cc}-d^{*}& c\\ c^{*}& d\\ b^{*}& a\\ -a^{*}& b\end{array}\right]$$

(11)

Note. 

The left column of the charge conjugate two-column spinor is exactly the same as the one for the usual charge conjugation operator $\widehat{\mathbf{C}}\mathrm{\Psi }=i{\gamma }^2{\mathrm{\Psi }}^{*}$, see [7]. Possible ambiguity in sign does not matter because DE is linear homogeneous.

Proposition 5.

Symmetry which reverses the signs for mass. Replacing the spinor $\mathrm{\Psi }$ with its charge conjugate ${\mathrm{\Psi }}_c$ in DE results in the same eight equations with the only difference that the signs in all equalities are reversed:

$$\widehat{D}\mathrm{\Psi }=mc\mathrm{\Psi }⟺\widehat{D}{\mathrm{\Psi }}_c=-mc{\mathrm{\Psi }}_c$$

(12)

Proof. 

The proof again consists or writing out the eight scalar equations below and comparing them to the eight equations of the original Dirac Equation (7):

$$\left[\begin{array}{cccc}\frac{\widehat{E}}{c}& 0& -{\widehat{p}}_3& i{\widehat{p}}_2-{\widehat{p}}_1\\ 0& \frac{\widehat{E}}{c}& -i{\widehat{p}}_2-{\widehat{p}}_1& {\widehat{p}}_3\\ {\widehat{p}}_3& -i{\widehat{p}}_2+{\widehat{p}}_1& -\frac{\widehat{E}}{c}& 0\\ i{\widehat{p}}_2+{\widehat{p}}_1& -{\widehat{p}}_3& 0& -\frac{\widehat{E}}{c}\end{array}\right]\left[\begin{array}{cc}-d^{*}& c\\ c^{*}& d\\ b^{*}& a\\ -a^{*}& b\end{array}\right]=-mc\left[\begin{array}{cc}-d^{*}& c\\ c^{*}& d\\ b^{*}& a\\ -a^{*}& b\end{array}\right]$$

(13)

For example, row 3, column 2:

$${\widehat{p}}_{3}c+(-i{\widehat{p}}_2+{\widehat{p}}_1)d-\frac{\widehat{E}}{c}=-mca$$

This equation is the same as the Equation (9) but the matrix Equation (13) has the sign of its equality reversed compared to the matrix Equation (8).

Other checks are similar and are omitted.

The reversal of the sign means the opposite sign of all energy and momentum operators, so this is the particle-antiparticle symmetry. □

3.3. Coupling to the Electromagnetic Field

The minimal substitution ${\widehat{p}}_{\mu }⟶{\widehat{p}}_{\mu }-\frac{e}{c}{A}_{\mu }={\widehat{P}}_{\mu }$ results in Hermitian operators whose sign is reversed by particle-antiparticle symmetry exactly as in the previous case of free Dirac Equation. But now the reversal of the sign results in opposite signs of charges as well as energy and momentum operators. Hence the charge conjugation and the particle–antiparticle symmetry are the same.

4. Biquaternionic Form of the Dirac Equation

Since the two-column forms of the standard Dirac Equation and its Hermitian conjugate equation are equivalent, we leave only the standard one and rewrite it first in block form:

$$\left[\begin{array}{cc}\frac{\widehat{E}}{c}& -\widehat{p}\\ \widehat{p}& -\frac{\widehat{E}}{c}\end{array}\right]\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]=mc\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]⟹\left[\begin{array}{cc}{\partial }_0& \nabla \\ -\nabla & -{\partial }_0\end{array}\right]\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]=-i\frac{mc}{\hslash }\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]$$

then as two coupled equations in Pauli algebra spinor variables:

$$\begin{array}{cc}{\partial }_0{\psi }_{upp}+\nabla {\psi }_{low}& =-i\frac{mc}{\hslash }{\psi }_{upp}i{\partial }_{0}v+\nabla u=\frac{mc}{\hslash }v\hfill \\ {\partial }_0{\psi }_{low}+\nabla {\psi }_{upp}& =i\frac{mc}{\hslash }{\psi }_{low}{\partial }_{0}u+i\nabla v=i\frac{mc}{\hslash }u\hfill \end{array}$$

Next, since the lower spinor is a quaternion and the upper spinor a product of a quaternion with the imaginary unit, their sum and difference will be related by the biquaternion conjugation, which is the same as the bar-star automorphism:

$$\psi ={\psi }_{low}+{\psi }_{upp}=u+iv\overline{\psi }{}^{*}={\psi }_{low}-{\psi }_{upp}=u-ivu,v\in \mathbb{H}$$

(14)

Now add and subtract the two equations. We obtain two coupled equations:

$$\begin{array}{cc}\hfill \left({\partial }_0+\nabla \right)\psi =& i\frac{mc}{\hslash }\overline{\psi }{}^{*}\hfill \\ \hfill \left({\partial }_0-\nabla \right)\overline{\psi }{}^{*}=& i\frac{mc}{\hslash }\psi \hfill \end{array}$$

(15)

Recalling that $\partial ={\partial }_0+\nabla ,\overline{\partial }={\partial }_0-\nabla$, we have a system of two equations:

$$\begin{array}{c}\hfill \partial \psi =i\frac{mc}{\hslash }\overline{\psi }{}^{*}\\ \hfill \overline{\partial }\overline{\psi }{}^{*}=i\frac{mc}{\hslash }\psi \end{array}$$

(16)

4.1. The Energy-Momentum Form of the Biquaternionic Dirac Equation

It is obtained from Equation (15) by multiplying both sides by $i\hslash$:

$$\begin{array}{cc}\hfill \left(\frac{\widehat{E}}{c}-\widehat{p}\right)\psi =& -mc\overline{\psi }{}^{*}\hfill \\ \hfill \left(\frac{\widehat{E}}{c}+\widehat{p}\right)\overline{\psi }{}^{*}=& -mc\psi \hfill \end{array}$$

(17)

4.2. Equivalence between the Standard and the Biquaternionic Forms of the Dirac Equation

Proposition 6.

The standard Dirac Equation (4) and the system of two biquaternionic Equations (16) are equivalent to each other.

Proof. 

We need to show one-to-one correspondence between the four-component spinors $\mathrm{\Psi }$ that are solutions of the Dirac Equation (4) and biquaternionic spinors that are solutions of the system Equation (16).

The procedure for converting four-component spinors that solve Equation (4) into biquaternionic spinors is as follows:

First, complete the four-component spinor with a right column so as to make the lower Pauli algebra spinor a quaternion and the upper Pauli algebra spinor a multiple of a quaternion by the imaginary unit (this is unambiguous).

$$\mathrm{\Psi }=\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]⟶\left[\begin{array}{cc}a& b^{*}\\ b& -a^{*}\\ c& -d^{*}\\ d& c^{*}\end{array}\right]u=\left[\begin{array}{cc}c& -d^{*}\\ d& c^{*}\end{array}\right]iv=\left[\begin{array}{cc}a& b^{*}\\ b& -a^{*}\end{array}\right]$$

Next form their sum and difference:

$$\psi =u+iv\psi =\left[\begin{array}{cc}a+c& b^{*}-d^{*}\\ b+d& -a^{*}+c^{*}\end{array}\right]\overline{\psi }{}^{*}=u-iv\overline{\psi }{}^{*}=\left[\begin{array}{cc}-a+c& -b^{*}-d^{*}\\ -b+d& a^{*}+c^{*}\end{array}\right]$$

(18)

The opposite procedure of converting biquaternionic spinors to four-component spinors is as follows:

$$\begin{array}{c}\hfill \psi =u+iv⟹\overline{\psi }{}^{*}=u-iv\\ \hfill u=\frac{1}{2}\left(\psi +\overline{\psi }{}^{*}\right)iv=\frac{1}{2}\left(\psi -\overline{\psi }{}^{*}\right)\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}\psi -\overline{\psi }{}^{*}\\ \psi +\overline{\psi }{}^{*}\end{array}\right]\end{array}$$

Then at the last step delete the right column to obtain a four-component spinor.

Since all transitions throughout the derivation are equivalences, the sets of solutions at the beginning and at the end will be in one-to-one correspondence with each other. □

Proposition 7.

Every entry of the biquaternionic spinors ψ and $\overline{\psi }{}^{*}$ satisfies the Klein–Gordon equation for the same mass as the Dirac Equation.

Proof. 

It follows from Equation (16) that both the biquaternionic spinor and its bar-star automorphic image can be calculated as differentials of each other:

$$\overline{\psi }{}^{*}=\frac{\hslash }{imc}\partial \psi \psi =\frac{\hslash }{imc}\overline{\partial }\overline{\psi }{}^{*}$$

(19)

Note. 

It is essential at this step that the mass is non-zero.

Next we compose these relations, using the fact that $\overline{\partial }\partial \psi =\left({\partial }_0^2-{\nabla }^2\right)\psi =\square \psi$:

$$\begin{array}{c}\hfill \overline{\partial }\overline{\psi }{}^{*}=\frac{\hslash }{imc}\overline{\partial }\partial \psi =i\frac{mc}{\hslash }\psi \\ \hfill \overline{\partial }\partial \psi =-\frac{m^2{c}^2}{{\hslash }^2}\psi \\ \hfill \overline{\partial }\partial \psi +\frac{m^2{c}^2}{{\hslash }^2}\psi =0\\ \hfill \left(\square +\frac{m^2{c}^2}{{\hslash }^2}\right)\psi =0\end{array}$$

However, the operator in brackets is scalar, which means it applies separately to all the four components of the spinor. So each of these components satisfy the Klein–Gordon equation. The same derivation works for the spinor’s bar-star image. □

5. Translation from the Four-Component to Biquaternionic Language

In this and subsequent sections the following formulas for conversion between the four-component and biquaternionic forms are used throughout:

$$\mathrm{\Psi }=\left[\begin{array}{cc}a& b^{*}\\ b& -a^{*}\\ c& -d^{*}\\ d& c^{*}\end{array}\right]=\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]=\left[\begin{array}{c}iv\\ u\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}\psi -\overline{\psi }{}^{*}\\ \psi +\overline{\psi }{}^{*}\end{array}\right]$$

(20)

$$\begin{array}{ccccccc}\hfill \psi & =\hfill & \hfill {\psi }_{upp}+{\psi }_{low}{\psi }_{upp}& =\hfill & \hfill \frac{1}{2}(\psi -\overline{\psi }{}^{*}){\psi }_{upp}{}^{*}& =\hfill & \hfill \frac{1}{2}(\psi {}^{*}-\overline{\psi })\\ \hfill \overline{\psi }{}^{*}& =\hfill & \hfill -{\psi }_{upp}+{\psi }_{low}{\psi }_{low}& =\hfill & \hfill \frac{1}{2}(\psi +\overline{\psi }{}^{*}){\psi }_{low}{}^{*}& =\hfill & \hfill \frac{1}{2}({\psi }^{*}+\overline{\psi })\end{array}$$

(21)

Keeping the standard choice of letters to represent the four complex functions of the four-component spinor, one also has the standard expressions for the biquaternionic spinor and its bar-star automorphism image:

$$\psi =\left[\begin{array}{cc}a+c& b^{*}-d^{*}\\ b+d& -a^{*}+c^{*}\end{array}\right]\overline{\psi }{}^{*}=\left[\begin{array}{cc}-a+c& -b^{*}-d^{*}\\ -b+d& a^{*}+c^{*}\end{array}\right]$$

(22)

5.1. Positive and Negative Energy

Rewrite the Equations (16) as:

$$\begin{array}{c}\hfill \partial (u+iv)=i\frac{mc}{\hslash }(u-iv)\overline{\partial }(u-iv)=i\frac{mc}{\hslash }(u+iv)\\ \hfill -\frac{\widehat{E}}{c}(u+iv)+\widehat{p}(u+iv)=mc(u-iv)\\ \hfill -\frac{\widehat{E}}{c}(u-iv)-\widehat{p}(u-iv)=mc(u+iv)\end{array}$$

5.1.1. Purely Real (Quaternionic) Biquaternion Spinor, Negative Energy

Let $v\equiv 0,\psi \equiv u$. Then both the imaginary component and its derivatives vanish and we have:

$$\begin{array}{c}\hfill -\frac{\widehat{E}}{c}\left(u\right)+\widehat{p}\left(u\right)=mc\left(u\right)\\ \hfill -\frac{\widehat{E}}{c}\left(u\right)-\widehat{p}\left(u\right)=mc\left(u\right)\end{array}$$

This implies that $\widehat{p}\left(u\right)=0$ and $\widehat{E}\left(u\right)=-m{c}^2\left(u\right)$. The conclusion is that states that are purely quaternionic are states of zero momentum and negative energy.

5.1.2. Purely Imaginary Biquaternion Spinor, Positive Energy

Let $u\equiv 0,\psi \equiv iv$. Then both the real component and its derivatives vanish and we have:

$$\begin{array}{c}\hfill -\frac{\widehat{E}}{c}\left(iv\right)+\widehat{p}\left(iv\right)=mc(-iv)\\ \hfill \frac{\widehat{E}}{c}\left(iv\right)+\widehat{p}\left(iv\right)=mc\left(iv\right)\end{array}$$

This implies that $\widehat{p}\left(iv\right)=0$ and $\widehat{E}\left(iv\right)=m{c}^2\left(iv\right)$. The conclusion is that states that are purely imaginary biquaternions are states of zero momentum and positive energy.

5.2. Chirality

The procedure for converting four-component spinors to biquaternionic spinors Equation (6) is used to obtain right-handed and left-handed projection operators for spinors in the biquaternionic picture.

First recall the procedure for converting from four-component to biquaternion spinor:

$$\mathrm{\Psi }=\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]⟶\psi =\left[\begin{array}{cc}a+c& b^{*}-d^{*}\\ b+d& -a^{*}+c^{*}\end{array}\right]\overline{\psi }{}^{*}=\left[\begin{array}{cc}-a+c& -b^{*}-d^{*}\\ -b+d& a^{*}+c^{*}\end{array}\right]$$

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5.2.1. Right-Handed Projection Operator

Let $\mathrm{\Psi }$ be a four-component spinor. First, the standard right-handed projector operator is applied to $\mathrm{\Psi }$; then it is completed to a two-column spinor, and then the result is converted to a biquaternion. Then we compare the result with the conversion of the original spinor to obtain the the right-handed projection operator in the biquaternionic picture.

$$\begin{array}{c}\hfill \mathrm{\Psi }=\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]P_r\mathrm{\Psi }=\frac{1}{2}(1+{\gamma }_5)\mathrm{\Psi }=\frac{1}{2}\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 0& 1& 0\\ 0& 1& 0& 1\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}a+c\\ b+d\\ a+c\\ b+d\end{array}\right]\end{array}$$

Then at the next step the spinor is completed with a right column as in Equation (3):

$$\frac{1}{2}\left[\begin{array}{c}a+c\\ b+d\\ a+c\\ b+d\end{array}\right]⟶\frac{1}{2}\left[\begin{array}{cc}a+c& b^{*}+d^{*}\\ b+d& -a^{*}-c^{*}\\ a+c& -b^{*}-d^{*}\\ b+d& a^{*}+c^{*}\end{array}\right]={\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]}_{right}$$

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Finally using the rules $\psi ={\psi }_{upp}+{\psi }_{low},\overline{\psi }{}^{*}=-{\psi }_{upp}+{\psi }_{low}$ we obtain:

$${\psi }_r=\left[\begin{array}{cc}a+c& 0\\ b+d& 0\end{array}\right]\overline{{\psi }_{right}}{}^{*}=\left[\begin{array}{cc}0& -b^{*}-d^{*}\\ 0& a^{*}+c^{*}\end{array}\right]$$

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5.2.2. Left-Handed Projection Operator

$$\begin{array}{c}\hfill \mathrm{\Psi }=\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]P_r\mathrm{\Psi }=\frac{1}{2}(1-{\gamma }_5)\mathrm{\Psi }=\frac{1}{2}\left[\begin{array}{cccc}1& 0& -1& 0\\ 0& 1& 0& -1\\ -1& 0& 1& 0\\ 0& -1& 0& 1\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}a-c\\ b-d\\ -a+c\\ -b+d\end{array}\right]\end{array}$$

Completing with the right column:

$$\frac{1}{2}\left[\begin{array}{c}a-c\\ b-d\\ -a+c\\ -b+d\end{array}\right]⟶\frac{1}{2}\left[\begin{array}{cc}a-c& b^{*}-d^{*}\\ b-d& -a^{*}+c^{*}\\ -a+c& b^{*}-d^{*}\\ -b+d& -a^{*}+c^{*}\end{array}\right]={\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]}_{left}$$

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Now again combine the upper and lower spinors:

$${\psi }_{left}=\left[\begin{array}{cc}0& b^{*}-d^{*}\\ 0& -a^{*}+c^{*}\end{array}\right]\overline{{\psi }_{left}}{}^{*}=\left[\begin{array}{cc}-a+c& 0\\ -b+d& 0\end{array}\right]$$

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As expected ${\psi }_{right}+{\psi }_{left}=\psi ,\overline{\psi }{\phantom{\int }}_{right}^{*}+\overline{\psi }{\phantom{\int }}_{left}^{*}=\overline{\psi }{\phantom{\int }}^{*}$.

The conclusion is that the right-handed projector nullifies the right column of a biquaternionic spinor, and leaves the left column unchanged. The left-handed projector nullifies the left column, and leaves the right column unchanged. They act the opposite way on the bar-star image of a spinor: right-handed projector nullifies its left column, and the left-handed projector nullifies its right column, leaving everything else unchanged.

5.3. Probability Density and Probability Current

The standard expressions using four-component spinors for probability density and for probability current are:

$$\rho ={\mathrm{\Psi }}^{*}\mathrm{\Psi }{j}^k={\mathrm{\Psi }}^{*}{\gamma }^0{\gamma }^k\mathrm{\Psi }={\mathrm{\Psi }}^{*}{\alpha }^k\mathrm{\Psi }$$

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If instead the spinors are completed to two column spinors as above the same construction results in scalar matrices instead of scalars, with the same values.

It is possible to use the biquaternion form of the spinors to calculate density and current, and the result is this

Proposition 8.

Let $\mathrm{\Psi }=\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]$ where $\psi ={\psi }_{upp}+{\psi }_{low}$. Then the expression for probability density and current is:

$$j^{\mu }=\frac{1}{2}\mathrm{Tr}\left({\psi }^{*}{\sigma }^{\mu }\psi \right)$$

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Proof. 

$\mu =0$:

$$\rho ={\mathrm{\Psi }}^{*}\mathrm{\Psi }=\left[\begin{array}{cc}{\psi }_{upp}{\phantom{\int }}^{*}& {\psi }_{low}{\phantom{\int }}^{*}\end{array}\right]\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]=\frac{1}{4}\left[({\psi }^{*}-\overline{\psi })(\psi -\overline{\psi }{}^{*})+({\psi }^{*}+\overline{\psi })(\psi +\overline{\psi }{}^{*})\right]=\frac{1}{2}\left({\psi }^{*}\psi +\overline{\psi }\overline{\psi }{}^{*}\right)$$

However, $\overline{\psi }\overline{\psi }{\phantom{\int }}^{*}=\left(\overline{\psi {\phantom{\int }}^{*}\psi }\right)$ so the last expression is a scalar matrix with the scalar being the $\frac{1}{2}\mathrm{Tr}\left(\psi {\phantom{\int }}^{*}\psi \right)$.

$\mu =k$:

$$\begin{array}{c}\hfill j^k={\mathrm{\Psi }}^{*}{\alpha }^k\psi =\left[\begin{array}{cc}{\psi }_{upp}{\phantom{\int }}^{*}& {\psi }_{low}{\phantom{\int }}^{*}\end{array}\right]\left[\begin{array}{cc}0& {\sigma }^k\\ {\sigma }^k& 0\end{array}\right]\left[\begin{array}{c}{\psi }_{upp}\\ {\psi }_{low}\end{array}\right]={\psi }_{low}{}^{*}{\sigma }^k{\psi }_{upp}+{\psi }_{upp}{}^{*}{\sigma }^k{\psi }_{low}=\\ \hfill =\frac{1}{4}\left[({\psi }^{*}+\overline{\psi }){\sigma }^k(\psi -\overline{\psi }{}^{*})+({\psi }^{*}-\overline{\psi }){\sigma }^k(\psi +\overline{\psi }{}^{*})\right]=\frac{1}{2}({\psi }^{*}{\sigma }^k\psi -\overline{\psi }{\sigma }^k\overline{\psi }{}^{*})=\frac{1}{2}\mathrm{Tr}\left({\psi }^{*}{\sigma }^k\psi \right)\end{array}$$

Note that $-\overline{\psi }{\sigma }^k\overline{\psi }{\psi }^{*}=\left(\overline{\psi {\phantom{\int }}^{*}{\sigma }^k\psi }\right)$ so that the sum above is a scalar. □

Next calculate the four components of the current using the $\psi$ and ${\psi }^{*}$ written in the standard shorthand for the four-component spinor:

$$\psi =\left[\begin{array}{cc}a+c& b^{*}-d^{*}\\ b+d& -a^{*}+c^{*}\end{array}\right]{\psi }^{*}=\left[\begin{array}{cc}{a}^{*}+c^{*}& b^{*}+d^{*}\\ b-d& -a+c\end{array}\right]$$

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$$\begin{array}{ccc}\hfill j^0=\frac{1}{2}\mathrm{Tr}\left({\psi }^{*}{\sigma }^0\psi \right)& =\hfill & \hfill a^{*}a+b^{*}b+c^{*}c+d^{*}d\\ \hfill j^1=\frac{1}{2}\mathrm{Tr}\left({\psi }^{*}{\sigma }^1\psi \right)& =\hfill & \hfill a{d}^{*}+a^{*}d+b{c}^{*}+b^{*}c\\ \hfill j^2=\frac{1}{2}\mathrm{Tr}\left({\psi }^{*}{\sigma }^2\psi \right)& =\hfill & \hfill i(a{d}^{*}-a^{*}d+b^{*}c-b{c}^{*})\\ \hfill j^3=\frac{1}{2}\mathrm{Tr}\left({\psi }^{*}{\sigma }^3\psi \right)& =\hfill & \hfill a^{*}c+a{c}^{*}-b^{*}d-b{d}^{*}\end{array}$$

These expressions coincide with ones calculated directly from the four-component spinor Equation (2) using the standard method.

5.4. Spin

The relation between four-component spinors and biquaternionic spinors is used to obtain the spin operator in biquaternionic form (this is not done from first principles). Beginning with the four-component form of the spin operator [8]:

$$\widehat{\mathbf{S}}=\frac{\hslash }{2}\left[\begin{array}{cc}\widehat{\sigma }& 0\\ 0& \widehat{\sigma }\end{array}\right]\widehat{\mathbf{S}}\mathrm{\Psi }=\frac{\hslash }{2}\left[\begin{array}{cc}\widehat{\sigma }& 0\\ 0& \widehat{\sigma }\end{array}\right]\mathrm{\Psi }{\widehat{\mathbf{S}}}_k\mathrm{\Psi }=\frac{\hslash }{2}\left[\begin{array}{cc}{\widehat{\sigma }}_k& 0\\ 0& {\widehat{\sigma }}_k\end{array}\right]\mathrm{\Psi }$$

the four-component spinor $\mathrm{\Psi }$ is replaced with its expression from the biquaternionic spinor $\psi$ Equation (20):

$$\mathrm{\Psi }=\frac{1}{2}\left[\begin{array}{c}\psi -\overline{\psi }{}^{*}\\ \psi +\overline{\psi }{}^{*}\end{array}\right]$$

then the spin operator takes the value:

$${\widehat{\mathbf{S}}}_k=\frac{\hslash }{4}\left[\begin{array}{cc}{\sigma }_k& 0\\ 0& {\sigma }_k\end{array}\right]\left[\begin{array}{c}\psi -\overline{\psi }{}^{*}\\ \psi +\overline{\psi }{}^{*}\end{array}\right]=\frac{\hslash }{4}{\sigma }_k\left[\begin{array}{c}\psi -\overline{\psi }{}^{*}\\ \psi +\overline{\psi }{}^{*}\end{array}\right]$$

and using Equation (20) again we get the familiar expression for the spin operator:

$$\widehat{\mathbf{S}}\psi =\frac{\hslash }{2}{\sigma }_k\psi$$

6. The Charge Conjugation Operator

The charge conjugation operator is denoted by a subscript as in $\widehat{\mathbf{C}}\mathrm{\Psi }={\mathrm{\Psi }}_c$.

The following summarises the action of the charge conjugation operator on a four-component spinor completed with a second column, written in standard shorthand, its biquaternion spinor, and the bar star image of its biquaternion spinor:

$$\begin{array}{cc}\hfill \mathrm{\Psi }=\left[\begin{array}{cc}a& b^{*}\\ b& -a^{*}\\ c& -d^{*}\\ d& c^{*}\end{array}\right]& {\mathrm{\Psi }}_c=\left[\begin{array}{cc}-d^{*}& c\\ c^{*}& d\\ b^{*}& a\\ -a^{*}& b\end{array}\right]\hfill \\ \hfill \psi =\left[\begin{array}{cc}a+c& b^{*}-d^{*}\\ b+d& -a^{*}+c^{*}\end{array}\right]& {\psi }_c=\left[\begin{array}{cc}{b}^{*}-d^{*}& a+c\\ -a^{*}+c^{*}& b+d\end{array}\right]\hfill \\ \hfill \overline{\psi }{}^{*}=\left[\begin{array}{cc}-a+c& -b^{*}-d^{*}\\ -b+d& a^{*}+c^{*}\end{array}\right]& \overline{\left({\psi }_c\right)}{}^{*}=\left[\begin{array}{cc}{b}^{*}+d^{*}& a-c\\ -a^{*}-c^{*}& b-d\end{array}\right]\hfill \end{array}$$

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6.1. Comparison of the Dirac Equation for a Spinor and for Its Charge Conjugate

Proposition 9.

Let ψ be a biquaternionic spinor which is a solution for the Equation (17):

$$\begin{array}{cc}\hfill \left(\frac{\widehat{E}}{c}-\widehat{p}\right)\psi =& -mc\overline{\psi }{}^{*}\hfill \\ \hfill \left(\frac{\widehat{E}}{c}+\widehat{p}\right)\overline{\psi }{}^{*}=& -mc\psi \hfill \end{array}$$

(32)

Then its charge conjugate spinor satisfies the equation with the signs reversed:

$$\begin{array}{cc}\hfill \left(\frac{\widehat{E}}{c}-\widehat{p}\right){\psi }_c=& mc\overline{\left({\psi }_c\right)}{}^{*}\hfill \\ \hfill \left(\frac{\widehat{E}}{c}+\widehat{p}\right)\overline{\left({\psi }_c\right)}{}^{*}=& mc{\psi }_c\hfill \end{array}$$

(33)

Proof. 

The proof follows the construction of the biquaternionic equation out of the Dirac equation for two-column spinors as in Equation (17).

When the four-component spinor $\mathrm{\Psi }$ is replaced with its charge conjugate ${\mathrm{\Psi }}_c$ the sign of the mass term in the matrix Equation (5) is reversed. This reversal follows through the derivation of the biquaternionic equation and results in Equation (33). □

6.2. Majorana Solutions of the Dirac Equation

If a spinor equals its charge conjugate, it is referred to as a Majorana solution:

$${\left({\mathrm{\Psi }}_M\right)}_c={\mathrm{\Psi }}_M{\left({\psi }_M\right)}_c={\psi }_M$$

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From the construction of the charge conjugation operator it is clear that

$${\psi }_M={\psi }_M{\sigma }^1\overline{{\psi }_M}{}^{*}=-\overline{{\psi }_M}{}^{*}{\sigma }^1$$

In fact one use the standard formulas of Equation (31), choose two parameters out of four and write the Majorana solution and its bar star image in detail:

$${\psi }_M=\left[\begin{array}{cc}a+b^{*}& a+b^{*}\\ -a^{*}+b& -a^{*}+b\end{array}\right]\overline{{\psi }_M}{}^{*}=\left[\begin{array}{cc}-a+b^{*}& a-b^{*}\\ -a^{*}-b& a^{*}+b\end{array}\right]$$

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Proposition 10.

Majorana solutions are massless.

Proof. 

It is enough to compare the two equations for a biquaternionic spinor and its charge conjugate:

$$\begin{array}{cc}\hfill \left(\frac{\widehat{E}}{c}-\widehat{p}\right){\psi }_M=& -mc\overline{\psi }{}_M^{*}\hfill \\ \hfill \left(\frac{\widehat{E}}{c}-\widehat{p}\right){\psi }_M=& mc\overline{\psi }{}_M^{*}\hfill \end{array}$$

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From this it immediately follows that the mass is zero. The same proof works with four-component spinors using Equation (12). □

7. When Multiplied by the Imaginary Unit, Charge Conjugation Is a Complex Lorentz Transformation

Definition 2.

$$L_c\psi :=i\widehat{\mathbf{C}}\psi =i{\psi }_c$$

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Proposition 11.

$L_c$ is a complex Lorentz transformation.

Proof. 

$$\psi =\left[\begin{array}{cc}{\psi }_0+{\psi }_3& -i{\psi }_2+{\psi }_1\\ i{\psi }_2+{\psi }_1& {\psi }_0-{\psi }_3\end{array}\right]\overrightarrow{\psi }=\left[\begin{array}{c}{\psi }_0\\ {\psi }_1\\ {\psi }_2\\ {\psi }_3\end{array}\right]$$

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$${\psi }_c=\left[\begin{array}{cc}-i{\psi }_2+{\psi }_1& {\psi }_0+{\psi }_3\\ {\psi }_0-{\psi }_3& i{\psi }_2+{\psi }_1\end{array}\right]L_c\psi =i{\psi }_c=\left[\begin{array}{cc}{\psi }_2+i{\psi }_1& i{\psi }_0+i{\psi }_3\\ i{\psi }_0-i{\psi }_3& -{\psi }_2+i{\psi }_1\end{array}\right]$$

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$$\overrightarrow{\left(L_c\psi \right)}=\left[\begin{array}{c}i{\psi }_1\\ i{\psi }_0\\ -{\psi }_3\\ {\psi }_2\end{array}\right]=\left[\begin{array}{cccc}0& i& 0& 0\\ i& 0& 0& 0\\ 0& 0& 0& -1\\ 0& 0& 1& 0\end{array}\right]\left[\begin{array}{c}{\psi }_0\\ {\psi }_1\\ {\psi }_2\\ {\psi }_3\end{array}\right]={\mathcal{L}}_c\overrightarrow{\psi }$$

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Check that the $4\times 4$ matrix ${\mathcal{L}}_c$ is Lorentz: ${\mathcal{L}}_c^{T}g{\mathcal{L}}_c\stackrel{?}{=}g$.

This check is a straightforward matrix calculation. □

Since ${\mathcal{L}}_c\in {\mathcal{L}}_{+}\left(\mathbb{C}\right)$ is a proper complex Lorentz transformation, there must exist unimodular matrices $A,B\in SL(2,\mathbb{C})$ so that $L_c\psi =A\psi B$ ([9]). They are easy to find: $L_c\psi ={\sigma }^0\psi i{\sigma }^1$.

8. Complex Lorentz Transformation of Charge Conjugation Commutes with the Bar-Star Automorphism

Note. 

This implies that the operator of charge conjugation anticommutes with the bar-star automorphism for the following reason:

$$\begin{array}{c}\hfill \overline{\left(L_c\psi \right)}{}^{*}=L_c\left(\overline{\psi }{}^{*}\right)\\ \hfill \overline{\left(i{\psi }_c\right)}{}^{*}=i{\left(\overline{\psi }{}^{*}\right)}_c\\ \hfill -i\overline{\left({\psi }_c\right)}{}^{*}=i{\left(\overline{\psi }{}^{*}\right)}_c\end{array}$$

Proposition 12.

Let $\widehat{\iota }$, $L_c$ and $\widehat{\mathbf{C}}$ be respectively the bar-star automorphism, the Lorentz charge conjugation transformation and the charge conjugation operator.

Then $[\widehat{\iota },L_c]=0$ and $\{\widehat{\iota },\widehat{\mathbf{C}}\}=0$.

Proof. 

We need to show that $\overline{\left(L_c\psi \right)}{}^{*}+L_c\left(\overline{\psi }{}^{*}\right)=0$ or equivalently $\widehat{\iota }{L}_c\psi +L_c\widehat{\iota }\psi =0$.

The proof proceeds by direct calculation, starting with an arbitrary Pauli algebra spinor $\psi$.

Write it side by side with its Lorentz charge conjugate:

$$\psi =\left[\begin{array}{cc}{\psi }_0+{\psi }_3& -i{\psi }_2+{\psi }_1\\ i{\psi }_2+{\psi }_1& {\psi }_0-{\psi }_3\end{array}\right]L_c\psi =i\left[\begin{array}{cc}-i{\psi }_2+{\psi }_1& {\psi }_0+{\psi }_3\\ {\psi }_0-{\psi }_3& i{\psi }_2+{\psi }_1\end{array}\right]=\left[\begin{array}{cc}{\psi }_2+i{\psi }_1& i{\psi }_0+i{\psi }_3\\ i{\psi }_0-i{\psi }_3& -{\psi }_2+i{\psi }_1\end{array}\right]$$

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Apply space inversion to both:

$$\overline{\psi }=\left[\begin{array}{cc}{\psi }_0-{\psi }_3& i{\psi }_2-{\psi }_1\\ -i{\psi }_2-{\psi }_1& {\psi }_0+{\psi }_3\end{array}\right]\overline{\left(L_c\psi \right)}=\left[\begin{array}{cc}-{\psi }_2+i{\psi }_1& -i{\psi }_0-i{\psi }_3\\ -i{\psi }_0+i{\psi }_3& {\psi }_2+i{\psi }_1\end{array}\right]$$

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Apply the Hermitian conjugation (“star”) to both:

$$\overline{\psi }{}^{*}=\left[\begin{array}{cc}{\psi }_0{}^{*}-{\psi }_3{}^{*}& i{\psi }_2{}^{*}-{\psi }_1{}^{*}\\ -i{\psi }_2{}^{*}-{\psi }_1{}^{*}& {\psi }_0{}^{*}+{\psi }_3{}^{*}\end{array}\right]\overline{\left(L_c\psi \right)}{}^{*}=\left[\begin{array}{cc}-{\psi }_2{}^{*}-i{\psi }_1{}^{*}& i{\psi }_0{}^{*}-i{\psi }_3{}^{*}\\ i{\psi }_0{}^{*}+i{\psi }_3{}^{*}& {\psi }_2{}^{*}-i{\psi }_1{}^{*}\end{array}\right]$$

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Apply Lorentz charge conjugation to the left:

$$L_c\left(\overline{\psi }{}^{*}\right)=i\left[\begin{array}{cc}i{\psi }_2{}^{*}-{\psi }_1{}^{*}& {\psi }_0{}^{*}-{\psi }_3{}^{*}\\ {\psi }_0{}^{*}+{\psi }_3{}^{*}& -i{\psi }_2{}^{*}-{\psi }_1{}^{*}\end{array}\right]=\left[\begin{array}{cc}-{\psi }_2{}^{*}-i{\psi }_1{}^{*}& i{\psi }_0{}^{*}-i{\psi }_3{}^{*}\\ i{\psi }_0{}^{*}+i{\psi }_3{}^{*}& {\psi }_2{}^{*}-i{\psi }_1{}^{*}\end{array}\right]$$

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which completes the proof that bar-star automorphism and the charge conjugation Lorentz transformation commute. □

9. Second Proof of Sign Reversal of the Mass by Charge Conjugation that Uses Its Anticommutation with Bar-Star Automorphism

The second proof starts with the biquaternionic Dirac Equations (16):

$$\partial \psi =i\frac{mc}{\hslash }\overline{\psi }{}^{*}$$

where $\partial \psi$ means:

$$\partial \psi =\left[\begin{array}{cc}{\partial }_0+{\partial }_3& -i{\partial }_2+{\partial }_1\\ i{\partial }_2+{\partial }_1& {\partial }_0-{\partial }_3\end{array}\right]\left[\begin{array}{cc}{\psi }_0+{\psi }_3& -i{\psi }_2+{\psi }_1\\ i{\psi }_2+{\psi }_1& {\psi }_0-{\psi }_3\end{array}\right]$$

Lemma 1.

$$\partial \left({\psi }_c\right)={(\partial \psi )}_c$$

Proof. 

This follows from associativity of matrix multiplication in $\partial \psi {\sigma }^1$. □

Proposition 13.

If ψ is a Majorana solution, i.e., ${\psi }_c=\psi$ then either $\psi =0$ or the mass $m=0$.

Proof. 

Let $\psi$ be a Majorana solution. We can rewrite Equation (16) as

$$\partial \left({\psi }_c\right)=i\frac{mc}{\hslash }\overline{\left({\psi }_c\right)}{}^{*}$$

Now use the Lemma (1) and the Proposition (12):

$$\begin{array}{c}\hfill {(\partial \psi )}_c=i\frac{mc}{\hslash }\overline{\left({\psi }_c\right)}{}^{*}\\ \hfill {(\partial \psi )}_c=-i\frac{mc}{\hslash }{\left(\overline{\psi }{}^{*}\right)}_c\\ \hfill {(\partial \psi )}_c={(-i\frac{mc}{\hslash }\overline{\psi }{}^{*})}_c\\ \hfill \partial \psi =-i\frac{mc}{\hslash }\overline{\psi }{}^{*}\end{array}$$

Combining this with the original $\partial \psi =i\frac{mc}{\hslash }\overline{\psi }{}^{*}$ proves the proposition (13). □

10. Dirac Mass vs. Majorana Mass

All the previous discussion concerned the solutions of the Dirac Equation. Now consider the Majorana equation which in full matrix form can be written as:

$$\left[\begin{array}{cccc}\frac{\widehat{E}}{c}& 0& -{\widehat{p}}_3& i{\widehat{p}}_2-{\widehat{p}}_1\\ 0& \frac{\widehat{E}}{c}& -i{\widehat{p}}_2-{\widehat{p}}_1& {\widehat{p}}_3\\ {\widehat{p}}_3& -i{\widehat{p}}_2+{\widehat{p}}_1& -\frac{\widehat{E}}{c}& 0\\ i{\widehat{p}}_2+{\widehat{p}}_1& -{\widehat{p}}_3& 0& -\frac{\widehat{E}}{c}\end{array}\right]\left[\begin{array}{cc}a& b^{*}\\ b& -a^{*}\\ c& -d^{*}\\ d& c^{*}\end{array}\right]=\pm mc\left[\begin{array}{cc}-d^{*}& c\\ c^{*}& d\\ b^{*}& a\\ -a^{*}& b\end{array}\right]$$

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The first four row by two column spinor is the same as in the Dirac Equation, and the second is its charge conjugate.

The minus sign cancels in the Dirac Equation, and both spinors belong to the same ray.

However, in the Majorana equation the minus sign does not cancel and is retained to account for possible ambiguity.

In compact form similar to Equation (6) Majorana equation can be written as:

$$\widehat{D}\mathrm{\Psi }=\pm mc{\mathrm{\Psi }}_c$$

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10.1. Biquaternionic Majorana Equation

Following the same procedure as before to convert the equation to biquaternionic form we obtain:

$$\left[\begin{array}{cc}\frac{\widehat{E}}{c}& -\widehat{p}\\ \widehat{p}& -\frac{\widehat{E}}{c}\end{array}\right]\left[\begin{array}{c}iv\\ u\end{array}\right]=\pm mc\left[\begin{array}{c}u{\sigma }^1\\ iv{\sigma }^1\end{array}\right]⟹\left[\begin{array}{cc}{\partial }_0& \nabla \\ -\nabla & -{\partial }_0\end{array}\right]\left[\begin{array}{c}iv\\ u\end{array}\right]=(\pm )(-i)\frac{mc}{\hslash }\left[\begin{array}{c}u{\sigma }^1\\ iv{\sigma }^1\end{array}\right]$$

Rewriting in Pauli algebra variables we obtain two coupled equations:

$$\begin{array}{cc}\hfill {\partial }_0\left(iv\right)+\nabla u& =(\pm )(-i)\frac{mc}{\hslash }u{\sigma }^1\\ \hfill {\partial }_{0}u+\nabla \left(iv\right)& =(\pm )\left(i\right)\frac{mc}{\hslash }iv{\sigma }^1\end{array}$$

and this gives the biquaternionic form of the Majorana equation:

$$\begin{array}{ccc}\hfill \partial \psi & =& (\pm )(-i)\frac{mc}{\hslash }\overline{\psi }{}^{*}{\sigma }^1\hfill \\ \hfill \overline{\partial }\overline{\psi }{}^{*}& =& (\pm )\left(i\right)\frac{mc}{\hslash }\psi {\sigma }^1\hfill \end{array}$$

10.2. Imaginary Mass in the Majorana Equation

Proposition 14.

If the mass of the solution to the biquaternionic Majorana equation is non-zero, then it is an imaginary number.

Proof. 

The proof is by calculating the Klein–Gordon equation satisfied by the scalar components of the Pauli algebra spinor which is the solution of the biquaternionic Majorana equation. It proceeds similarly to that for components of a solution to the DE.

First find the bar-star image of the spinor from the first biquaternionic equation: $\overline{\psi }{}^{*}=(\pm )\left(i\frac{\hslash }{mc}\right)\partial \psi {\sigma }^1$.

Then apply the parity conjugate of the differential: $\overline{\partial }\overline{\psi }{}^{*}=(\pm )\left(i\frac{\hslash }{mc}\right)\overline{\partial }\partial \psi {\sigma }^1$.

Recalling that $\overline{\partial }\partial =\square$ we have: $\overline{\partial }\overline{\psi }{}^{*}=(\pm )\left(i\frac{\hslash }{mc}\right)\square \psi {\sigma }^1=(\pm )\left(i\right)\frac{mc}{\hslash }\psi {\sigma }^1$ from the second equation.

After cancelling we obtain: $\left(\square -\frac{m^2{c}^2}{{\hslash }^2}\right)\psi =0$.

However, the sign in the Klein–Gordon equation should be plus! This is only possible if the mass m is an imaginary number. □

11. Relation to Other Work

There is similarity with the treatment of the Dirac Equation in [6] in the section “The Dirac Equation in space-time algebra” but the essential difference is the use of Pauli algebra and the bar-star automorphism in this article.

12. Conclusions

The main assumption in this article is that it is possible to use the multiplicative structure of the Pauli algebra when working with solutions of the Dirac and Majorana equations. Another assumption is that masses are scalars, not matrices.

With these assumptions, the following conclusions are obtained:

• If a solution of the Dirac Equation coincides with its charge conjugate, then its mass is necessarily zero;
• If a solution of the Majorana equation has non-zero mass, then the mass is an imaginary number.

The first conclusion implies that if the mass of neutrinos are indeed non-zero real scalars, as experimental evidence indicates [10], then the result of the double beta decay experiment will be negative. The second conclusion implies that Majorana particles of non-zero mass are either non-physical or tachions, if such exist.

In other words, a negative result of the double beta decay experiment would be supporting evidence for the multiplicative structure of the spinor space being physical, and not only a mathematical artifact.

Data sharing is not applicable to this article as no new data were created or analyzed in this study.