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Description of the Electron in the Electromagnetic Field: The Dirac Type Equation and the Equation for the Wave Function in Spinor Coordinate Space

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30 July 2024

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31 July 2024

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Abstract
Physical processes are usually described using four-dimensional vector quantities - coordinate vector, momentum vector, current vector. But at the fundamental level they are characterized by spinors - coordinate spinors, momentum spinors, spinor wave functions. The propagation of fields and their interaction takes place at the spinor level, and since each spinor uniquely corresponds to a certain vector, the results of physical processes appear before us in vector form. For example, the relativistic Schrödinger equation and the Dirac equation are formulated by means of coordinate vectors, momentum vectors and quantum operators corresponding to them. In the Schrödinger equation the wave function is represented by a single complex quantity, in the Dirac equation a step forward is taken and the wave function is a spinor with complex components, but still coordinates and momentum are vectors. For a closed description of nature using only spinor quantities, it is necessary to have an equation similar to the Dirac equation in which momentum, coordinates and operators are spinors. It is such an equation that is presented in this paper. Using the example of the interaction between an electron and an electromagnetic field, we can see that the spinor equation contains more detailed information about the interaction than the vector equations. This is not new for quantum mechanics, since it describes interactions using complex wave functions, which cannot be observed directly, and only when measured goes to probabilities in the form of squares of the moduli of the wave functions. In the same way spinor quantities are not observable, but they completely determine observable vectors.In Section 2 of the paper, we analyze the quadratic form for an arbitrary four-component complex vector based on Pauli matrices. The form is invariant with respect to Lorentz transformations including any rotations and boosts. The invariance of the form allows us to construct on its basis an equation for a free particle combining the properties of the relativistic wave equation and the Dirac equation. For an electron in the presence of an electromagnetic potential it is shown that taking into account the commutation relations between the momentum and coordinate components allows us to obtain from this equation the known results describing the interactions of the electron spin with the electric and magnetic field. In section 3 of the paper this quadratic form is expressed through momentum spinors, which makes it possible to obtain an equation for the spinor wave function in spinor coordinate space by replacing the momentum spinor components by partial derivative operators on the corresponding coordinate spinor component. Section 4 presents a modification of the theory of the path integral, which consists in considering the path integral in the spinor coordinate space. The Lagrangian densities for the scalar field and for the electron field, along with their corresponding propagators, are presented. An equation of motion for the electron is proposed that is relativistically invariant, in contrast to the Dirac equation, which lacks this invariance. This novel equation permitted the construction of a relativistically invariant procedure for the second quantization of the fermion field in spinor coordinate space.
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1. Introduction

Nowadays, the interest to study applications of the Dirac equation to different situations and to find out the conditions of its generalization is not weakening. In particular, in [1] new versions of an extended Dirac equation and the associated Clifford algebra are presented. In [2] a study of the Schrödinger-Dirac covariant equation in the presence of gravity, where the non-commuting gamma matrices become space-time-dependent, is carried out. In [3] an idea is discussed that the visible properties of the electron, including rest mass and magnetic moment, are determined by a massless charge spinning at light speed within a Compton domain. In [4] some aspects of conformal rescaling in detail are explored and the role of the "quantum" potential is discussed as a natural consequence of non-inertial motion and is not exclusive to the quantum domain. Author establishes the fundamental importance of conformal symmetry, in which rescaling of the rest mass plays a vital role. Thus, the basis for a radically new theory of quantum phenomena based on the process of mass-energy flow is proposed. In [5] author have derived the covariant fourth-order/one-function equivalent of the Dirac equation for the general case of an arbitrary set of γ-matrices.
Supporting these search aspirations, in our work we propose a deeper understanding of the Dirac equation with an emphasis on the direct use of the principles of symmetry and invariance to Lorentz transformations. For the first time we present a formulation of the Dirac and Schrödinger equations in spinor coordinate space.

2. Generalized Dirac Type Equation

Let us introduce notations, which will be used further on. The speed of light and the rationalized Planck’s constant will be considered as unity.
Pauli matrices
σ 0 = 1 0 0 1           σ 1 = 0 1 1 0           σ 2 = 0 i i 0           σ 3 = 1 0 0 1
Matrices constructed from Pauli matrices
S 0 = σ 0 0 0 σ 0       S 1 = σ 1 0 0 σ 1       S 2 = σ 2 0 0 σ 2       S 3 = σ 3 0 0 σ 3
A vector of matrices
S T ( S 1 , S 2 , S 3 )
A set of arbitrary complex numbers and a vector of its three components
X T ( X 0 , X 1 , X 2 , X 3 )
X T ( X 1 , X 2 , X 3 )
Let us define a 2×2 matrix of Lorentz transformations given by the set of real rotation angles ( α 1 , α 2 , α 3 )  and boosts ( β 1 , β 2 , β 3 )
n = e x p 1 2 i α 1 σ 1 e x p 1 2 β 1 σ 1 e x p 1 2 i α 2 σ 2 e x p 1 2 β 2 σ 2 e x p 1 2 i α 3 σ 3 e x p 1 2 β 3 σ 3
and a similar 4×4 transformation matrix
N = e x p 1 2 i α 1 S 1 e x p 1 2 β 1 S 1 e x p 1 2 i α 2 S 2 e x p 1 2 β 2 S 2 e x p 1 2 i α 3 S 3 e x p 1 2 β 3 S 3
We also define a 4×4 matrix of Lorentz transformations Λ , where μ and ν take values 0,1,2,3
Λ   ν μ = 1 2 T r [ σ μ n σ ν n ]  
which can also be written explicitly using the 4×4 matrices of turn generators ( R 1 , R 2 , R 3 ) and boosts ( K 1 , K 2 , K 3 )
Λ = e x p α 1 R 1 e x p β 1 K 1 e x p α 2 R 2 e x p β 2 K 2 e x p α 3 R 3 e x p β 3 K 3
Let's define a 4×4 matrix
M 2 = S 0 X 0 S 1 X 1 S 2 X 2 S 3 X 3 S 0 X 0 + S 1 X 1 + S 2 X 2 + S 3 X 3 = S 0 X 0 S T X S 0 X 0 + S T X = S 0 X 0 S 0 X 0 S 1 X 1 S 1 X 1 S 2 X 2 S 2 X 2 S 3 X 3 S 3 X 3 + S 0 X 0 S 1 X 1 + S 2 X 2 + S 3 X 3 S 1 X 1 S 0 X 0 + S 2 X 2 + S 3 X 3 S 2 X 2 S 0 X 0 + S 1 X 1 + S 3 X 3 S 3 X 3 S 0 X 0 + S 1 X 1 + S 2 X 2
In fact, we consider a quaternion with complex coefficients, which we multiply by its conjugate quaternion (due to the complexity of the coefficients, these are biquaternions, but we still use quaternionic conjugation, without complex conjugation).
Let us subject the set of complex numbers to the Lorentz transformation
X = Λ X
Let us write a relation whose validity for an arbitrary set of complex numbers can be checked directly
S 0 X 0 S 1 X 1 S 2 X 2 S 3 X 3 S 0 X 0 + S 1 X 1 + S 2 X 2 + S 3 X 3 = S 0 X 0 S 1 X 1 S 2 X 2 S 3 X 3 S 0 X 0 + S 1 X 1 + S 2 X 2 + S 3 X 3 = M 2
The matrix M 2  in the simplest case is diagonal with equal complex elements on the diagonal equal to the square of the length of the vector X  in the metric of Minkowski space, which we denote m 2 . Both M 2  and m 2  do not change under any rotations and boosts, in physical applications the invariance of m 2  is usually used, in particular, for the four-component momentum vector this quantity is called the square of mass.
Since the matrices S  anticommutate with each other, for a vector X  whose components commute with each other, we have just the simplest case with a diagonal matrix with m 2  on the diagonal. But if the components of vector X  do not commute, the matrix M 2  already has a more complex structure and carries additional physical information compared to m 2 . For example, the vector X  may include the electron momentum vector and the electromagnetic potential vector. The four-component potential vector is a function of the four-dimensional coordinates of Minkowski space. The components of the four-component momentum do not commute with the components of the coordinate vector, respectively, and the coordinate function does not commute with the momentum components, and their commutator is expressed through the partial derivative of this function by the corresponding coordinate. If the components of the vector X  do not commute, the matrix M 2  will no longer be invariant with respect to Lorentz transformations.
Suppose that the complex numbers we consider commute with all matrices, and note that the squares of all matrices are equal to the unit 4×4 matrix I
M 2 = ( X 0 X 0 X 1 X 1 X 2 X 2 X 3 X 3 ) I + S 1 X 0 X 1 + S 2 X 0 X 2 + S 3 X 0 X 3 S 1 X 1 X 0 + S 1 S 2 X 1 X 2 + S 1 S 3 X 1 X 3 S 2 X 2 X 0 + S 2 S 1 X 2 X 1 + S 2 S 3 X 2 X 3 S 3 X 3 X 0 + S 3 S 1 X 3 X 1 + S 3 S 2 X 3 X 2 = ( X 0 X 0 X 1 X 1 X 2 X 2 X 3 X 3 ) I + S 1 ( X 0 X 1 X 1 X 0 ) + S 2 ( X 0 X 2 X 2 X 0 ) + S 3 ( X 0 X 3 X 3 X 0 ) S 1 S 2 X 1 X 2 + S 1 S 3 X 1 X 3 S 2 S 1 X 2 X 1 + S 2 S 3 X 2 X 3 S 3 S 1 X 3 X 1 + S 3 S 2 X 3 X 2 = ( X 0 X 0 X 1 X 1 X 2 X 2 X 3 X 3 ) I + S 1 ( X 0 X 1 X 1 X 0 ) + S 2 ( X 0 X 2 X 2 X 0 ) + S 3 ( X 0 X 3 X 3 X 0 ) S 1 S 2 X 1 X 2 + S 2 S 1 X 2 X 1 S 2 S 3 X 2 X 3 + S 3 S 2 X 3 X 2 S 3 S 1 X 3 X 1 + S 1 S 3 X 1 X 3 = ( X 0 X 0 X 1 X 1 X 2 X 2 X 3 X 3 ) I + S 1 ( X 0 X 1 X 1 X 0 ) + S 2 ( X 0 X 2 X 2 X 0 ) + S 3 ( X 0 X 3 X 3 X 0 ) ( S 1 S 2 X 1 X 2 + S 2 S 1 X 1 X 2 + S 2 S 1 ( X 2 X 1 X 1 X 2 ) ) S 2 S 3 X 2 X 3 + S 3 S 2 X 2 X 3 + S 3 S 2 ( X 3 X 2 X 2 X 3 ) S 3 S 1 X 3 X 1 + S 1 S 3 X 3 X 1 + S 1 S 3 ( X 1 X 3 X 3 X 1 )
Taking into account anticommutative properties of matrices and expressions for their pairwise products we obtain
M 2 = ( X 0 X 0 X 1 X 1 X 2 X 2 X 3 X 3 ) I + S 1 ( X 0 X 1 X 1 X 0 ) + S 2 ( X 0 X 2 X 2 X 0 ) + S 3 ( X 0 X 3 X 3 X 0 ) S 2 S 1 ( X 2 X 1 X 1 X 2 ) S 3 S 2 ( X 3 X 2 X 2 X 3 ) S 1 S 3 ( X 1 X 3 X 3 X 1 ) = ( X 0 X 0 X 1 X 1 X 2 X 2 X 3 X 3 ) I + S 1 ( X 0 X 1 X 1 X 0 ) + S 2 ( X 0 X 2 X 2 X 0 ) + S 3 ( X 0 X 3 X 3 X 0 ) + i S 3 ( X 2 X 1 X 1 X 2 ) + i S 1 ( X 3 X 2 X 2 X 3 ) + i S 2 ( X 1 X 3 X 3 X 1 ) = ( X 0 X 0 X 1 X 1 X 2 X 2 X 3 X 3 ) I + S 1 ( X 0 X 1 X 1 X 0 ) + i S 1 ( X 3 X 2 X 2 X 3 ) + S 2 ( X 0 X 2 X 2 X 0 ) + i S 2 ( X 1 X 3 X 3 X 1 ) + S 3 ( X 0 X 3 X 3 X 0 ) + i S 3 ( X 2 X 1 X 1 X 2 )
Consider the case when X  is the sum of the momentum vector and the electromagnetic potential vector, which is a function of coordinates
X = +
T ( P 0 , P 1 , P 2 , P 3 )
T ( A 0 , A 1 , A 2 , A 3 )
P T ( P 1 , P 2 , P 3 )
A T ( A 1 , A 2 , A 3 )
M 2 = I [ ( P 0 + A 0 ) ( P 0 + A 0 ) ( P 1 + A 1 ) ( P 1 + A 1 ) ( P 2 + A 2 ) ( P 2 + A 2 ) ( P 3 + A 3 ) ( P 3 + A 3 ) ] + S 1 [ ( P 0 + A 0 ) ( P 1 + A 1 ) ( P 1 + A 1 ) ( P 0 + A 0 ) ] + i S 1 [ ( P 3 + A 3 ) ( P 2 + A 2 ) ( P 2 + A 2 ) ( P 3 + A 3 ) ] + S 2 [ ( P 0 + A 0 ) ( P 2 + A 2 ) ( P 2 + A 2 ) ( P 0 + A 0 ) ] + i S 2 [ ( P 1 + A 1 ) ( P 3 + A 3 ) ( P 3 + A 3 ) ( P 1 + A 1 ) ] + S 3 [ ( P 0 + A 0 ) ( P 3 + A 3 ) ( P 3 + A 3 ) ( P 0 + A 0 ) ] + i S 3 [ ( P 2 + A 2 ) ( P 1 + A 1 ) ( P 1 + A 1 ) ( P 2 + A 2 ) ]
For now, we'll stick with the Heisenberg approach, that is, we will consider the components of the momentum vector P 0 , P 1 , P 2 , P 3  as operators for which there are commutation relations with coordinates or coordinate functions such as A 0 , A 1 , A 2 , A 3 . In this approach, the operators do not have to act on any wave function.
Taking into account the commutation relations of the components of the momentum vector and the coordinate vector, the commutator of the momentum component and the coordinate function is expressed through the derivative of this function by the corresponding coordinate, e.g.
P 2 + A 2 P 1 + A 1 P 1 + A 1 P 2 + A 2 = P 2 A 1 A 1 P 2 P 1 A 2 A 2 P 1 = i A 1 x 2 i A 2 x 1
As a result, we obtain
M 2 = I P 0 + A 0 P 0 + A 0 P 1 + A 1 P 1 + A 1 P 2 + A 2 P 2 + A 2 P 3 + A 3 P 3 + A 3 + S 1 i A 1 x 0 + i A 0 x 1 + i S 1 i A 2 x 3 + i A 3 x 2 + S 2 i A 2 x 0 + i A 0 x 2 + i S 2 i A 3 x 1 + i A 1 x 3 + S 3 i A 3 x 0 + i A 0 x 3 + i S 3 i A 1 x 2 + i A 2 x 1 = I P 0 + A 0 P 0 + A 0 P 1 + A 1 P 1 + A 1 P 2 + A 2 P 2 + A 2 P 3 + A 3 P 3 + A 3 i S 1 A 1 x 0 A 0 x 1 + S 1 A 2 x 3 A 3 x 2 i S 2 A 2 x 0 A 0 x 2 + S 2 A 3 x 1 A 1 x 3 i S 3 A 3 x 0 A 0 x 3 + S 3 A 1 x 2 A 2 x 1 = I P 0 + A 0 P 0 + A 0 P 1 + A 1 P 1 + A 1 P 2 + A 2 P 2 + A 2 P 3 + A 3 P 3 + A 3 i S 1 F 01 + S 1 F 32 i S 2 F 02 + S 2 F 13 i S 3 F 03 + S 3 F 21 = I P 0 + A 0 P 0 + A 0 P 1 + A 1 P 1 + A 1 P 2 + A 2 P 2 + A 2 P 3 + A 3 P 3 + A 3 i S 1 E x + S 1 B x i S 2 E y + S 2 B y i S 3 E z + S 3 B z
where
F μ ν = μ A ν ν A μ
μ x μ
F μ ν = 0     E x E x     0 E y E z B z B y E y     B z E z   B y 0 B x B x 0
As a result, we have the expression
M 2 = I P 0 + A 0 P 0 + A 0 P 1 + A 1 P 1 + A 1 P 2 + A 2 P 2 + A 2 P 3 + A 3 P 3 + A 3 + S T B i S T E
B T ( B x , B y , B z ) ( B 1 , B 2 , B 3 )
E T ( E x , E y , E z )   ( E 1 , E 2 , E 3 )
Similarly, it can be shown that
S 0 P 0 S 1 P 1 S 2 P 2 S 3 P 3 S 0 A 0 + S 1 A 1 + S 2 A 2 + S 3 A 3 + S 0 A 0 S 1 A 1 S 2 A 2 S 3 A 3 S 0 P 0 + S 1 P 1 + S 2 P 2 + S 3 P 3 = 2 I ( P 0 A 0 P 1 A 1 P 2 A 2 P 3 A 3 ) + S T B i S T E
The matrix
M 2 S T B i S T E = I P 0 + A 0 P 0 + A 0 P 1 + A 1 P 1 + A 1 P 2 + A 2 P 2 + A 2 P 3 + A 3 P 3 + A 3 I d 2
does not change under Lorentz transformations involving any rotations and boosts.
I d 2 = S 0 P 0 + A 0 S 1 P 1 + A 1 S 2 P 2 + A 2 S 3 P 3 + A 3 S 0 P 0 + A 0 + S 1 P 1 + A 1 + S 2 P 2 + A 2 + S 3 P 3 + A 3 S T B i S T E = S 0 P 0 + A 0 S T P + A S 0 P 0 + A 0 + S T P + A S T B i S T E
Taking into account the electron charge we have
X = e
I d 2 = S 0 P 0 e A 0 S T P e A S 0 P 0 e A 0 + S T P e A + e S T B i S T E
Let us summarize our consideration. There is a correlation
I d 2 = M 2 + e S T B i S T E
where
M 2 S 0 P 0 e A 0 S T P e A S 0 P 0 e A 0 + S T P e A
I d 2 I P 0 e A 0 2 P 1 e A 1 2 P 2 e A 2 2 P 3 e A 3 2 = I P 0 e A 0 P 0 e A 0 P 1 e A 1 P 1 e A 1 P 2 e A 2 P 2 e A 2 P 3 e A 3 P 3 e A 3 = I P 0 e A 0 P 0 e A 0 P e A T P e A = I P 0 e A 0 2 P e A 2
Let's analyze the obtained equality
M 2 = I d 2 e S T B i S T E
Note that the quantity d 2  is invariant to the Lorentz transformations irrespective of whether the momentum and field components commute or not. To solve this equation, we have to make additional simplifications. For example, to arrive at an equation similar to the Dirac equation, we must equate M 2  with the matrix I m 2 , where m 2   is the square of the mass of a free electron. Then
I m 2 = I d 2 e S T B i S T E
I d 2 I m 2 e S T B i S T E = 0
I P 0 e A 0 2 P e A 2 I m 2 e S T B i S T E = 0
With this substitution the generalized equation almost coincides with the equation [6], formula (43.25), the difference is that there is a plus sign before   e S T B , and instead of i S T E  there is i α T E , in which the matrices α  have the following form
α T ( α 1 , α 2 , α 3 )
α 1 = 0 σ 1 σ 1 0       α 2 = 0 σ 2 σ 2 0       α 3 = 0 σ 3 σ 3 0
A similar equation is given by Dirac in [7], Para. 76, Equation 24]; he does not use the matrices α , only the matrices S , but the signs of the contributions of the magnetic and electric fields are the same.
Along with the original form
M 2 = S 0 P 0 e A 0 S T P e A S 0 P 0 e A 0 + S T P e A = d 2 e S T B i S T E
it is possible to consider the form with a different order of the factors. It can be shown that this leads to a change in the sign of the electric field contribution
M 2 = S 0 P 0 e A 0 + S T P e A S 0 P 0 e A 0 S T P e A = d 2 e S T B + i S T E
Since I d 2 , unlike M 2 , is invariant to Lorentz transformations, it would be logical to replace it by I m 2 . At least both these matrices are diagonal, and in the case of a weak field their diagonal elements are close. Nevertheless, the approach based on the Dirac equation leads to solutions consistent with experiment.
The matrix M 2  in the general case has complex elements and is not diagonal, and in the Dirac equations instead of it is substituted the product of the unit matrix by the square of mass m 2 , the physical meaning of such a substitution is not obvious. Apparently it is implied that it is the square of the mass of a free electron. But the square of the length of the sum of the lengths of the electron momentum vectors and the electromagnetic potential vector is not equal to the sum of the squares of the lengths of these vectors, that is, it is not equal to the square of the mass of the electron, even if the square of the length of the potential vector were zero. But, for example, in the case of an electrostatic central field, even the square of the length of one potential vector is not equal to zero. Therefore, it is difficult to find a logical justification for using the mass of a free electron in the Dirac equation in the presence of an electromagnetic field. Due to the noted differences, the solutions of the generalized equation can differ from the solutions arising from the Dirac equation.
In the case when there is a constant magnetic field directed along the z-axis, we can write down
A 0 = 0                 A 1 = 1 2 B 3 x 2                   A 2 = 1 2 B 3 x 1             A 3 = 0
S 0 P 0 2 M 2 P e A T P e A I e S 3 B 3 = 0
S 0 P 0 2 M 2 P 1 e A 1 P 1 e A 1 I P 2 e A 2 P 2 e A 2 I e S 3 B 3 = 0
S 0 P 0 2 M 2 P 0 2 I P 3 2 I P 1 2 e A 1 2 P 2 2 e A 2 2 + e 1 2 B 3 x 1   P 2 x 2 P 1 + x 1   P 2 x 2 P 1 e S 3 B 3 = 0
P 0 2 I M 2 P 0 2 I P 3 2 I P 1 2 I e A 1 2 I P 2 2 I e A 2 2 I + e B 3 x 1   P 2 x 2 P 1 I e S 3 B 3 = 0
I P 1 2 P 2 2 P 3 2 e A 1 2 e A 2 2 I M 2 e B 3 L 3 + 1 0 0 L 3 1 0                     0 0                     0 0                       0 0                       0 L 3 + 1 0 0 L 3 1 = 0
Here x 1 P 2 x 2 P 1 L 3 . Only when the field is directed along the z-axis, the matrix M 2  is diagonal and real because the third Pauli matrix is diagonal and real. And if the field is weak, M 2  can be approximated by the m 2 I  matrix. This is probably why it is customary to illustrate the interaction of electron spin with the magnetic field by choosing its direction along the z-axis. In any other direction M 2  is not only non-diagonal, but also complex, so that it is difficult to justify the use of m 2 I .
When the influence of the electromagnetic field was taken into account, no specific characteristics of the electron were used. When deriving a similar result using the Dirac equation, it is assumed that since the electron equation is used, the result is specific to the electron. In our case Pauli matrices and commutation relations are used, apparently these two assumptions or only one of them characterize the properties of the electron, distinguishing it from other particles with non-zero masses.
The proposed equation echoes the Dirac equation, at least from it one can obtain the same formulas for the interaction of spin and electromagnetic field as with the Dirac equation, and in the absence of a field the proposed equation is invariant to the Lorentz transformations. In contrast, to prove the invariance of the Dirac equation even in the absence of a field, the infinitesimal Lorentz transformations are used, but the invariance at finite angles of rotations and boosts is not demonstrated. The proof of invariance of the Dirac equation is based on the claim that a combination of rotations at finite angles can be represented as a combination of infinitesimal rotations. But this is true only for rotations or boosts around one axis, and if there are at least two axes, this statement is not true because of non-commutability of Pauli matrices, which are generators of rotations, so that the exponent of the sum is not equal to the product of exponents if the sum includes generators of rotations or boosts around different axes. By a direct check we can verify that the invariance of the Dirac equation takes place at any combination of rotations, but only under the condition of zero boosts, i.e., only in a rest frame of reference, any boost violates the invariance.
A test case for any theory is the model of the central electrostatic field used in the description of the hydrogen atom, in which the components of the vector potential are zero
S 0 P 0 e A 0 S T P S 0 P 0 e A 0 + S T P = I P 0 e A 0 2 P 1 2 P 2 2 P 3 2 + i e S T E
If again we equate the left part with I m 2 , we obtain
I P 0 e A 0 2 P 1 2 P 2 2 P 3 2 I m 2 + i e S T E = 0
I P 0 e A 0 2 P 1 2 P 2 2 P 3 2 m 2 i e S 1 A 0 x 1 + S 2 A 0 x 2 + S 3 A 0 x 3 = 0
Introducing the notations ( A 0 φ r = Q / r , P 0 E , r = 1 / x 1 2 + x 2 2 + x 3 2 ) ,  we obtain
I E e Q r 2 P 1 2 P 2 2 P 3 2 m 2 i e S 1 φ r x 1 + S 2 φ r x 2 + S 3 φ r x 3 = 0
I E e Q r 2 P 1 2 P 2 2 P 3 2 m 2 + i e Q r 3 S 1 x 1   + S 2 x 2   + S 3 x 3   = 0
If we substitute operators acting on the wave function instead of momentum components into the equation, we obtain a generalized analog of the relativistic Schrödinger equation, in which the wave function has four components and changes as a spinor under Lorentz transformations. Using the substitutions
P 0 i t               P 1 i x 1           P 2 i x 2             P 3 i x 3
the equation for the four-component wave function ψ  before all transformations has the form
S 0 t e A 0 + S T e A S 0 t e A 0 S T e A ψ + M 2 ψ = 0
and after transformations
S 0 P 0 e A 0 2 P e A 2 I e S T B + i e S T E ψ = M 2 ψ
Once again, note that the matrix M 2  is not diagonal and real.
All the above deductions are also valid when replacing 4×4 matrices S μ  by 2×2 matrices σ μ , since their commutative and anticommutative properties are the same. The corresponding generalized equation is of the form
σ 0 P 0 e A 0 2 M 2 P e A 2 I e σ T B + i e σ T E = 0
where
σ T ( σ 1 , σ 2 , σ 3 )
and the equation for the now two-component wave function looks like
σ 0 t e A 0 + σ T e A σ 0 t e A 0 σ T e A ψ + M 2 ψ = 0
In deriving his equation, Dirac [7], paragraph 74 noted that as long as we are dealing with matrices with two rows and columns, we cannot obtain a representation of more than three anticommuting quantities; to represent four anticommuting quantities, he turned to matrices with four rows and columns. In our case, however, three anticommuting matrices are sufficient, so the wave function can also be two-component. Dirac also explains that the presence of four components results in twice as many solutions, half of which have negative energy. In the case of a two-component wave function, however, no negative energy solutions are obtained. Particles with negative energy in this case also exist, but they are described by the same equation in which the signs of all four matrices S  or σ  are reversed.
One would seem to expect similar results from other representations of the momentum operator, e.g., [6] , formula (24.15)]
ω 0 = 1 0 0 0 1 0 0 0 1           ω 1 = 1 2 0 1 0 1 0 1 0 1 0           ω 2 = 1 2 0 i 0 i 0 i 0 i 0           ω 3 = 1 0 0 0 0 0 0 0 1
under the assumption that this representation can describe a particle with spin one. But this expectation is not justified, since the last three matrices do not anticommutate, and therefore the quadratic form constructed on their basis is not invariant under Lorentz transformations.
Let's see what happens to M 2  when we change the sign of the matrices. When changing the sign of S 0  we have
M 2 = S 0 P 0 S P S 0 P 0 + S P = S 0 P 0 + S P S 0 P 0 S P
swapping the places of the multipliers. The multipliers do not necessarily commute, so M 2  is not invariant with respect to the change of sign of S 0 , which can be interpreted as a reflection in time. The same picture of invariance absence we have at the sign change of matrices S , i.e. at spatial reflection
M 2 = S 0 P 0 + S P S 0 P 0 S P
If we change the signs of all matrices at once, we have
M 2 = S 0 P 0 + S P S 0 P 0 S P = S 0 P 0 S P S 0 P 0 + S P
i.e. invariance. The physical interpretation of this case can be given by taking into account the change of signs of the matrices in equation
S 0 P 0 e A 0 2 M 2 P e A 2 I e S T B + i e S T E = 0
which can be rewritten as
S 0 ( P 0 ) ( e ) A 0 2 M 2 P e A 2 I ( e ) S T B + i ( e ) S T E = 0
it can be interpreted as an equation for a particle with negative energy and positive charge, i.e. for the positron. Thus, the generalized equation with matrices S 0 , S 1 , S 2 , S 3 describes a particle, and with matrices S 0 , S 1 , S 2 , S 3  an antiparticle. However, in the next section of the paper another approach is described in which we do not have to deal with negative energy, it is positive for all particles, and the difference between particles and antiparticles consist in a different sign of their mass. And the mass in the generally accepted sense is simply equal to its modulus.
If one consistently adheres to the Heisenberg approach and does not involve the notion of wave function, it is not very clear how to search for solutions of the presented equations. The Schrödinger approach with finding the eigenvalues of the M 2 matrix and their corresponding eigenfunctions can help here.
S 0 P 0 e A 0 2 P e A 2 I e S T B + i e S T E ψ = M 2 ψ
In the left-hand side are the operators acting on the wave function, and in the right-hand side is a constant matrix on which the wave function is simply multiplied. This equality must be satisfied for all values of the four-dimensional coordinates t , x 1 , x 2 , x 3  at once. Then M 2  is not fixed but can take a set of possible values, finding all these values is the goal of solving the equation.
Thus, we have arrived at an equation containing a matrix M 2  which is non-diagonal, complex and in general depends on the coordinates t , x 1 , x 2 , x 3 . After the standard procedure of separating the time and space variables, we can go to a stationary equation in which there will be no time dependence, but the dependence the matrix M 2  on the coordinates will remain. It is possible to ignore the dependence of M 2  on the coordinates and its non-diagonality and simply replace this matrix by a unit matrix with a coefficient in the form of the square of the free electron mass. Then the equation will give solutions coinciding with those of the Dirac equation. But this solution can be considered only approximate and the question remains how far we depart from strict adherence to the principle of invariance with respect to Lorentz transformations and how far we deviate from the hypothetical true solution, which is fully consistent with this principle. To find this solution, we need to approach this equation without simplifying assumptions and look for a set of solutions, each of which represents an eigenvalue matrix M 2  of arbitrary form and its corresponding four-component eigenfunction.
When searching for solutions, one can try to use two equations
S 0 t e A 0 + S T e A S 0 t e A 0 S T e A ψ + M 2 ψ = 0
S 0 t e A 0 S T e A S 0 t e A 0 + S T e A ψ + M 2 ψ = 0
successively applying the operators with first order derivatives included in them to the eigenfunctions already found, similarly as described in Schrödinger's work [8].

3. Equation for the Spinor Coordinate Space

Let us return to the set of arbitrary complex numbers, for simplicity we will call it a vector
X T ( X 0 , X 1 , X 2 , X 3 )
Let us consider in connection with it arbitrary four-component complex spinors
p T ( p 0 , p 1 , p 2 , p 3 )
x 1 T ( x 1 0 , x 1 1 , x 1 2 , x 1 3 )
x 2 T ( x 2 0 , x 2 1 , x 2 2 , x 2 3 )
Among all possible vectors, let us select a set of such vectors for which there is a representation of components through arbitrary complex spinors
X μ = 1 2 x 1 S μ x 2  
and there is another way to calculate them
X μ = 1 2 T r [ x 1 x 2 S μ ]  
Further we will assume that both spinors are identical, then the vector constructed from them is
P T ( P 0 , P 1 , P 2 , P 3 )
has real components, and we will assume that this is the electron momentum vector constructed from the complex momentum spinor p
P μ = 1 2 p S μ p  
P μ = 1 2 T r [ p p S μ ]  
Consider the complex quantity
p T Σ M M x = p 0 , p 1 , p 2 , p 3 0 1 1 0   0 0 0   0     0     0 0   0 0 1 1 0 x 0 x 1 x 2 x 3 = p 0 , p 1 , p 2 , p 3 x 1 x 0 x 3 x 2 = p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
where we introduce one more complex spinor, which in the future we will give the meaning of the complex coordinate spinor
x T ( x 0 , x 1 , x 2 , x 3 )
and
Σ M M = σ M 0 0 σ M = 0 1 1 0   0 0 0   0     0     0 0   0 0 1 1 0       σ M = 0 1 1 0
Coordinate vector of the four-dimensional Minkowski space
X T ( X 0 , X 1 , X 2 , X 3 )
is obtained from the coordinate spinor by the same formulas
X μ = 1 2 x S μ x  
X μ = 1 2 T r [ x x S μ ]  
Thus, the vector in the Minkowski space is not a set of four arbitrary real numbers, but only such that are the specified bilinear combinations of components of completely arbitrary complex spinors
X 0 = 1 2 x 0 ¯ x 0 + x 1 ¯ x 1 + x 2 ¯ x 2 + x 3 ¯ x 3
X 1 = 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 + x 2 ¯ x 3 + x 3 ¯ x 2
X 2 = 1 2 i x 0 ¯ x 1 + i x 1 ¯ x 0 i x 2 ¯ x 3 + i x 3 ¯ x 2
X 3 = 1 2 x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3
Accordingly, the components of the vector in Minkowski space are interdependent, from this dependence automatically follow the relations of the special theory of relativity between space and time. For the same reason, the coordinates of Minkowski space cannot serve as independent variables in the equations.  From the commutative properties of S μ  matrices, which are generators of rotations and boosts with respect to which the length of vectors is invariant, quantum mechanics automatically follows. Indeed, the commutation relations between the components of momenta are related to the noncommutativity of rotations in some way, and from them the commutation relations between the components of coordinates and momenta are directly deduced. And from these relations the differential equations are derived.
And since we do not doubt the truth of the theory of relativity and quantum mechanics, we cannot doubt the reality of spinor space, which by means of the simplest arithmetic operations generates our space and time.
The quantity p T Σ M M x  is invariant under the Lorentz transformation simultaneously applied to the momentum and coordinate spinor, which automatically transforms both corresponding vectors as well
p = N p
P μ = 1 2 T r [ p p S μ ]  
P μ = 1 2 p S μ p  
P = Λ P
x = N x
X μ = 1 2 T r [ x x S μ ]  
X μ = 1 2 x S μ x  
X = Λ X
This quantity does not change for any combination of turns and boosts
p T Σ M M x = p T Σ M M x
Accordingly, the exponent   
exp p T Σ M M x = exp p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
characterizes the propagation process of a plane wave in spinor space with phase invariant to Lorentz transformations.
Let us apply the differential operator to the spinor analog of a plane wave
x 1 x 2 x 0 x 3 exp p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 = p 0 ( p 3 ) ( p 1 ) p 2 exp p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 = = p 1 p 2 p 0 p 3 exp p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
Applying this operator at another definition of the phase gives the same eigenvalue
x 1 x 2 x 0 x 3 exp p 0 x 0 + p 1 x 1 + p 2 x 2 + p 3 x 3 = p 1 p 2 p 0 p 3 exp p 0 x 0 + p 1 x 1 + p 2 x 2 + p 3 x 3
that is, two different eigenfunctions correspond to this eigenvalue, but in the second case the phase in the exponent is not invariant with respect to the Lorentz transformation, so we will use the first definition.
Since  
( p 0 , p 1 ) T a n d ( p 2 , p 3 ) T
are complex spinors, which, under the transformation
p = N p = n 0 0 n p
is affected by the same matrix n , then the complex quantity
m p 1 p 2 p 0 p 3
is invariant under the action on the momentum spinor p  of the transformation N . m  is an eigenvalue of the differential operator, and the plane wave is the corresponding m eigenfunction, which is a solution of the equation
x 1 x 2 x 0 x 3 ψ x 0 , x 1 , x 2 , x 3 = m   ψ ( x 0 , x 1 , x 2 , x 3 )
Here ψ ( x 0 , x 1 , x 2 , x 3 )  denotes the complex function of complex spinor coordinates.
When substantiating the Schrödinger equation for a plane wave in four-dimensional vector space, an assumption is made (further confirmed in the experiment) about its applicability to an arbitrary wave function. Let us make a similar assumption about the applicability of the reduced spinor equation to an arbitrary function of spinor coordinates, that is, we will consider this equation as universal and valid for all physical processes.
Let us clarify that by the derivative on a complex variable from a complex function we here understand the derivative from an arbitrary stepped complex function using the formula that is valid at least for any integer degrees
z k z = k z k 1
In particular, this is true for the exponential function, which is an infinite power series.
It is not by chance that we denote the eigenvalue by the symbol m, because if we form the momentum vector from the momentum spinor p  included in the expression for the plane wave
P μ = 1 2 p S μ p  
then for the square of its length the following equality will be satisfied
P 0 2 P 1 2 P 2 2 P 3 2 = m ¯ m = m 2
That is the square of the modulus m has the sense of the square of the mass of a free particle, which is described by a plane wave in spinor space as well as by a plane wave in vector space. For the momentum spinor of a fermionic type particle having in the rest frame the following form
p T = ( p 0 , p 1 , p 1 ¯ , p 0 ¯ )
Quantity
m = p 1 p 2 p 0 p 3 = p 1 p 1 ¯ + p 0 p 0 ¯
is real and not equal to zero, and for the bosonic-type momentum spinor having in the rest frame the following form
p T = ( p 0 , p 1 , p 0 , p 1 )
it is zero
m = p 1 p 2 p 0 p 3 = p 1 p 0 p 0 p 1 = 0
i.e., the boson satisfies the plane wave equation in spinor space with zero eigenvalue.
For the momentum spinor of a fermion-type particle we can consider another form in the rest system
p T = ( p 0 , p 1 , p 1 ¯ , p 0 ¯ )
then the mass will be real and negative
m = p 1 p 2 p 0 p 3 = p 1 p 1 ¯ p 0 p 0 ¯
This particle with negative mass can be treated as an antiparticle, and in the rest frame its energy is equal to its mass modulo, but it is always positive
P 0 = 1 2 p S 0 p = 1 2 p 0 ¯ p 0 + p 1 ¯ p 1 + ( p 1 ) ( p 1 ¯ ) + p 0 p 0 ¯ = 1 2 p 0 ¯ p 0 + p 1 ¯ p 1 + p 1 p 1 ¯ + p 0 p 0 ¯
To describe the behavior of an electron in the presence of an external electromagnetic field, it is common practice to add the electromagnetic potential vector to its momentum vector. We use the same approach at the spinor level and to each component of the momentum spinor of the electron we add the corresponding component of the electromagnetic potential spinor. For simplicity, the electron charge is equal to unity.
Further we need an expression for the commutation relation between the components of the momentum spinor, to which is added the corresponding component of the electromagnetic potential spinor, which is a function of the spinor coordinates
p 0 + a 0 x 1 , x 2 p 1 + a 1 x 1 , x 2 p 1 + a 1 x 1 , x 2 p 0 + a 0 x 1 , x 2
Let us replace the momenta by differential operators
p 0 x 1             p 1 x 0             p 2 x 3               p 3 x 2
and find the commutation relation
x 1 + a 0 x 0 , x 1 , x 2 , x 3 x 0 + a 1 x 0 , x 1 , x 2 , x 3 x 0 + a 1 x 0 , x 1 , x 2 , x 3 x 1 + a 0 x 0 , x 1 , x 2 , x 3 ψ x 0 , x 1 , x 2 , x 3 = x 1 a 1 ψ a 0 ψ x 0 + x 0 ( a 0 ψ ) a 1 ψ x 1 = a 1 x 1 ψ + a 1 ψ x 1 a 0 ψ x 0 + a 0 x 0 ψ + a 0 ψ x 0 a 1 ψ x 1 = a 1 x 1 ψ + a 0 x 0 ψ = a 1 x 0 , x 1 , x 2 , x 3 x 1 + a 0 x 0 , x 1 , x 2 , x 3 x 0   ψ x 0 , x 1 , x 2 , x 3
Thus
p 0 + a 0 p 1 + a 1 p 1 + a 1 p 0 + a 0 = a 1 x 1 + a 0 x 0
Let us apply the proposed equation to analyze the wave function of the electron in a centrally symmetric electric field, this model is used to describe the hydrogen-like atom. For the components of the vector potential of a centrally symmetric electric field it is true that
A 0 = 1 2 a S 0 a = 1 2 a 0 ¯ a 0 + a 1 ¯ a 1 + a 2 ¯ a 2 + a 3 ¯ a 3 = 1 R
A 1 = 1 2 a S 1 a = 1 2 a 0 ¯ a 1 + a 1 ¯ a 0 + a 2 ¯ a 3 + a 3 ¯ a 2 = 0
A 2 = 1 2 a S 2 a = 1 2 i a 0 ¯ a 1 + i a 1 ¯ a 0 i a 2 ¯ a 3 + i a 3 ¯ a 2 = 0
A 3 = 1 2 a S 3 a = 1 2 a 0 ¯ a 0 a 1 ¯ a 1 + a 2 ¯ a 2 a 3 ¯ a 3 = 0
a 0 ¯ a 0 + a 2 ¯ a 2 = a 1 ¯ a 1 + a 3 ¯ a 3
a 0 ¯ a 0 + a 2 ¯ a 2 = 1 R
a 0 ¯ a 1 + a 2 ¯ a 3 = a 1 ¯ a 0 + a 3 ¯ a 2
1 2 a 0 ¯ a 1 + a 1 ¯ a 0 + a 2 ¯ a 3 + a 3 ¯ a 2 = a 0 ¯ a 1 + a 2 ¯ a 3 = 0
a 0 ¯ a 1 = a 2 ¯ a 3
a 0 ¯ = i a 2 ¯
a 0 = i a 2
a 0 ¯ a 0 + a 2 ¯ a 2 = i a 2 ¯ * i a 2 + a 2 ¯ a 2 = 2 a 2 ¯ a 2 = 2 a 2 2 = 1 R
As a result, it is possible to accept
a 0 = i 2 R                     a 1 = 1 2 R                     a 2 = 1 2 R                       a 3 = i 2 R
a 0 ¯ a 1 = i 1 2 R 1 2 R = i 2 R
a 2 ¯ a 3 = 1 2 R i 1 2 R = i 2 R
R = X 1 2 + X 2 2 + X 3 2 = 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 + x 2 ¯ x 3 + x 3 ¯ x 2 2 + 1 2 i x 0 ¯ x 1 + i x 1 ¯ x 0 i x 2 ¯ x 3 + i x 3 ¯ x 2 2 + 1 2 x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3 2 = 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 + x 2 ¯ x 3 + x 3 ¯ x 2 2 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 x 2 ¯ x 3 + x 3 ¯ x 2 2 + 1 2 x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3 2
We are looking for a solution of the spinor equation; we do not consider the electron's spin yet
x 1 x 2 x 0 x 3 φ x 0 , x 1 , x 2 , x 3 = m   φ x 0 , x 1 , x 2 , x 3
This equation can be interpreted in another way.
Let us take the invariant expression
( p 1 p 2 p 0 p 3 ) = m
And let's do the substitution
p 0 x 1 + a 0 x 0 , x 1 , x 2 , x 3               p 1 x 0 + a 1 x 0 , x 1 , x 2 , x 3
p 2 x 3 + a 2 x 0 , x 1 , x 2 , x 3                 p 3 x 2 + a 3 x 0 , x 1 , x 2 , x 3
x 0 + a 1 x 3 + a 2 x 1 + a 0 x 2 + a 3 φ = m φ
We will consider this equation as an equation for determining the eigenvalues of m and the corresponding eigenfunctions
x 0 x 3 φ + x 1 x 2 φ + a 2 x 0 a 3 x 1 φ a 2 φ x 0 + a 1 φ x 3 a 3 φ x 1 + a 0 φ x 2 + a 1 a 2 a 0 a 3 φ = m φ
a 0 = i 2 R                     a 1 = 1 2 R                     a 2 = 1 2 R                       a 3 = i 2 R
a 1 a 2 a 0 a 3 = 1 2 R + 1 2 R = 1 R
a 2 x 0 a 3 x 1 = 1 2 x 0 1 R + i 1 2 x 1 1 R = 1 2 x 0 1 R 2 4 + i 1 2 x 1 1 R 2 4 = 1 2 1 4 1 R 2 5 4 x 0 R 2 + i 1 2 1 4 1 R 2 5 4 x 1 R 2 = = 1 2 1 4 1 R 2 5 4 x 0 R 2 i x 1 R 2 = 1 2 R 5 x 0 R 2 i x 1 R 2
R = X 1 2 + X 2 2 + X 3 2 = 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 + x 2 ¯ x 3 + x 3 ¯ x 2 2 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 x 2 ¯ x 3 + x 3 ¯ x 2 2 + 1 2 x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3 2
x 0 R 2 = x 0 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 + x 2 ¯ x 3 + x 3 ¯ x 2 2 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 x 2 ¯ x 3 + x 3 ¯ x 2 2 + 1 2 x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3 2 = 1 4 2 x 0 ¯ x 1 + x 1 ¯ x 0 + x 2 ¯ x 3 + x 3 ¯ x 2 x 0 x 0 ¯ x 1 + x 1 ¯ x 0 + x 2 ¯ x 3 + x 3 ¯ x 2 2 x 0 ¯ x 1 + x 1 ¯ x 0 x 2 ¯ x 3 + x 3 ¯ x 2 x 0 x 0 ¯ x 1 + x 1 ¯ x 0 x 2 ¯ x 3 + x 3 ¯ x 2 + 2 x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3 x 0 x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3 = 1 4 2 x 0 ¯ x 1 + x 1 ¯ x 0 + x 2 ¯ x 3 + x 3 ¯ x 2 x 1 ¯ 2 x 0 ¯ x 1 + x 1 ¯ x 0 x 2 ¯ x 3 + x 3 ¯ x 2 x 1 ¯ + 2 x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3 x 0 ¯ = 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 + x 2 ¯ x 3 + x 3 ¯ x 2 x 1 ¯ x 0 ¯ x 1 + x 1 ¯ x 0 x 2 ¯ x 3 + x 3 ¯ x 2 x 1 ¯ + x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3 x 0 ¯ = 1 2 x 0 ¯ x 1 + x 2 ¯ x 3 x 1 ¯ x 0 ¯ x 1 x 2 ¯ x 3 x 1 ¯ + x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3 x 0 ¯ = 1 2 x 0 ¯ x 1 + x 2 ¯ x 3 x 1 ¯ + x 0 ¯ x 1 + x 2 ¯ x 3 x 1 ¯ + x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3 x 0 ¯ = 1 2 x 0 ¯ x 1 + x 2 ¯ x 3 x 1 ¯ + x 2 ¯ x 3 x 1 ¯ + x 0 ¯ x 0 + x 2 ¯ x 2 x 3 ¯ x 3 x 0 ¯ = 1 2 x 0 ¯ x 1 x 1 ¯ + 2 x 2 ¯ x 3 x 1 ¯ + x 0 ¯ x 0 + x 2 ¯ x 2 x 3 ¯ x 3 x 0 ¯ = 1 2 2 x 2 ¯ x 3 x 1 ¯ 2 x 3 ¯ x 3 x 0 ¯ + x 0 ¯ x 0 + x 1 x 1 ¯ + x 2 ¯ x 2 + x 3 ¯ x 3 x 0 ¯ = 1 2 2 x 3 ( x 2 ¯ x 1 ¯ x 3 ¯ x 0 ¯ ) + x 0 ¯ x 0 + x 1 x 1 ¯ + x 2 ¯ x 2 + x 3 ¯ x 3 x 0 ¯ = 1 2 2 x 3 ( x 2 ¯ x 1 ¯ x 3 ¯ x 0 ¯ ) + x 0 ¯ x 0 + x 1 x 1 ¯ + x 2 ¯ x 2 + x 3 ¯ x 3 x 0 ¯
x 1 R 2 = 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 + x 2 ¯ x 3 + x 3 ¯ x 2 x 0 ¯ + x 0 ¯ x 1 + x 1 ¯ x 0 x 2 ¯ x 3 + x 3 ¯ x 2 x 0 ¯ x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3 x 1 ¯ = 1 2 x 1 ¯ x 0 + x 3 ¯ x 2 x 0 ¯ + x 1 ¯ x 0 + x 3 ¯ x 2 x 0 ¯ x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3 x 1 ¯ = 1 2 x 1 ¯ x 0 x 0 ¯ + 2 x 3 ¯ x 2 x 0 ¯ + x 1 ¯ x 1 x 2 ¯ x 2 + x 3 ¯ x 3 x 1 ¯ = 1 2 2 x 2 ( x 3 ¯ x 0 ¯ x 2 ¯ x 1 ¯ ) + x 0 x 0 ¯ + x 1 ¯ x 1 + x 2 ¯ x 2 + x 3 ¯ x 3 x 1 ¯
Let's introduce the notations
x 2 ¯ x 1 ¯ x 3 ¯ x 0 ¯ l
this quantity does not change under rotations and boosts and is some analog of the interval defined for Minkowski space and
1 2 x 0 x 0 ¯ + x 1 ¯ x 1 + x 2 ¯ x 2 + x 3 ¯ x 3 t
this quantity represents time in four-dimensional vector space.
An interesting fact is that time is always a positive quantity. As an assumption it can be noted that since we observe that time value goes forward, i.e. the value of t grows, and it is possible only due to scaling of all components of spinor space, such scaling leads to increase of distance between any two points of Minkowski space. As a result, with the passage of time the Minkowski space should expand, herewith at first relatively quickly, and then more and more slowly.
x 0 R 2 i x 1 R 2 = 1 2 2 x 3 ( x 2 ¯ x 1 ¯ x 3 ¯ x 0 ¯ ) + x 0 ¯ x 0 + x 1 x 1 ¯ + x 2 ¯ x 2 + x 3 ¯ x 3 x 0 ¯ i 1 2 2 x 2 x 3 ¯ x 0 ¯ x 2 ¯ x 1 ¯ + x 0 x 0 ¯ + x 1 ¯ x 1 + x 2 ¯ x 2 + x 3 ¯ x 3 x 1 ¯ = x 3 x 2 ¯ x 1 ¯ x 3 ¯ x 0 ¯ + 1 2 x 0 ¯ x 0 + x 1 x 1 ¯ + x 2 ¯ x 2 + x 3 ¯ x 3 x 0 ¯ i x 2 x 3 ¯ x 0 ¯ x 2 ¯ x 1 ¯ i 1 2 x 0 x 0 ¯ + x 1 ¯ x 1 + x 2 ¯ x 2 + x 3 ¯ x 3 x 1 ¯ = x 3 l + t x 0 ¯ + i x 2 l i t x 1 ¯ = l ( x 3 + i x 2 ) + t ( x 0 ¯ i x 1 ¯ )
As a result, we have an equation for determining the eigenvalues of m and their corresponding eigenfunctions   φ x 0 , x 1 , x 2 , x 3
x 0 x 3 + x 1 x 2 φ + 1 2 R φ x 0 + φ x 3 + i φ x 1 i φ x 2 + 1 2 R 5 l x 3 + i x 2 + t x 0 ¯ i x 1 ¯ φ + 1 R φ = m φ
Instead of looking for solutions to this equation directly, we can first try substituting already known solutions to the Schrödinger equation for the hydrogen-like atom. If φ ( X 0 ,   X 1 , X 2 , X 3 ) is one of these solutions, we need to find its derivatives over all spinor components
φ x µ = φ X ν X ν x µ
X 0 = 1 2 x 0 ¯ x 0 + x 1 ¯ x 1 + x 2 ¯ x 2 + x 3 ¯ x 3
X 1 = 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 + x 2 ¯ x 3 + x 3 ¯ x 2
X 2 = 1 2 i x 0 ¯ x 1 + i x 1 ¯ x 0 i x 2 ¯ x 3 + i x 3 ¯ x 2
X 3 = 1 2 x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3
For example
φ x 0 = φ X 0 x 0 ¯ 2 + φ X 1 x 1 ¯ 2 + φ X 2 i x 1 ¯ 2 + φ X 3 x 0 ¯ 2
Let’s pay attention to the shift in priorities. In the Schrödinger equation one looks for energy eigenvalues, while here it is proposed to look for mass eigenvalues, it seem more natural to us. The mass of a free particle is an invariant of the Lorentz transformations, and in the bound state the mass of the particle has a discrete series of allowed values, each of which corresponds to an energy eigenvalue, and the eigenfunction of these eigenvalues is the same. But these energy eigenvalues are not the same as the energy eigenvalues of the Schrödinger equation, because the equations are different. When an electron absorbs a photon, their spinors sum up and the mass of the electron changes. If the new mass coincides with some allowed value, the electron enters a new state. The kay idea here is the assumption that the interaction of spinors occurs simply by summing them.
The advantages of considering physical processes in spinor coordinate space may not be limited to electrodynamics. It may turn out, for example, that the spinor space is not subject to curvature under the influence of matter, as it takes place in the general theory of relativity for the vector coordinate space. On the contrary, it can be assumed that it is when the components of vector coordinate space are computed from the coordinate spinor that the momentum spinor with a multiplier of the order of the gravitational constant is added to this spinor. This results in a warp that affects other massive bodies.
To account for the electron spin, we will further represent the electron wave function as a four-component spinor function of four-component spinor coordinates
ψ x 0 , x 1 , x 2 , x 3 = ψ 0 x 0 , x 1 , x 2 , x 3 ψ 1 x 0 , x 1 , x 2 , x 3 ψ 2 x 0 , x 1 , x 2 , x 3 ψ 3 x 0 , x 1 , x 2 , x 3 = u 0 u 1 u 2 u 3 φ x 0 , x 1 , x 2 , x 3
where the coefficients u μ are complex quantities independent of coordinates. In fact, as shown at the end of the paper, the wave function is a linear combination of such right-hand sides with operator coefficients.
We will search for the solution of the wave equation considered in the first part of this paper
S 0 P 0 S 1 P 1 S 2 P 2 S 3 P 3 ( S 0 P 0 + S 1 P 1 + S 2 P 2 + S 3 P 3 ) ψ = M 2 ψ
Let's express the left part through the components of the momentum spinor
P μ = 1 2 p S μ p  
P 0 = 1 2 p S 0 p = 1 2 p 0 ¯ , p 1 ¯ , p 2 ¯ , p 3 ¯ 1       0 0     1   0       0 0       0     0       0 0       0 1       0 0     1   p 0 p 1 p 2 p 3 = 1 2 p 0 ¯ , p 1 ¯ , p 2 ¯ , p 3 ¯   p 0 p 1 p 2 p 3 = 1 2 p 0 ¯ p 0 + p 1 ¯ p 1 + p 2 ¯ p 2 + p 3 ¯ p 3
P 1 = 1 2 p S 1 p = 1 2 p 0 ¯ , p 1 ¯ , p 2 ¯ , p 3 ¯ 0       1 1         0   0       0 0       0     0       0 0       0 0       1 1         0   p 0 p 1 p 2 p 3 = 1 2 p 0 ¯ , p 1 ¯ , p 2 ¯ , p 3 ¯   p 1 p 0 p 3 p 2 = 1 2 p 0 ¯ p 1 + p 1 ¯ p 0 + p 2 ¯ p 3 + p 3 ¯ p 2
P 2 = 1 2 p S 2 p = 1 2 p 0 ¯ , p 1 ¯ , p 2 ¯ , p 3 ¯ 0   i i         0   0       0 0       0     0       0 0       0 0   i i         0   p 0 p 1 p 2 p 3 = 1 2 p 0 ¯ , p 1 ¯ , p 2 ¯ , p 3 ¯   i p 1 i p 0 i p 3 i p 2 = 1 2 i p 0 ¯ p 1 + i p 1 ¯ p 0 i p 2 ¯ p 3 + i p 3 ¯ p 2
P 3 = 1 2 p S 3 p = 1 2 p 0 ¯ , p 1 ¯ , p 2 ¯ , p 3 ¯ 1       0 0     1   0       0 0       0     0       0 0       0 1       0 0 1   p 0 p 1 p 2 p 3 = 1 2 p 0 ¯ , p 1 ¯ , p 2 ¯ , p 3 ¯   p 0 p 1 p 2 p 3 = 1 2 p 0 ¯ p 0 p 1 ¯ p 1 + p 2 ¯ p 2 p 3 ¯ p 3
P 0 P 3 = 1 2 p 0 ¯ p 0 + p 1 ¯ p 1 + p 2 ¯ p 2 + p 3 ¯ p 3 1 2 p 0 ¯ p 0 p 1 ¯ p 1 + p 2 ¯ p 2 p 3 ¯ p 3 = 1 2 p 0 ¯ p 0 + p 1 ¯ p 1 + p 2 ¯ p 2 + p 3 ¯ p 3 p 0 ¯ p 0 + p 1 ¯ p 1 p 2 ¯ p 2 + p 3 ¯ p 3 = 1 2 p 1 ¯ p 1 + p 3 ¯ p 3 + p 1 ¯ p 1 + p 3 ¯ p 3 = p 1 ¯ p 1 + p 3 ¯ p 3
P 0 + P 3 = 1 2 p 0 ¯ p 0 + p 1 ¯ p 1 + p 2 ¯ p 2 + p 3 ¯ p 3 + 1 2 p 0 ¯ p 0 p 1 ¯ p 1 + p 2 ¯ p 2 p 3 ¯ p 3 = p 0 ¯ p 0 + p 2 ¯ p 2
P 1 + i P 2 = 1 2 p 0 ¯ p 1 + p 1 ¯ p 0 + p 2 ¯ p 3 + p 3 ¯ p 2 + i 1 2 i p 0 ¯ p 1 + i p 1 ¯ p 0 i p 2 ¯ p 3 + i p 3 ¯ p 2 = 1 2 p 0 ¯ p 1 + p 1 ¯ p 0 + p 2 ¯ p 3 + p 3 ¯ p 2 + p 0 ¯ p 1 p 1 ¯ p 0 + p 2 ¯ p 3 p 3 ¯ p 2 = 1 2 p 0 ¯ p 1 + p 2 ¯ p 3 + p 0 ¯ p 1 + p 2 ¯ p 3 = p 0 ¯ p 1 + p 2 ¯ p 3
P 1 i P 2 = 1 2 p 0 ¯ p 1 + p 1 ¯ p 0 + p 2 ¯ p 3 + p 3 ¯ p 2 i 1 2 i p 0 ¯ p 1 + i p 1 ¯ p 0 i p 2 ¯ p 3 + i p 3 ¯ p 2 = 1 2 p 0 ¯ p 1 + p 1 ¯ p 0 + p 2 ¯ p 3 + p 3 ¯ p 2 p 0 ¯ p 1 + p 1 ¯ p 0 p 2 ¯ p 3 + p 3 ¯ p 2 = 1 2 p 1 ¯ p 0 + p 3 ¯ p 2 + p 1 ¯ p 0 + p 3 ¯ p 2 = p 1 ¯ p 0 + p 3 ¯ p 2
S 0 P 0 S 1 P 1 S 2 P 2 S 3 P 3 = 1       0 0         1   0       0 0       0     0       0 0       0 1       0 0     1 P 0 0       1 1         0   0       0 0       0     0       0 0       0 0       1 1         0 P 1 0   i i         0   0       0 0       0     0       0 0       0 0   i i         0 P 2 1       0 0     1   0       0 0       0     0       0 0       0 1       0 0 1 P 3 = P 0 P 3 P 1 + i P 2 P 1 i P 2         P 0 + P 3   0       0 0       0     0       0 0       0 P 0 P 3 P 1 + i P 2 P 1 i P 2     P 0 + P 3 = = p 1 ¯ p 1 + p 3 ¯ p 3 p 0 ¯ p 1 + p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2         p 0 ¯ p 0 + p 2 ¯ p 2   0       0 0       0     0       0 0       0 p 1 ¯ p 1 + p 3 ¯ p 3 p 0 ¯ p 1 + p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2         p 0 ¯ p 0 + p 2 ¯ p 2
S 0 P 0 + S 1 P 1 + S 2 P 2 + S 3 P 3 = 1       0 0         1   0       0 0       0     0       0 0       0 1       0 0     1 P 0 + 0       1 1         0   0       0 0       0     0       0 0       0 0       1 1         0 P 1 + 0   i i         0   0       0 0       0     0       0 0       0 0   i i         0 P 2 + 1       0 0     1   0       0 0       0     0       0 0       0 1       0 0 1 P 3 = P 0 + P 3 P 1 i P 2 P 1 + i P 2         P 0 P 3   0       0 0       0     0       0 0       0 P 0 + P 3 P 1 i P 2 P 1 + i P 2         P 0 P 3 = p 0 ¯ p 0 + p 2 ¯ p 2 p 0 ¯ p 1 p 2 ¯ p 3 p 1 ¯ p 0 p 3 ¯ p 2         p 1 ¯ p 1 + p 3 ¯ p 3   0       0 0       0     0       0 0       0 p 0 ¯ p 0 + p 2 ¯ p 2 p 0 ¯ p 1 p 2 ¯ p 3 p 1 ¯ p 0 p 3 ¯ p 2         p 1 ¯ p 1 + p 3 ¯ p 3
Let's distinguish the direct products of vectors in these matrices
S 0 P 0 + S 1 P 1 + S 2 P 2 + S 3 P 3 = p 0 ¯ p 0 + p 2 ¯ p 2 p 0 ¯ p 1 p 2 ¯ p 3 p 1 ¯ p 0 p 3 ¯ p 2         p 1 ¯ p 1 + p 3 ¯ p 3   0       0 0       0     0       0 0       0 P 0 + P 3 P 1 i P 2 P 1 + i P 2       P 0 P 3 = p 0 ¯ p 0 p 0 ¯ p 1 p 1 ¯ p 0         p 1 ¯ p 1   0       0 0       0     0       0 0       0 p 0 ¯ p 0 p 0 ¯ p 1 p 1 ¯ p 0         p 1 ¯ p 1 + p 2 ¯ p 2 p 2 ¯ p 3 p 3 ¯ p 2         p 3 ¯ p 3   0       0 0       0     0       0 0       0 p 2 ¯ p 2 p 2 ¯ p 3 p 3 ¯ p 2         p 3 ¯ p 3
= p 0 ¯ p 1 ¯ 0 0 p 0 , p 1 , 0 , 0 + 0   0 p 0 ¯ p 1 ¯ 0,0 , p 0 , p 1 + p 2 ¯ p 3 ¯ 0 0 p 2 , p 3 , 0 , 0 + 0   0 p 2 ¯ p 3 ¯ 0,0 , p 2 , p 3
S 0 P 0 S 1 P 1 S 2 P 2 S 3 P 3 = p 1 ¯ p 1 + p 3 ¯ p 3   p 0 ¯ p 1 + p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2         p 0 ¯ p 0 + p 2 ¯ p 2   0       0 0       0     0       0 0       0 p 1 ¯ p 1 + p 3 ¯ p 3   p 0 ¯ p 1 + p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2         p 0 ¯ p 0 + p 2 ¯ p 2 = p 1 ¯ p 1   p 0 ¯ p 1 p 1 ¯ p 0       p 0 ¯ p 0   0       0 0       0     0       0 0       0 p 1 ¯ p 1   p 0 ¯ p 1 p 1 ¯ p 0       p 0 ¯ p 0 + p 3 ¯ p 3   p 2 ¯ p 3 p 3 ¯ p 2       p 2 ¯ p 2   0       0 0       0     0       0 0       0 p 3 ¯ p 3   p 2 ¯ p 3 p 3 ¯ p 2       p 2 ¯ p 2 = p 1 p 1 ¯ p 1 p 1 ¯ p 1 ¯ p 1 p 1 p 0 ¯ p 1 p 0 ¯ p 0 ¯ p 1 p 0 p 1 ¯ p 0 p 1 ¯ p 1 ¯ p 0     p 0 p 0 ¯ p 0 p 0 ¯ p 0 ¯ p 0   0       0 0       0     0       0 0       0 p 1 p 1 ¯ p 1 p 1 ¯ p 1 ¯ p 1 p 1 p 0 ¯ p 1 p 0 ¯ p 0 ¯ p 1 p 0 p 1 ¯ p 0 p 1 ¯ p 1 ¯ p 0     p 0 p 0 ¯ p 0 p 0 ¯ p 0 ¯ p 0 + p 3 p 3 ¯ p 3 p 3 ¯ p 3 ¯ p 3 p 3 p 2 ¯ p 3 p 2 ¯ p 2 ¯ p 3 p 2 p 3 ¯ p 2 p 3 ¯ p 3 ¯ p 2     p 2 p 2 ¯ p 2 p 2 ¯ p 2 ¯ p 2   0       0 0       0     0       0 0       0 p 3 p 3 ¯ p 3 p 3 ¯ p 3 ¯ p 3 p 3 p 2 ¯ p 3 p 2 ¯ p 2 ¯ p 3 p 2 p 3 ¯ p 2 p 3 ¯ p 3 ¯ p 2     p 2 p 2 ¯ p 2 p 2 ¯ p 2 ¯ p 2
= p 1   p 0 0 0 p 1 ¯ , p 0 ¯ , 0 , 0 + 0   0 p 1 p 0 0,0 , p 1 ¯ , p 0 ¯ p 1 p 1 ¯ p 1 ¯ p 1 p 1 p 0 ¯ p 0 ¯ p 1 p 0 p 1 ¯ p 1 ¯ p 0     p 0 p 0 ¯ p 0 ¯ p 0   0       0 0       0     0       0 0       0 p 1 p 1 ¯ p 1 ¯ p 1 p 1 p 0 ¯ p 0 ¯ p 1 p 0 p 1 ¯ p 1 ¯ p 0     p 0 p 0 ¯ p 0 ¯ p 0 + p 3   p 2 0 0 p 3 ¯ , p 2 ¯ , 0 , 0 + 0   0 p 3 p 2 0,0 , p 3 ¯ , p 2 ¯ p 3 p 3 ¯ p 3 ¯ p 3 p 3 p 2 ¯ p 2 ¯ p 3 p 2 p 3 ¯ p 3 ¯ p 2     p 2 p 2 ¯ p 2 ¯ p 2   0       0 0       0     0       0 0       0 p 3 p 3 ¯ p 3 ¯ p 3 p 3 p 2 ¯ p 2 ¯ p 3 p 2 p 3 ¯ p 3 ¯ p 2     p 2 p 2 ¯ p 2 ¯ p 2
Let's introduce the notations
p 0 ¯ p 1 ¯ 0 0 p 0 , p 1 , 0 , 0 + 0   0 p 0 ¯ p 1 ¯ 0,0 , p 0 , p 1 + p 2 ¯ p 3 ¯ 0 0 p 2 , p 3 , 0 , 0 + 0   0 p 2 ¯ p 3 ¯ 0,0 , p 2 , p 3 S +
p 1   p 0 0 0 p 1 ¯ , p 0 ¯ , 0 , 0 + 0   0 p 1 p 0 0,0 , p 1 ¯ , p 0 ¯ + p 3   p 2 0 0 p 3 ¯ , p 2 ¯ , 0 , 0 + 0   0 p 3 p 2 0,0 , p 3 ¯ , p 2 ¯ S
p 1 p 1 ¯ p 1 ¯ p 1 p 1 p 0 ¯ p 0 ¯ p 1 p 0 p 1 ¯ p 1 ¯ p 0     p 0 p 0 ¯ p 0 ¯ p 0   0       0 0       0     0       0 0       0 p 1 p 1 ¯ p 1 ¯ p 1 p 1 p 0 ¯ p 0 ¯ p 1 p 0 p 1 ¯ p 1 ¯ p 0     p 0 p 0 ¯ p 0 ¯ p 0 + p 3 p 3 ¯ p 3 ¯ p 3 p 3 p 2 ¯ p 2 ¯ p 3 p 2 p 3 ¯ p 3 ¯ p 2     p 2 p 2 ¯ p 2 ¯ p 2   0       0 0       0     0       0 0       0 p 3 p 3 ¯ p 3 ¯ p 3 p 3 p 2 ¯ p 2 ¯ p 3 p 2 p 3 ¯ p 3 ¯ p 2     p 2 p 2 ¯ p 2 ¯ p 2 K
Let us substitute differential operators instead of spinor components
p 0 x 1   1           p 1 x 0 0             p 2 x 3   3             p 3 x 2 2
p 0 ¯ [   ] ¯ x 1 ¯ 1 ¯             p 1 ¯   ¯ x 0 ¯ 0 ¯             p 2 ¯   ¯ x 3 ¯ 3 ¯               p 3 ¯   ¯ x 2 ¯ 2 ¯
Then the quantities included in the wave equation
( S K ) S + ψ x 0 , x 1 , x 2 , x 3 = M 2 ψ x 0 , x 1 , x 2 , x 3
will have the form
S = 0 1 0 0 0 ¯ , 1 ¯ , 0 , 0 + 0   0 0 1 0,0 , 0 ¯ , 1 ¯ + 2   3 0 0 2 ¯ , 3 ¯ , 0 , 0 + 0   0 2 3 0,0 , 2 ¯ , 3 ¯
S + = 1 ¯ 0 ¯ 0 0 1 , 0 , 0 , 0 + 0   0 1 ¯ 0 ¯ 0,0 , 1 , 0 + 3 ¯ 2 ¯ 0 0 3 , 2 , 0 , 0 + 0   0 3 ¯ 2 ¯ 0,0 , 3 , 2
K = = 0 0 ¯ 0 ¯ 0 0 1 ¯ 1 ¯ 0 1 0 ¯ 0 ¯ 1   1 1 ¯ 1 ¯ 1   0       0 0       0     0       0 0       0 0 0 ¯ 0 ¯ 0 0 1 ¯ 1 ¯ 0 1 0 ¯ 0 ¯ 1   1 1 ¯ 1 ¯ 1 + 2 2 ¯ 2 ¯ 2 2 3 ¯ 3 ¯ 2 3 2 ¯ 2 ¯ 3   3 3 ¯ 3 ¯ 3   0       0 0       0     0       0 0       0 2 2 ¯ 2 ¯ 2 2 3 ¯ 3 ¯ 2 3 2 ¯ 2 ¯ 3   3 3 ¯ 3 ¯ 3
Let us consider the case of a free particle and represent the electron field as a four-component spinor function of four-component spinor coordinates
ψ x 0 , x 1 , x 2 , x 3 = ψ 0 x 0 , x 1 , x 2 , x 3 ψ 1 x 0 , x 1 , x 2 , x 3 ψ 2 x 0 , x 1 , x 2 , x 3 ψ 3 x 0 , x 1 , x 2 , x 3 = u 0 u 1 u 2 u 3 φ x 0 , x 1 , x 2 , x 3
For a free particle, the components of the momentum spinor commute with each other, so all components of the matrix K are zero.
Let us use the model of a plane wave in spinor space
φ x 0 , x 1 , x 2 , x 3 = exp p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
Substituting the plane wave solution into the differential equation, we obtain the algebraic equation
S S + u 0 u 1 u 2 u 3 φ x 0 , x 1 , x 2 , x 3 = M 2 u 0 u 1 u 2 u 3 φ x 0 , x 1 , x 2 , x 3
S p 0 ¯ p 1 ¯ 0 0 p 0 u 0 + p 1 u 1 + 0   0 p 0 ¯ p 1 ¯ p 0 u 2 + p 1 u 3 + p 2 ¯ p 3 ¯ 0 0 p 2 u 0 + p 3 u 1 + 0   0 p 2 ¯ p 3 ¯ p 2 u 2 + p 3 u 3 φ x 0 , x 1 , x 2 , x 3 = m 2 u 0 u 1 u 2 u 3 φ x 0 , x 1 , x 2 , x 3
p 1   p 0 0 0 p 1 ¯ , p 0 ¯ , 0 , 0 + 0   0 p 1 p 0 0,0 , p 1 ¯ , p 0 ¯ + p 3   p 2 0 0 p 3 ¯ , p 2 ¯ , 0 , 0 + 0   0 p 3 p 2 0,0 , p 3 ¯ , p 2 ¯
p 0 ¯ p 1 ¯ 0 0 p 0 u 0 + p 1 u 1 + 0   0 p 0 ¯ p 1 ¯ p 0 u 2 + p 1 u 3 + p 2 ¯ p 3 ¯ 0 0 p 2 u 0 + p 3 u 1 + 0   0 p 2 ¯ p 3 ¯ p 2 u 2 + p 3 u 3 φ x 0 , x 1 , x 2 , x 3 = M 2 u 0 u 1 u 2 u 3 φ x 0 , x 1 , x 2 , x 3
p 1   p 0 0 0 p 1 ¯ , p 0 ¯ , 0 , 0 + 0   0 p 1 p 0 0,0 , p 1 ¯ , p 0 ¯ + p 3   p 2 0 0 p 3 ¯ , p 2 ¯ , 0 , 0 + 0   0 p 3 p 2 0,0 , p 3 ¯ , p 2 ¯
p 0 ¯ p 1 ¯ 0 0 p 0 u 0 + p 1 u 1 + 0   0 p 0 ¯ p 1 ¯ p 0 u 2 + p 1 u 3 + p 2 ¯ p 3 ¯ 0 0 p 2 u 0 + p 3 u 1 + 0   0 p 2 ¯ p 3 ¯ p 2 u 2 + p 3 u 3 = M 2 u 0 u 1 u 2 u 3
p 1   p 0 0 0 p 1 ¯ p 0 ¯ + p 0 ¯ p 1 ¯ p 0 u 0 + p 1 u 1 + p 3   p 2 0 0 p 3 ¯ p 0 ¯ + p 2 ¯ p 1 ¯ p 0 u 0 + p 1 u 1 + 0   0 p 1 p 0 p 1 ¯ p 0 ¯ + p 0 ¯ p 1 ¯ p 0 u 2 + p 1 u 3 + 0   0 p 3 p 2 p 3 ¯ p 0 ¯ + p 2 ¯ p 1 ¯ p 0 u 2 + p 1 u 3 + p 1   p 0 0 0 p 1 ¯ p 2 ¯ + p 0 ¯ p 3 ¯ p 2 u 0 + p 3 u 1 + p 3   p 2 0 0 p 3 ¯ p 2 ¯ + p 2 ¯ p 3 ¯ p 2 u 0 + p 3 u 1 + 0   0 p 1 p 0 p 1 ¯ p 2 ¯ + p 0 ¯ p 3 ¯ p 2 u 2 + p 3 u 3 + 0   0 p 3 p 2 p 3 ¯ p 2 ¯ + p 2 ¯ p 3 ¯ p 2 u 2 + p 3 u 3 = M 2 u 0 u 1 u 2 u 3
Let us take into account the commutativity of the momentum components, besides, let us introduce the notations
p 3 ¯ p 0 ¯ + p 2 ¯ p 1 ¯ m ¯             p 1 ¯ p 2 ¯ + p 0 ¯ p 3 ¯ m ¯            
for the quantities which are invariant under any rotations and boosts, then we obtain
p 3   p 2 0 0 m ¯ p 0 u 0 + p 1 u 1 + 0   0 p 3 p 2 m ¯ p 0 u 2 + p 1 u 3 + p 1   p 0 0 0 ( m ¯ ) p 2 u 0 + p 3 u 1 + 0   0 p 1 p 0 ( m ¯ ) p 2 u 2 + p 3 u 3 = M 2 u 0 u 1 u 2 u 3
u 0 p 3   p 2 0 0 m ¯ p 0 + p 1   p 0 0 0 m ¯ p 2 + u 1 p 3   p 2 0 0 m ¯ p 1 p 1   p 0 0 0 m ¯ p 3 + u 2 0   0 p 3 p 2 m ¯ p 0 + 0   0 p 1 p 0 m ¯ p 2 + u 3 0   0 p 3 p 2 m ¯ p 1 0   0 p 1 p 0 m ¯ p 3 = M 2 u 0 u 1 u 2 u 3
u 0 m ¯ p 1 p 2 p 3 p 0   p 0 p 2 p 2 p 0 0 0 + u 1 m ¯ p 3 p 1 p 1 p 3   p 2 p 1 p 0 p 3 0 0 + u 2 m ¯ 0   0 p 1 p 2 p 3 p 0 p 0 p 2 p 2 p 0 + u 3 m ¯ 0   0 p 3 p 1 p 1 p 3 p 2 p 1 p 0 p 3 = M 2 u 0 u 1 u 2 u 3
Additionally, introducing notation for Lorentz invariant quantities
p 1 p 2 p 3 p 0 m             p 2 p 1 p 0 p 3 m
we obtain
u 0 m ¯ m 0 0 0 + u 1 m ¯ 0   m 0 0 + u 2 m ¯ 0   0 m 0 + u 3 m ¯ 0   0 0 m = m 2 u 0 u 1 u 2 u 3
u 0 m 2 0 0 0 + u 1 0   m 2 0 0 + u 2 0   0 m 2 0 + u 3 0   0 0 m 2 = m 2 u 0 u 1 u 2 u 3
m 2   0   0     m 2   0           0   0                 0     0         0   0             0   m 2     0   0     m 2 u 0 u 1 u 2 u 3 = M 2 u 0 u 1 u 2 u 3
We see that in the case of a plane wave in spinor space, the matrix in the left part of the equation is diagonal and remains so at any rotations and boosts, the diagonal element also does not change.
In this case we can consider the matrix M 2 in the right part to be diagonal with the same elements on the diagonal m 2 , then the equation can be rewritten as an equation for the problem of finding eigenvalues and eigenfunctions
S S + ψ x 0 , x 1 , x 2 , x 3 = m 2 I ψ x 0 , x 1 , x 2 , x 3
S S + ψ x 0 , x 1 , x 2 , x 3 = m 2 ψ x 0 , x 1 , x 2 , x 3
Let us compare our equation with the Dirac equation [6], formula (43.16)]
P 0 + M   0   0       P 0 + M   P 3       P 1 i P 2 P 1 + i P 2       P 3     P 3       P 1 i P 2 P 1 + i P 2       P 3 P 0 M   0 0     P 0 M u 0 u 1 u 2 u 3 = 0
In the rest frame of reference, the three components of momentum are zero and the equation is simplified
P 0 + M           0   0       P 0 + M   0                             0 0                               0     0                             0 0                               0   P 0 M           0 0     P 0 M u 0 u 1 u 2 u 3 = 0
That is, in the rest frame the Dirac equation and the spinor equation analyzed by us look identically and contain a diagonal matrix. The corresponding problem on eigenvalues and eigenvectors of these matrices has degenerate eigenvalues, which correspond to the linear space of eigenfunctions. In this space, one can choose an orthogonal basis of linearly independent functions, and this choice is quite arbitrary. For example, in [9], formula (2.127)], solutions in the form of plane waves in the vector space have been proposed for the Dirac equation in the rest frame
u i ( 0 ) exp i M t
v i ( 0 ) exp + i M t
and the following spinors are chosen as basis vectors
u 1 0 = 1 0 0 0         u 2 0 = 0 1 0 0           v 1 0 = 0 0 1 0             v 2 0 = 0 0 0 1
For transformation to a moving coordinate system in [9], formula (2.133)] the following formula is used
ψ i ( X ) = u i ( P ) exp i P X
ψ i ( X ) = v i ( P ) exp + i P X
Where
u 1 P = P 0 + M 2 M 1 0 P 3 P 0 + M P 1 + i P 2 P 0 + M       u 2 P = P 0 + M 2 M 1 0 P 1 i P 2 P 0 + M P 3 P 0 + M       v 1 P = P 0 + M 2 M P 3 P 0 + M P 1 + i P 2 P 0 + M 1 0        
v 2 P = P 0 + M 2 M P 1 i P 2 P 0 + M P 3 P 0 + M 0 1
The basis spinors form a complete system, that is, any four-component complex spinor can be represented as their linear combination and this arbitrary spinor will be a solution to the problem on eigenvalues and eigenfunctions in a resting coordinate system. The choice of the given particular basis has disadvantages, because if to find a four-dimensional current vector from any of these basis functions
j μ = 1 2 u 1 0 S μ u 1 0  
then this current in the rest frame of reference
j T = 1 2 , 0 , 0 , 1 2
has non-zero components, and the square of the length of the current vector is zero. It turns out that a resting electron creates a current, which contradicts physical common sense. Since we have freedom of choice of the basis, it is reasonable to choose the spinor for the wave function as some set of momentum spinor components, for example
u 0 = e m p 2 p 3 p 0 p 1
The proportionality factor is chosen so that in the rest frame the zero component of the current is equal to the charge of, for example, an electron or a positron. If the momentum spinor in the rest frame has the form
p T = ( p 0 , p 1 , p 1 ¯ , p 0 ¯ )
then the momentum vector in this rest frame of reference will be
P T = ( m , 0 , 0 , 0 )
and the current vector
j T = ( e , 0 , 0 , 0 )
The same momentum vector in the rest frame of reference can be obtained from different spinors, e.g,
p 1 = 1 0 0 1           p 2 = 0 1 1 0           p 3 = 1 0 0 1         p 4 = 0 1 1 0 P = 1 0 0 0
after a 30-degree boost along the z-axis we get
p 1 = 1.299 0 0 0.77           p 2 = 0 0.77 1.299 0           p 3 = 1.299 0 0 0.77         p 4 = 0 0.77 1.299 0 P = 1.14 0 0 0.548
After scaling the spinors by the factor e m , similar relations are true for the current vector. Thus, electrons can have the same momentum and current vector but different spinors, i.e., they are characterized by different spins. As it is supposed, the electron here has two physical degrees of freedom, since in a rest frame of reference one can choose the components p 0 and p 1 to be real.
Thus defined spinor wave function for a free particle is invariant to Lorentz transformations, since in this case the mass of electron m = p 1 p 2 p 3 p 0 , its charge and the phase of the plane spinor wave
exp p T Σ M M x = exp p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
do not change at rotations and boosts. The matrix on the left side of the equation does not change either, remaining diagonal with m 2 on the diagonal.
It is logical to use the same considerations when choosing the basis for the wave function of the photon, whose mass, i.e., the eigenvalues of the wave function equation, are also degenerate and thus equal to zero. In this case, the choice of the proportionality factor between the spinor of the wave function and the momentum spinor is not so obvious, one can, for example, consider the option of
u 0 = e p 2 p 3 p 2 p 3
For a fermion, which can be an electron or a positron in the rest frame takes place p T = ( p 0 , p 1 , p 1 ¯ , p 0 ¯ ) , so the quantity
m = p 1 p 2 p 3 p 0 = p 1 p 1 ¯ + p 0 ¯ p 0
which, unlike the mass M in the Dirac equation, is complex in the general case, is also real for the fermion and can be positive for the electron or negative for the positron. The charge is proportional to the mass with a minus sign, since the electron charge is considered negative and the positron positive. For simplicity it is possible to consider the mass of the electron as negative and that of the positron as positive, then the charge will be proportional to the mass without changing the sign.
For the momentum spinor of a boson, such as a photon, it is true that p T = ( p 0 , p 1 , p 0 , p 1 ) , so its mass is zero
m = p 1 p 2 p 3 p 0 = p 1 p 0 p 1 p 0 = 0
The given constructions are not abstract, but describe the physical reality, since the results of the processes occurring in the spinor space are displayed in the Minkowski vector space. In particular, the momentum vector corresponding to the momentum spinor has the following parameters
P μ = 1 2 T r [ p p S μ ]  
the square of the length is equal to the square of the mass of the electron or positron
P 0 2 P 1 2 P 2 2 P 3 2 = m 2
And to the spinor wave function ψ x 0 , x 1 , x 2 , x 3 at some point in spinor space corresponds the vector wave function Ψ ( X 0 , X 1 , X 2 , X 3 )
Ψ μ = 1 2 T r [ ψ ψ S μ ]   (which for a plane wave coincides with the current vector), taking its value in the corresponding point of physical space with coordinates
X μ = 1 2 T r [ x x S μ ]   The vector wave function Ψ can be compared in meaning to the square of the modulus of the conventional scalar wave function, in particular Ψ 0 is equal to this square and has the meaning of probability. The conventional scalar wave function itself is closer in meaning to the spinor wave function considered here, they both have complex values, and the four-component wave functions of the electron have in both cases the same meaning.
The arbitrary choice of the basis of the linear space of the eigenvectors of the matrix takes place only for a free particle. In the general case the matrix K is not zero, the wave equation has no solution in the form of plane waves in spinor space and ceases to be invariant with respect to Lorentz transformations, and the eigenvalues become nondegenerate.
We propose to extend the scope of applicability of the presented equation consisting of differential operators in the form of partial derivatives on the components of coordinate spinors with a nonzero matrix K
( S K ) S + ψ x 0 , x 1 , x 2 , x 3 = M 2 ψ x 0 , x 1 , x 2 , x 3
not only to the case of a plane wave, but to any situation in general. This transition is analogous to the transition from the application of the Schrödinger equation to a plane wave in vector space to its application in a general situation. The legitimacy of such transitions should be confirmed by the results of experiments.
This equation will be called the equation for the spinor wave function defined on the spinor coordinate space. Here the matrix M 2 is, generally speaking, neither diagonal nor real, but it does not depend on the coordinates and is determined solely by the parameters of the electromagnetic field. Only in the case of a plane wave it is diagonal and has on the diagonal the square of the mass of the free particle. We can try to simplify the problem and require that the matrix M 2   is diagonal with the same elements on the diagonal m 2 , then the equation can be rewritten in the form of the equation for the problem of search of eigenvalues and eigenfunctions for any quantum states
( S K ) S + ψ x 0 , x 1 , x 2 , x 3 = m 2 ψ x 0 , x 1 , x 2 , x 3
This approach is pleasant in the Dirac equation, where the mass is fixed and equated to the mass of a free particle, and at the same time results giving good agreement with experiment are obtained.
We are of the opinion that the spinor equation is more fundamental than the relativistic Schrödinger and Dirac equations, it is not a generalization of them, it is a refinement of them, because it describes nature at the spinor level, and hence is more precise and detailed than the equations for the wave function defined on the vector space.
Let us consider the proposed equation for the special case when the particle is in an external electromagnetic field, which we will also represent by a four-component spinor function at a point of the spinor coordinate space
a x 0 , x 1 , x 2 , x 3 = a 0 x 0 , x 1 , x 2 , x 3 a 1 x 0 , x 1 , x 2 , x 3 a 2 x 0 , x 1 , x 2 , x 3 a 3 x 0 , x 1 , x 2 , x 3
We will apply to the wave function of the electron the operators corresponding to the components of the momentum spinor, putting for simplicity the electron charge equal to unity
p 0 x 1 + a 0 x 0 , x 1 , x 2 , x 3                               p 1 x 0 + a 1 x 0 , x 1 , x 2 , x 3
p 2 x 3 + a 2 x 0 , x 1 , x 2 , x 3                               p 3 x 2 + a 3 x 0 , x 1 , x 2 , x 3
p 0 ¯ [   ] ¯ x 1 ¯ + a 0 x 0 , x 1 , x 2 , x 3 ¯                                         p 1 ¯ [   ] ¯ x 0 ¯ + a 1 x 0 , x 1 , x 2 , x 3 ¯            
p 2 ¯ [   ] ¯ x 3 ¯ + a 2 x 0 , x 1 , x 2 , x 3 ¯                                           p 3 ¯ [   ] ¯ x 2 ¯ + a 3 x 0 , x 1 , x 2 , x 3 ¯            
Note that the electromagnetic potential vector can be calculated from the electromagnetic potential spinor by the standard formula
A μ = 1 2 a S μ a  
The advantage of the spinor description over the vector description is that instead of summing up the components of the momentum and electromagnetic potential vectors as is usually done
P μ + A μ = 1 2 p S μ p + 1 2 a S μ a  
now we sum the spinor components and then the resulting vector is
1 2 p + a S μ p + a = 1 2 p S μ p + 1 2 p S μ a + 1 2 a S μ p + 1 2 a S μ a
in addition to the usual momentum and field vectors, contains an additional term
1 2 p S μ a + 1 2 a S μ p
taking real values and describing the mutual influence of the fields of the electron and photon.
After the addition of the electromagnetic field the components of the momentum spinor do not commute, the corresponding commutators are found above
x 1 + a 0 x 0 + a 1 x 0 + a 1 x 1 + a 0 φ = a 1 x 1 + a 0 x 0 φ
x 3 + a 2 x 2 + a 3 x 2 + a 3 x 3 + a 2 φ = a 3 x 3 + a 2 x 2 φ
Let's find commutators for other operators
[   ] ¯ x 1 ¯ + a 0 ¯ [   ] ¯ x 0 ¯ + a 1 ¯ [   ] ¯ x 0 ¯ + a 1 ¯ [   ] ¯ x 1 ¯ + a 0 ¯ φ = [   ] ¯ x 1 ¯ + a 0 ¯ [   ] ¯ x 0 ¯ + a 1 ¯ φ   ¯ x 0 ¯ + a 1 ¯   ¯ x 1 ¯ + a 0 ¯ φ = =   ¯ x 1 ¯ + a 0 ¯ φ ¯ x 0 ¯ + a 1 ¯ φ   ¯ x 0 ¯ + a 1 ¯ φ ¯ x 1 ¯ + a 0 ¯ φ =   ¯ x 1 ¯ φ ¯ x 0 ¯ + a 0 ¯ a 1 ¯ φ +   ¯ x 1 ¯ a 1 ¯ φ + a 0 ¯ φ ¯ x 0 ¯ +   ¯ x 0 ¯ φ ¯ x 1 ¯ a 1 ¯ a 0 ¯ φ +   ¯ x 0 ¯ a 0 ¯ φ a 1 ¯ φ ¯ x 1 ¯ =   ¯ x 1 ¯ a 1 ¯ φ + a 0 ¯ φ ¯ x 0 ¯ +   ¯ x 0 ¯ a 0 ¯ φ a 1 ¯ φ ¯ x 1 ¯ = φ ¯ x 1 ¯ a 1 ¯ + a 1 ¯ ¯ x 1 ¯ φ + a 0 ¯ φ ¯ x 0 ¯ + φ ¯ x 0 ¯ a 0 ¯ + a 0 ¯ ¯ x 0 ¯ φ a 1 ¯ φ ¯ x 1 ¯ = a 1 ¯ ¯ x 1 ¯ φ + a 0 ¯ ¯ x 0 ¯ φ = a 1 x 1 ¯ + a 0 x 0 ¯ φ
x 1 + a 0 [   ] ¯ x 0 ¯ + a 1 ¯ [   ] ¯ x 0 ¯ + a 1 ¯ x 1 + a 0 φ = x 1 + a 0 [   ] ¯ x 0 ¯ + a 1 ¯ φ   ¯ x 0 ¯ + a 1 ¯ x 1 + a 0 φ = = x 1 + a 0 φ ¯ x 0 ¯ + a 1 ¯ φ   ¯ x 0 ¯ + a 1 ¯ φ x 1 + a 0 φ = x 1 φ ¯ x 0 ¯ + a 0 a 1 ¯ φ + x 1 a 1 ¯ φ + a 0 φ ¯ x 0 ¯ +   ¯ x 0 ¯ φ x 1 a 1 ¯ a 0 φ +   ¯ x 0 ¯ a 0 φ a 1 ¯ φ x 1 = x 1 a 1 ¯ φ + a 0 φ ¯ x 0 ¯ +   ¯ x 0 ¯ a 0 φ a 1 ¯ φ x 1 = φ x 1 a 1 ¯ + a 1 ¯ x 1 φ + a 0 φ ¯ x 0 ¯ + φ ¯ x 0 ¯ a 0 + a 0 ¯ x 0 ¯ φ a 1 ¯ φ ¯ x 1 ¯ = a 1 ¯ x 1 φ + a 0 ¯ x 0 ¯ φ = a 1 ¯ x 1 + a 0 ¯ x 0 ¯ φ
Further we will use these and analogous relations
x 1 + a 0 x 0 + a 1 x 0 + a 1 x 1 + a 0 φ = a 1 x 1 + a 0 x 0 φ
x 1 + a 0 [   ] ¯ x 0 ¯ + a 1 ¯ [   ] ¯ x 0 ¯ + a 1 ¯ x 1 + a 0 φ = a 1 ¯ x 1 + a 0 ¯ x 0 ¯ φ
x 0 + a 1 [   ] ¯ x 1 ¯ + a 0 ¯ [   ] ¯ x 1 ¯ + a 0 ¯ x 0 + a 1 φ = a 0 ¯ x 0 a 1 ¯ x 1 ¯ φ
x 0 + a 1 [   ] ¯ x 0 ¯ + a 1 ¯ [   ] ¯ x 0 ¯ + a 1 ¯ x 0 + a 1 φ = a 1 ¯ x 0 + a 1 ¯ x 0 ¯ φ
x 1 + a 0 [   ] ¯ x 1 ¯ + a 0 ¯ [   ] ¯ x 1 ¯ + a 0 ¯ x 1 + a 0 φ = a 0 ¯ x 1 a 0 ¯ x 1 ¯ φ
[   ] ¯ x 1 ¯ + a 0 ¯ [   ] ¯ x 0 ¯ + a 1 ¯ [   ] ¯ x 0 ¯ + a 1 ¯ [   ] ¯ x 1 ¯ + a 0 ¯ φ = a 1 x 1 ¯ + a 0 x 0 ¯ φ
[   ] ¯ x 0 ¯ + a 1 ¯ [   ] ¯ x 1 ¯ + a 0 ¯ + [   ] ¯ x 1 ¯ + a 0 ¯ [   ] ¯ x 0 ¯ + a 1 ¯ φ = a 1 x 1 ¯ ¯ + a 0 x 0 ¯ ¯ φ
Let's solve the equation
( S K ) S + ψ x 0 , x 1 , x 2 , x 3 = M 2 ψ x 0 , x 1 , x 2 , x 3
( S K ) S + ψ 0 x 0 , x 1 , x 2 , x 3 ψ 1 x 0 , x 1 , x 2 , x 3 ψ 2 x 0 , x 1 , x 2 , x 3 ψ 3 x 0 , x 1 , x 2 , x 3 = M 2 ψ 0 x 0 , x 1 , x 2 , x 3 ψ 1 x 0 , x 1 , x 2 , x 3 ψ 2 x 0 , x 1 , x 2 , x 3 ψ 3 x 0 , x 1 , x 2 , x 3
S = 0 + a 1   1 + a 0 0 0 0 ¯ + a 1 ¯ , 1 ¯ + a 0 ¯ , 0 , 0 + 0   0 0 + a 1 1 + a 0 0,0 , 0 ¯ + a 1 ¯ , 1 ¯ + a 0 ¯ + 2 + a 3   3 + a 2 0 0 2 ¯ + a 3 ¯ , 3 ¯ + a 2 ¯ , 0 , 0 + 0   0 2 + a 3 3 + a 2 0,0 2 ¯ + a 3 ¯ 3 ¯ + a 2 ¯
S + = 1 ¯ + a 0 ¯ 0 ¯ + a 1 ¯ 0 0 1 + a 0 , 0 + a 1 , 0 , 0 + 0   0 1 ¯ + a 0 ¯ 0 ¯ + a 1 ¯ 0,0 , 1 + a 0 0 , 0 + a 1 + 3 ¯ + a 2 ¯ 2 ¯ + a 3 ¯ 0 0 3 + a 2 , 2 + a 3 , 0 , 0 + 0   0 3 ¯ + a 2 ¯ 2 ¯ + a 3 ¯ 0,0 , 3 + a 2 , 2 + a 3
K = = 0 a 1 0 ¯ a 1 ¯ 0 ¯ a 1 ¯ 0 a 1 0 + a 1 1 ¯ + a 0 ¯ 1 ¯ + a 0 ¯ 0 + a 1 1 + a 0 0 ¯ + a 1 ¯ 0 ¯ + a 1 ¯ 1 + a 0     1 + a 0 1 ¯ + a 0 ¯ 1 ¯ + a 0 ¯ 1 + a 0   0       0 0       0     0       0 0       0 p 1 p 1 ¯ p 1 ¯ p 1 p 1 p 0 ¯ p 0 ¯ p 1 p 0 p 1 ¯ p 1 ¯ p 0 p 0 p 0 ¯ p 0 ¯ p 0 + 2 a 3 2 ¯ a 3 ¯ 2 ¯ a 3 ¯ 2 a 3 2 + a 3 3 ¯ + a 2 ¯ 3 ¯ + a 2 ¯ 2 + a 3 3 + a 2 2 ¯ + a 3 ¯ 2 ¯ + a 3 ¯ 3 + a 2     3 + a 2 3 ¯ + a 2 ¯ 3 ¯ + a 2 ¯ 3 + a 2   0       0 0       0     0       0 0       0 p 3 p 3 ¯ p 3 ¯ p 3 p 3 p 2 ¯ p 2 ¯ p 3 p 2 p 3 ¯ p 3 ¯ p 2     p 2 p 2 ¯ p 2 ¯ p 2
= a 1 ¯ x 0 + a 1 ¯ x 0 ¯ a 0 ¯ x 0 a 1 ¯ x 1 ¯ a 1 ¯ x 1 + a 0 ¯ x 0 ¯   a 0 ¯ x 1 a 0 ¯ x 1 ¯   0       0 0       0     0       0 0       0 a 1 ¯ x 0 + a 1 ¯ x 0 ¯ a 0 ¯ x 0 a 1 ¯ x 1 ¯ a 1 ¯ x 1 + a 0 ¯ x 0 ¯   a 0 ¯ x 1 a 0 ¯ x 1 ¯
+ a 3 ¯ x 2 + a 3 ¯ x 2 ¯ a 2 ¯ x 2 a 3 ¯ x 3 ¯ a 3 ¯ x 3 + a 2 ¯ x 2 ¯   a 2 ¯ x 3 a 2 ¯ x 3 ¯   0       0 0       0     0       0 0       0 a 3 ¯ x 2 + a 3 ¯ x 2 ¯ a 2 ¯ x 2 a 3 ¯ x 3 ¯ a 3 ¯ x 3 + a 2 ¯ x 2 ¯   a 2 ¯ x 3 a 2 ¯ x 3 ¯
S + ψ = 1 ¯ + a 0 ¯ 0 ¯ + a 1 ¯ 0 0 1 + a 0 ψ 0 + 0 + a 1 ψ 1 + 0   0 1 ¯ + a 0 ¯ 0 ¯ + a 1 ¯ 1 + a 0 ψ 2 + 0 + a 1 ψ 3 + 3 ¯ + a 2 ¯ 2 ¯ + a 3 ¯ 0 0 3 + a 2 ψ 0 + 2 + a 3 ψ 1 + 0   0 3 ¯ + a 2 ¯ 2 ¯ + a 3 ¯ 3 + a 2 ψ 2 + 2 + a 3 ψ 3
Since the second factor S + in the left-hand side of the equation has a simpler structure than the first factor, perhaps as a first step we should find the eigenvalues and eigenfunctions of the equation
S + ψ x 0 , x 1 , x 2 , x 3 = M 2 ψ x 0 , x 1 , x 2 , x 3
and use them when solving the equation as a whole.
S S + ψ = 0 + a 1   1 + a 0 0 0 0 ¯ + a 1 ¯ , 1 ¯ + a 0 ¯ , 0 , 0 + 0   0 0 + a 1 1 + a 0 0,0 , 0 ¯ + a 1 ¯ , 1 ¯ + a 0 ¯ + 2 + a 3   3 + a 2 0 0 2 ¯ + a 3 ¯ , 3 ¯ + a 2 ¯ , 0 , 0 + 0   0 2 + a 3 3 + a 2 0,0 , 2 ¯ + a 3 ¯ 3 ¯ + a 2 ¯
1 ¯ + a 0 ¯ 0 ¯ + a 1 ¯ 0 0 ψ 0 x 1 + a 0 ψ 0 + ψ 1 x 0 + a 1 ψ 1 + 0   0 1 ¯ + a 0 ¯ 0 ¯ + a 1 ¯ ψ 2 x 1 + a 0 ψ 2 + ψ 3 x 0 + a 1 ψ 3 + 3 ¯ + a 2 ¯ 2 ¯ + a 3 ¯ 0 0 ψ 0 x 3 + a 2 ψ 0 + ψ 1 x 2 + a 3 ψ 1 + 0   0 3 ¯ + a 2 ¯ 2 ¯ + a 3 ¯ ψ 2 x 3 + a 2 ψ 2 + ψ 3 x 2 + a 3 ψ 3
= 0 + a 1   1 + a 0 0 0 0 ¯ + a 1 ¯ 1 ¯ + a 0 ¯ + 1 ¯ + a 0 ¯ 0 ¯ + a 1 ¯ ψ 0 x 1 + a 0 ψ 0 + ψ 1 x 0 + a 1 ψ 1 + 2 + a 3   3 + a 2 0 0 2 ¯ + a 3 ¯ 1 ¯ + a 0 ¯ + 3 ¯ + a 2 ¯ 0 ¯ + a 1 ¯ ψ 0 x 1 + a 0 ψ 0 + ψ 1 x 0 + a 1 ψ 1 + 0   0 0 + a 1 1 + a 0 0 ¯ + a 1 ¯ 1 ¯ + a 0 ¯ + 1 ¯ + a 0 ¯ 0 ¯ + a 1 ¯ ψ 2 x 1 + a 0 ψ 2 + ψ 3 x 0 + a 1 ψ 3 + 0   0 2 + a 3   3 + a 2 2 ¯ + a 3 ¯ 1 ¯ + a 0 ¯ + 3 ¯ + a 2 ¯ 0 ¯ + a 1 ¯ ψ 2 x 1 + a 0 ψ 2 + ψ 3 x 0 + a 1 ψ 3 + 0 + a 1 1 + a 0 0 0 0 ¯ + a 1 ¯ 3 ¯ + a 2 ¯ + 1 ¯ + a 0 ¯ 2 ¯ + a 3 ¯ ψ 0 x 3 + a 2 ψ 0 + ψ 1 x 2 + a 3 ψ 1 + 2 + a 3   3 + a 2 0 0 2 ¯ + a 3 ¯ 3 ¯ + a 2 ¯ + 3 ¯ + a 2 ¯ 2 ¯ + a 3 ¯ ψ 0 x 3 + a 2 ψ 0 + ψ 1 x 2 + a 3 ψ 1 + 0   0 0 + a 1 1 + a 0 0 ¯ + a 1 ¯ 3 ¯ + a 2 ¯ + 1 ¯ + a 0 ¯ 2 ¯ + a 3 ¯ ψ 2 x 3 + a 2 ψ 2 + ψ 3 x 2 + a 3 ψ 3 + 0   0 2 + a 3   3 + a 2 2 ¯ + a 3 ¯ 3 ¯ + a 2 ¯ + 3 ¯ + a 2 ¯ 2 ¯ + a 3 ¯ ψ 2 x 3 + a 2 ψ 2 + ψ 3 x 2 + a 3 ψ 3
S S + ψ = 0 + a 1 1 + a 0 0 0 a 1 x 1 ¯ + a 0 x 0 ¯ ψ 0 x 1 + a 0 ψ 0 + ψ 1 x 0 + a 1 ψ 1 + 2 + a 3   3 + a 2 0 0 2 ¯ + a 3 ¯ 1 ¯ + a 0 ¯ + 3 ¯ + a 2 ¯ 0 ¯ + a 1 ¯ ψ 0 x 1 + a 0 ψ 0 + ψ 1 x 0 + a 1 ψ 1 + 0   0 0 + a 1 1 + a 0 a 1 x 1 ¯ + a 0 x 0 ¯ ψ 2 x 1 + a 0 ψ 2 + ψ 3 x 0 + a 1 ψ 3 + 0   0 2 + a 3   3 + a 2 2 ¯ + a 3 ¯ 1 ¯ + a 0 ¯ + 3 ¯ + a 2 ¯ 0 ¯ + a 1 ¯ ψ 2 x 1 + a 0 ψ 2 + ψ 3 x 0 + a 1 ψ 3 + 0 + a 1 1 + a 0 0 0 0 ¯ + a 1 ¯ 3 ¯ + a 2 ¯ + 1 ¯ + a 0 ¯ 2 ¯ + a 3 ¯ ψ 0 x 3 + a 2 ψ 0 + ψ 1 x 2 + a 3 ψ 1 + 2 + a 3   3 + a 2 0 0 a 3 x 3 ¯ + a 2 x 2 ¯ ψ 0 x 3 + a 2 ψ 0 + ψ 1 x 2 + a 3 ψ 1 + 0   0 0 + a 1 1 + a 0 0 ¯ + a 1 ¯ 3 ¯ + a 2 ¯ + 1 ¯ + a 0 ¯ 2 ¯ + a 3 ¯ ψ 2 x 3 + a 2 ψ 2 + ψ 3 x 2 + a 3 ψ 3 + 0   0 2 + a 3   3 + a 2 a 3 x 3 ¯ + a 2 x 2 ¯ ψ 2 x 3 + a 2 ψ 2 + ψ 3 x 2 + a 3 ψ 3
Let's calculate the expressions included in the equation
2 ¯ + a 3 ¯ 1 ¯ + a 0 ¯ + 3 ¯ + a 2 ¯ 0 ¯ + a 1 ¯ φ = 3 ¯ + a 2 ¯ 0 ¯ + a 1 ¯ φ 2 ¯ + a 3 ¯ 1 ¯ + a 0 ¯ φ = 3 ¯ + a 2 ¯ φ ¯ x 0 ¯ + a 1 ¯ φ 2 ¯ + a 3 ¯ φ ¯ x 1 ¯ + a 0 ¯ φ
= 3 ¯ φ ¯ x 0 ¯ + 3 ¯ a 1 ¯ φ + a 2 ¯ φ ¯ x 0 ¯ + a 2 ¯ a 1 ¯ φ + 2 ¯ φ ¯ x 1 ¯ 2 ¯ a 0 ¯ φ a 3 ¯ φ ¯ x 1 ¯ a 3 ¯ a 0 ¯ φ
= 3 ¯ φ ¯ x 0 ¯ + a 1 x 3 ¯ φ a 2 ¯ φ ¯ x 0 ¯ + a 2 ¯ a 1 ¯ φ + 2 ¯ φ ¯ x 1 ¯ + a 0 ¯ φ ¯ x 2 ¯ + a 0 x 2 ¯ φ a 3 ¯ φ ¯ x 1 ¯ a 3 ¯ a 0 ¯ φ = 2 ¯ φ ¯ x 1 ¯ 3 ¯ φ ¯ x 0 ¯ + a 1 x 3 ¯ + a 0 x 2 ¯ φ + a 1 ¯ φ ¯ x 3 ¯ a 2 ¯ φ ¯ x 0 ¯ + a 0 ¯ φ ¯ x 2 ¯ a 3 ¯ φ ¯ x 1 ¯ + a 2 ¯ a 1 ¯ a 3 ¯ a 0 ¯ φ
Let us consider the situation when the electromagnetic potential can be described by a plane wave in spinor space
a x 0 , x 1 , x 2 , x 3 = a 0 x 0 , x 1 , x 2 , x 3 a 1 x 0 , x 1 , x 2 , x 3 a 2 x 0 , x 1 , x 2 , x 3 a 3 x 0 , x 1 , x 2 , x 3 = u a 0 u a 1 u a 2 u a 3 φ a = u a 0 u a 1 u a 2 u a 3 exp p a 0 x 1 p a 1 x 0 + p a 2 x 3 p a 3 x 2
x 0 + a 1 x 3 + a 2 x 1 + a 0 x 2 + a 3 φ = x 0 x 3 φ + x 1 x 2 φ + a 2 x 0 a 3 x 1 φ a 2 φ x 0 + a 1 φ x 3 a 3 φ x 1 + a 0 φ x 2 + a 1 a 2 a 0 a 3 φ = x 0 x 3 φ + x 1 x 2 φ + u a 2 p a 1 u a 3 p a 0 φ a φ u a 2 φ a φ x 0 u a 1 φ a φ x 2 + u a 3 φ a φ x 0 + u a 0 φ a φ x 3 + u a 1 u a 2 u a 0 u a 3 φ a 2 φ
x 1 + a 0 x 3 + a 2 x 3 + a 2 x 1 + a 0 φ = a 2 x 1 a 0 x 3 φ = ( u a 2 p a 0 u a 0 p a 2 ) φ a φ
When the electromagnetic potential is represented by a plane wave, the field created by a charged particle is not taken into account, so this model adequately describes only the situation when the electromagnetic field is strong enough and the influence of the particle charge can be neglected.
It would be interesting in this context to consider for the presented spinor model the case of a centrally symmetric electric field and to find solutions of the spinor wave equation for the hydrogen-like atom, taking into account the presence of spin at the electron. For such a model we can take
a 0 = i 1 2 R a 1 = 1 2 R               a 2 = 1 2 R               a 3 = i 1 2 R
R = 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 + x 2 ¯ x 3 + x 3 ¯ x 2 2 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 x 2 ¯ x 3 + x 3 ¯ x 2 2 + 1 2 x 0 ¯ x 0 x 1 ¯ x 1 + x 2 ¯ x 2 x 3 ¯ x 3 2
As mentioned above, we can substitute into the equation the already known exact solutions of the Dirac equation for the hydrogen-like atom by expressing the components of the coordinate vector and derivatives on them through the components of the coordinate spinor and derivatives on them. It is likely that the solution of the Dirac equation would not make the spinor equation an identity; it would be evidence that more arbitrary assumptions are made in the Dirac equation than in the spinor equation, and that the latter claims to be a better description of nature.
We can also consider the case of a constant magnetic field directed along the z-axis
A 0 = 0                 A 1 = 1 2 B 3 X 2                   A 2 = 1 2 B 3 X 1             A 3 = 0
X 1 = 1 2 x 0 ¯ x 1 + x 1 ¯ x 0 + x 2 ¯ x 3 + x 3 ¯ x 2
X 2 = 1 2 i x 0 ¯ x 1 + i x 1 ¯ x 0 i x 2 ¯ x 3 + i x 3 ¯ x 2
A 1 = 1 2 a 0 ¯ a 1 + a 1 ¯ a 0 + a 2 ¯ a 3 + a 3 ¯ a 2
A 2 = 1 2 i a 0 ¯ a 1 + i a 1 ¯ a 0 i a 2 ¯ a 3 + i a 3 ¯ a 2
A 0 = 1 2 a 0 ¯ a 0 + a 1 ¯ a 1 + a 2 ¯ a 2 + a 3 ¯ a 3
A 3 = 1 2 a 0 ¯ a 0 a 1 ¯ a 1 + a 2 ¯ a 2 a 3 ¯ a 3
Let's say
a 0 = i x 1 ¯ B 3 / 2                     a 1 = x 0 ¯ B 3 / 2
a 2 = i x 3 ¯ B 3 / 2                     a 3 = x 2 ¯ B 3 / 2
A 1 = 1 4 B 3 i x 1 x 0 ¯ i x 0 ¯ x 1 + i x 3 ¯ x 2 i x 2 ¯ x 3 = 1 2 B 3 X 2
A 2 = 1 4 B 3 x 1 ¯ x 0 + x 0 ¯ x 1 + x 3 ¯ x 2 + x 2 ¯ x 3 = 1 2 B 3 X 1  
A 0 = 1 4 B 3 x 1 x 1 ¯ + x 0 x 0 ¯ + x 3 x 3 ¯ + x 2 x 2 ¯ = 1 2 B 3 t
A 3 = 1 4 B 3 x 1 x 1 ¯ x 0 x 0 ¯ + x 3 x 3 ¯ x 2 x 2 ¯ = 1 2 B 3 X 3  
We see that the scalar potential A 0 grows with time, but does not depend on spatial coordinates, and the vector potential does not depend on time, so that there is no electric field. In this case
K = a 1 ¯ x 0 + a 1 ¯ x 0 ¯ a 0 ¯ x 0 a 1 ¯ x 1 ¯ a 1 ¯ x 1 + a 0 ¯ x 0 ¯   a 0 ¯ x 1 a 0 ¯ x 1 ¯   0       0 0       0     0       0 0       0 a 1 ¯ x 0 + a 1 ¯ x 0 ¯ a 0 ¯ x 0 a 1 ¯ x 1 ¯ a 1 ¯ x 1 + a 0 ¯ x 0 ¯   a 0 ¯ x 1 a 0 ¯ x 1 ¯ + a 3 ¯ x 2 + a 3 ¯ x 2 ¯ a 2 ¯ x 2 a 3 ¯ x 3 ¯ a 3 ¯ x 3 + a 2 ¯ x 2 ¯   a 2 ¯ x 3 a 2 ¯ x 3 ¯   0       0 0       0     0       0 0       0 a 3 ¯ x 2 + a 3 ¯ x 2 ¯ a 2 ¯ x 2 a 3 ¯ x 3 ¯ a 3 ¯ x 3 + a 2 ¯ x 2 ¯   a 2 ¯ x 3 a 2 ¯ x 3 ¯ =
= B 3 2 1 0 0 i   0       0 0       0     0       0 0       0 1       0 0 i + B 3 / 2 1 0 0 i   0       0 0       0     0       0 0       0 1 0 0 i = 2 B 3 1 0 0 i   0       0 0       0     0       0 0       0 1       0 0 i
The equation considered up to now is rather cumbersome, therefore we would like to have a simpler and compact relativistic invariant equation for the fermion, taking into account the presence of a half-integer spin. Such equation really exists, its derivation is given in section 4 of the paper. Here we will give its form for the electron in the presence of the electromagnetic field
S R + S R ¯ + S R + S R ¯ 4 ( m + m ¯ ) I φ x = 0
where
S R = ( 2 + a 3 ) ( 3 + a 2 ) ( 0 + a 1 ) ( 1 + a 0 ) ( 1 + a 0 ) , 0 + a 1 , 3 + a 2 , ( 2 + a 3 ) ( 0 + a 1 ) ( 1 + a 0 ) ( 2 + a 3 ) ( 3 + a 2 ) 3 + a 2 , ( 2 + a 3 ) , ( 1 + a 0 ) , ( 0 + a 1 )
+ ( 0 + a 1 ) ( 1 + a 0 ) ( 2 + a 3 ) ( 3 + a 2 ) 3 + a 2 , ( 2 + a 3 ) , ( 1 + a 0 ) , ( 0 + a 1 ) ( 2 + a 3 ) ( 3 + a 2 ) ( 0 + a 1 ) ( 1 + a 0 ) 1 + a 0 , ( 0 + a 1 ) , ( 3 + a 2 ) , ( 2 + a 3 )
S R = ( 1 + a 0 ) ( 0 + a 1 ) ( 3 + a 2 ) ( 2 + a 3 ) ( 2 + a 3 ) , ( 3 + a 2 ) , ( 0 + a 1 ) , ( 1 + a 0 ) ( 3 + a 2 ) ( 2 + a 3 ) ( 1 + a 0 ) ( 0 + a 1 ) ( 0 + a 1 ) , ( 1 + a 0 ) , ( 2 + a 3 ) , ( 3 + a 2 )
+ ( 3 + a 2 ) ( 2 + a 3 ) ( 1 + a 0 ) ( 0 + a 1 ) ( 0 + a 1 ) , ( 1 + a 0 ) , ( 2 + a 3 ) , ( 3 + a 2 ) ( 1 + a 0 ) ( 0 + a 1 ) ( 3 + a 2 ) ( 2 + a 3 ) ( 2 + a 3 ) , ( 3 + a 2 ) , ( 0 + a 1 ) , ( 1 + a 0 )
In general case electric and magnetic fields are expressed through partial derivatives of components of the vector potential by components of the space vector. We also can find the expression through these fields for the derivatives of the spinor components of the electromagnetic potential by the components of the coordinate spinor. To do this, we first find all derivatives
A γ x µ = A γ X ν X ν x µ
then express the components of the vector potential through the components of the spinor potential, substitute the components of the electric and magnetic fields instead of the derivatives of the components of the vector potential by the components of the coordinate vector, and then find the required derivatives from the resulting system of linear equations.
From general considerations taking into account the substitutions
p 0 ¯ [   ] ¯ x 1 ¯                     p 1 ¯ [   ] ¯ x 0 ¯
it is possible to write the commutation relations for the components of the momentum spinor and functions from the components of the coordinate spinor
[ φ ] ¯ x 1 ¯ = 1 c φ , p 0 ¯ = 1 c ( φ p 0 ¯ p 0 ¯ φ )
x 1 , p 0 ¯ = ( x 1 p 0 ¯ p 0 ¯ x 1 ) = c x 1 ¯ x 1 ¯ = c
x 1 ¯ , p 0 = ( x 1 ¯ p 0 p 0 x 1 ¯ ) = c ¯
[ φ ] ¯ x 0 ¯ = 1 d φ , p 1 ¯ = 1 d ( φ p 1 ¯ p 1 ¯ φ )
x 0 , p 1 ¯ = x 0 p 1 ¯ p 1 ¯ x 0 = d x 0 ¯ x 0 ¯ = d
x 0 ¯ , p 1 = x 0 ¯ p 1 p 1 x 0 ¯ = d ¯
All other combinations commute with each other. The constant coefficients c and d possibly include a minus sign, an imaginary unit and some degree of the rationalized Planck’s constant. Their values can be determined using known commutation relations for vector components, e.g.
X 1 P 1 P 1 X 1 = i ћ
and using expressions of vector components through spinor components
X 1 P 1 P 1 X 1 = x 0 ¯ , x 1 ¯ 0 1 1 0 x 0 x 1 p 0 ¯ , p 1 ¯ 0 1 1 0 p 0 p 1 p 0 ¯ , p 1 ¯ 0 1 1 0 p 0 p 1 x 0 ¯ , x 1 ¯ 0 1 1 0 x 0 x 1 = x 0 ¯ , x 1 ¯ x 1 x 0 p 0 ¯ , p 1 ¯ p 1 p 0 p 0 ¯ , p 1 ¯ p 1 p 0 x 0 ¯ , x 1 ¯ x 1 x 0 = x 0 ¯ x 1 + x 1 ¯ x 0 p 0 ¯ p 1 + p 1 ¯ p 0 p 0 ¯ p 1 + p 1 ¯ p 0 x 0 ¯ x 1 + x 1 ¯ x 0 = x 0 ¯ x 1 p 0 ¯ p 1 + x 0 ¯ x 1 p 1 ¯ p 0 + x 1 ¯ x 0 p 0 ¯ p 1 + x 1 ¯ x 0 p 1 ¯ p 0 p 0 ¯ p 1 x 0 ¯ x 1 p 0 ¯ p 1 x 1 ¯ x 0 p 1 ¯ p 0 x 0 ¯ x 1 p 1 ¯ p 0 x 1 ¯ x 0 = ( x 0 ¯ x 1 p 0 ¯ p 1 p 0 ¯ p 1 x 0 ¯ x 1 ) + ( x 0 ¯ x 1 p 1 ¯ p 0 p 1 ¯ p 0 x 0 ¯ x 1 ) + ( x 1 ¯ x 0 p 0 ¯ p 1 p 0 ¯ p 1 x 1 ¯ x 0 ) + ( x 1 ¯ x 0 p 1 ¯ p 0 p 1 ¯ p 0 x 1 ¯ x 0 )
x 0 ¯ x 1 p 0 ¯ p 1 p 0 ¯ p 1 x 0 ¯ x 1 = x 0 ¯ p 0 ¯ x 1 + c p 1 p 0 ¯ x 0 ¯ p 1 + d ¯ x 1 = x 0 ¯ p 0 ¯ x 1 p 1 + c x 0 ¯ p 1 p 0 ¯ x 0 ¯ p 1 x 1 d ¯ p 0 ¯ x 1 = x 0 ¯ p 0 ¯ x 1 p 1 x 0 ¯ p 0 ¯ x 1 p 1 + c x 0 ¯ p 1 d ¯ p 0 ¯ x 1 = c x 0 ¯ p 1 d ¯ p 0 ¯ x 1
x 1 ¯ x 0 p 1 ¯ p 0 p 1 ¯ p 0 x 1 ¯ x 0 = x 1 ¯ p 1 ¯ x 0 d p 0 p 1 ¯ x 1 ¯ p 0 c ¯ x 0 = x 1 ¯ p 1 ¯ x 0 p 0 d x 1 ¯ p 0 p 1 ¯ x 1 ¯ p 0 x 0 + c ¯ p 1 ¯ x 0 = x 1 ¯ p 1 ¯ x 0 p 0 x 1 ¯ p 1 ¯ x 0 p 0 + c ¯ p 1 ¯ x 0 d x 1 ¯ p 0 = c ¯ p 1 ¯ x 0 d x 1 ¯ p 0
x 0 ¯ x 1 p 1 ¯ p 0 p 1 ¯ p 0 x 0 ¯ x 1 = x 0 ¯ p 1 ¯ x 1 p 0 p 1 ¯ p 0 x 0 ¯ x 1 = p 1 ¯ x 0 ¯ x 1 p 0 p 1 ¯ p 0 x 0 ¯ x 1 = p 1 ¯ x 0 ¯ p 0 x 1 p 1 ¯ p 0 x 0 ¯ x 1 = p 1 ¯ p 0 x 0 ¯ x 1 p 1 ¯ p 0 x 0 ¯ x 1 = 0
x 1 ¯ x 0 p 0 ¯ p 1 p 0 ¯ p 1 x 1 ¯ x 0 = x 1 ¯ x 0 p 0 ¯ p 1 p 0 ¯ p 1 x 1 ¯ x 0 = p 0 ¯ x 1 ¯ x 0 p 1 p 0 ¯ p 1 x 1 ¯ x 0 = p 0 ¯ p 1 x 1 ¯ x 0 p 0 ¯ p 1 x 1 ¯ x 0 = 0
X 1 P 1 P 1 X 1 = ( x 0 ¯ x 1 p 0 ¯ p 1 p 0 ¯ p 1 x 0 ¯ x 1 ) + ( x 1 ¯ x 0 p 1 ¯ p 0 p 1 ¯ p 0 x 1 ¯ x 0 ) = c x 0 ¯ p 1 d ¯ p 0 ¯ x 1 + c ¯ p 1 ¯ x 0 d x 1 ¯ p 0 = c p 1 x 0 ¯ d ¯ d ¯ x 1 p 0 ¯ c + c ¯ x 0 p 1 ¯ + d d p 0 x 1 ¯ + c ¯ = c p 1 x 0 ¯ c d ¯ d ¯ x 1 p 0 ¯ + d ¯ c + c ¯ x 0 p 1 ¯ + c ¯ d d p 0 x 1 ¯ d c ¯ = c p 1 x 0 ¯ d ¯ x 1 p 0 ¯ + c ¯ x 0 p 1 ¯ d p 0 x 1 ¯  
Let's return to the relations
P 0 2 P 1 2 P 2 2 P 3 2 = m ¯ m = m 2
p 1 p 2 p 0 p 3 = m  
p 1 ¯ p 2 ¯ p 0 ¯ p 3 ¯ = m ¯
p 1 ¯ p 2 ¯ p 0 ¯ p 3 ¯ ( p 1 p 2 p 0 p 3 ) = m ¯ m = m 2
In this form they are equivalent, but if an external field is added, a difference arises, since in one case the field is added at the vector level and in the other at the spinor level
( P 0 A 0 ) 2 ( P 1 A 1 ) 2 ( P 2 A 2 ) 2 ( P 3 A 3 ) 2 = m 2
p 1 a 1 ¯ p 2 a 2 ¯ p 0 a 0 ( p 3 a 3 ) ¯ ( ( p 1 a 1 ) ( p 2 a 2 ) ( p 0 a 0 ) ( p 3 a 3 ) ) = m 2
These relations correspond to differential equations including the relativistic Schrödinger equation
2 X 0 2 2 X 1 2 2 X 2 2 2 X 3 2 φ X 0 , X 1 , X 2 , X 3 = m 2 φ X 0 , X 1 , X 2 , X 3
x 1 x 2 x 0 x 3 φ x 0 , x 1 , x 2 , x 3 = m   φ x 0 , x 1 , x 2 , x 3
[   ] ¯ x 1 ¯ [   ] ¯ x 2 ¯ [   ] ¯ x 0 ¯ [   ] ¯ x 3 ¯ φ x 0 , x 1 , x 2 , x 3 = m ¯   φ x 0 , x 1 , x 2 , x 3
[   ] ¯ x 1 ¯ [   ] ¯ x 2 ¯ [   ] ¯ x 0 ¯ [   ] ¯ x 3 ¯ x 1 x 2 x 0 x 3 φ x 0 , x 1 , x 2 , x 3 = m 2 φ x 0 , x 1 , x 2 , x 3
For a free particle the eigenfunctions and eigenvalues solving these equations should coincide, but in the presence of an external field the eigenvalues and the corresponding eigenfunctions will differ because of the above mentioned difference in summation in one case of vector components and in the other case of spinor components.
While the Dirac equation is sometimes referred to as extracting the square root of the Klein-Gordon equation, here we see a different way of doing it.
Let us describe in more detail the square of the length of the momentum vector
4 P 0 P 0 P 1 P 1 P 2 P 2 P 3 P 3 = = p 0 ¯ p 0 + p 1 ¯ p 1 + p 2 ¯ p 2 + p 3 ¯ p 3 p 0 ¯ p 0 + p 1 ¯ p 1 + p 2 ¯ p 2 + p 3 ¯ p 3 p 0 ¯ p 1 + p 1 ¯ p 0 + p 2 ¯ p 3 + p 3 ¯ p 2 p 0 ¯ p 1 + p 1 ¯ p 0 + p 2 ¯ p 3 + p 3 ¯ p 2 + p 0 ¯ p 1 + p 1 ¯ p 0 p 2 ¯ p 3 + p 3 ¯ p 2 p 0 ¯ p 1 + p 1 ¯ p 0 p 2 ¯ p 3 + p 3 ¯ p 2 p 0 ¯ p 0 p 1 ¯ p 1 + p 2 ¯ p 2 p 3 ¯ p 3 p 0 ¯ p 0 p 1 ¯ p 1 + p 2 ¯ p 2 p 3 ¯ p 3
p 0 ¯ p 0 + p 1 ¯ p 1 + p 2 ¯ p 2 + p 3 ¯ p 3 p 0 ¯ p 0 + p 1 ¯ p 1 + p 2 ¯ p 2 + p 3 ¯ p 3 p 0 ¯ p 0 p 1 ¯ p 1 + p 2 ¯ p 2 p 3 ¯ p 3 p 0 ¯ p 0 p 1 ¯ p 1 + p 2 ¯ p 2 p 3 ¯ p 3 = p 0 ¯ p 0 p 1 ¯ p 1 + p 2 ¯ p 2 + p 3 ¯ p 3 + p 1 ¯ p 1 p 0 ¯ p 0 + p 2 ¯ p 2 + p 3 ¯ p 3 + p 2 ¯ p 2 p 0 ¯ p 0 + p 1 ¯ p 1 + p 3 ¯ p 3 + p 3 ¯ p 3 p 0 ¯ p 0 + p 1 ¯ p 1 + p 2 ¯ p 2 p 0 ¯ p 0 p 1 ¯ p 1 + p 2 ¯ p 2 p 3 ¯ p 3 + p 1 ¯ p 1 p 0 ¯ p 0 + p 2 ¯ p 2 p 3 ¯ p 3 p 2 ¯ p 2 p 0 ¯ p 0 p 1 ¯ p 1 p 3 ¯ p 3 + p 3 ¯ p 3 p 0 ¯ p 0 p 1 ¯ p 1 + p 2 ¯ p 2 = p 0 ¯ p 0 p 1 ¯ p 1 + p 3 ¯ p 3 + p 1 ¯ p 1 p 0 ¯ p 0 + p 2 ¯ p 2 + p 2 ¯ p 2 p 1 ¯ p 1 + p 3 ¯ p 3 + p 3 ¯ p 3 p 0 ¯ p 0 + p 2 ¯ p 2 p 0 ¯ p 0 p 1 ¯ p 1 p 3 ¯ p 3 + p 1 ¯ p 1 p 0 ¯ p 0 + p 2 ¯ p 2 p 2 ¯ p 2 p 1 ¯ p 1 p 3 ¯ p 3 + p 3 ¯ p 3 p 0 ¯ p 0 + p 2 ¯ p 2 = p 0 ¯ p 0 p 1 ¯ p 1 + p 3 ¯ p 3 + p 1 ¯ p 1 p 0 ¯ p 0 + p 2 ¯ p 2 + p 2 ¯ p 2 p 1 ¯ p 1 + p 3 ¯ p 3 + p 3 ¯ p 3 p 0 ¯ p 0 + p 2 ¯ p 2 + p 0 ¯ p 0 p 1 ¯ p 1 + p 3 ¯ p 3 + p 1 ¯ p 1 p 0 ¯ p 0 + p 2 ¯ p 2 + p 2 ¯ p 2 p 1 ¯ p 1 + p 3 ¯ p 3 + p 3 ¯ p 3 p 0 ¯ p 0 + p 2 ¯ p 2
p 0 ¯ p 1 + p 1 ¯ p 0 + p 2 ¯ p 3 + p 3 ¯ p 2 p 0 ¯ p 1 + p 1 ¯ p 0 + p 2 ¯ p 3 + p 3 ¯ p 2 + p 0 ¯ p 1 + p 1 ¯ p 0 p 2 ¯ p 3 + p 3 ¯ p 2 p 0 ¯ p 1 + p 1 ¯ p 0 p 2 ¯ p 3 + p 3 ¯ p 2 = p 0 ¯ p 1 p 1 ¯ p 0 + p 2 ¯ p 3 + p 3 ¯ p 2 p 1 ¯ p 0 p 0 ¯ p 1 + p 2 ¯ p 3 + p 3 ¯ p 2 p 2 ¯ p 3 p 0 ¯ p 1 + p 1 ¯ p 0 + p 3 ¯ p 2 p 3 ¯ p 2 p 0 ¯ p 1 + p 1 ¯ p 0 + p 2 ¯ p 3 p 0 ¯ p 1 p 1 ¯ p 0 p 2 ¯ p 3 + p 3 ¯ p 2 + p 1 ¯ p 0 p 0 ¯ p 1 p 2 ¯ p 3 + p 3 ¯ p 2 p 2 ¯ p 3 p 0 ¯ p 1 + p 1 ¯ p 0 + p 3 ¯ p 2 + p 3 ¯ p 2 p 0 ¯ p 1 + p 1 ¯ p 0 p 2 ¯ p 3 = p 0 ¯ p 1 p 1 ¯ p 0 + p 3 ¯ p 2 p 1 ¯ p 0 p 0 ¯ p 1 + p 2 ¯ p 3 p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2 p 3 ¯ p 2 p 0 ¯ p 1 + p 2 ¯ p 3 p 0 ¯ p 1 p 1 ¯ p 0 + p 3 ¯ p 2 + p 1 ¯ p 0 p 0 ¯ p 1 p 2 ¯ p 3 p 2 ¯ p 3 + p 1 ¯ p 0 + p 3 ¯ p 2 + p 3 ¯ p 2 p 0 ¯ p 1 p 2 ¯ p 3
4 P 0 P 0 P 1 P 1 P 2 P 2 P 3 P 3 = p 0 ¯ p 0 p 1 ¯ p 1 + p 3 ¯ p 3 + p 1 ¯ p 1 p 0 ¯ p 0 + p 2 ¯ p 2 + p 2 ¯ p 2 p 1 ¯ p 1 + p 3 ¯ p 3 + p 3 ¯ p 3 p 0 ¯ p 0 + p 2 ¯ p 2 + p 0 ¯ p 0 p 1 ¯ p 1 + p 3 ¯ p 3 + p 1 ¯ p 1 p 0 ¯ p 0 + p 2 ¯ p 2 + p 2 ¯ p 2 p 1 ¯ p 1 + p 3 ¯ p 3 + p 3 ¯ p 3 p 0 ¯ p 0 + p 2 ¯ p 2 p 0 ¯ p 1 p 1 ¯ p 0 + p 3 ¯ p 2 p 1 ¯ p 0 p 0 ¯ p 1 + p 2 ¯ p 3 p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2 p 3 ¯ p 2 p 0 ¯ p 1 + p 2 ¯ p 3 p 0 ¯ p 1 p 1 ¯ p 0 + p 3 ¯ p 2 + p 1 ¯ p 0 p 0 ¯ p 1 p 2 ¯ p 3 p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2 + p 3 ¯ p 2 p 0 ¯ p 1 p 2 ¯ p 3
To obtain this result, we did not have to make assumptions about commutability of the spinor components among themselves. Accordingly, a similar expression takes place for the phase of a plane wave in vector space
4 P 0 X 0 P 1 X 1 P 2 X 2 P 3 X 3 = p 0 ¯ p 0 x 1 ¯ x 1 + x 3 ¯ x 3 + p 1 ¯ p 1 x 0 ¯ x 0 + x 2 ¯ x 2 + p 2 ¯ p 2 x 1 ¯ x 1 + x 3 ¯ x 3 + p 3 ¯ p 3 x 0 ¯ x 0 + x 2 ¯ x 2 + p 0 ¯ p 0 x 1 ¯ x 1 + x 3 ¯ x 3 + p 1 ¯ p 1 x 0 ¯ x 0 + x 2 ¯ x 2 + p 2 ¯ p 2 x 1 ¯ x 1 + x 3 ¯ x 3 + p 3 ¯ p 3 x 0 ¯ x 0 + x 2 ¯ x 2 p 0 ¯ p 1 x 1 ¯ x 0 + x 3 ¯ x 2 p 1 ¯ p 0 x 0 ¯ x 1 + x 2 ¯ x 3 p 2 ¯ p 3 x 1 ¯ x 0 + x 3 ¯ x 2 p 3 ¯ p 2 x 0 ¯ x 1 + x 2 ¯ x 3 p 0 ¯ p 1 x 1 ¯ x 0 + x 3 ¯ x 2 + p 1 ¯ p 0 x 0 ¯ x 1 x 2 ¯ x 3 p 2 ¯ p 3 x 1 ¯ x 0 + x 3 ¯ x 2 + p 3 ¯ p 2 x 0 ¯ x 1 x 2 ¯ x 3
Further we assume that the components of the momentum spinor commute, which takes place for a free particle, then we obtain
4 P 0 P 0 P 1 P 1 P 2 P 2 P 3 P 3 = p 0 ¯ p 0 p 1 ¯ p 1 + p 3 ¯ p 3 + p 1 ¯ p 1 p 0 ¯ p 0 + p 2 ¯ p 2 + p 2 ¯ p 2 p 1 ¯ p 1 + p 3 ¯ p 3 + p 3 ¯ p 3 p 0 ¯ p 0 + p 2 ¯ p 2 + p 0 ¯ p 0 p 1 ¯ p 1 + p 3 ¯ p 3 + p 1 ¯ p 1 p 0 ¯ p 0 + p 2 ¯ p 2 + p 2 ¯ p 2 p 1 ¯ p 1 + p 3 ¯ p 3 + p 3 ¯ p 3 p 0 ¯ p 0 + p 2 ¯ p 2 p 0 ¯ p 1 p 1 ¯ p 0 + p 3 ¯ p 2 p 1 ¯ p 0 p 0 ¯ p 1 + p 2 ¯ p 3 p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2 p 3 ¯ p 2 p 0 ¯ p 1 + p 2 ¯ p 3 p 0 ¯ p 1 p 1 ¯ p 0 + p 3 ¯ p 2 + p 1 ¯ p 0 p 0 ¯ p 1 p 2 ¯ p 3 p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2 + p 3 ¯ p 2 p 0 ¯ p 1 p 2 ¯ p 3 = 2 p 0 ¯ p 0 p 1 ¯ p 1 + p 3 ¯ p 3 + 2 p 1 ¯ p 1 p 0 ¯ p 0 + p 2 ¯ p 2 + 2 p 2 ¯ p 2 p 1 ¯ p 1 + p 3 ¯ p 3 + 2 p 3 ¯ p 3 p 0 ¯ p 0 + p 2 ¯ p 2 p 0 ¯ p 1 p 1 ¯ p 0 + p 3 ¯ p 2 p 1 ¯ p 0 p 0 ¯ p 1 + p 2 ¯ p 3 p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2 p 3 ¯ p 2 p 0 ¯ p 1 + p 2 ¯ p 3 p 0 ¯ p 1 p 1 ¯ p 0 + p 3 ¯ p 2 + p 1 ¯ p 0 p 0 ¯ p 1 p 2 ¯ p 3 p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2 + p 3 ¯ p 2 p 0 ¯ p 1 p 2 ¯ p 3 = 2 p 0 ¯ p 0 p 1 ¯ p 1 + p 3 ¯ p 3 + 2 p 1 ¯ p 1 p 0 ¯ p 0 + p 2 ¯ p 2 + 2 p 2 ¯ p 2 p 1 ¯ p 1 + p 3 ¯ p 3 + 2 p 3 ¯ p 3 p 0 ¯ p 0 + p 2 ¯ p 2 2 p 0 ¯ p 1 p 1 ¯ p 0 + p 3 ¯ p 2 2 p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2 2 p 0 ¯ p 1 p 1 ¯ p 0 + p 3 ¯ p 2 2 p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2
= 2 p 0 ¯ p 0 p 1 ¯ p 1 + p 3 ¯ p 3 + 2 p 1 ¯ p 1 p 0 ¯ p 0 + p 2 ¯ p 2 + 2 p 2 ¯ p 2 p 1 ¯ p 1 + p 3 ¯ p 3 + 2 p 3 ¯ p 3 p 0 ¯ p 0 + p 2 ¯ p 2 2 p 0 ¯ p 1 p 1 ¯ p 0 + p 3 ¯ p 2 2 p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2 2 p 0 ¯ p 1 p 1 ¯ p 0 + p 3 ¯ p 2 2 p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2 = 2 p 0 ¯ p 0 p 3 ¯ p 3 + 2 p 1 ¯ p 1 p 2 ¯ p 2 + 2 p 2 ¯ p 2 p 1 ¯ p 1 + 2 p 3 ¯ p 3 p 0 ¯ p 0 2 p 0 ¯ p 1 p 3 ¯ p 2 2 p 2 ¯ p 3 p 1 ¯ p 0 2 p 0 ¯ p 1 p 3 ¯ p 2 2 p 2 ¯ p 3 p 1 ¯ p 0 = 4 p 0 ¯ p 0 p 3 ¯ p 3 + 4 p 1 ¯ p 1 p 2 ¯ p 2 4 p 0 ¯ p 1 p 3 ¯ p 2 4 p 2 ¯ p 3 p 1 ¯ p 0
On the other hand, we can write
m ¯ m = p 1 p 2 p 0 p 3 ¯ p 1 p 2 p 0 p 3 = p 1 p 2 ¯ p 1 p 2 p 1 p 2 ¯ p 0 p 3 p 0 p 3 ¯ p 1 p 2 + p 0 p 3 ¯ p 0 p 3
Thus, the results of calculations coincide.
Let us compare the phases of plane waves in vector and spinor spaces. Let us hypothesize that the plane wave in spinor space has a more complicated form than it was supposed earlier in the paper, namely, it contains an additional conjugate multiplier
exp i   p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 ¯ p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
On the one hand, this assumption does not cancel the reasoning given in the paper concerning the equations and derivatives, since the derivatives on the components of the spatial spinor from the first conjugate factor are equal to zero, and on the other hand, the phase of the wave in this form is closer to the generally accepted phase of a plane wave in vector space.
For the boson the momentum spinor has a special form
p T = ( p 0 , p 1 , p 0 , p 1 )
Thus phases calculated by two methods do not coincide with each other, though both of them are invariant under Lorentz transformations
p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 ¯ p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 P 0 X 0 P 1 X 1 P 2 X 2 P 3 X 3
However, if to impose restrictions also on the form of the coordinate spinor
x T = ( x 0 , x 1 , x 0 , x 1 )
then
p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 ¯ p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 = p 0 x 1 + x 3 ¯ p 0 x 1 + x 3 + p 1 x 0 + x 2 ¯ p 1 x 0 + x 2 p 0 x 1 + x 3 ¯ p 1 x 0 + x 2 p 1 ( x 0 + x 2 ) ¯ p 0 ( x 1 + x 3 ) = 4 p 0 x 1 ¯ p 0 x + 4 p 1 x 0 ¯ p 1 x 0 4 p 0 x 1 ¯ p 1 x 0 4 p 1 x 0 ¯ p 0 x 1
4 P 0 X 0 P 1 X 1 P 2 X 2 P 3 X 3 = 2 p 0 ¯ p 0 x 1 ¯ x 1 + 2 p 1 ¯ p 1 x 0 ¯ x 0 + 2 p 2 ¯ p 2 x 1 ¯ x 1 + 2 p 3 ¯ p 3 x 0 ¯ x 0 + 2 p 0 ¯ p 0 x 1 ¯ x 1 + 2 p 1 ¯ p 1 x 0 ¯ x 0 + 2 p 2 ¯ p 2 x 1 ¯ x 1 + 2 p 3 ¯ p 3 x 0 ¯ x 0 2 p 0 ¯ p 1 x 1 ¯ x 0 2 p 1 ¯ p 0 x 0 ¯ x 1 2 p 2 ¯ p 3 x 1 ¯ x 0 2 p 3 ¯ p 2 x 0 ¯ x 1 2 p 0 ¯ p 1 x 1 ¯ x 0 2 p 1 ¯ p 0 x 0 ¯ x 1 2 p 2 ¯ p 3 x 1 ¯ x 0 2 p 3 ¯ p 2 x 0 ¯ x 1 = 4 p 0 ¯ p 0 x 1 ¯ x 1 + 4 p 1 ¯ p 1 x 0 ¯ x 0 + 4 p 0 ¯ p 0 x 1 ¯ x 1 + 4 p 1 ¯ p 1 x 0 ¯ x 0 4 p 0 ¯ p 1 x 1 ¯ x 0 4 p 1 ¯ p 0 x 0 ¯ x 1 4 p 0 ¯ p 1 x 1 ¯ x 0 4 p 1 ¯ p 0 x 0 ¯ x 1 = 4 p 0 ¯ p 0 x 1 ¯ x 1 + 4 p 1 ¯ p 1 x 0 ¯ x 0 + 4 p 0 ¯ p 0 x 1 ¯ x 1 + 4 p 1 ¯ p 1 x 0 ¯ x 0 4 p 0 ¯ p 1 x 1 ¯ x 0 4 p 1 ¯ p 0 x 0 ¯ x 1 4 p 0 ¯ p 1 x 1 ¯ x 0 4 p 1 ¯ p 0 x 0 ¯ x 1 = 8 p 0 ¯ p 0 x 1 ¯ x 1 + 8 p 1 ¯ p 1 x 0 ¯ x 0 8 8 p 0 ¯ p 1 x 1 ¯ x 0 8 p 1 ¯ p 0 x 0 ¯ x 1
That is, there is an equality
p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 ¯ p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 = 2 ( P 0 X 0 P 1 X 1 P 2 X 2 P 3 X 3 )
The form of the coordinate spinor defined in this way leads to the zero length of the coordinate vector, but the boson is characterized by this property. That is, it turns out that the requirement of coincidence of two definitions of the phase allows revealing the form of the coordinate spinor of the particle.
If a fermion is at rest in some coordinate system, then all components of its momentum vector, except the zero component, i.e., energy, are zero. Therefore, the phase of the corresponding plane wave in vector space depends on time but does not depend on spatial coordinates. It turns out that oscillations in time occur synchronously throughout space, and there is no wave propagation in the usual sense. On the contrary, the phase of the wave in spinor space depends on the spatial coordinates in such a situation. In addition, two fermions with different spins correspond to the same momentum vector, so the phases of the corresponding waves do not differ. But the phases of the wave in spinor space for fermions with different spins are different even when they are stationary.
If we accept the proposed hypothesis, then we need to change the equation for which the plane wave is an eigenfunction
x 1 x 2 x 0 x 3 exp i   p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 ¯ p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 = = p 1 p 2 p 0 p 3   f x ¯ f x ¯ exp i   p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 ¯ p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
where
f x p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
As a result, we have the equation
x 1 x 2 x 0 x 3 exp i   f x ¯ f x = m f x ¯ f x ¯ exp i   f x ¯ f x
Although the complex multiplier in front of the exponent in the right-hand side does not change with rotations and boosts, it now depends on the coordinates.
Let's consider the equation
x 1 x 2 x 0 x 3 φ x = m f x ¯ f x ¯ φ x
then the function corresponding to the free particle
φ x = exp i   p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 ¯ p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 f x ¯ f x ¯
is its solution with
m = p 1 p 2 p 0 p 3
This equation can be ascribed a universal character and its solutions can be sought at different possible values of m for real physical conditions, for example, in the presence of an electromagnetic field. Note that the function with imaginary unit under the exponent
exp i f x ¯ f x f x ¯ f x ¯
tends to zero at removal from the origin, i.e. the wave function of the plane wave is localized in space.
It was an expression for the amplitude of probability; the probability itself has the form
exp i f x ¯ f x f x ¯ f x ¯ exp i f x ¯ f x f x ¯ f x ¯ ¯ = 1 f x ¯ f x ¯   f x f x
This quantity will not be infinite even at zero coordinates, since the coordinate components enter the denominator only as a product with the momentum components, and such a product cannot be zero, since this is forbidden by commutation relations and the uncertainty relation.
The photon has a mass equal to zero, so the right side of the equation is also zero, and it does not have a multiplier in the denominator in the solution, so the photon is not localized in space.

4. Path Integral and Second Quantization in Spinor Coordinate Space

Based on the above, we can modify the theory of the integral over trajectories. We will consider it in the notations in which it is presented in [10]. For a free scalar field with sources J(X) the trajectory integral has the form
Z ( J ) = D φ X   e x p i S ( φ X ) = D φ X   e x p i d 4 X L ( φ X ) + J X φ X ) = D φ X   e x p i d 4 X   1 2 φ X 0 2 φ X 1 2 φ X 2 2 φ X 3 2 m 2 φ X 2 + J X φ X
It includes the action of
S ( φ X = d 4 X L ( φ X ) + J X φ X
and the Lagrangian density for the free field
L ( φ X = 1 2 φ X 0 2 φ X 1 2 φ X 2 2 φ X 3 2 m 2 φ X 2
For convenience and clarity, the following notations are introduced
( φ ) 2 = μ φ μ φ = η μ ν μ φ ν φ = 0 φ 2 1 φ 2 2 φ 2 3 φ 2 = φ X 0 2 φ X 1 2 φ X 2 2 φ X 3 2
μ X μ
For the general case the Lagrangian density has the form
L φ X = 1 2 φ X 2 V φ X
where V φ X -polynomial over the field φ X .
Substituting the Lagrangian into the Euler equation
μ δ L δ ( μ φ ) δ L δ φ = 0
the field equation of motion is obtained.
The free field theory is developed for a special kind of polynomial
V φ X = 1 2 m 2 φ 2
L φ = 1 2 [ φ 2 m 2 φ 2 ]  
δ L δ ( μ φ ) = 1 2 δ φ 2 δ ( μ φ ) = 1 2 δ [ 0 φ 2 1 φ 2 2 φ 2 3 φ 2 ] δ ( μ φ ) = ± 1 2 δ μ φ 2 δ ( μ φ ) = ± μ φ
δ L δ φ = 1 2 m 2 δ φ 2 δ φ = m 2 φ
In summary, Euler's equation defines the equation of motion
0 0 φ 0 0 φ 0 0 φ 0 0 φ + m 2 φ = 0
0 2 φ 1 2 φ 2 2 φ 3 2 φ + m 2 φ = 0
2 φ 0 2 φ 1 2 φ 2 2 φ 3 2 φ
2 φ + m 2 φ = 0
( 2 + m 2 ) φ = 0
The notations used here are
2 φ 0 2 φ 1 2 φ 2 2 φ 3 2 φ
2 0 2 1 2 2 2 3 2
Thus, there is a correspondence of the Lagrangian and the equation of motion for the free field
L φ X = 1 2 0 φ X 2 1 φ X 2 2 φ X 2 3 φ X 2 m 2 φ X 2
L φ = 1 2 [ φ 2 m 2 φ 2 ]
L φ = 1 2 [ 0 φ 2 1 φ 2 2 φ 2 3 φ 2 m 2 φ 2 ]
0 2 φ X 1 2 φ X 2 2 φ X 3 2 φ X + m 2 φ X = 0
Our proposal is to replace the Lagrangian in vector coordinate space by the Lagrangian in spinor coordinate space. For this purpose we use the equation of motion in spinor coordinate space and we want to find the Lagrangian for which the Euler equation defines this equation of motion
x 1 x 2 x 0 x 3 φ x + m φ x = 0
1 2 0 3 φ x + m φ x = 0
μ δ L δ ( μ φ x ) δ L δ φ x = 0
For the sake of clarity, we use the same notation for the spinor coordinate derivative as for the vector coordinate derivative; the context allows us to distinguish between them
μ x μ
Let us write the Lagrangian plus sources in the form
L φ x = 1 2 1 φ x 2 φ x 0 φ x 3 φ x ) V ( φ x ) + j x φ x
And let's substitute the Lagrangian into the Euler equation
0 δ L δ ( 0 ) + 1 δ L δ ( 1 ) + 2 δ L δ ( 2 ) + 3 δ L δ ( 3 ) δ L δ φ = 0
1 2 0 3 φ x + 1 ( 2 φ x ) + 2 ( 1 φ x ) 3 ( 0 φ x ) δ L δ φ = 0
For the case of a free field the derivative operators commute, so we can write
1 2 φ x 0 3 φ x δ L δ φ = 0
x 1 x 2 x 0 x 3 φ x δ L δ φ = 0
x 1 x 2 x 0 x 3 φ x δ V φ δ φ = 0
It is pleasant that the Euler equation in invariant form works also in this situation, so that we obtain the desired form of the equation of motion in the spinor coordinate space. It is important that the proposed Lagrangian has a relativistically invariant form, even in the general case, and not only at commuting derivatives. The polynomial has the form
V φ = 1 2 m φ x 2 + g 3 ! φ x 3 + λ 4 ! φ x 4 +
In the case of a free field we restrict ourselves to the first term of the polynomial
V φ = 1 2   m φ x 2
Then the Lagrangian density and the equation of motion for the scalar field in spinor coordinate space have the form
L φ x = 1 2 1 φ x 2 φ x 0 φ x 3 φ x ) 1 2   m φ x 2
1 2 1 2 + 2 1 0 3 3 0 φ x + m φ x = 0
For a free field when the derivative operators commute, we obtain
1 2 0 3 φ x + m φ x = 0
In the spinor equation of motion there is a plus sign before the mass, although in the rest of the paper there was a minus sign. To return to the minus sign it is enough to put a plus sign in front of the polynomial V φ in the Lagrangian.
Now we have to find the integral over the trajectories, which, along with the Lagrangian, includes the sources
Z j = D φ x   e x p i d 4 x L ( φ x ) + j x φ x = D φ x   e x p i d 4 x   1 2 1 φ x 2 φ x 0 φ x 3 φ x ) 1 2 m φ x 2 + j x φ x
The components of spinors are complex, and we have already noted that the derivatives on complex variables are applied to the degree functions, which, most likely, can describe physical fields, respectively, and at integration the finding of a first-form for the degree function of a complex variable can be treated similarly, i.e. as a first-form from the degree function.
It is possible to recover Planck's constant, which provides a transition to the classical limit
Z ( j ) = D φ x   e x p i d 4 x L ( φ x
One of the steps in computing the integral over trajectories in [10] is to find the free propagator from Eq.
( 2 + m 2 ) D ( X Y ) = δ ( X Y )
the solution of which has the form
D X Y = d 4 P ( 2 π ) 4 e i P ( X Y )     P 2 m 2 + i ε    
herewith
δ ( X Y ) = d 4 P ( 2 π ) 4 e i P ( X Y )
In our case, we want to find
Z j = D φ x   e x p i d 4 x   1 2 1 φ x 2 φ x 0 φ x 3 φ x ) 1 2 m φ x 2 + j x φ x
After integration by parts by analogy with [10], Chapter 1.3] we obtain for the special case of a free field
Z j = D φ x   e x p i d 4 x 1 2 φ x ( 1 2 0 3 ) + m φ x + j x φ x
In the process of calculation, it is necessary to find the solution of the equation
1 2 0 3 + m D ( x y ) = δ ( x y )
For this purpose, we pass to the momentum space by means of the integral transformation
φ x = d 4 p ( 2 π ) 4   φ p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
The assumed propagator has the form
D x y = d 4 p ( 2 π ) 4         e i p 0 ( x 1 y 1 ) p 1 ( x 0 y 0 ) + p 2 ( x 3 y 3 ) p 3 ( x 2 y 2 )       p 1 p 2 p 0 p 3 m + i ε
which is verified by substitution into Eq. Here it is assumed that the representation of the delta function
δ x y = d 4 p ( 2 π ) 4   e i p 0 ( x 1 y 1 ) p 1 ( x 0 y 0 ) + p 2 ( x 3 y 3 ) p 3 ( x 2 y 2 )
One can see the difference between the propagators, since in one case m 2 is real and positive, while in spinor space m is complex in general.
We can use the relation
1       p 1 p 2 p 0 p 3 m = p 1 p 2 p 0 p 3 ¯ + m ¯       p 1 p 2 p 0 p 3 ¯ + m ¯     p 1 p 2 p 0 p 3 m     = p 1 p 2 p 0 p 3 ¯ + m ¯         P 2 m 2 + ( m ¯ m ) p 1 p 2 p 0 p 3 = p 1 p 2 p 0 p 3 ¯ + m ¯         P 2 m 2
where
    P 2 P 0 2 P 1 2 P 2 2 P 3 2
in which it is taken into account that the fermion mass is real. Now the propagator has the form
D x = d 4 p ( 2 π ) 4 p 1 p 2 p 0 p 3 ¯ + m ¯         P 2 m 2 + i ε e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
The derivatives of the scalar field on spinor coordinates can be expressed through the derivatives on vector coordinates
0 φ x = φ x x 0 = φ X x x 0 = φ X x X 0 X 0 x x 0 + φ X x X 1 X 1 x x 0 + φ X x X 2 X 2 x x 0 + φ X x X 3 X 3 x x 0 = φ X 0 x 0 ¯ 2 + φ X 1 x 1 ¯ 2 + φ X 2 i x 1 ¯ 2 + φ X 3 x 0 ¯ 2
0 φ x = φ X 0 x 0 ¯ 2 + φ X 1 x 1 ¯ 2 + φ X 2 i x 1 ¯ 2 + φ X 3 x 0 ¯ 2
1 φ x = φ X 0 x 1 ¯ 2 + φ X 1 x 0 ¯ 2 φ X 2 i x 0 ¯ 2 φ X 3 x 1 ¯ 2
2 φ x = φ X 0 x 2 ¯ 2 + φ X 1 x 3 ¯ 2 + φ X 2 i x 3 ¯ 2 + φ X 3 x 2 ¯ 2
3 φ x = φ X 0 x 3 ¯ 2 + φ X 1 x 2 ¯ 2 φ X 2 i x 2 ¯ 2 φ X 3 x 3 ¯ 2
If in the right part to represent the wave function as a plane wave in vector space
φ X = exp P 0 X 0 P 1 X 1 P 2 X 2 P 3 X 3
then in the left part it should be represented as a plane wave of a special form in spinor space
φ x = exp p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 ¯ p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
Only in this case the left and right parts will be dimensionally consistent, e.g.
1 φ x = p 0 x 1 ¯ p 1 x 0 ¯ + p 2 x 3 ¯ p 3 x 2 ¯ p 0
φ X 0 x 0 ¯ 2 = P 0 x 0 ¯ 2 = 1 4 p 0 ¯ p 0 + p 1 ¯ p 1 + p 2 ¯ p 2 + p 3 ¯ p 3 x 0 ¯
In any case, a complete coincidence will not be obtained due to the mismatch of dimensionless exponents of the exponents
p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 ¯ p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 P 0 X 0 P 1 X 1 P 2 X 2 P 3 X 3
Since we call the field under consideration a scalar field, we expect its value to be invariant to Lorentz transformations. But how to formalize this statement and to what exactly does this transformation apply? We propose to consider that the value of a scalar field is the scalar product of the representatives of a spinor field, which is the most fundamental field in nature, and vectors, tensors and, among others, scalars are formed from the spinors representing it. The scalar product is defined by means of the metric tensor of the spinor space. From any two spinors we can obtain a scalar, in general the complex case. But if we want to obtain a scalar with real values, we must impose some restrictions on the original spinors. For example, to any spinor u we can correspond a scalar U taking real values, whose value does not change under the action of the Lorentz transformation on the spinor and the action of the same transformation on the conjugate spinor
U = i u T Σ M M u ¯ = u T S 2 u ¯ = N * u T S 2 ( N * u ¯ )
U = i u 0 * u 1 ¯ u 1 * u 0 ¯ + u 2 * u 3 ¯ u 3 * u 2 ¯
When a spinor and its conjugate spinor are simultaneously rotated or boosted by some angle, the scalar undergoes a rotation or boost by zero angle.
We can find the derivatives of the scalar by the components of the coordinate spinor
U x x μ = u x x μ T S 2 u ¯ + u x T S 2 u x ¯ x μ
The components of the coordinate spinor are complex quantities, the derivative on them is taken virtually formally, since physical fields can be represented by power functions of the components of the coordinate spinor and its conjugate.
What advantages does the transition from the integral over trajectories in vector space to the integral over trajectories in spinor space provide? A possible answer is that it may be easier to compute this integral, or the integration may not lead to divergences. If the spinor coordinate space is indeed more fundamental, and the vector coordinate space is an offspring of it, then we may benefit from this transition in any case.
Now let us move from the scalar field to the field of an electron, that is, the field of a particle with half-integer spin. We will use gamma matrices in the Weyl basis
γ 0 V = 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0                                 γ 1 V = 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0
    γ 2 V = 0 0 0 0 0 i i 0 0 i i 0 0 0 0 0                           γ 3 V = 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0
Let us consider the linear combination of these matrices with components of the momentum vector as coefficients, substituting the expressions of the vector components through the components of the momentum spinor
γ 0 V P 0 + γ 1 V P 1 + γ 2 V P 2 + γ 3 V P 3 = 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 P 0 + 0       0 0       0 0 1 1 0 0 1 1       0 0 0 0 0 P 1 + 0       0 0       0 0 i i 0 0 i i       0 0 0 0 0 P 2 + 0     0 0     0 1     0 0 1 1 0 0 1 0       0 0       0 P 3 = 0       0 0         0   P 0 + P 3       P 1 i P 2 P 1 + i P 2       P 0 P 3     P 0 P 3       P 1 + i P 2 P 1 i P 2       P 0 + P 3 0       0 0         0 = 0       0 0       0         p 0 ¯ p 0 + p 2 ¯ p 2 p 0 ¯ p 1 p 2 ¯ p 3 p 1 ¯ p 0 p 3 ¯ p 2       p 1 ¯ p 1 + p 3 ¯ p 3     p 1 ¯ p 1 + p 3 ¯ p 3       p 0 ¯ p 1 + p 2 ¯ p 3 p 1 ¯ p 0 + p 3 ¯ p 2       p 0 ¯ p 0 + p 2 ¯ p 2 0       0 0       0
= 0       0 0       0   p 0 ¯ p 0       p 0 ¯ p 1 p 1 ¯ p 0       p 1 ¯ p 1     p 1 ¯ p 1       p 0 ¯ p 1 p 1 ¯ p 0       p 0 ¯ p 0 0       0 0       0 + 0       0 0       0   p 2 ¯ p 2       p 2 ¯ p 3 p 3 ¯ p 2       p 3 ¯ p 3     p 3 ¯ p 3       p 2 ¯ p 3 p 3 ¯ p 2       p 2 ¯ p 2 0       0 0       0
= 0       0 0       0   p 0 ¯ p 0       p 0 ¯ p 1 p 1 ¯ p 0       p 1 ¯ p 1     p 1 p 1 ¯ p 1 p 1 ¯ p 1 ¯ p 1       p 1 p 0 ¯ p 1 p 0 ¯ p 0 ¯ p 1 p 0 p 1 ¯ p 0 p 1 ¯ p 1 ¯ p 0       p 0 p 0 ¯ p 0 p 0 ¯ p 0 ¯ p 0 0       0 0       0 + 0       0 0       0   p 2 ¯ p 2       p 2 ¯ p 3 p 3 ¯ p 2       p 3 ¯ p 3     p 3 p 3 ¯ p 3 p 3 ¯ p 3 ¯ p 3       p 3 p 2 ¯ p 3 p 2 ¯ p 2 ¯ p 3 p 2 p 3 ¯ p 2 p 3 ¯ p 3 ¯ p 2       p 2 p 2 ¯ p 2 p 2 ¯ p 2 ¯ p 2 0       0 0       0
= 0       0 0       0   p 0 ¯ p 0       p 0 ¯ p 1 p 1 ¯ p 0       p 1 ¯ p 1     p 1 p 1 ¯       p 1 p 0 ¯ p 0 p 1 ¯       p 0 p 0 ¯ 0       0 0       0 + 0       0 0       0   p 2 ¯ p 2       p 2 ¯ p 3 p 3 ¯ p 2       p 3 ¯ p 3     p 3 p 3 ¯       p 3 p 2 ¯ p 2 p 3 ¯       p 2 p 2 ¯ 0       0 0       0 0       0 0       0   0       0 0       0     p 1 p 1 ¯ p 1 ¯ p 1       p 1 p 0 ¯ p 0 ¯ p 1 p 0 p 1 ¯ p 1 ¯ p 0       p 0 p 0 ¯ p 0 ¯ p 0 0       0 0       0 0       0 0       0   0       0 0       0     p 3 p 3 ¯ p 3 ¯ p 3       p 3 p 2 ¯ p 2 ¯ p 3 p 2 p 3 ¯ p 3 ¯ p 2       p 2 p 2 ¯ p 2 ¯ p 2 0       0 0       0 S V ( p ) K V ( p )
Let us represent the matrix S V ( p ) as a sum of direct products of spinors
S V ( p ) = 0   0 p 1 p 0 p 1 ¯ , p 0 ¯ , 0,0 + p 0 ¯ p 1 ¯ 0 0 0,0 , p 0 , p 1 + 0   0 p 3 p 2 p 3 ¯ , p 2 ¯ , 0,0 + p 2 ¯ p 3 ¯ 0 0 0,0 , p 2 , p 3
For a free field the components of the momentum spinor commute, therefore
γ 0 V P 0 + γ 1 V P 1 + γ 2 V P 2 + γ 3 V P 3 = S V ( p )
Complex mass
m = p 1 p 2 p 0 p 3
does not change at rotations and boosts for an arbitrary complex spinor. Moreover, by a direct check it is possible to check that for an arbitrary spinor
S V p S V p = m ¯ m I = m 2 I
For a free field, when all components of the momentum spinor commute, we can write the relativistic equation of motion of the fermionic field
S V S V φ x = m ¯ m I φ x
Where the matrix of derivatives S V is obtained from the matrix S V p by substitutions
p 1 0                 p 0 1               p 3 2                   p 2 3
p 1 ¯ 0 ¯                 p 0 ¯ 1 ¯               p 3 ¯ 2 ¯                 p 2 ¯ 3 ¯
μ ¯ φ x φ x ¯ x μ ¯
S V = 0   0 0 1 0 ¯ , 1 ¯ , 0 , 0 + 1 ¯ 0 ¯ 0 0 0,0 , 1 , 0 + 0   0 2 3 2 ¯ , 3 ¯ , 0 , 0 + 3 ¯ 2 ¯ 0 0 0,0 , 3 , 2
However, it is generally accepted to write for this field another equation, the Dirac equation, which does not possess the invariance property anymore
( S V m I ) φ x = 0
And for the more general case, when the momentum components do not commute, we need to write the equation
( S V K V m I ) φ x = 0
K V p = 0       0 0       0   0       0 0       0     p 1 p 1 ¯ p 1 ¯ p 1       p 1 p 0 ¯ p 0 ¯ p 1 p 0 p 1 ¯ p 1 ¯ p 0       p 0 p 0 ¯ p 0 ¯ p 0 0       0 0       0 + 0       0 0       0   0       0 0       0     p 3 p 3 ¯ p 3 ¯ p 3       p 3 p 2 ¯ p 2 ¯ p 3 p 2 p 3 ¯ p 3 ¯ p 2       p 2 p 2 ¯ p 2 ¯ p 2 0       0 0       0
K V = 0       0 0       0   0       0 0       0     0 0 ¯ 0 ¯ 0       0 1 ¯ + 1 ¯ 0 1 0 ¯ + 0 ¯ 1       1 1 ¯ 1 ¯ 1 0       0 0       0 + 0       0 0       0   0       0 0       0     2 2 ¯ 2 ¯ 2       2 3 ¯ + 3 ¯ 2 3 2 ¯ + 2 ¯ 3       3 3 ¯ 3 ¯ 3 0       0 0       0
Further we will consider the equation of motion for a free field
( S V m I ) φ x = 0
We again want to find the integral over trajectories
Z j = D φ x   e x p i d 4 x L ( φ x ) + j x φ x
for which we need the Lagrangian, from which the Euler equation is derived equation of motion
( S V m I ) φ x = 0
It is suggested to use the Lagrangian
L = 1 2 φ x T S V φ x 1 2 m φ x T φ x
Let us substitute the Lagrangian into the Euler equation and obtain the equation of motion
0 δ L δ ( 0 ) + 1 δ L δ ( 1 ) + 2 δ L δ ( 2 ) + 3 δ L δ ( 3 ) δ L δ φ = 0
1 2 S V φ x + m φ x = 0
Since the Lagrangian includes, along with the derivatives of μ , the derivatives of μ ¯ , it is logical to use a different definition of Euler's equation
0 δ L δ ( 0 ) + 0 ¯ δ L δ ( 0 ¯ ) + 1 δ L δ ( 1 ) + 1 ¯ δ L δ ( 1 ¯ ) + 2 δ L δ ( 2 ) + 2 ¯ δ L δ ( 2 ¯ ) + 3 δ L δ ( 3 ) + 3 ¯ δ L δ ( 3 ¯ ) δ L δ φ = 0
Then for the free field case when the derivative operators commute with each other, we obtain the equation of motion
S V φ x + m φ x = 0
If the derivative operators do not commute, additional terms will appear in the equation of motion in the form of matrices similar to the K V matrix, and these additional terms will not necessarily coincide with K V . In this connection it is necessary to consider the Lagrangian as more fundamental notion than the equation of motion and to derive the equation of motion from the Lagrangian, i.e. to take as a basis not the derivation of the equation of motion in momentum space, with what we started, but to take as an axiom the form of the Lagrangian in the form of field derivatives in the relativistically invariant form. Then, if to follow the invariance principle quite strictly, we should start from the product of two matrices, i.e. to use the Lagrangian
L = 1 2 φ x T S V S V φ x m 2 φ x T φ x
Or, not limited to fermions,
L = 1 2 φ x T S V S V φ x m m ¯ φ x T φ x
Nevertheless, further we will search for the integral over trajectories in the simplest case with the originally proposed Lagrangian and in addition assume commutativity of all derivative operators
Z j = D φ x   e x p i d 4 x   1 2 φ x T S V φ x 1 2 m φ x T φ x + j x T φ x
After integration by parts, we presumably obtain
Z j = D φ x   e x p i d 4 x 1 2 φ x T S V + m I φ x + j x φ x
Then it is necessary to find the solution of the equation
S V + m I D ( x ) = I δ ( x )
For this purpose, we pass to the momentum space by means of the integral transformation
φ x = d 4 p ( 2 π ) 4   φ p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
We get the equation
S V p m I D V ( p ) = I
with the decision
D V p = S V p + m ¯ I     P 2 m ¯ m
Indeed
S V p m I ( S V p + m ¯ I )     P 2 m ¯ m = ( P 2 m ¯ m ) I     P 2 m ¯ m = I
Here we use the equality, which is valid for an arbitrary complex spinor p
S V ( p ) m I S V ( p ) + m ¯ I = P 2 I m m ¯ S V p m ¯ m I = ( P 2 m 2 ) I
P μ = 1 2 p S μ p  
    P 2 = P 0 2 P 1 2 P 2 2 P 3 2
It is based on the correlation verified earlier in our work
p 1 p 2 p 0 p 3 p 1 ¯ p 2 ¯ p 0 ¯ p 3 ¯ = P 0 2 P 1 2 P 2 2 P 3 2
it is also taken into account that we consider fermions whose mass is real.
As a result, the propagator has the form
D V x = d 4 p ( 2 π ) 4   S V p + m ¯ I     P 2 m ¯ m e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
here we assume the validity of the relation
δ x = d 4 p ( 2 π ) 4   e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
In the case of a fermion, the mass in integration is a fixed real quantity, and it can be considered negative for the electron and positive for the positron. Theoretically, the mass can be complex or purely imaginary. If we put mass equal to zero, it may be possible to apply this Lagrangian to describe massless particles. I wonder if there are particles with complex or purely imaginary mass. In the latter case, the square of the mass will still be positive and the particle will satisfy the Klein-Gordon equation. Such particles can interact among themselves, but not with particles whose mass is real.
Let's return to the question about the use of completely relativistically invariant Lagrangian
L = 1 2 φ x T S V S V φ x m 2 φ x T φ x
Let's find the product of matrices
S V p S V p =
0   0 p 1 p 0 p 1 ¯ , p 0 ¯ , 0 , 0 + p 0 ¯ p 1 ¯ 0 0 0,0 , p 0 , p 1 + 0   0 p 3 p 2 p 3 ¯ , p 2 ¯ , 0 , 0 + p 2 ¯ p 3 ¯ 0 0 0,0 , p 2 , p 3
0   0 p 1 p 0 p 1 ¯ , p 0 ¯ , 0 , 0 + p 0 ¯ p 1 ¯ 0 0 0,0 , p 0 , p 1 + 0   0 p 3 p 2 p 3 ¯ , p 2 ¯ , 0 , 0 + p 2 ¯ p 3 ¯ 0 0 0,0 , p 2 , p 3 =
p 1 ¯ p 2 ¯ p 0 ¯ p 3 ¯ 0   0 p 1 p 0 0,0 , p 2 , p 3 + p 0 p 3 p 1 p 2 p 0 ¯ p 1 ¯ 0 0 p 3 ¯ , p 2 ¯ , 0 , 0 +
p 3 ¯ p 0 ¯ p 2 ¯ p 1 ¯ 0   0 p 3 p 2 0,0 , p 0 , p 1 + p 2 p 1 p 3 p 0 p 2 ¯ p 3 ¯ 0 0 p 1 ¯ , p 0 ¯ , 0 , 0 =
m ¯ 0   0 p 1 p 0 0,0 , p 2 , p 3 m p 0 ¯ p 1 ¯ 0 0 p 3 ¯ , p 2 ¯ , 0 , 0 m ¯ 0   0 p 3 p 2 0,0 , p 0 , p 1 + m p 2 ¯ p 3 ¯ 0 0 p 1 ¯ , p 0 ¯ , 0 , 0 =
m 0   0 p 1 p 0 0,0 , p 2 , p 3 p 0 ¯ p 1 ¯ 0 0 p 3 ¯ , p 2 ¯ , 0 , 0 0   0 p 3 p 2 0,0 , p 0 , p 1 + p 2 ¯ p 3 ¯ 0 0 p 1 ¯ , p 0 ¯ , 0 , 0
m S V V p
The assumption that the following equalities hold is used
p 1 p 2 p 0 p 3 = p 2 p 1 p 3 p 0 = m
p 1 ¯ p 2 ¯ p 0 ¯ p 3 ¯ = p 2 ¯ p 1 ¯ p 3 ¯ p 0 ¯ = m ¯
m ¯ = m
Further we find the product of matrices
S V V p S V V p =
0   0 p 1 p 0 0,0 , p 2 , p 3 p 0 ¯ p 1 ¯ 0 0 p 3 ¯ , p 2 ¯ , 0 , 0 0   0 p 3 p 2 0,0 , p 0 , p 1 + p 2 ¯ p 3 ¯ 0 0 p 1 ¯ , p 0 ¯ , 0 , 0
0   0 p 1 p 0 0,0 , p 2 , p 3 p 0 ¯ p 1 ¯ 0 0 p 3 ¯ , p 2 ¯ , 0 , 0 0   0 p 3 p 2 0,0 , p 0 , p 1 + p 2 ¯ p 3 ¯ 0 0 p 1 ¯ , p 0 ¯ , 0 , 0
= p 2 p 1 p 3 p 0 0   0 p 1 p 0 0,0 , p 2 , p 3 + p 3 ¯ p 0 ¯ p 2 ¯ p 1 ¯ p 0 ¯ p 1 ¯ 0 0 p 3 ¯ , p 2 ¯ , 0 , 0
+ p 0 p 3 p 1 p 2 0   0 p 3 p 2 0,0 , p 0 , p 1 + p 2 ¯ p 1 ¯ p 3 ¯ p 0 ¯ p 2 ¯ p 3 ¯ 0 0 p 1 ¯ , p 0 ¯ , 0 , 0 = m 0   0 p 1 p 0 0,0 , p 2 , p 3 0   0 p 3 p 2 0,0 , p 0 , p 1 + m ¯ p 2 ¯ p 3 ¯ 0 0 p 1 ¯ , p 0 ¯ , 0 , 0 p 0 ¯ p 1 ¯ 0 0 p 3 ¯ , p 2 ¯ , 0 , 0
= m 0 0 0 0 0 0 0 0 0 0 0 0 p 1 p 2 p 1 p 3 p 0 p 2 p 0 p 3 0 0 0 0 0 0 0 0 0 0 0 0 p 3 p 0 p 3 p 1 p 2 p 0 p 2 p 1
+ m ¯ p 2 ¯ p 1 ¯ p 2 ¯ p 0 ¯ p 3 ¯ p 1 ¯ p 3 ¯ p 0 ¯ 0 0 0 0 0 0 0 0 0 0 0 0 p 0 ¯ p 3 ¯ p 0 ¯ p 2 ¯ p 1 ¯ p 3 ¯ p 1 ¯ p 2 ¯ 0 0 0 0 0 0 0 0 0 0 0 0
= m 0 0 0 0 0 0 0 0 0 0 0 0 p 1 p 2 p 3 p 0 0 0 p 0 p 3 + p 2 p 1 + m ¯ p 2 ¯ p 1 ¯ p 0 ¯ p 3 ¯ 0 0 p 3 ¯ p 0 ¯ + p 1 ¯ p 2 ¯ 0 0 0 0 0 0 0 0 0 0 0 0
= m ¯ m ¯ 0 0 m ¯ m ¯ 0 0 0 0 0 0 0 0 m m 0 0 m m
Again we use the equality
p 1 p 2 p 0 p 3 p 1 ¯ p 2 ¯ p 0 ¯ p 3 ¯ = P 0 2 P 1 2 P 2 2 P 3 2 = P 2
and consider that the mass of the fermion is real, i.e.
p 1 p 2 p 0 p 3 = p 1 ¯ p 2 ¯ p 0 ¯ p 3 ¯
p 1 p 2 p 0 p 3 p 1 p 2 p 0 p 3 = p 1 ¯ p 2 ¯ p 0 ¯ p 3 ¯ p 1 ¯ p 2 ¯ p 0 ¯ p 3 ¯ = P 2
therefore, the relations are valid
S V V p S V V p = P 2 0 0 P 2 0 0 0 0 0 0 0 0 P 2 0 0 P 2 = P 2 I
S V V p m I S V V p + m I = P 2 I m 2 I = P 2 m 2 I
S V V p m I ( S V V p + m I ) P 2 m 2 = I
But the main advantage of the obtained matrix is the following
S V V p = 0   0 p 1 p 0 0,0 , p 2 , p 3 p 0 ¯ p 1 ¯ 0 0 p 3 ¯ , p 2 ¯ , 0 , 0 0   0 p 3 p 2 0,0 , p 0 , p 1 + p 2 ¯ p 3 ¯ 0 0 p 1 ¯ , p 0 ¯ , 0 , 0 =
0 0 0 0 0 0 0 0 0 0 0 0 p 1 p 2 p 1 p 3 p 0 p 2 p 0 p 3 0 0 0 0 0 0 0 0 0 0 0 0 p 3 p 0 p 3 p 1 p 2 p 0 p 2 p 1
+ p 2 ¯ p 1 ¯ p 2 ¯ p 0 ¯ p 3 ¯ p 1 ¯ p 3 ¯ p 0 ¯ 0 0 0 0 0 0 0 0 0 0 0 0 p 0 ¯ p 3 ¯ p 0 ¯ p 2 ¯ p 1 ¯ p 3 ¯ p 1 ¯ p 2 ¯ 0 0 0 0 0 0 0 0 0 0 0 0
= 0 0 0 0 0 0 0 0 0 0 0 0 p 1 p 2 p 3 p 0 0 0 p 0 p 3 + p 2 p 1 + p 2 ¯ p 1 ¯ p 0 ¯ p 3 ¯ 0 0 p 3 ¯ p 0 ¯ + p 1 ¯ p 2 ¯ 0 0 0 0 0 0 0 0 0 0 0 0
= m ¯ 0 0 m ¯ 0 0 0 0 0 0 0 0 m 0 0 m
This matrix does not change at rotations and boosts, so it can be stated that the equation of motion, e.g., in the form of
S V V m ¯ 0 0 m ¯ 0 0 0 0 0 0 0 0 m 0 0 m φ x = 0
where
S V V = 0   0 0 1 0,0 , 3 , 2 1 ¯ 0 ¯ 0 0 2 ¯ , 3 ¯ , 0 , 0 0   0 2 3 0,0 , 1 , 0 + 3 ¯ 2 ¯ 0 0 0 ¯ , 1 ¯ , 0 , 0
is truly relativistically invariant, respectively we can use the invariant Lagrangian
L = 1 2 φ x T S V V φ x m φ x T φ x
to which corresponds the relativistically invariant propagator of the boson having a real mass, which is negative for the electron and positive for the positron
D V V x = d 4 p ( 2 π ) 4   S V V p + m I     P 2 m 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
Let us compare the propagator in spinor space with the propagator of the fermion given in [10], formula II.2.22 and formula II.5.18]
S X = d 4 P ( 2 π ) 4 e i P X     γ μ P μ m I = d 4 P ( 2 π ) 4 γ μ P μ + m I         P 2 m 2 e i P X
In [10] this formula is obtained by applying the second quantization procedure or using Grassmann integrals. The results are similar, but the integration here is performed in the vector momentum space. The Dirac equation and the corresponding Lagrangian are not relativistically invariant. Besides, here the mass is considered always real and positive, but then it is not clear how electron and positron differ from the point of view of this formula.
Let us consider in detail the derivation of the expression for the fermion propagator in [10], Sec. II.2. It is based on the assumption of relativistic invariance of the Dirac equation and therefore the calculations are carried out in the rest frame, and then the result is extended to an arbitrary frame of reference. Thus for the field spinor u the spinor u_   u γ 0 is defined and it is asserted that the value of
u γ 0 u = u 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 u
is a Lorentz scalar. But it is not so, since in the spinor space the scalar is formed exclusively by the scalar product of two spinors, where the metric tensor of the spinor space is included
u Σ M M u = u 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 u
there are no other ways to construct a scalar in the spinor space.
Nevertheless, this fact and the fact of non-invariance of the Dirac equation itself do not cancel the value of the second quantization procedure and the final form of the fermion propagator, which allows to make accurate predictions of the experimental results.
We hope that the proposed Lagrangian for the spinor coordinate space can find application in the calculation of the integral over trajectories, but already in the spinor space. Whether such a calculation in spinor space has an advantage over the calculation of the integral over trajectories in vector space can be shown by their real comparison.
By analogy with the propagator of a photon, more precisely of a massive vector meson, given in [10], formula I.5.3
D ν λ X = d 4 P ( 2 π ) 4     η ν λ + P ν P λ / m 2     P 2 m 2 e i P X
we can assume the propagator form in the spinor space without revealing for compactness the expression of the momentum vector components through the momentum spinor components
D ν λ x = d 4 p ( 2 π ) 4 η ν λ + P ν P λ / m 2         P 2 m ¯ m     e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
Among other things, the equation
S V V m ¯ 0 0 m ¯ 0 0 0 0 0 0 0 0 m 0 0 m φ x = 0
can be modified to take into account the electromagnetic potential, the electron charge is taken as a unit
p 0 1 + a 0                       p 1 0 + a 1                       p 2 3 + a 2                     p 3 2 + a 3
p 0 ¯ 1 ¯ + a 0 ¯                   p 1 ¯ 0 ¯ + a 1 ¯                     p 2 ¯ 3 ¯ + a 2 ¯                   p 3 ¯ 2 ¯ + a 3 ¯
S V V = 0   0 0 + a 1 1 + a 0 0,0 , 3 + a 2 , 2 a 3 1 ¯ + a 0 ¯ 0 ¯ a 1 ¯ 0 0 2 ¯ + a 3 ¯ , 3 ¯ + a 2 ¯ , 0 , 0 0   0 2 + a 3 3 + a 2 0,0 , 1 + a 0 , 0 a 1 + 3 ¯ + a 2 ¯ 2 ¯ a 3 ¯ 0 0 0 ¯ + a 1 ¯ , 1 ¯ + a 0 ¯ , 0 , 0
and apply, in particular, to analyze the radiation spectrum of a hydrogen-like atom.
Let us formulate again the difference between the equations, the second of which is derived from the Dirac equation with gamma matrices in the Weyl basis
S V V m ¯ 0 0 m ¯ 0 0 0 0 0 0 0 0 m 0 0 m φ x = 0
S V m I φ x = 0
The difference is, the matrix S V V p (p) remains unchanged under any rotations and boosts applied to the spinor p , while the matrix S V p (p) changes under any rotations and boosts.
S V = 0   0 0 1 0 ¯ , 1 ¯ , 0 , 0 + 1 ¯ 0 ¯ 0 0 0,0 , 1 , 0 + 0   0 2 3 2 ¯ , 3 ¯ , 0 , 0 + 3 ¯ 2 ¯ 0 0 0,0 , 3 , 2
S V V = 0   0 0 1 0,0 , 3 , 2 1 ¯ 0 ¯ 0 0 2 ¯ , 3 ¯ , 0 , 0 0   0 2 3 0,0 , 1 , 0 + 3 ¯ 2 ¯ 0 0 0 ¯ , 1 ¯ , 0 , 0
Equally radically different are the corresponding Lagrangians and propagators.
By analogy with [10], Chapter II.2] we will carry out the procedure of second quantization of the fermion field. Let us write the equation
S V V m ¯ 0 0 m ¯ 0 0 0 0 0 0 0 0 m 0 0 m φ x = 0
in the momentum space, for which we apply the integral transformation
φ x = d 4 p ( 2 π ) 4   φ p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
Let's substitute the wave function into the equation and obtain
S V V p m ¯ 0 0 m ¯ 0 0 0 0 0 0 0 0 m 0 0 m φ p = 0
S V V p = 0   0 p 1 p 0 0,0 , p 2 , p 3 p 0 ¯ p 1 ¯ 0 0 p 3 ¯ , p 2 ¯ , 0 , 0 0   0 p 3 p 2 0,0 , p 0 , p 1 + p 2 ¯ p 3 ¯ 0 0 p 1 ¯ , p 0 ¯ , 0 , 0
Let us define two sets of four reference spinors
          u 1 = 0   0 p 1 p 0           u 2 = p 0 ¯ p 1 ¯ 0 0 u 3 = 0   0 p 3 p 2           u 4 = p 2 ¯ p 3 ¯ 0 0
          v 1 = p 1 p 0 0 0           v 2 = 0 0 p 0 ¯ p 1 ¯ v 3 = p 3 p 2 0 0           v 4 = 0 0 p 2 ¯ p 3 ¯
v 1 = γ 0 V u 1 v 2 = γ 0 V u 2 v 3 = γ 0 V u 3 v 4 = γ 0 V u 4
where
γ 0 V = 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0
And let's express the matrix through them
S V V p = 0   0 p 1 p 0 0,0 , p 2 , p 3 p 0 ¯ p 1 ¯ 0 0 p 3 ¯ , p 2 ¯ , 0 , 0 0   0 p 3 p 2 0,0 , p 0 , p 1 + p 2 ¯ p 3 ¯ 0 0 p 1 ¯ , p 0 ¯ , 0 , 0 = u 1 p v 4 + p u 2 p v 3 + p u 3 p v 2 + p + u 4 p v 1 + p
Developing the idea of invariance, we pass to the set of reference spinors with wider filling, but continuing to form matrices possessing the invariance property
          u 1 = p 3 p 2 p 1 p 0           u 2 = p 2 p 3 p 0 p 1 u 3 = p 1 p 0 p 3 p 2           u 4 = p 0 p 1 p 2 p 3
          v 1 = p 1 p 0 p 3 p 2               v 2 = p 0 p 1 p 2 p 3 v 3 = p 3 p 2 p 1 p 0               v 4 = p 2 p 3 p 0 p 1
Let's express through the reference spinors the matrix
S R p = p 3 p 2 p 1 p 0 p 0 , p 1 , p 2 , p 3 p 1 p 0 p 3 p 2 p 2 , p 3 , p 0 , p 1
+ p 1 p 0 p 3 p 2 p 2 , p 3 , p 0 , p 1 p 3 p 2 p 1 p 0 p 0 , p 1 , p 2 , p 3
= u 1 p u 4 T p u 3 p u 2 T p + v 1 p v 4 T p v 3 p v 2 T p
S R p = p 3 p 2 p 1 p 0 p 0 , p 1 , p 2 , p 3 p 1 p 0 p 3 p 2 p 2 , p 3 , p 0 , p 1
+ p 1 p 0 p 3 p 2 p 2 , p 3 , p 0 , p 1 p 3 p 2 p 1 p 0 p 0 , p 1 , p 2 , p 3
= p 3 p 0 p 3 p 1 p 2 p 0 p 2 p 1 p 3 p 2 p 3 p 3 p 2 p 2 p 2 p 3 p 1 p 0 p 1 p 1 p 0 p 0 p 0 p 1 p 1 p 2 p 1 p 3 p 0 p 2 p 0 p 3 p 1 p 2 p 1 p 3 p 0 p 2 p 0 p 3 p 1 p 0 p 1 p 1 p 0 p 0 p 0 p 1 p 3 p 2 p 3 p 3 p 2 p 2 p 2 p 3 p 3 p 0 p 3 p 1 p 2 p 0 p 2 p 1
+ p 1 p 2 p 1 p 3 p 0 p 2 p 0 p 3 p 1 p 0 p 1 p 1 p 0 p 0 p 0 p 1 p 3 p 2 p 3 p 3 p 2 p 2 p 2 p 3 p 3 p 0 p 3 p 1 p 2 p 0 p 2 p 1 p 3 p 0 p 3 p 1 p 2 p 0 p 2 p 1 p 3 p 2 p 3 p 3 p 2 p 2 p 2 p 3 p 1 p 0 p 1 p 1 p 0 p 0 p 0 p 1 p 1 p 2 p 1 p 3 p 0 p 2 p 0 p 3
= p 3 p 0 + p 1 p 2 0 0 p 2 p 1 p 0 p 3 0 0 0 0 0 0 0 0 p 1 p 2 p 3 p 0 0 0 p 0 p 3 + p 2 p 1
+ p 1 p 2 p 3 p 0 0 0 p 0 p 3 + p 2 p 1 0 0 0 0 0 0 0 0 p 3 p 0 + p 1 p 2 0 0 p 2 p 1 p 0 p 3
= m + m 0 0 m + m 0 0 0 0 0 0 0 0 m + m 0 0 m + m
and matrix
S R p = p 0 p 1 p 2 p 3 p 3 , p 2 , p 1 , p 0 p 2 p 3 p 0 p 1 p 1 , p 0 , p 3 , p 2
+ p 2 p 3 p 0 p 1 p 1 , p 0 , p 3 , p 2 p 0 p 1 p 2 p 3 p 3 , p 2 , p 1 , p 0
= u 4 p u 1 T p u 2 p u 3 T p + v 4 p v 1 T p v 2 p v 3 T p
S R p = p 0 p 1 p 2 p 3 p 3 , p 2 , p 1 , p 0 p 2 p 3 p 0 p 1 p 1 , p 0 , p 3 , p 2
+ p 2 p 3 p 0 p 1 p 1 , p 0 , p 3 , p 2 p 0 p 1 p 2 p 3 p 3 , p 2 , p 1 , p 0 = = p 0 p 3 p 0 p 2 p 1 p 3 p 1 p 2 p 0 p 1 p 0 p 0 p 1 p 1 p 1 p 0 p 2 p 3 p 2 p 2 p 3 p 3 p 3 p 2 p 2 p 1 p 2 p 0 p 3 p 1 p 3 p 0 p 2 p 1 p 2 p 0 p 3 p 1 p 3 p 0 p 2 p 3 p 2 p 2 p 3 p 3 p 3 p 2 p 0 p 1 p 0 p 0 p 1 p 1 p 1 p 0 p 0 p 3 p 0 p 2 p 1 p 3 p 1 p 2
+ p 2 p 1 p 2 p 0 p 3 p 1 p 3 p 0 p 2 p 3 p 2 p 2 p 3 p 3 p 3 p 2 p 0 p 1 p 0 p 0 p 1 p 1 p 1 p 0 p 0 p 3 p 0 p 2 p 1 p 3 p 1 p 2 p 0 p 3 p 0 p 2 p 1 p 3 p 1 p 2 p 0 p 1 p 0 p 0 p 1 p 1 p 1 p 0 p 2 p 3 p 2 p 2 p 3 p 3 p 3 p 2 p 2 p 1 p 2 p 0 p 3 p 1 p 3 p 0 = p 0 p 3 + p 2 p 1 0 0 p 1 p 2 p 3 p 0 0 0 0 0 0 0 0 0 p 2 p 1 p 0 p 3 0 0 p 3 p 0 + p 1 p 2 + p 2 p 1 p 0 p 3 0 0 p 3 p 0 + p 1 p 2 0 0 0 0 0 0 0 0 p 0 p 3 + p 2 p 1 0 0 p 1 p 2 p 3 p 0 = m + m 0 0 m + m 0 0 0 0 0 0 0 0 m + m 0 0 m + m
here
m = p 1 p 2 p 0 p 3
Let us deviate from the canonical approach to the definition of antisymmetry of the field and consider the following scheme of reasoning. Let us decompose the fermion field into plane waves with operator coefficients
φ x = d 4 p 2 π 2  
d 1 p u 1 p + i d 2 p u 3 p + i b 2 p u 2 ¯ p +   b 1 p u 4 ¯ p + d 4 p v 1 p + i d 3 p v 3 p + i b 3 p v 2 ¯ p + b 4 p v 4 ¯ p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
+ b 1 * p u 1 ¯ p + i b 2 * p u 3 ¯ p + i d 2 * p u 2 p + d 1 * p u 4 p + b 4 * p v 1 ¯ p + i b 3 * p v 3 ¯ p + i d 3 * p v 2 p + d 4 * p v 4 p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
φ T x = d 4 p 2 π 2  
d 1 p u 1 T p + i d 2 p u 3 T p + i b 2 p u 2 + p + b 1 p u 4 + p + d 4 p v 1 T p + i d 3 p v 3 T p + i b 3 p v 2 + p + b 4 p v 4 + p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 + b 1 * p u 1 + p + i b 2 * p u 3 + p + i d 2 * p u 2 T p + d 1 * p u 4 T p + b 4 * p v 1 + p + i b 3 * p v 3 + p + i d 3 * p v 2 T p + d 4 * p v 4 T p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
φ + x = d 4 p 2 π 2  
d 1 * p u 1 + p i d 2 * p u 3 + p i b 2 * p u 2 T p + b 1 * p u 4 T p + d 4 * p v 1 + p i d 3 * p v 3 + p i b 3 * p v 2 T p + b 4 * p v 4 T p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
+ b 1 p u 1 T p i b 2 p u 3 T p i d 2 p u 2 + p + d 1 p u 4 + p + b 4 p v 1 T p i b 3 p v 3 T p i d 3 p v 2 + p + d 4 p v 4 + p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
Let's impose the anticommutation conditions on the operator coefficients
  b 1 p b 1 * p + b 1 p b 1 * p = b 1 * p b 1 p + b 1 * p b 1 p = δ p p
d 1 p d 1 * p + d 1 p d 1 * p = d 1 * p d 1 p + d 1 * p d 1 p = δ p p
  d 2 p d 2 * p + d 2 p d 2 * p = b 2 * p b 2 p + b 2 * p b 2 p = δ p p
b 2 p b 2 * p + b 2 p b 2 * p = d 2 * p d 2 p + d 2 * p d 2 p = δ p p
  d 3 p d 3 * p + d 3 p d 3 * p = b 3 * p b 3 p + b 3 * p b 3 p = δ p p
b 3 p b 3 * p + b 3 p b 3 * p = d 3 * p d 3 p + d 3 * p d 3 p = δ p p
  b 4 p b 4 * p + b 4 p b 4 * p = b 4 * p b 4 p + b 4 * p b 4 p = δ p p
d 4 p d 4 * p + d 4 p d 4 * p = d 4 * p d 4 p + d 4 * p d 4 p = δ p p
We consider the rest anticommutators to be equal to zero. Then we can write the expression for the anticommutator of the field with its transposed version but without its complex conjugation
φ x , φ T x =   d 4 p 2 π 2   d 4 p 2 π 2 =
d 1 p u 1 p + i d 2 p u 3 p + i b 2 p u 2 ¯ p +   b 1 p u 4 ¯ p + d 4 p v 1 p + i d 3 p v 3 p + i b 3 p v 2 ¯ p + b 4 p v 4 ¯ p
b 1 * p u 1 + p + i b 2 * p u 3 + p + i d 2 * p u 2 T p + d 1 * p u 4 T p + b 4 * p v 1 + p + i b 3 * p v 3 + p + i d 3 * p v 2 T p + d 4 * p v 4 T p
e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
+
d 1 p u 1 p + i d 2 p u 3 p + i b 2 p u 2 ¯ p +   b 1 p u 4 ¯ p + d 4 p v 1 p + i d 3 p v 3 p + i b 3 p v 2 ¯ p + b 4 p v 4 ¯ p
b 1 * p u 1 + p + i b 2 * p u 3 + p + i d 2 * p u 2 T p + d 1 * p u 4 T p + d 4 * p v 1 + p + i d 3 * p v 3 + p + i d 3 * p v 2 T p + d 4 * p v 4 T p
e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
+
b 1 * p u 1 ¯ p + i b 2 * p u 3 ¯ p + i d 2 * p u 2 p + d 1 * p u 4 p + b 4 * p v 1 ¯ p + i b 3 * p v 3 ¯ p + i d 3 * p v 2 p + d 4 * p v 4 p
d 1 p u 1 T p + i d 2 p u 3 T p + i b 2 p u 2 + p + b 1 p u 4 + p + d 4 p v 1 T p + i d 3 p v 3 T p + i b 3 p v 2 + p + b 4 p v 4 + p
e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
+
b 1 * p u 1 ¯ p + i b 2 * p u 3 ¯ p + i d 2 * p u 2 p + d 1 * p u 4 p + b 4 * p v 1 ¯ p + i b 3 * p v 3 ¯ p + i d 3 * p v 2 p + d 4 * p v 4 p
d 1 p u 1 T p + i d 2 p u 3 T p + i b 2 p u 2 + p + b 1 p u 4 + p + d 4 p v 1 T p + i d 3 p v 3 T p + i b 3 p v 2 + p + b 4 p v 4 + p
e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
=   d 4 p 2 π 2   d 4 p 2 π 2 d 1 p u 1 p d 1 * p u 4 T p + d 1 p u 1 p d 1 * p u 4 T p d 2 p u 3 p d 2 * p u 2 T p d 2 p u 3 p d 2 * p u 2 T p + e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 +   b 1 p u 4 ¯ p b 1 * p u 1 + p +   b 1 p u 4 ¯ p b 1 * p u 1 + p b 2 p u 2 ¯ p b 2 * p u 3 + p b 2 p u 2 ¯ p b 2 * p u 3 + p + e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
+   d 4 p 2 π 2   d 4 p 2 π 2 b 1 * p u 1 ¯ p b 1 p u 4 + p + b 1 * p u 1 ¯ p b 1 p u 4 + p b 2 * p u 3 ¯ p b 2 p u 2 + p b 2 * p u 3 ¯ p b 2 p u 2 + p + e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 + d 1 * p u 4 p d 1 p u 1 T p + d 1 * p u 4 p d 1 p u 1 T p d 2 * p u 2 p d 2 p u 3 T p d 2 * p u 2 p d 2 p u 3 T p + e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
=   d 4 p 2 π 2 u 1 p u 4 T p + u 3 p u 2 T p + e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2 + u 4 ¯ p u 1 + p + u 2 ¯ p u 3 + p + e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
+   d 4 p 2 π 2 u 1 ¯ p u 4 + p + u 3 ¯ p u 2 + p + e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2 + u 4 p u 1 T p + u 2 p u 3 T p + e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
=   d 4 p 2 π 2 u 1 p u 4 T p u 3 p u 2 T p + + u 4 p u 1 T p u 2 p u 3 T p + + e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2 + u 4 ¯ p u 1 + p u 2 ¯ p u 3 + p + + u 1 ¯ p u 4 + p u 3 ¯ p u 2 + p + + e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
=   d 4 p 2 π 2 u 1 p u 4 T p u 3 p u 2 T p + v 1 p v 4 T p v 3 p v 2 T p + u 4 p u 1 T p u 2 p u 3 T p + v 4 p v 1 T p v 2 p v 3 T p e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2 + u 4 ¯ p u 1 + p u 2 ¯ p u 3 + p + v 4 ¯ p v 1 + p v 2 ¯ p v 3 + p + u 1 ¯ p u 4 + p u 3 ¯ p u 2 + p + v 1 ¯ p v 4 + p v 3 ¯ p v 2 + p e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
d 4 p 2 π 4   S R p + S R p e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2 +
d 4 p 2 π 4   S R ¯ p + S R ¯ p e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
=
d 4 p 2 π 4   4 m 0 0 m 0 0 0 0 0 0 0 0 m 0 0 m e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2 +
d 4 p 2 π 4   4 m ¯ 0 0 m ¯ 0 0 0 0 0 0 0 0 m ¯ 0 0 m ¯ e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
= 4 m I δ x x + 4 m ¯ I δ x x
We will consider this relation as a proof of the anti-symmetry of the fermion wave function under the stipulated anticommutation relations.
Let us calculate the total energy of the fermion field
E = P 0 = d 4 x φ + x S 0 φ x
= d 4 x       d 4 p 2 π 2 d 4 p 2 π 2
d 1 * p u 1 + p i d 2 * p u 3 + p i b 2 * p u 2 T p + b 1 * p u 4 T p + d 4 * p v 1 + p i d 3 * p v 3 + p i b 3 * p v 2 T p + b 4 * p v 4 T p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 + b 1 p u 1 T p i b 2 p u 3 T p i d 2 p u 2 + p + d 1 p u 4 + p + b 4 p v 1 T p i b 3 p v 3 T p i d 3 p v 2 + p + d 4 p v 4 + p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
d 1 p u 1 T p + i d 2 p u 3 T p + i b 2 p u 2 + p + b 1 p u 4 + p + d 4 p v 1 T p + i d 3 p v 3 T p + i b 3 p v 2 + p + b 4 p v 4 + p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 + b 1 * p u 1 + p + i b 2 * p u 3 + p + i d 2 * p u 2 T p + d 1 * p u 4 T p + b 4 * p v 1 + p + i b 3 * p v 3 + p + i d 3 * p v 2 T p + d 4 * p v 4 T p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
= d 4 x     d 4 p 2 π 2   d 4 p 2 π 2
d 1 * p u 1 + p i d 2 * p u 3 + p i b 2 * p u 2 T p + b 1 * p u 4 T p + d 4 * p v 1 + p i d 3 * p v 3 + p i b 3 * p v 2 T p + b 4 * p v 4 T p d 1 p u 1 p + i d 2 p u 3 p + i b 2 p u 2 ¯ p +   b 1 p u 4 ¯ p + d 4 p v 1 p + i d 3 p v 3 p + i b 3 p v 2 ¯ p +   b 4 p v 4 ¯ p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 + b 1 p u 1 T p i b 2 p u 3 T p i d 2 p u 2 + p + d 1 p u 4 + p + b 4 p v 1 T p i b 3 p v 3 T p i d 3 p v 2 + p + d 4 p v 4 + p b 1 * p u 1 ¯ p + i b 2 * p u 3 ¯ p + i d 2 * p u 2 p + d 1 * p u 4 p + b 4 * p v 1 ¯ p + i b 3 * p v 3 ¯ p + i d 3 * p v 2 p + d 4 * p v 4 p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
=   d 4 p 2 π 2   d 4 p 2 π 2
d 1 * p u 1 + p i d 2 * p u 3 + p i b 2 * p u 2 T p + b 1 * p u 4 T p + d 4 * p v 1 + p i d 3 * p v 3 + p i b 3 * p v 2 T p + b 4 * p v 4 T p d 1 p u 1 p + i d 2 p u 3 p + i b 2 p u 2 ¯ p +   b 1 p u 4 ¯ p + d 4 p v 1 p + i d 3 p v 3 p + i b 3 p v 2 ¯ p +   b 4 p v 4 ¯ p δ p p + b 1 p u 1 T p i b 2 p u 3 T p i d 2 p u 2 + p + d 1 p u 4 + p + b 4 p v 1 T p i b 3 p v 3 T p i d 3 p v 2 + p + d 4 p v 4 + p b 1 * p u 1 ¯ p + i b 2 * p u 3 ¯ p + i d 2 * p u 2 p + d 1 * p u 4 p + b 4 * p v 1 ¯ p + i b 3 * p v 3 ¯ p + i d 3 * p v 2 p + d 4 * p v 4 p δ p p
=   d 4 p 2 π 2 d 1 * p d 1 p u 1 + p u 1 p + d 1 p d 1 * p u 4 + p u 4 p + b 1 p b 1 * p u 1 T p u 1 ¯ p + b 1 * p b 1 p u 4 T p u 4 ¯ p + b 2 p b 2 * p u 3 T p u 3 ¯ p + b 2 * p b 2 p u 2 T p u 2 ¯ p + d 2 * p d 2 p u 3 + p u 3 p + d 2 p d 2 * p u 2 + p u 2 p + d 4 * p d 4 p v 1 + p v 1 p + d 4 p d 4 * p v 4 + p v 4 p + b 4 p b 4 * p v 1 T p v 1 ¯ p + b 4 * p b 4 p v 4 T p v 4 ¯ p + b 3 p b 3 * p v 3 T p v 3 ¯ p + b 3 * p b 3 p v 2 T p v 2 ¯ p + d 3 * p d 3 p v 3 + p v 3 p + d 3 p d 3 * p v 2 + p v 2 p
= d 4 p 2 π 4   e 0 p b 1 p b 1 * p + b 1 * p b 1 p + d 1 * p d 1 p + d 1 p d 1 * p + b 2 p b 2 * p + d 2 * p d 2 p + b 2 * p b 2 p + d 2 p d 2 * p + b 4 p b 4 * p + b 4 * p b 4 p + d 4 * p d 4 p + d 4 p d 4 * p + b 3 p b 3 * p + d 3 * p d 3 p + b 3 * p b 3 p + d 3 p d 3 * p
= 8 d 4 p 2 π 4   e 0 p δ 0 = 8 d 4 x 2 π 4 d 4 p 2 π 4   e 0 p  
here
e 0 p = p 0 ¯ p 0 + p 1 ¯ p 1 + p 2 ¯ p 2 + p 3 ¯ p 3
Each summand in brackets represents the operator of the number of particles with a certain reference spinor. The operator's action consists of consecutive application of the annihilation operator and the operator of the birth of a particle. On initial examination, it would appear that the energy associated with zero-point fluctuations in the vacuum has been overlooked. However, an examination of the final expression reveals that the field always possesses a constant energy, regardless of the particles that contribute to it. This constant energy of the field can be interpreted as the energy of zero-point fluctuations of the vacuum.
It is important that all the above deductions are valid in any frame of reference, while the proof of anticommutativity of the fermion field in [10] is carried out for the rest frame.
The following relations were taken into account in the derivation
  d 1 p d 1 * p + d 1 p d 1 * p = b 1 * p b 1 p + b 1 * p b 1 p = δ 0
b 1 p b 1 * p + b 1 p d 1 * p = d 1 * p d 1 p + d 1 * p b 1 p = δ 0
  d 2 p b 2 * p + d 2 p b 2 * p = b 2 * p d 2 p + b 2 * p d 2 p = δ 0
b 2 p d 2 * p + b 2 p d 2 * p = d 2 * p b 2 p + d 2 * p b 2 p = δ 0
  d 3 p b 3 * p + d 3 p b 3 * p = b 3 * p d 3 p + b 3 * p d 3 p = δ 0
b 3 p d 3 * p + b 3 p d 3 * p = d 3 * p b 3 p + d 3 * p b 3 p = δ 0
  d 4 p d 4 * p + d 4 p d 4 * p = b 4 * p b 4 p + b 4 * p b 4 p = δ 0
b 4 p b 4 * p + b 4 p b 4 * p = d 4 * p d 4 p + d 4 * p d 4 p = δ 0
  d 1 p d 1 * p = 1 2 δ 0                   b 1 * p b 1 p = 1 2 δ 0                     b 1 p b 1 * p = 1 2 δ 0                     d 1 * p d 1 p = 1 2 δ 0
  d 2 p b 2 * p = 1 2 δ 0                   b 2 * p d 2 p = 1 2 δ 0                     b 2 p d 2 * p = 1 2 δ 0                       d 2 * p b 2 p = 1 2 δ 0
  d 3 p b 3 * p = 1 2 δ 0                   b 3 * p d 3 p = 1 2 δ 0                     b 3 p d 3 * p = 1 2 δ 0                       d 3 * p b 3 p = 1 2 δ 0
  d 4 p d 4 * p = 1 2 δ 0                   b 4 * p b 4 p = 1 2 δ 0                     b 4 p b 4 * p = 1 2 δ 0                     d 4 * p d 4 p = 1 2 δ 0
δ 0 = d 4 x ( 2 π ) 4  
Other components of the total field momentum are calculated by the formula
P μ = d 4 x φ + x S μ φ x
Total momentum
P T ( P 0 , P 1 , P 2 , P 3 )
is a vector in Minkowski space. Multiplied by the ratio of the electron or positron charge to their mass, it turns into the vector of the total current of the fermionic field
I = ± e m e P
But the real interest is not the total current of the fermionic field, but the density of this current as a function of coordinates
J μ = ± e m e φ + x S μ φ x = ± e m e F μ x
where
F μ x = φ + x S μ φ x
is a four-dimensional probability density current, which is transformed as a four-dimensional vector by Lorentz transformations. Multiplication by ± e m e transforms it into a four-dimensional current density. Let us perform a series of transformations analogous to those presented by Dirac in [11], Lecture 11].
P 0 = d 4 x 2 π 2 φ + x S 0 φ x =
1 2 m d 4 x 2 π 2 φ + x S 0 S R φ x
= 1 2 m d 4 x 2 π 2 φ + x S 0 d 4 p 2 π 2   S R p φ p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 = 1 2 m d 4 p 2 π 2   d 4 x 2 π 2 φ + x S 0 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 S R p φ p = 1 2 m d 4 p 2 π 2   φ + p S 0 S R p φ p = d 4 p 2 π 2   φ + p S 0 φ p = d 4 p 2 π 2   φ + p φ p = d 4 p 2 π 2   φ 0 + p φ 0 p + φ 1 + p φ 1 p + φ 2 + p φ 2 p + φ 3 + p φ 3 p
For an arbitrary component of the total momentum we have
P μ = d 4 p 2 π 2   φ + p S μ φ p
Following Dirac's argument in [11], the value of
P 0 = H = d 4 p 2 π 2   φ 0 + p φ 0 p + φ 1 + p φ 1 p + φ 2 + p φ 2 p + φ 3 + p φ 3 p
can be treated as either a Hamiltonian or a total energy operator, with φ 0 + p representing the birth operator and φ μ p representing the annihilation operator.
In [11] the quantization procedure includes the use of one definite Lorentzian reference frame, i.e. it is not invariant. In our case all deductions are valid in any reference frame in the spinor space, and it means invariance to change of reference frames in the Minkowski space also.
The following relations are used in the transformations
S R φ x = 2 m φ x
φ x = 1 2 m S R φ x
φ x = d 4 p 2 π 2   φ p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2  
φ p = d 4 x ( 2 π ) 2   φ x e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
δ p = d 4 x ( 2 π ) 2   e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
δ x = d 4 p ( 2 π ) 2   e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
φ + p = d 4 x ( 2 π ) 2   φ + x e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
S R φ x = d 4 p ( 2 π ) 2   S R p φ p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
S R p = 2 m I
S R = 2 3 0 1 1 , 0 , 3 , 2 0 1 2 3 3 , 2 , 1 , 0
+ 0 1 2 3 3 , 2 , 1 , 0 2 3 0 1 1 , 0 , 3 , 2
S R p = p 3 p 2 p 1 p 0 p 0 , p 1 , p 2 , p 3 p 1 p 0 p 3 p 2 p 2 , p 3 , p 0 , p 1
+ p 1 p 0 p 3 p 2 p 2 , p 3 , p 0 , p 1 p 3 p 2 p 1 p 0 p 0 , p 1 , p 2 , p 3
The chain of reasoning can be organized in a slightly different way as well
P 0 = d 4 x 2 π 2 φ + x S 0 φ x = 1 2 m ¯ 1 2 m d 4 x 2 π 2 S R φ x + S R φ x = 1 2 m ¯ 1 2 m d 4 x 2 π 2 d 4 p 2 π 2   S R p φ p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 +
d 4 p 2 π 2   S R p φ p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 = 1 2 m ¯ 1 2 m d 4 x 2 π 2 d 4 p 2 π 2   S R p φ p + e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
d 4 p 2 π 2   S R p φ p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
= 1 2 m ¯ 1 2 m d 4 p 2 π 2   S R p φ p + d 4 p 2 π 2   S R p φ p δ p p
= 1 2 m ¯ 1 2 m d 4 p 2 π 2   d 4 p 2 π 2   S R p φ p + S R p φ p δ p p
= 1 2 m ¯ 1 2 m d 4 p 2 π 2   S R p φ p + S R p φ p
= 1 2 m ¯ 1 2 m d 4 p 2 π 2   φ p + S R p + S R p φ p
= 1 2 m ¯ 1 2 m d 4 p 2 π 2   φ p + S R ¯ p T S R p φ p
= 1 2 m ¯ 1 2 m d 4 p 2 π 2   φ p + 2 ( p 1 p 2 p 3 p 0 ) ¯ I T 2 ( p 1 p 2 p 3 p 0 ) I φ p
= 1 m ¯ 1 m d 4 p 2 π 2   φ p + p 1 p 2 p 3 p 0 ¯ p 1 p 2 p 3 p 0 φ p
= 1 m ¯ 1 m d 4 p 2 π 2 P 0 2 P 1 2 P 2 2 P 3 2 φ p + φ p
= d 4 p 2 π 2 φ p + φ p
Here it is taken into account that
S R p = 2 ( p 1 p 2 p 3 p 0 ) I
p 1 p 2 p 0 p 3 ¯ p 1 p 2 p 0 p 3 = P 0 2 P 1 2 P 2 2 P 3 2 = m ¯ m =
= S 0 P 0 S 1 P 1 S 2 P 2 S 3 P 3 S 0 P 0 + S 1 P 1 + S 2 P 2 + S 3 P 3
Let us draw an analogy between our approach and the relations given in [12], Volume 1, Chapter 3, Section 3.3.1]. There it is noted that the birth and annihilation operators of the fermionic field must satisfy such commutation relations that the equality expressing translational invariance is satisfied
φ X + A = e i P T A φ X e i P T A
which in differential form is written as
μ φ X = i [ P μ , φ X ]
On the basis of these relations the anticommutation relations between the birth and annihilation operators are derived. The coordinates here are the components of the Minkowski vector space.
We can perform a similar consideration in the spinor coordinate space, describing for it the translational invariance by the relations
φ x + a = e i p 0 a 1 p 1 a 0 + p 2 a 3 p 3 a 2 φ x e i p 0 a 1 p 1 a 0 + p 2 a 3 p 3 a 2
0 φ x = i [ p 1 , φ x ]
1 φ x = i [ p 0 , φ x ]
2 φ x = i [ p 3 , φ x ]
3 φ x = i p 2 , φ x
Or we can try to approximate the formulation of translational invariance in Minkowski space by means of the formula
φ x + a =
e i p 0 a 1 p 1 a 0 + p 2 a 3 p 3 a 2 ¯ p 0 a 1 p 1 a 0 + p 2 a 3 p 3 a 2 φ x e i p 0 a 1 p 1 a 0 + p 2 a 3 p 3 a 2 ¯ p 0 a 1 p 1 a 0 + p 2 a 3 p 3 a 2
Let us calculate the total mass of the fermion field
M = d 4 x φ T x φ x =
d 4 x   d 4 p 2 π 2 d 4 p 2 π 2  
d 1 p u 1 T p + i d 2 p u 3 T p + i b 2 p u 2 + p + b 1 p u 4 + p + d 4 p v 1 T p + i d 3 p v 3 T p + i b 3 p v 2 + p + b 4 p v 4 + p
b 1 * p u 1 ¯ p + i b 2 * p u 3 ¯ p + i d 2 * p u 2 p + d 1 * p u 4 p + b 4 * p v 1 ¯ p + i b 3 * p v 3 ¯ p + i d 3 * p v 2 p + d 4 * p v 4 p
e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
+ d 4 x   d 4 p 2 π 2 d 4 p 2 π 2  
b 1 * p u 1 + p + i b 2 * p u 3 + p + i d 2 * p u 2 T p + d 1 * p u 4 T p + b 4 * p v 1 + p + i b 3 * p v 3 + p + i d 3 * p v 2 T p + d 4 * p v 4 T p
d 1 p u 1 p + i d 2 p u 3 p + i b 2 p u 2 ¯ p +   b 1 p u 4 ¯ p + d 4 p v 1 p + i d 3 p v 3 p + i b 3 p v 2 ¯ p + b 4 p v 4 ¯ p
e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
=   d 4 p 2 π 2 d 4 p 2 π 2  
d 1 p u 1 T p + i d 2 p u 3 T p + i b 2 p u 2 + p + b 1 p u 4 + p + d 4 p v 1 T p + i d 3 p v 3 T p + i b 3 p v 2 + p + b 4 p v 4 + p
b 1 * p u 1 ¯ p + i b 2 * p u 3 ¯ p + i d 2 * p u 2 p + d 1 * p u 4 p + b 4 * p v 1 ¯ p + i b 3 * p v 3 ¯ p + i d 3 * p v 2 p + d 4 * p v 4 p
δ p p
+   d 4 p 2 π 2 d 4 p 2 π 2  
b 1 * p u 1 + p + i b 2 * p u 3 + p + i d 2 * p u 2 T p + d 1 * p u 4 T p + b 4 * p v 1 + p + i b 3 * p v 3 + p + i d 3 * p v 2 T p + d 4 * p v 4 T p
d 1 p u 1 p + i d 2 p u 3 p + i b 2 p u 2 ¯ p +   b 1 p u 4 ¯ p + d 4 p v 1 p + i d 3 p v 3 p + i b 3 p v 2 ¯ p + b 4 p v 4 ¯ p
δ p p
=   d 4 p 2 π 2 d 1 p d 1 * p u 1 T p u 4 p + b 1 p b 1 * p u 4 + p u 1 ¯ p d 2 p d 2 * p u 3 T p u 2 p b 2 p b 2 * p u 2 + p u 3 ¯ p + d 4 p d 4 * p v 1 T p v 4 p + b 4 p b 4 * p v 4 + p v 1 ¯ p d 3 p d 3 * p v 3 T p v 2 p b 3 p b 3 * p v 2 + p v 3 ¯ p
+   d 4 p 2 π 2 b 1 * p   b 1 p u 1 + p u 4 ¯ p + d 1 * p d 1 p u 4 T p u 1 p b 2 * p b 2 p u 3 + p u 2 ¯ p d 2 * p d 2 p u 2 p u 3 p + b 4 * p b 4 p v 1 + p v 4 ¯ p + d 4 * p d 4 p v 4 T p v 1 p b 3 * p b 3 p v 3 + p v 2 ¯ p d 3 * p d 3 p v 2 T p v 3 p
= d 4 p 2 π 4   ( m + m ¯ ) d 1 p d 1 * p + b 1 p b 1 * p + d 4 p d 4 * p + b 4 p b 4 * p + b 2 * p b 2 p + d 2 * p d 2 p + d 3 p d 3 * p + b 3 p b 3 * p + b 1 * p   b 1 p + d 1 * p d 1 p + b 4 * p b 4 p + d 4 * p d 4 p + b 2 * p b 2 p + d 2 * p d 2 p + b 3 * p b 3 p + d 3 * p d 3 p
= d 4 p 2 π 4 8 ( m + m ¯ ) δ 0 = d 4 x 2 π 4 d 4 p 2 π 4   8 ( m + m ¯ )
The ratios used in the derivation are
u 1 T p u 4 p = p 3 p 0 + p 2 p 1 + p 1 p 2 p 0 p 3 = 2 m
u 4 T p u 1 p = p 0 p 3 + p 1 p 2 + p 2 p 1 p 3 p 0 = 2 m
u 3 T p u 2 p = p 1 p 2 + p 0 p 3 + p 3 p 0 p 2 p 1 = 2 m
u 2 T p u 3 p = p 2 p 1 + p 3 p 0 + p 0 p 3 p 1 p 2 = 2 m
u 1 T p u 4 p = p 3 p 0 + p 2 p 1 + p 1 p 2 p 0 p 3 = 2 m
v 1 T p v 4 p = p 1 p 2 p 3 p 0 p 0 p 3 + p 2 p 1 = 2 m
u 1 + p u 4 ¯ p = p 3 p 0 + p 2 p 1 + p 1 p 2 p 0 p 3 ¯ = 2 m ¯
  d 1 p d 1 * p + d 1 p d 1 * p = b 1 * p b 1 p + b 1 * p b 1 p = δ 0
b 1 p b 1 * p + b 1 p d 1 * p = d 1 * p d 1 p + d 1 * p b 1 p = δ 0
  d 2 p b 2 * p + d 2 p b 2 * p = b 2 * p d 2 p + b 2 * p d 2 p = δ 0
b 2 p d 2 * p + b 2 p d 2 * p = d 2 * p b 2 p + d 2 * p b 2 p = δ 0
  d 3 p b 3 * p + d 3 p b 3 * p = b 3 * p d 3 p + b 3 * p d 3 p = δ 0
b 3 p d 3 * p + b 3 p d 3 * p = d 3 * p b 3 p + d 3 * p b 3 p = δ 0
  d 4 p d 4 * p + d 4 p d 4 * p = b 4 * p b 4 p + b 4 * p b 4 p = δ 0
b 4 p b 4 * p + b 4 p b 4 * p = d 4 * p d 4 p + d 4 * p d 4 p = δ 0
  d 1 p d 1 * p = 1 2 δ 0                   b 1 * p b 1 p = 1 2 δ 0                     b 1 p b 1 * p = 1 2 δ 0                     d 1 * p d 1 p = 1 2 δ 0
  d 2 p b 2 * p = 1 2 δ 0                   b 2 * p d 2 p = 1 2 δ 0                     b 2 p d 2 * p = 1 2 δ 0                       d 2 * p b 2 p = 1 2 δ 0
  d 3 p b 3 * p = 1 2 δ 0                   b 3 * p d 3 p = 1 2 δ 0                     b 3 p d 3 * p = 1 2 δ 0                       d 3 * p b 3 p = 1 2 δ 0
  d 4 p d 4 * p = 1 2 δ 0                   b 4 * p b 4 p = 1 2 δ 0                     b 4 p b 4 * p = 1 2 δ 0                     d 4 * p d 4 p = 1 2 δ 0
δ 0 = d 4 x ( 2 π ) 4   Let us give an interpretation of the operator coefficients for this approach
          u 1 = p 3 p 2 p 1 p 0           u 2 = p 2 p 3 p 0 p 1 u 3 = p 1 p 0 p 3 p 2           u 4 = p 0 p 1 p 2 p 3
m u 1 = p 2 p 1 + p 3 p 0 = m
m u 2 = p 3 p 0 + p 2 p 1 = m
m u 3 = p 0 p 3 + p 1 p 2 = m
m u 4 = p 1 p 2 + p 0 p 3 = m
                v 1 = p 1 p 0 p 3 p 2           v 2 = p 0 p 1 p 2 p 3 v 3 = p 3 p 2 p 1 p 0           v 4 = p 2 p 3 p 0 p 1
m v 1 = p 0 p 3 p 1 p 2 = m
m v 2 = p 1 p 2 p 0 p 3 = m
m v 3 = p 2 p 1 p 3 p 0 = m
m v 4 = p 3 p 0 p 2 p 1 = m
φ x = d 4 p 2 π 2  
d 1 p u 1 p + i d 2 p u 3 p + i b 2 p u 2 ¯ p +   b 1 p u 4 ¯ p + d 4 p v 1 p + i d 3 p v 3 p + i b 3 p v 2 ¯ p + b 4 p v 4 ¯ p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
+ b 1 * p u 1 ¯ p + i b 2 * p u 3 ¯ p + i d 2 * p u 2 p + d 1 * p u 4 p + b 4 * p v 1 ¯ p + i b 3 * p v 3 ¯ p + i d 3 * p v 2 p + d 4 * p v 4 p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
d 1 * p creates and d 1 p destroys a particle u 1 p = p 3 p 2 p 1 p 0 with mass m , spin up and momentum in the interval d 4 p , d 1 * p d 1 p is the operator of the number of such particles
b 1 p creates and b 1 * p destroys a particle u 1 ¯ p = p 3 ¯ p 2 ¯ p 1 ¯ p 0 ¯ with mass m ¯ , spin up and momentum in the interval d 4 p , b 1 p b 1 * p is the operator of the number of such particles
d 1 p creates and d 1 * p destroys a particle u 4 p = p 0 p 1 p 2 p 3 with mass m , spin up and momentum in the interval d 4 p , d 1 p d 1 * p is the operator of the number of such particles
b 1 * p creates and b 1 p destroys a particle u 4 ¯ p = p 0 ¯ p 1 ¯ p 2 ¯ p 3 ¯ with mass m ¯ , spin up and momentum in the interval d 4 p , b 1 * p b 1 p is the operator of the number of such particles
Note that u 1 p and u 4 p are translated into each other by a linear transformation, this is also true for other pairs of spinors
u 4 = 0     0 0     0   0     1 1     0     0     1 1     0   0   0 0 0 u 1
u 1 = 0     0 0     0   0     1 1     0     0     1 1     0   0   0 0 0 u 4
It is known [10], formula II.1.30] that the charge conjugation operation transforms an electron into a positron with a change of the sign of the charge. Let us apply the charge conjugation to the reference spinor
  0 0   0 0   0 i i 0     0 i i 0   0 0 0 0 u 1 =   0 0   0 0   0 i i 0     0 i i 0   0 0 0 0 p 3 p 2 p 1 p 0 = i p 0 p 1 p 2 p 3 = i u 4
As a result u 1 not only transforms to u 4 , but also changes a sign of mass due to the imaginary unit in the charge conjugation matrix. This confirms our thesis that the charge conjugation synchronously changes signs of charge and mass.
The properties of all particles and operators are summarized in a table
creates destroys particle number mass spin wave sign
d 1 * p d 1 p u 1 p = p 3 p 2 p 1 p 0 d 1 * p d 1 p m up +
d 1 p d 1 * p u 4 p = p 0 p 1 p 2 p 3 d 1 p d 1 * p m up -
b 1 p b 1 * p u 1 ¯ p = p 3 ¯ p 2 ¯ p 1 ¯ p 0 ¯ b 1 p b 1 * p m ¯ up -
b 1 * p b 1 p u 4 ¯ p = p 0 ¯ p 1 ¯ p 2 ¯ p 3 ¯ b 1 * p b 1 p m ¯ up +
d 4 * p d 4 p v 1 p = p 1 p 0 p 3 p 2 d 4 * p d 4 m down +
d 4 p d 4 * p v 4 p = p 2 p 3 p 0 p 1 d 4 p d 4 * p m down -
b 4 p b 4 * p v 1 ¯ p = p 1 ¯ p 0 ¯ p 3 ¯ p 2 ¯ b 4 p b 4 * p m ¯ down -
b 4 * p b 4 p v 4 ¯ p = p 2 ¯ p 3 ¯ p 0 ¯ p 1 ¯ b 4 * p b 4 p m ¯ down +
d 2 * p d 2 p i u 3 p = i p 1 p 0 p 3 p 2 d 2 * p d 2 p m up +
d 2 p d 2 * p i u 2 p = i p 2 p 3 p 0 p 1 d 2 p d 2 * p m up -
b 2 p b 2 * p i u 3 ¯ p = i p 1 ¯ p 0 ¯ p 3 ¯ p 2 ¯ b 2 p b 2 * p m ¯ up -
b 2 * p b 2 p i u 2 ¯ p = i p 2 ¯ p 3 ¯ p 0 ¯ p 1 ¯ b 2 * p b 2 p m ¯ up +
d 3 * p d 3 p i v 3 p = i p 3 p 2 p 1 p 0 d 3 * p d 3 p m down +
d 3 p d 3 * p i v 2 p = i p 0 p 1 p 2 p 3 d 3 p d 3 * p m down -
b 3 p b 3 * p i v 3 ¯ p = i p 3 ¯ p 2 ¯ p 1 ¯ p 0 ¯   b 3 p b 3 * p m ¯ down -
b 3 * p b 3 p i v 2 ¯ p = i p 0 ¯ p 1 ¯ p 2 ¯ p 3 ¯ b 3 * p b 3 p m ¯ down +
By the words d 1 p destroys the particle u 1 p it should be understood that this operator transforms this particle into the particle u 4 p , and the operator d 1 * p performs the reverse transformation of u 4 p into u 1 p . Since both of these particles have the same mass, the total mass of the fermionic field does not change from these transformations. The mass m itself can have any sign.
If the operator d 1 p acts on the particle u 1 p , it transforms it into the particle u 4 p , the action on any other particle gives zero.
Let us see what result we get if we apply the canonical definition of anticommutativity of the fermionic field.
φ x , φ + x =   d 4 p 2 π 2   d 4 p 2 π 2
d 1 p u 1 p + i d 2 p u 3 p + i b 2 p u 2 ¯ p +   b 1 p u 4 ¯ p + d 4 p v 1 p + i d 3 p v 3 p + i b 3 p v 2 ¯ p + b 4 p v 4 ¯ p
d 1 * p u 1 + p i d 2 * p u 3 + p i b 2 * p u 2 T p + b 1 * p u 4 T p + d 4 * p v 1 + p i d 3 * p v 3 + p i b 3 * p v 2 T p + b 4 * p v 4 T p
e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
+
d 1 p u 1 p + i d 2 p u 3 p + i b 2 p u 2 ¯ p +   b 1 p u 4 ¯ p + d 4 p v 1 p + i d 3 p v 3 p + i b 3 p v 2 ¯ p + b 4 p v 4 ¯ p
d 1 * p u 1 + p i d 2 * p u 3 + p i b 2 * p u 2 T p + b 1 * p u 4 T p + d 4 * p v 1 + p i d 3 * p v 3 + p i b 3 * p v 2 T p + b 4 * p v 4 T p
e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
+
b 1 * p u 1 ¯ p + i b 2 * p u 3 ¯ p + i d 2 * p u 2 p + d 1 * p u 4 p + b 4 * p v 1 ¯ p + i b 3 * p v 3 ¯ p + i d 3 * p v 2 p + d 4 * p v 4 p
b 1 p u 1 T p i b 2 p u 3 T p i d 2 p u 2 + p + d 1 p u 4 + p + b 4 p v 1 T p i b 3 p v 3 T p i d 3 p v 2 + p + d 4 p v 4 + p
e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
+
b 1 * p u 1 ¯ p + i b 2 * p u 3 ¯ p + i d 2 * p u 2 p + d 1 * p u 4 p + b 4 * p v 1 ¯ p + i b 3 * p v 3 ¯ p + i d 3 * p v 2 p + d 4 * p v 4 p
b 1 p u 1 T p i b 2 p u 3 T p i d 2 p u 2 + p + d 1 p u 4 + p + b 4 p v 1 T p i b 3 p v 3 T p i d 3 p v 2 + p + d 4 p v 4 + p
e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
=   d 4 p 2 π 2   d 4 p 2 π 2 d 1 p u 1 p d 1 * p u 1 + p + d 1 p u 1 p d 1 * p u 1 + p + + d 2 p u 3 p d 2 * p u 3 + p + d 2 p u 3 p d 2 * p u 3 + p + e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 +   b 1 p u 4 ¯ p b 1 * p u 4 T p +   b 1 p u 4 ¯ p b 1 * p u 4 T p + + b 2 p u 2 ¯ p b 2 * p u 2 T p + b 2 p u 2 ¯ p i b 2 * p u 2 T p + e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
+
  d 4 p 2 π 2   d 4 p 2 π 2 b 1 * p u 1 ¯ p b 1 p u 1 T p + b 1 * p u 1 ¯ p b 1 p u 1 T p + + b 2 * p u 3 ¯ p b 2 p u 3 T p + b 2 * p u 3 ¯ p b 2 p u 3 T p + e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 + d 1 * p u 4 p d 1 p u 4 + p + d 1 * p u 4 p d 1 p u 4 + p + + d 2 * p u 2 p d 2 p u 2 + p + d 2 * p u 2 p d 2 p u 2 + p + e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
=   d 4 p 2 π 2 u 1 p u 1 + p + + u 3 p u 3 + p + e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2 + u 4 ¯ p u 4 T p + + u 2 ¯ p u 2 T p + e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
+   d 4 p 2 π 2 u 1 ¯ p u 1 T p + + u 3 ¯ p u 3 T p + e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2 + u 4 p u 4 + p + + u 2 p u 2 + p + e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
=   d 4 p 2 π 2 u 1 p u 1 + p + u 3 p u 3 + p + u 4 p u 4 + p + u 2 p u 2 + p + e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2 + u 4 ¯ p u 4 T p + u 2 ¯ p u 2 T p + u 1 ¯ p u 1 T p + u 3 ¯ p u 3 T p + e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
=   d 4 p 2 π 2 u 1 p u 1 + p + u 2 p u 2 + p + u 3 p u 3 + p + u 4 p u 4 + p + v 1 p v 1 + p + v 2 p v 2 + p + v 3 p v 3 + p + v 4 p v 4 + p e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2 + u 1 ¯ p u 1 + p + u 2 ¯ p u 2 + p + u 3 ¯ p u 3 + p + u 4 ¯ p u 4 + p + v 1 ¯ p v 1 + p + v 2 ¯ p v 2 + p + v 3 ¯ p v 3 + p + v 4 ¯ p v 4 + p e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
=
d 4 p 2 π 4   T R p + T R p e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
+
d 4 p 2 π 4   T R ¯ p + T R ¯ p e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
=
d 4 p 2 π 4   4 e ( p ) 0 0 e ( p ) 0 0 0 0 0 0 0 0 e ( p ) 0 0 e ( p ) e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
+
d 4 p 2 π 4   4 e ( p ) 0 0 e ( p ) 0 0 0 0 0 0 0 0 e ( p ) 0 0 e ( p ) e i p 0 x 1 x 1 p 1 x 0 x 0 + p 2 x 3 x 3 p 3 x 2 x 2
= 4 e ( p ) I δ x x + 4 e ( p ) I δ x x
where
T R p = u 1 p u 1 + p + u 2 p u 2 + p + u 3 p u 3 + p + u 4 p u 4 + p
T R p = v 1 p v 1 + p + v 2 p v 2 + p + v 3 p v 3 + p + v 4 p v 4 + p
T R p + T R p + T R ¯ p + T R ¯ p =
4 p 0 p 0 ¯ + p 1 p 1 ¯ + p 2 p 2 ¯ + p 3 p 3 ¯ 1       0 0       1   0       0 0       0     0       0 0       0 1       0 0       1 = 4 e ( p ) I
In deriving this result, the following relations are taken into account
T R p + T R p = u 1 p u 1 + p + u 2 p u 2 + p + u 3 p u 3 + p + u 4 p u 4 + p
+ v 1 p v 1 + p + v 2 p v 2 + p + v 3 p v 3 + p + v 4 p v 4 + p =
p 3 p 2 p 1 p 0 p 3 ¯ , p 2 ¯ , p 1 ¯ , p 0 ¯ + p 2 p 3 p 0 p 1 p 2 ¯ , p 3 ¯ , p 0 ¯ , p 1 ¯
+ p 1 p 0 p 3 p 2 p 1 ¯ , p 0 ¯ , p 3 ¯ , p 2 ¯ + p 0 p 1 p 2 p 3 p 0 ¯ , p 1 ¯ , p 2 ¯ , p 3 ¯ +
p 1 p 0 p 3 p 2 p 1 ¯ , p 0 ¯ , p 3 ¯ , p 2 ¯ + p 0 p 1 p 2 p 3 p 0 ¯ , p 1 ¯ , p 2 ¯ , p 3 ¯
+ p 3 p 2 p 1 p 0 p 3 ¯ , p 2 ¯ , p 1 ¯ , p 0 ¯ + p 2 p 3 p 0 p 1 p 2 ¯ , p 3 ¯ , p 0 ¯ , p 1 ¯ =
p 3 p 3 ¯       p 3 p 2 ¯ p 2 p 3 ¯       p 2 p 2 ¯   p 3 p 1 ¯       p 3 p 0 ¯ p 2 p 1 ¯       p 2 p 0 ¯     p 1 p 3 ¯     p 1 p 2 ¯ p 0 p 3 ¯       p 0 p 2 ¯ p 1 p 1 ¯       p 1 p 0 ¯ p 0 p 1 ¯       p 0 p 0 ¯ + p 2 p 2 ¯   p 2 p 3 ¯ p 3 p 2 ¯       p 3 p 3 ¯   p 2 p 0 ¯       p 2 p 1 ¯ p 3 p 0 ¯       p 3 p 1 ¯     p 0 p 2 ¯     p 0 p 3 ¯ p 1 p 2 ¯       p 1 p 3 ¯ p 0 p 0 ¯       p 0 p 1 ¯ p 1 p 0 ¯       p 1 p 1 ¯
+ p 1 p 1 ¯       p 1 p 0 ¯ p 0 p 1 ¯       p 0 p 0 ¯   p 1 p 3 ¯       p 1 p 2 ¯ p 0 p 3 ¯       p 0 p 2 ¯     p 3 p 1 ¯     p 3 p 0 ¯ p 2 p 1 ¯       p 2 p 0 ¯ p 3 p 3 ¯       p 3 p 2 ¯ p 2 p 3 ¯       p 2 p 2 ¯ + p 0 p 0 ¯   p 0 p 1 ¯ p 1 p 0 ¯       p 1 p 1 ¯   p 0 p 2 ¯       p 0 p 3 ¯ p 1 p 2 ¯       p 1 p 3 ¯     p 2 p 0 ¯     p 2 p 1 ¯ p 3 p 0 ¯       p 3 p 1 ¯ p 2 p 2 ¯       p 2 p 3 ¯ p 3 p 2 ¯       p 3 p 3 ¯
+ p 1 p 1 ¯       p 1 p 0 ¯ p 0 p 1 ¯       p 0 p 0 ¯   p 1 p 3 ¯       p 1 p 2 ¯ p 0 p 3 ¯       p 0 p 2 ¯     p 3 p 1 ¯     p 3 p 0 ¯ p 2 p 1 ¯       p 2 p 0 ¯ p 3 p 3 ¯       p 3 p 2 ¯ p 2 p 3 ¯       p 2 p 2 ¯ + p 0 p 0 ¯   p 0 p 1 ¯ p 1 p 0 ¯       p 1 p 1 ¯   p 0 p 2 ¯       p 0 p 3 ¯ p 1 p 2 ¯       p 1 p 3 ¯     p 2 p 0 ¯     p 2 p 1 ¯ p 3 p 0 ¯       p 3 p 1 ¯ p 2 p 2 ¯       p 2 p 3 ¯ p 3 p 2 ¯       p 3 p 3 ¯
+ p 3 p 3 ¯       p 3 p 2 ¯ p 2 p 3 ¯       p 2 p 2 ¯   p 3 p 1 ¯       p 3 p 0 ¯ p 2 p 1 ¯       p 2 p 0 ¯     p 1 p 3 ¯     p 1 p 2 ¯ p 0 p 3 ¯       p 0 p 2 ¯ p 1 p 1 ¯       p 1 p 0 ¯ p 0 p 1 ¯       p 0 p 0 ¯ + p 2 p 2 ¯   p 2 p 3 ¯ p 3 p 2 ¯       p 3 p 3 ¯   p 2 p 0 ¯       p 2 p 1 ¯ p 3 p 0 ¯       p 3 p 1 ¯     p 0 p 2 ¯     p 0 p 3 ¯ p 1 p 2 ¯       p 1 p 3 ¯ p 0 p 0 ¯       p 0 p 1 ¯ p 1 p 0 ¯       p 1 p 1 ¯
= p 3 p 3 ¯       p 3 p 2 ¯ p 2 p 3 ¯       p 2 p 2 ¯   0       0 0       0     0       0 0       0 p 1 p 1 ¯       p 1 p 0 ¯ p 0 p 1 ¯       p 0 p 0 ¯ + p 2 p 2 ¯   p 2 p 3 ¯ p 3 p 2 ¯       p 3 p 3 ¯   0       0 0       0     0       0 0       0 p 0 p 0 ¯       p 0 p 1 ¯ p 1 p 0 ¯       p 1 p 1 ¯
+ p 1 p 1 ¯       p 1 p 0 ¯ p 0 p 1 ¯       p 0 p 0 ¯   0       0 0       0     0       0 0       0 p 3 p 3 ¯       p 3 p 2 ¯ p 2 p 3 ¯       p 2 p 2 ¯ + p 0 p 0 ¯   p 0 p 1 ¯ p 1 p 0 ¯       p 1 p 1 ¯   0       0 0       0     0       0 0       0 p 2 p 2 ¯       p 2 p 3 ¯ p 3 p 2 ¯       p 3 p 3 ¯
+ p 1 p 1 ¯       p 1 p 0 ¯ p 0 p 1 ¯       p 0 p 0 ¯   0       0 0       0     0       0 0       0 p 3 p 3 ¯       p 3 p 2 ¯ p 2 p 3 ¯       p 2 p 2 ¯ + p 0 p 0 ¯   p 0 p 1 ¯ p 1 p 0 ¯       p 1 p 1 ¯   0       0 0       0     0       0 0       0 p 2 p 2 ¯       p 2 p 3 ¯ p 3 p 2 ¯       p 3 p 3 ¯
+ p 3 p 3 ¯       p 3 p 2 ¯ p 2 p 3 ¯       p 2 p 2 ¯   0       0 0       0     0       0 0       0 p 1 p 1 ¯       p 1 p 0 ¯ p 0 p 1 ¯       p 0 p 0 ¯ + p 2 p 2 ¯   p 2 p 3 ¯ p 3 p 2 ¯       p 3 p 3 ¯   0       0 0       0     0       0 0       0 p 0 p 0 ¯       p 0 p 1 ¯ p 1 p 0 ¯       p 1 p 1 ¯
T R p + T R p + T R ¯ p + T R ¯ p =
4 p 0 p 0 ¯ + p 1 p 1 ¯ + p 2 p 2 ¯ + p 3 p 3 ¯ 1       0 0       1   0       0 0       0     0       0 0       0 1       0 0       1 = 4 e ( p ) I
The last operation of taking the value p 0 p 0 ¯ + p 1 p 1 ¯ + p 2 p 2 ¯ + p 3 p 3 ¯ out from under the sign of the integral seems doubtful because of its dependence on the momentum over which the integration is performed. If one closes one's eyes to this, as is generally accepted in the literature, in particular in [10], this relation is taken to be interpreted as a proof of the anti-symmetry of the fermion wave function under the stipulated anticommutation relations. The only situation where this is unquestionably true is when considering in a rest system where boosts are excluded, energy is equal to mass, and invariant to rotations.
It is noteworthy that the antisymmetric treatment, whether or not complex conjugation is considered, yields a diagonal matrix that is invariant in one case but not in the other. It is encouraging to observe that the set of reference spinors remain consistent.
It is crucial to note that the proposed invariant approach cannot be realized within the Minkowski vector space. To achieve this, it is necessary to transition to the spinor space. This reiterates the secondary role of the Minkowski space in comparison to the spinor space.
Dirac's equation can be expressed in both spinor and vector spaces, a fact that led Dirac to discover it. In contrast, the invariant equation can be written in spinor space but not in vector space, which explains why it was unknown.
Let us write down the propagator of the fermionic field and the fermionic field invariant equation of motion using the proposed matrices
S R p = p 3 p 2 p 1 p 0 p 0 , p 1 , p 2 , p 3 p 1 p 0 p 3 p 2 p 2 , p 3 , p 0 , p 1
+ p 1 p 0 p 3 p 2 p 2 , p 3 , p 0 , p 1 p 3 p 2 p 1 p 0 p 0 , p 1 , p 2 , p 3
S R p = p 0 p 1 p 2 p 3 p 3 , p 2 , p 1 , p 0 p 2 p 3 p 0 p 1 p 1 , p 0 , p 3 , p 2
+ p 2 p 3 p 0 p 1 p 1 , p 0 , p 3 , p 2 p 0 p 1 p 2 p 3 p 3 , p 2 , p 1 , p 0
The equation of motion has the form
S R + S R ¯ + S R + S R ¯ 4 ( m + m ¯ ) I φ x = 0
where
p 0 x 1   1           p 1 x 0 0             p 2 x 3   3             p 3 x 2 2
p 0 ¯ [   ] ¯ x 1 ¯ 1 ¯             p 1 ¯   ¯ x 0 ¯ 0 ¯             p 2 ¯   ¯ x 3 ¯ 3 ¯               p 3 ¯   ¯ x 2 ¯ 2 ¯
S R = 2 3 0 1 1 , 0 , 3 , 2 0 1 2 3 3 , 2 , 1 , 0
+ 0 1 2 3 3 , 2 , 1 , 0 2 3 0 1 1 , 0 , 3 , 2
S R = 1 0 3 2 2 , 3 , 0 , 1 3 2 1 0 0 , 1 , 2 , 3
+ 3 2 1 0 0 , 1 , 2 , ( 3 ) 1 0 3 2 2 , 3 , 0 , 1
The equation is relativistically invariant, respectively we can use the invariant Lagrangian
L = 1 2 φ x T ( S R + S R ¯ + S R + S R ¯ ) φ x 4 ( m + m ¯ ) φ x T φ x
to which corresponds the relativistically invariant fermion propagator
D R x = d 4 p ( 2 π ) 4   S R p + S R ¯ p + S R p + S R ¯ p + 4 ( m + m ¯ ) I     P 2 m 2 e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
The equation can be modified to take into account the electromagnetic potential, the electron charge is taken as a unit
p 0 1 + a 0                       p 1 0 + a 1                       p 2 3 + a 2                     p 3 2 + a 3
S R = 2 + a 3 3 + a 2 0 + a 1 1 + a 0 1 + a 0 , 0 + a 1 , 3 + a 2 , 2 + a 3
( 0 + a 1 ) ( 1 + a 0 ) ( 2 + a 3 ) ( 3 + a 2 ) 3 + a 2 , ( 2 + a 3 ) , ( 1 + a 0 ) , ( 0 + a 1 )
+ 0 + a 1 1 + a 0 2 + a 3 3 + a 2 3 + a 2 , 2 + a 3 , 1 + a 0 , 0 + a 1
( 2 + a 3 ) ( 3 + a 2 ) ( 0 + a 1 ) ( 1 + a 0 ) 1 + a 0 , ( 0 + a 1 ) , ( 3 + a 2 ) , ( 2 + a 3 )
S R = 1 + a 0 0 + a 1 3 + a 2 2 + a 3 2 + a 3 , 3 + a 2 , 0 + a 1 , 1 + a 0
( 3 + a 2 ) ( 2 + a 3 ) ( 1 + a 0 ) ( 0 + a 1 ) ( 0 + a 1 ) , ( 1 + a 0 ) , ( 2 + a 3 ) , ( 3 + a 2 )
+ 3 + a 2 2 + a 3 1 + a 0 0 + a 1 0 + a 1 , 1 + a 0 , 2 + a 3 , 3 + a 2
( 1 + a 0 ) ( 0 + a 1 ) ( 3 + a 2 ) ( 2 + a 3 ) ( 2 + a 3 ) , ( 3 + a 2 ) , ( 0 + a 1 ) , ( 1 + a 0 )
and apply, in particular, to analyze the emission spectrum of the hydrogen-like atom.
It is our intention to extend the proposed approach to the description of the bosonic field. In order to transition to such a description, it would be prudent to consider the existence of the fermion annihilation process as a potential indicator. It is reasonable to posit a relationship between electrons and photons, given their ability to transform into one another. It is therefore proposed that the photon field be expressed by means of analogous reference spinors, which can be obtained from the fermion spinors but which have zero mass. Let us utilize the table of properties of the reference spinors to identify suitable pairs for describing the annihilation process. The two spinors should exhibit identical properties, with the exception of their respective masses, which should have opposite signs. For example,
          u 1 + u 3 = p 3 p 2 p 1 p 0 + p 1 p 0 p 3 p 2 = p 3 p 1 p 2 i p 0 p 1 + p 3 p 0 + p 2 = p 3 p 2 p 3 p 2 + p 1 p 0 p 1 p 0  
          u 4 + u 2 = p 0 p 1 p 2 p 3 + p 2 p 3 p 0 p 1 = p 0 p 2 p 1 p 3 p 2 + p 0 p 3 + p 1 = p 0 p 1 p 0 p 1 + p 2 p 3 p 2 p 3  
f u 1 = p 3 p 2 p 3 p 2           f u 3 = p 1 p 0 p 1 p 0           f u 4 = p 2 p 3 p 2 p 3           f u 2 = p 0 p 1 p 0 p 1  
m f u 1 = p 2 p 3 + p 3 p 2 = 0
m f u 3 = p 0 p 1 + p 1 p 0 = 0
m f u 4 = p 3 p 2 + p 2 p 3 = 0
m f u 3 = p 1 p 0 + p 0 p 1 = 0
          v 1 + i v 3 = p 1 p 0 p 3 p 2 + p 3 p 2 p 1 p 0 = p 1 + p 3 p 0 + p 2 p 3 + p 1 p 2 + p 0 = p 1 p 0 p 1 p 0 + p 3 p 2 p 3 p 2  
          v 4 + i v 2 = p 2 p 3 p 0 p 1 + p 0 i p 1 i p 2 p 3 = p 2 + p 0 p 3 p 1 p 0 p 2 p 1 + p 3 = p 2 p 3 p 2 p 3 + p 0 p 1 p 0 p 1  
f v 1 = p 1 p 0 i p 1 i p 0           f v 3 = p 3 p 2 p 3 p 2             f v 4 = p 2 p 3 p 2 p 3           f v 2 = p 0 p 1 p 0 p 1  
m f v 1 = p 0 p 1 p 1 p 0 = 0
m f v 3 = p 2 p 3 p 3 p 2 = 0
m f v 4 = p 3 p 2 p 2 p 3 = 0
m f v 2 = p 1 p 0 p 0 p 1 = 0
Let us represent the photon field as a decomposition
ψ x = d 4 p 2 π 2  
f d 1 p f u 1 p + f d 2 p f u 3 p + f b 2 p f u 2 ¯ p + f b 1 p f u 4 ¯ p + f d 4 p v 1 p + f d 3 p f v 3 p + f b 3 p f v 2 ¯ p + f b 4 p f v 4 ¯ p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
+ f b 1 * p f u 1 ¯ p + f b 2 * p f u 3 ¯ p + f d 2 * p f u 2 p + f d 1 * p f u 4 p + f b 4 * p f v 1 ¯ p + f b 3 * p f v 3 ¯ p + f d 3 * p f v 2 p + f d 4 * p f v 4 p e i p 0 x 1 p 1 x 0 + p 2 x 3 p 3 x 2
We are interested in the coordinate function
ψ + x S μ ψ x
It represents a four-dimensional probability density current, which is transformed by the Lorentz transformations as a four-dimensional vector. It is proposed that this vector, possibly with some dimensional coefficient, be treated as a vector potential of the electromagnetic field
A μ x = e m e ψ + x S μ ψ x
The final step is to verify that, under the specified commutation relations between the operator coefficients, the commutation relations of the following type are satisfied for the vector potential
A + x , A x = A + x A x A + x A x = C I δ x x
or
A T x , A x = A + x A x A + x A x = C I δ x x
with a certain constant C .

Conclusions

An alternative approach to analyze relativistic and quantum effects inherent in charged particles in the presence of an electromagnetic field is proposed. Two ways of describing the electron behavior in the electromagnetic field are considered: by means of the vector equation, which is based on the plane wave model for a free electron, and the spinor equation, which is based on the representation of the electron as a plane wave in spinor space. For both equations, which are valid for a free particle, their applicability to an arbitrary physical situation is postulated, in particular to describe the behavior of a particle in the presence of an electromagnetic field. The presented equations are intended to fulfill the same role as the Schrödinger equation and the Dirac equation. At the same time, in our opinion, the spinor equations more accurately describe the details of the interaction between fields and particles.

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