One-to-One Map of Dirac Equation between Matrix Algebra and Geometric Algebra Cl3,0

Jesús Sánchez

doi:10.4236/jamp.2025.139169

Journal of Applied Mathematics and Physics > Vol.13 No.9, September 2025

One-to-One Map of Dirac Equation between Matrix Algebra and Geometric Algebra Cl_3,0

Jesús Sánchez

Bilbao, Spain.
DOI: 10.4236/jamp.2025.139169 PDF HTML XML 65 Downloads 274 Views

Abstract

In this paper, it is created a one-to-one map between the Dirac Equation in Matrix Algebra and the Dirac Equation in Geometric Algebra Cl_3,0. The eight free parameters of the spinors in matrix representation correspond to the coefficients of the eight elements of the Geometric Algebra Cl_3,0 (scalars, three vectors, three bivectors and the trivector). This way, a one-to-one mapping between matrix representation and Geometric Algebra Cl_3,0 can be obtained.

Keywords

Geometric Algebra, Matrix Algebra, Dirac Equation, Quantum Mechanics, Wavefunction, Basis Vectors

Share and Cite:

Sánchez, J. (2025) One-to-One Map of Dirac Equation between Matrix Algebra and Geometric Algebra Cl_3,0. Journal of Applied Mathematics and Physics, 13, 2945-2975. doi: 10.4236/jamp.2025.139169.

1. Introduction

In this paper, we will use derive the Dirac Equation in Geometric Algebra Cl_3,0. Then, after some transformations, we will create a one-to-one map of the Dirac Equation and the wavefunction between Matrix Algebra and Geometric Algebra.

2. Introduction to Geometric Algebra

If some background is needed regarding geometric algebra, the reference [1] is recommended. Also, you may find a complete study of Geometric Algebra in [2].

We will use Geometric Algebra Cl_3,0. This means it has three basis vectors with positive signature and zero basis vectors with negative signature. We will explain this in a minute.

In Geometric Algebra Cl_3,0, we have the vectors:

$\hat{x} \hat{y} \hat{z}$

In this case, we will consider them orthonormal, and in the Appendix A1, I will explain what the difference would be in the calculations if they were not orthonormal. (Figure 1)

Figure 1. Orthonormal basis vectors in Cl_3,0.

The square of these vectors in Geometric Algebra is its norm to the square. The norm of a vector is a scalar (not a vector anymore). As we have considered the basis as orthonormal, its square is scalar 1.

${\hat{x}}^{2} = \hat{x} \hat{x} = {‖ \hat{x} ‖}^{2} = 1$ (1)

${\hat{y}}^{2} = \hat{y} \hat{y} = {‖ \hat{y} ‖}^{2} = 1$ (2)

${\hat{z}}^{2} = \hat{z} \hat{z} = {‖ \hat{z} ‖}^{2} = 1$ (3)

In the nomenclature Cl_3,0, the 3 stands for the number of vectors which square is positive and the 0 for the number of basis vectors which squares is negative. In this case, no basis vectors have a negative square (also known as negative signature), so all of them have a positive square (positive signature), that equals +1 in an orthonormal basis.

The basis vectors can be multiplied by each other (this operation is called Geometric Product). For orthonormal or orthogonal bases, this product follows the anticommutative property, this is:

$\hat{x} \hat{y} = - \hat{y} \hat{x}$ (4)

$\hat{y} \hat{z} = - \hat{z} \hat{y}$ (5)

$\hat{z} \hat{x} = - \hat{x} \hat{z}$ (6)

This combination of two vectors via this product is called a bivector. The bivector instead of representing a vector (an oriented segment), it represents an oriented plane. So $\hat{x} \hat{y}$ represents the plane xy with its normal in a certain direction. And $\hat{y} \hat{x}$ represents the same plane xy but with its normal in the opposite direction.

In Geometric Algebra we do not talk about normal vectors anymore. Instead, we talk about the orientation of a theoretical rotation in that plane. See Figure 2 for a visual explanation. Also, in [1]-[3], you can find more information about the meaning or interpretation of the bivectors.

Figure 2. Representation of the bivectors $\hat{x} \hat{y}$ and $\hat{y} \hat{x}$ . They represent the same plane with opposite orientation. In fact, $\hat{x} \hat{y} = - \hat{y} \hat{x}$ .

If we multiply the three vectors, we obtain the trivector (also called pseudoscalar in the literature [1] [2]):

$\hat{x} \hat{y} \hat{z} = - \hat{y} \hat{x} \hat{z} = \hat{y} \hat{z} \hat{x} = - \hat{z} \hat{y} \hat{x} = \hat{z} \hat{x} \hat{y} = - \hat{x} \hat{z} \hat{y}$ (7)

You can check that the same relations as in Equations (4) - (6) apply. So, every time you swap the position of two vectors you have to put a minus sign (or multiply by −1, as you prefer).

The meaning of the trivector is an oriented volume. The same as the bivector is a plane with two possible orientations. The trivector is a volume with two possible orientations. You can see visual representation in Figure 3. Again in [1]-[3] you can find a more information regarding trivectors.

Figure 3. Representation of the two possible orientations of the trivector. We can check that $\hat{x} \hat{y} \hat{z} = - \hat{y} \hat{x} \hat{z}$ .

3. Operations in Geometric Algebra

One of the most surprising characteristics of Geometric Algebra is that you can mix scalars with vectors, bivectors and trivectors. You represent this a sum. For example, a typical element in Geometric Algebra could have the form:

$A = 3 + 5 \hat{x} + 3 \hat{y} + 4 \hat{y} \hat{z} - 2 \hat{x} \hat{y} \hat{z}$

The same that is done with polynomials or complex numbers, that is to leave the sum among different components indicated, it is done in Geometric Algebra. This type of element in Geometric Algebra that has different components as scalars, vectors, bivectors, etc. is called a multivector. So, the A element in the example above is a multivector.

In a multivector, the vectors and the trivector are called odd-grade elements. The reason is because they are composed by one vector or by three vectors (odd grade number).

In a multivector, the scalars and the bivectors are called even grade elements. The reason is because the elements have 0 vectors (the scalars) or 2 vectors (the bivectors). We consider the 0 and 2 even for this purpose.

And if you want to make a product between two multivectors in Geometric Algebra you just have to follow the laws (1) to (6). For example:

$(2 + 3 \hat{x}) (5 \hat{y} + 7 \hat{x} + \hat{y} \hat{z} + \hat{z} \hat{x})$

The first thing we have to do is to multiply component by component as we would do in a polynomial for example. But the very important thing is that you have to keep the order of the product as we have seen that it is not commutative, so:

$(2 + 3 \hat{x}) (5 \hat{y} + 7 \hat{x} + \hat{y} \hat{z} + \hat{z} \hat{x}) = 10 \hat{y} + 14 \hat{x} + 2 \hat{y} \hat{z} + 2 \hat{z} \hat{x} + 15 \hat{x} \hat{y} + 21 \hat{x} \hat{x} + 3 \hat{x} \hat{y} \hat{z} + 3 \hat{x} \hat{z} \hat{x} =$

Now, with the relations (1) to (6) we will operate the square of $\hat{x}$ and we will swap the $\hat{x}$ and the $\hat{z}$ in the last component:

$= 10 \hat{y} + 14 \hat{x} + 2 \hat{y} \hat{z} + 2 \hat{z} \hat{x} + 15 \hat{x} \hat{y} + 21 (+ 1) + 3 \hat{x} \hat{y} \hat{z} - 3 \hat{x} \hat{x} \hat{z} =$

Now, we have again a square of $\hat{x}$ in the last component, so we can operate:

$\begin{array}{l} = 10 \hat{y} + 14 \hat{x} + 2 \hat{y} \hat{z} + 2 \hat{z} \hat{x} + 15 \hat{x} \hat{y} + 21 + 3 \hat{x} \hat{y} \hat{z} - 3 (+ 1) \hat{z} \\ = 10 \hat{y} + 14 \hat{x} + 2 \hat{y} \hat{z} + 2 \hat{z} \hat{x} + 15 \hat{x} \hat{y} + 21 + 3 \hat{x} \hat{y} \hat{z} - 3 \hat{z} = \end{array}$

If we order the terms, starting by the scalar, vectors, bivectors and finally the trivector we have:

$= 21 + 14 \hat{x} + 10 \hat{y} - 3 \hat{z} + 15 \hat{x} \hat{y} + 2 \hat{y} \hat{z} + 2 \hat{z} \hat{x} + 3 \hat{x} \hat{y} \hat{z}$

Let’s see another example:

$(3 + \hat{x} + 2 \hat{y}) (5 \hat{x} + 7 \hat{y}) =$

We start multiplying the components but keeping always the order of the vectors.

$(3 + \hat{x} + 2 \hat{y}) (5 \hat{x} + 7 \hat{y}) = 15 \hat{x} + 21 \hat{y} + 5 \hat{x} \hat{x} + 7 \hat{x} \hat{y} + 10 \hat{y} \hat{x} + 14 \hat{y} \hat{y} =$

Now we apply the (1) to (6) to the squares:

$15 \hat{x} + 21 \hat{y} + 5 (+ 1) + 7 \hat{x} \hat{y} + 10 \hat{y} \hat{x} + 14 (+ 1) = 15 \hat{x} + 21 \hat{y} + 5 + 7 \hat{x} \hat{y} + 10 \hat{y} \hat{x} + 14 =$

We can see that now we have two scalars (5 and 14) that have appeared coming from vector products that can be summed, so:

$= 15 \hat{x} + 21 \hat{y} + 19 + 7 \hat{x} \hat{y} + 10 \hat{y} \hat{x} =$

Also, we see that we have the same bivector $\hat{x} \hat{y}$ in two different forms, so we apply (1) to (6) to get:

$= 15 \hat{x} + 21 \hat{y} + 19 + 7 \hat{x} \hat{y} - 10 \hat{x} \hat{y} = 15 \hat{x} + 21 \hat{y} + 19 - 3 \hat{x} \hat{y} =$

Ordering the terms:

$= 19 + 15 \hat{x} + 21 \hat{y} - 3 \hat{x} \hat{y}$

To sum up, we can say that the geometric product keeps the associative and the distributive properties but not the commutative property. In an orthonormal basis the commutative property is substituted by the anticommutative property as can be seen in (4) to (6). For n on orthonormal basis, the thing is not so simple but we will not treat this case in this paper. You can see a hint about it in Appendix A1.

4. Square of the Bivectors and the Trivector

If we multiply a bivector by itself (applying (1) to (6)):

$\hat{x} \hat{y} \hat{x} \hat{y} = - \hat{x} \hat{y} \hat{y} \hat{x} = - \hat{x} (1) \hat{x} = - \hat{x} \hat{x} = - 1$

$\hat{y} \hat{z} \hat{y} \hat{z} = - 1$

$\hat{z} \hat{x} \hat{z} \hat{x} = - 1$

We see that the result is −1. The same happens with the trivector:

$\hat{x} \hat{y} \hat{z} \hat{x} \hat{y} \hat{z} = - \hat{x} \hat{y} \hat{z} \hat{x} \hat{z} \hat{y} = \hat{x} \hat{y} \hat{z} \hat{z} \hat{x} \hat{y} = \hat{x} \hat{y} (1) \hat{x} \hat{y} = \hat{x} \hat{y} \hat{x} \hat{y} = - \hat{x} \hat{y} \hat{y} \hat{x} = - \hat{x} (1) \hat{x} = - 1$

In Geometric Algebra the imaginary or complex numbers are not used and are not necessary. The reason is that there are elements that are already in fact the square root of −1, as the bivectors or the trivector. We will see the importance of this in Quantum Mechanics. Instead of using imaginary numbers, these will be substituted by bivectors and trivectors with geometric meaning.

The imaginary unit i was defined as “something unknown” (whatever it is) that is the square root of −1. Now that we have elements that are in fact, known, and are the square root of −1 (the bivectors and the trivector), we can be more specific and use these elements to play this role. We will use this conversion from the i imaginary unit into bivectors and trivector, mainly in Quantum Mechanics.

When the i does not have any preferred spatial direction, it will be related to the trivector $\hat{x} \hat{y} \hat{z}$ . This happens, for example, when the i appears related to mass, energy or time.

If the i is related to something with a preferred direction like speed or momentum, normally the i is related to a bivector. Do not worry, we will see how to work with this in the next chapters.

Summing up, even if we are in Cl_3,0 with the three basis vectors with positive signature (positive square), the algebra itself has created two types of elements more (bivectors and trivectors) whose squares are negative.

In a multivector, we will have these two types of elements depending on their square, the scalars and the vectors whose square is +1 (positive signature) and the bivectors and the trivector which square is −1 (negative signature).

5. Inverse of a Vector in Geometric Algebra

Another interesting property in Geometric Algebra is that you can take the inverse of a vector. We can calculate its value for a basis vector in the following way. We start with Equation (1):

$\hat{x} \hat{x} = 1$

We premultiply both equations by the inverse of $\hat{x}$ :

${\hat{x}}^{- 1} \hat{x} \hat{x} = {\hat{x}}^{- 1} (1)$

By definition, the product of the inverse of an element by itself is equal to 1.

$(1) \hat{x} = {\hat{x}}^{- 1}$

So,

$\hat{x} = {\hat{x}}^{- 1}$

${\hat{x}}^{- 1} = \hat{x}$

The inverse of a basis vector in an orthonormal basis is the vector itself. So:

${\hat{x}}^{- 1} = \hat{x}$

${\hat{y}}^{- 1} = \hat{y}$

${\hat{z}}^{- 1} = \hat{z}$

When we have to take the inverse of a product of vectors (bivectors or trivectors) you can check in [2] that apart from inverting each element you have to reverse the order of them, this way:

${(\hat{x} \hat{y})}^{- 1} = {\hat{y}}^{- 1} {\hat{x}}^{- 1} = \hat{y} \hat{x} = - \hat{x} \hat{y}$

${(\hat{x} \hat{y} \hat{z})}^{- 1} = {\hat{z}}^{- 1} {\hat{y}}^{- 1} {\hat{x}}^{- 1} = \hat{z} \hat{y} \hat{x} = - \hat{x} \hat{y} \hat{z}$

Remember that every time you swap two vectors, you add a minus sign (4) to (6). To convert $\hat{z} \hat{y} \hat{x}$ into $- \hat{x} \hat{y} \hat{z}$ you have two make three swaps, that is the reason of the final negative sign.

We will use the convention that the division by a vector is to postmultiply by the inverse of that vector. This means, for example, if we want to do the following operation, this will be the result:

$\frac{\hat{x}}{\hat{y}} = \hat{x} {(\hat{y})}^{- 1} = \hat{x} \hat{y}$

Remind that we are always talking about orthonormal bases. To have a hint about non-orthonormal bases, you check Annex A1.

6. Reverse Operation and Reverse Product

There is another operation we can make in Geometric Algebra, which is the reversion of a multivector. I will represent this with a line above the multivector. This operation reverses all the internal order of bivectors and trivectors. As an example:

$A = 3 + 5 \hat{x} + 3 \hat{y} + 4 \hat{y} \hat{z} - 2 \hat{x} \hat{y} \hat{z}$

$\begin{matrix} \bar{A} = \bar{(3 + 5 \hat{x} + 3 \hat{y} + 4 \hat{y} \hat{z} - 2 \hat{x} \hat{y} \hat{z})} \\ = 3 + 5 \hat{x} + 3 \hat{y} + 4 \hat{z} \hat{y} - 2 \hat{z} \hat{y} \hat{x} \\ = 3 + 5 \hat{x} + 3 \hat{y} - 4 \hat{y} \hat{z} + 2 \hat{x} \hat{y} \hat{z} \end{matrix}$

You can see that it is similar to a conjugate in complex numbers. It changes the sign of the elements whose square is −1 (in this case, the bivectors and the trivector).

With this we can define the reverse product. It consists of the product of a multivector by the reverse of itself.

The main characteristic of the reverse product is that if the multivector only has one type of elements with positive square and only one type of elements with negative square, the result of this product is a scalar.

This means, if the multivector only has scalars (positive square) and bivectors (negative square) the reverse product of the multivector will be a scalar. The same is true if it only has vectors (positive square) and bivectors (negative square). Or vectors (positive square) and trivector (negative square).

But when the multivector has scalars and vectors (both positive squares) and a bivector for example, the result could not be a scalar. The same if it has scalars and both bivectors and the trivector (both negative squares).

This is, the multivector has to have only scalars or vectors (not both) mixed with only bivectors or trivectors (not both).

Let’s see some examples. B only has vectors (positive square) and the trivector (negative square), the result must be scalar:

$B = 5 \hat{x} + 3 \hat{y} + 4 x \hat{y} \hat{z}$

$\bar{B} = \bar{(5 \hat{x} + 3 \hat{y} + 4 x \hat{y} \hat{z})} = 5 \hat{x} + 3 \hat{y} - 4 x \hat{y} \hat{z}$

$\begin{matrix} B \bar{B} = (5 \hat{x} + 3 \hat{y} + 4 \hat{x} \hat{y} \hat{z}) (5 \hat{x} + 3 \hat{y} - 4 \hat{x} \hat{y} \hat{z}) \\ = 25 + 15 \hat{x} \hat{y} - 20 \hat{x} \hat{x} \hat{y} \hat{z} + 15 \hat{y} \hat{x} + 9 - 12 \hat{y} \hat{x} \hat{y} \hat{z} + 20 \hat{x} \hat{y} \hat{z} \hat{x} + 12 \hat{x} \hat{y} \hat{z} \hat{y} - 16 \hat{x} \hat{y} \hat{z} \hat{x} \hat{y} \hat{z} \\ = 25 + 15 \hat{x} \hat{y} - 20 \hat{y} \hat{z} - 15 \hat{x} \hat{y} + 9 + 12 \hat{x} \hat{y} \hat{y} \hat{z} - 20 \hat{x} \hat{y} \hat{x} \hat{z} - 12 \hat{x} \hat{y} \hat{y} \hat{z} + 16 \hat{x} \hat{y} \hat{x} \hat{z} \hat{y} \hat{z} = \end{matrix}$

We sum the scalars, we see that the elements in xy sum zero, we square to +1 the vectors that are the same and consecutive and we continue swapping vectors to try to simplify:

$\begin{array}{l} = 34 - 20 \hat{y} \hat{z} + 12 \hat{x} \hat{z} + 20 \hat{x} \hat{x} \hat{y} \hat{z} - 12 \hat{x} \hat{z} - 16 \hat{x} \hat{x} \hat{y} \hat{z} \hat{y} \hat{z} \\ = 34 - 20 \hat{y} \hat{z} + 12 \hat{x} \hat{z} + 20 \hat{y} \hat{z} - 12 \hat{x} \hat{z} + 16 \hat{x} \hat{x} \hat{y} \hat{y} \hat{z} \hat{z} = 34 + 16 = 50 \end{array}$

The result, 50, is a scalar as we expected.

Another example. C has only scalars and the trivector; the result should be scalar:

$C = 5 + 4 \hat{x} \hat{y} \hat{z}$

$\bar{C} = \bar{(5 + 4 \hat{x} \hat{y} \hat{z})} = 5 - 4 \hat{x} \hat{y} \hat{z}$

$C \bar{C} = (5 + 4 \hat{x} \hat{y} \hat{z}) (5 - 4 \hat{x} \hat{y} \hat{z}) = 25 - 20 \hat{x} \hat{y} \hat{z} + 20 \hat{x} \hat{y} \hat{z} - 16 \hat{x} \hat{y} \hat{z} \hat{x} \hat{y} \hat{z} =$

The elements in xyz sum to zero. Swapping vectors in the last element, we get:

$= 25 + 16 \hat{x} \hat{y} \hat{x} \hat{z} \hat{y} \hat{z} = 25 - 16 \hat{x} \hat{x} \hat{y} \hat{z} \hat{y} \hat{z} = 25 + 16 \hat{x} \hat{x} \hat{y} \hat{y} \hat{z} \hat{z} = 25 + 16 = 31$

31 is a scalar as expected.

New example. D has scalars but has a mix of bivectors and trivectors (the result could not be a scalar):

$D = 5 + 2 \hat{x} \hat{y} + 3 \hat{x} \hat{y} \hat{z}$

$\bar{D} = \bar{(5 + 2 \hat{x} \hat{y} + 4 \hat{x} \hat{y} \hat{z})} = 5 - 2 \hat{x} \hat{y} - 3 \hat{x} \hat{y} \hat{z}$

$\begin{matrix} D \bar{D} = (5 + 2 \hat{x} \hat{y} + 3 \hat{x} \hat{y} \hat{z}) (5 - 2 \hat{x} \hat{y} - 3 \hat{x} \hat{y} \hat{z}) \\ = 25 - 10 \hat{x} \hat{y} - 15 \hat{x} \hat{y} \hat{z} + 10 \hat{x} \hat{y} - 4 \hat{x} \hat{y} \hat{x} \hat{y} - 6 \hat{x} \hat{y} \hat{x} \hat{y} \hat{z} + 15 \hat{x} \hat{y} \hat{z} - 6 \hat{x} \hat{y} \hat{z} \hat{x} \hat{y} - 9 \hat{x} \hat{y} \hat{z} \hat{x} \hat{y} \hat{z} = \end{matrix}$

We see that the terms in $\hat{x} \hat{y}$ and $\hat{x} \hat{y} \hat{z}$ vanish. Also, we swap some vectors:

$\begin{array}{l} = 25 - 4 \hat{x} \hat{y} \hat{x} \hat{y} - 6 \hat{x} \hat{y} \hat{x} \hat{y} \hat{z} - 6 \hat{x} \hat{y} \hat{z} \hat{x} \hat{y} - 9 \hat{x} \hat{y} \hat{z} \hat{x} \hat{y} \hat{z} \\ = 25 + 4 \hat{x} \hat{x} \hat{y} \hat{y} + 6 \hat{x} \hat{x} \hat{y} \hat{y} \hat{z} + 6 \hat{x} \hat{y} \hat{z} \hat{y} \hat{x} \\ = 25 + 4 + 6 \hat{z} - 6 \hat{x} \hat{y} \hat{y} \hat{z} \hat{x} = 29 + 6 \hat{z} - 6 \hat{x} \hat{z} \hat{x} \\ = 29 + 6 \hat{z} + 6 \hat{x} \hat{x} \hat{z} = 29 + 6 \hat{z} + 6 \hat{z} = 29 + 12 \hat{z} \end{array}$

We see that the result is not a scalar as we had both bivectors and the trivector (both of negative signature) in the same multivector.

One important thing to comment about the reverse product is that it acts very similarly to the scalar product of an element with itself (the square) in the bra-ket notation of Dirac Algebra.

In the bra-ket notation of Dirac algebra, when you want to calculate the square of a complex function or vector, you multiply this function or vector by the conjugate of itself, so you always get a real scalar result. This reverse product makes the same, you multiply a multivector by a version of itself where the sign of different elements of this multivector has changed with the aim of obtaining a real scalar as a result.

7. Summary of Geometric Algebra Cl_3,0

We have seen that the Geometric Algebra has some elements called multivectors that are composed of scalars, vectors, bivectors and a trivector. In fact, although the Geometric Algebra Cl_3,0 has only three basis vectors, it has really 8 degrees of freedom. A general multivector in Geometric Algebra could have the form (being all the coefficients $α_{i}$ real scalars):

$A = α_{0} + α_{x} \hat{x} + α_{y} \hat{y} + α_{z} \hat{z} + α_{x y} \hat{x} \hat{y} + α_{y z} \hat{y} \hat{z} + α_{z x} \hat{z} \hat{x} + α_{x y z} \hat{x} \hat{y} \hat{z}$

This means, although we have only three special dimensions (x, y and z) we have really 8 degrees of freedom (or 8 expanded dimensions in a meta sense) coming from these original three special dimensions.

These eight degrees of freedom are represented by these 8 $α_{i}$ scalars. These scalars are always real. As commented, we do not need imaginary numbers in Geometric Algebra as we have two types of elements (the bivectors and the trivector) whose squares are −1 and fulfill this necessity.

One comment for people who have some experience in Geometric Algebra used in Physics: If you are new to Geometric Algebra, please do not read it, so you do not start running.

In most of the literature regarding the use of Geometric Algebra both in Quantum Mechanics and General Relativity the Cl_1,3 or Cl_3,1 is used. This means there are three basis vectors (the spatial dimensions) with one signature and another one (the time) with opposite signature.

The issue is that these 4 dimensions expand to 16 degrees of freedom. However, in reality, only the sub-even algebra of these 16 degrees of freedom is used (only 8 degrees of the 16 possible are used). So why is this Cl_1,3 or Cl_3,1 used in the first place? We know that with Cl_3,0 we already have the 8 degrees of freedom we need.

The need for Cl_1,3 and Cl_3,1 is to accommodate the time dimension in Geometric Algebra. But we will explain in the next chapter why this is not necessary anymore.

8. So, Where Is the Time?

If you have worked with Quantum Mechanics or with the Dirac Equation (whether you have done it using Geometric Algebra or not) you might be asking where the time is.

We have $\hat{x}$ , $\hat{y}$ and $\hat{z}$ . But where is the $\hat{t}$ ? As I have commented in some papers already [3] [4] we can use the trivector as the basis vector of the dimension of time. Does this mean that the dimension of time does not exist? No, the dimension of time has its own freedom (its own scalar coefficient t) but the basis vector $\hat{t}$ that accompanies this coefficient is a combination of the space vectors.

I know, it is very difficult to believe but if you continue reading the next chapters, you will see that this works perfectly. This led to a one-to-one map of the Dirac equation solutions in standard algebra and Geometric Algebra without the need for a specific vector of time.

In fact, we will work with the following definition:

${\hat{t}}^{- 1} = \hat{x} \hat{y} \hat{z}$ (7)

The reason why we define the inverse of the basis vector instead of the basis vector itself will be seen later. Anyhow, following the rules in Chapter 5 you can see that for an orthonormal basis (not in general for other bases):

$\hat{t} = {(\hat{x} \hat{y} \hat{z})}^{- 1} = {\hat{z}}^{- 1} {\hat{y}}^{- 1} x^{- 1} = \hat{z} \hat{y} \hat{x} = - \hat{x} \hat{y} \hat{z}$ (7.1)

So:

$\hat{x} \hat{y} \hat{z} = - \hat{t}$

So, a general multivector will be of the type:

$\begin{matrix} A = α_{0} + α_{x} \hat{x} + α_{y} \hat{y} + α_{z} \hat{z} + α_{x y} \hat{x} \hat{y} + α_{y z} \hat{y} \hat{z} + α_{z x} \hat{z} \hat{x} + α_{x y z} \hat{x} \hat{y} \hat{z} \\ = α_{0} + α_{x} \hat{x} + α_{y} \hat{y} + α_{z} \hat{z} + α_{x y} \hat{x} \hat{y} + α_{y z} \hat{y} \hat{z} + α_{z x} \hat{z} \hat{x} - α_{x y z} \hat{t} \end{matrix}$

Even, we reorder putting the time consecutive to the spatial dimensions we would have:

$A = α_{0} + α_{x} \hat{x} + α_{y} \hat{y} + α_{z} \hat{z} - α_{x y z} \hat{t} + α_{x y} \hat{x} \hat{y} + α_{y z} \hat{y} \hat{z} + α_{z x} \hat{z} \hat{x}$

We can even recall the $α_{x y z}$ as $α_{t}$ :

$α_{t} = α_{x y z}$

This leads to,

$\begin{matrix} A = α_{0} + α_{x} \hat{x} + α_{y} \hat{y} + α_{z} \hat{z} + α_{x y} \hat{x} \hat{y} + α_{y z} \hat{y} \hat{z} + α_{z x} \hat{z} \hat{x} - α_{t} \hat{t} \\ = α_{0} + α_{x} \hat{x} + α_{y} \hat{y} + α_{z} \hat{z} - α_{t} \hat{t} + α_{x y} \hat{x} \hat{y} + α_{y z} \hat{y} \hat{z} + α_{z x} \hat{z} \hat{x} \end{matrix}$

You can see that the $α_{t}$ assures that the time has its own freedom compared to the spatial dimensions. But we do not need an original dimension more to accommodate it, it appears naturally in Geometric Algebra. In fact, we will not use it is it is above, we will use the more convenient definition we put in the beginning for a multivector:

$A = α_{0} + α_{x} \hat{x} + α_{y} \hat{y} + α_{z} \hat{z} + α_{x y} \hat{x} \hat{y} + α_{y z} \hat{y} \hat{z} + α_{z x} \hat{z} \hat{x} + α_{x y z} \hat{x} \hat{y} \hat{z}$

And we will explain how to work with these multivectors when time is involved.

If you have worked with Geometric Algebra before in Cl_1,3 or Cl_3,1, I will give you the following relations we will use in advance. If you do not know what we are talking about, just skip the following equations and continue reading:

$i = I = σ_{1} σ_{2} σ_{3} = γ_{0} γ_{1} γ_{2} γ_{3} = {\hat{t}}^{- 1} = \hat{x} \hat{y} \hat{z}$

$σ_{1} = γ_{1} γ_{0} = \hat{x}$

$σ_{2} = γ_{2} γ_{0} = \hat{y}$

$σ_{3} = γ_{3} γ_{0} = \hat{z}$

$γ_{0} \to {\hat{t}}^{- 1} = \hat{x} \hat{y} \hat{z}$

As commented before, not always the i will be equal to the trivector, but sometimes to the bivectors also. But we will check this on a case-by-case basis.

You can check more things regarding $\hat{t}$ as a composition of special vectors in Annex A2. More information regarding non-orthonormal bases is in Annex A1.

9. The Dirac Equation in Geometric Algebra Cl_3,0

We will do exactly the same thing that Dirac did to discover his famous equation. We will start from this relativistic equation that relates energy with momentum and mass [5].

$E^{2} = p^{2} c^{2} + m^{2} c^{4}$

For simplicity, we will do what is commonly done in these cases: we will consider a system of units where the speed of light c = 1 and reduced constant Planck ħ = 1. This is commonly done, in [6], which is already done de facto, with no loss of generality:

$E^{2} = p^{2} + m^{2}$ (8)

Now, if we use the common operator for Energy used in Quantum Mechanics, see [6], we have:

$E = i ℏ \frac{\partial}{\partial t}$

As we have commented before when we see an i imaginary unit in an equation with no preferred spatial direction (like Energy, mass, time…) we can directly convert it to the trivector so we have:

$E = \hat{x} \hat{y} \hat{z} ℏ \frac{\partial}{\partial t}$

And, as we have said considering ħ = 1 we have:

$E = \hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t}$ (9)

If you want more details of how to obtain this equation instead of using the conversion of i, deriving it directly from the wave equation, you can see [4].

That’s it for the Energy. Let’s go with the momentum. For simplicity we will consider only the direction of x and later we will generalize it. We start from the momentum operator in Quantum Mechanics [6]:

$p_{x} = - i ℏ \frac{\partial}{\partial x}$

Here, we cannot convert the i directly to the trivector as it has a preferred direction (x). So we will do it in another way. We know that the momentum units are mass multiplied by distance divided by time (in SI this is kg, m, s)

$p_{x_u n i t s} = kg \cdot \frac{m}{s}$

Considering mass, a scalar, we have that the vectors applying should be for the distance the direction x (the $\hat{x}$ vector) and be divided by time ( $\hat{t}$ vector).

$p_{x_v e c t o r s} = \frac{\hat{x}}{\hat{t}} = \hat{x} {(\hat{t})}^{- 1}$

where for the last step we have used the convention of division by vectors commented in Chapter 5. Now we use Equation (7) to convert the inverse of the time vector in the trivector:

$p_{v e c t o r s} = \hat{x} {(\hat{t})}^{- 1} = \hat{x} \hat{x} \hat{y} \hat{z} = (1) \hat{y} \hat{z} = \hat{y} \hat{z}$ (10)

So, we have seen that for Equation (10) where x is a preferred direction, the vectors that apply are $\hat{y} \hat{z}$ (not $\hat{x}$ ). This is logic as we need to have an element (in this case the bivector) whose square is −1 to substitute the i.

So, we would have:

$p_{x} = \hat{y} \hat{z} ℏ \frac{\partial}{\partial x}$

And the minus sign? The minus sign is a convention about which direction of the wave is considered positive. We could add it directly or we could reverse the $\hat{y} \hat{z}$ bivector to $- \hat{z} \hat{y}$ so we get it anyhow. But we will not add it at this stage, as it will appear naturally at a later step. We will see it later. So again, considering ħ is equal to 1, we have:

$p_{x} = \hat{y} \hat{z} \frac{\partial}{\partial x}$

Doing the same operation for $p_{y}$ and $p_{z}$ we will obtain:

$p_{y} = \hat{z} \hat{x} \frac{\partial}{\partial y}$

$p_{z} = \hat{x} \hat{y} \frac{\partial}{\partial z}$

So, summing all, we have:

$p = \hat{y} \hat{z} \frac{\partial}{\partial x} + \hat{z} \hat{x} \frac{\partial}{\partial y} + \hat{x} \hat{y} \frac{\partial}{\partial z}$ (11)

Again, do not worry about the minus sign not appearing; it will appear naturally later.

Regarding Equation (8) the only pending element is the mass that we consider as a scalar and we will not apply any vector to it. So, we have:

$E^{2} = p^{2} + m^{2}$

${(\hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t})}^{2} = {(\hat{y} \hat{z} \frac{\partial}{\partial x} + \hat{z} \hat{x} \frac{\partial}{\partial y} + \hat{x} \hat{y} \frac{\partial}{\partial z})}^{2} + m^{2}$ ? (12)

The above should be a kind of Klein-Gordon equation [6] in Geometric Algebra in Cl_3,0. But first thing to comment on is that in Equation (8) we have scalars to the square (E, P and m). But if you make the squares directly as they are in the geometric algebra Equation (12) you would not obtain scalars. So that Equation (12) is not properly ok.

We have to convert those squares to an operation that gives as results scalars and continue being true to its nature. Something similar and when you conjugate a vector or wavefunction to perform the bra-ket square in Dirac notation and obtain a scalar.

In Geometric Algebra, you guessed it, this operation is the reverse product.

In fact, if we do:

$(\hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t}) (\hat{z} \hat{y} \hat{x} \frac{\partial}{\partial t}) = (\hat{y} \hat{z} \frac{\partial}{\partial x} + \hat{z} \hat{x} \frac{\partial}{\partial y} + \hat{x} \hat{y} \frac{\partial}{\partial z} + m) (\hat{z} \hat{y} \frac{\partial}{\partial x} + \hat{x} \hat{z} \frac{\partial}{\partial y} + \hat{y} \hat{x} \frac{\partial}{\partial z} + m)$

$(\hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t}) (- \hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t}) = (\hat{y} \hat{z} \frac{\partial}{\partial x} + \hat{z} \hat{x} \frac{\partial}{\partial y} + \hat{x} \hat{y} \frac{\partial}{\partial z} + m) (- \hat{y} \hat{z} \frac{\partial}{\partial x} - \hat{z} \hat{x} \frac{\partial}{\partial y} - \hat{x} \hat{y} \frac{\partial}{\partial z} + m)$ (13)

You will see that all the cross products vanish (the dream of Dirac) and you will get:

$\frac{\partial^{2}}{\partial t^{2}} = \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}} + m^{2}$ (14)

Which yes, we obtain all scalars (vectors have disappeared) as we wanted to be in line with Equation (8). But what we have obtained (14) is not the Dirac equation. The reason is that we are playing with operators, but we are missing the wavefunction.

We will use the wavefunction in Geometric Algebra that has this form:

$ψ = ψ_{0} + \hat{x} ψ_{x} + \hat{y} ψ_{y} + \hat{z} ψ_{z} + \hat{x} \hat{y} ψ_{x y} + \hat{y} \hat{z} ψ_{y z} + \hat{z} \hat{x} ψ_{z x} + \hat{x} \hat{y} \hat{z} ψ_{x y z}$

You can see that it has 8 degrees of freedom that are represented by the 8 factors $ψ_{i}$ that multiply each of the elements that exist in Geometric Algebra Cl_3,0 (the scalars, 3 vectors, 3 bivectors and the trivector). These factors $ψ_{i}$ are real functions. So, no generality is lost, as the solution to Dirac equation is four complex functions (that if they could be divided into real and imaginary parts, would map to 8 functions).

Now, we will include the wavefunction in Equation (13). In Geometric Algebra, when an operator has two parts and one is the reverse of the other, the function that is affected by the operator is always between both. You can check this in [1] [2]. A typical operator like this is the rotors, but this is another story. We introduce our wave function between the two parts of the operator:

$(\hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t}) ψ (- \hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t}) = (\hat{y} \hat{z} \frac{\partial}{\partial x} + \hat{z} \hat{x} \frac{\partial}{\partial y} + \hat{x} \hat{y} \frac{\partial}{\partial z} + m) ψ (- \hat{y} \hat{z} \frac{\partial}{\partial x} - \hat{z} \hat{x} \frac{\partial}{\partial y} - \hat{x} \hat{y} \frac{\partial}{\partial z} + m)$

If we take everything to the left side of the equation, we have:

$(\hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t}) ψ (- \hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t}) - (\hat{y} \hat{z} \frac{\partial}{\partial x} + \hat{z} \hat{x} \frac{\partial}{\partial y} + \hat{x} \hat{y} \frac{\partial}{\partial z} + m) ψ (- \hat{y} \hat{z} \frac{\partial}{\partial x} - \hat{z} \hat{x} \frac{\partial}{\partial y} - \hat{x} \hat{y} \frac{\partial}{\partial z} + m) = 0$

As the geometric product is associative and distributive, we can take the wavefunction as a common factor of both sides of the operator and get:

$(\hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t} - (\hat{y} \hat{z} \frac{\partial}{\partial x} + \hat{z} \hat{x} \frac{\partial}{\partial y} + \hat{x} \hat{y} \frac{\partial}{\partial z} + m)) ψ (- \hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t} - (- \hat{y} \hat{z} \frac{\partial}{\partial x} - \hat{z} \hat{x} \frac{\partial}{\partial y} - \hat{x} \hat{y} \frac{\partial}{\partial z} + m)) = 0$

Operating the signs, we have:

$(\hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t} - \hat{y} \hat{z} \frac{\partial}{\partial x} - \hat{z} \hat{x} \frac{\partial}{\partial y} - \hat{x} \hat{y} \frac{\partial}{\partial z} - m) ψ (- \hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t} + \hat{y} \hat{z} \frac{\partial}{\partial x} + \hat{z} \hat{x} \frac{\partial}{\partial y} + \hat{x} \hat{y} \frac{\partial}{\partial z} - m) = 0$ (15)

As the equation is zero, it could be that the product of the first two elements of the triple product is zero:

Or it could be that the product of the last two elements is zero:

$ψ (- \hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t} + \hat{y} \hat{z} \frac{\partial}{\partial x} + \hat{z} \hat{x} \frac{\partial}{\partial y} + \hat{x} \hat{y} \frac{\partial}{\partial z} - m) = 0$ (17)

As convention and what Dirac did was to take the first half of the Equation (16) (but in standard algebra not in Geometric Algebra).

The first comment on that Equation (16) is that the element that represents energy (the partial derivative with respect to time) is positive while the ones regarding momentum (the partial derivatives with respect to spatial coordinates) are negative. Remember the minus sign of the momentum? It appears naturally here. Even if we chose Equation (17), you can check that the sign of the momentum would be the opposite of the sign of the energy again.

Continuing, if we put the complete equation with all the elements of the wavefunction we have:

$(\hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t} - \hat{y} \hat{z} \frac{\partial}{\partial x} - \hat{z} \hat{x} \frac{\partial}{\partial y} - \hat{x} \hat{y} \frac{\partial}{\partial z} - m) (ψ_{0} + \hat{x} ψ_{x} + \hat{y} ψ_{y} + \hat{z} ψ_{z} + \hat{x} \hat{y} ψ_{x y} + \hat{y} \hat{z} ψ_{y z} + \hat{z} \hat{x} ψ_{z x} + \hat{x} \hat{y} \hat{z} ψ_{x y z}) = 0$ (18)

If we operate it, we have:

$(\hat{x} \hat{y} \hat{z} \frac{\partial}{\partial t} - \hat{y} \hat{z} \frac{\partial}{\partial x} - \hat{z} \hat{x} \frac{\partial}{\partial y} - \hat{x} \hat{y} \frac{\partial}{\partial z} - m) (ψ_{0} + \hat{x} ψ_{x} + \hat{y} ψ_{y} + \hat{z} ψ_{z} + \hat{x} \hat{y} ψ_{x y} + \hat{y} \hat{z} ψ_{y z} + \hat{z} \hat{x} ψ_{z x} + \hat{x} \hat{y} \hat{z} ψ_{x y z}) = 0$

$\begin{array}{l} \hat{x} \hat{y} \hat{z} \frac{\partial ψ_{0}}{\partial t} + \hat{y} \hat{z} \frac{\partial ψ_{x}}{\partial t} + \hat{z} \hat{x} \frac{\partial ψ_{y}}{\partial t} + \hat{x} \hat{y} \frac{\partial ψ_{z}}{\partial t} - \hat{z} \frac{\partial ψ_{x y}}{\partial t} - \hat{x} \frac{\partial ψ_{y z}}{\partial t} - \hat{y} \frac{\partial ψ_{z x}}{\partial t} - \frac{\partial ψ_{x y z}}{\partial t} \\ - \hat{y} \hat{z} \frac{\partial ψ_{0}}{\partial x} - \hat{x} \hat{y} \hat{z} \frac{\partial ψ_{x}}{\partial x} + \hat{z} \frac{\partial ψ_{y}}{\partial x} - \hat{y} \frac{\partial ψ_{z}}{\partial x} - \hat{z} \hat{x} \frac{\partial ψ_{x y}}{\partial x} + \frac{\partial ψ_{y z}}{\partial x} + \hat{x} \hat{y} \frac{\partial ψ_{z x}}{\partial x} + \hat{x} \frac{\partial ψ_{x y z}}{\partial x} \\ - \hat{z} \hat{x} \frac{\partial ψ_{0}}{\partial y} - \hat{z} \frac{\partial ψ_{x}}{\partial y} - \hat{x} \hat{y} \hat{z} \frac{\partial ψ_{y}}{\partial y} + \hat{x} \frac{\partial ψ_{z}}{\partial y} + \hat{y} \hat{z} \frac{\partial ψ_{x y}}{\partial y} - \hat{x} \hat{y} \frac{\partial ψ_{y z}}{\partial y} + \frac{\partial ψ_{z x}}{\partial y} + \hat{y} \frac{\partial ψ_{x y z}}{\partial y} \\ - \hat{x} \hat{y} \frac{\partial ψ_{0}}{\partial z} + \hat{y} \frac{\partial ψ_{x}}{\partial z} - \hat{x} \frac{\partial ψ_{y}}{\partial z} - \hat{x} \hat{y} \hat{z} \frac{\partial ψ_{z}}{\partial z} + \frac{\partial ψ_{x y}}{\partial z} + \hat{z} \hat{x} \frac{\partial ψ_{y z}}{\partial z} - \hat{y} \hat{z} \frac{\partial ψ_{z x}}{\partial z} + \hat{z} \frac{\partial ψ_{x y z}}{\partial z} \\ - m ψ_{0} - \hat{x} m ψ_{x} - \hat{y} m ψ_{y} - \hat{z} m ψ_{z} - \hat{x} \hat{y} m ψ_{x y} - \hat{y} \hat{z} m ψ_{y z} - \hat{z} \hat{x} m ψ_{z x} - \hat{x} \hat{y} \hat{z} m ψ_{x y z} = 0 \end{array}$ (19)

Now, for this equation to be zero, it has to be zero the sum of all the elements that multiply the same vector, bivector or trivector. This is, if we take all the elements that multiply the trivector $\hat{x} \hat{y} \hat{z}$ we get this equation:

$\frac{\partial ψ_{0}}{\partial t} - \frac{\partial ψ_{x}}{\partial x} - \frac{\partial ψ_{y}}{\partial y} - \frac{\partial ψ_{z}}{\partial z} - m ψ_{x y z} = 0$

If we take the elements that multiply the $\hat{y} \hat{z}$ bivector, we have:

$\frac{\partial ψ_{x}}{\partial t} - \frac{\partial ψ_{0}}{\partial x} + \frac{\partial ψ_{x y}}{\partial y} - \frac{\partial ψ_{z x}}{\partial z} - m ψ_{y z} = 0$

And so on. Doing this for the scalars, the vectors, bivectors and the trivector, we get these eight equations:

$\frac{\partial ψ_{0}}{\partial t} - \frac{\partial ψ_{x}}{\partial x} - \frac{\partial ψ_{y}}{\partial y} - \frac{\partial ψ_{z}}{\partial z} - m ψ_{x y z} = 0$ (20)

$\frac{\partial ψ_{x}}{\partial t} - \frac{\partial ψ_{0}}{\partial x} + \frac{\partial ψ_{x y}}{\partial y} - \frac{\partial ψ_{z x}}{\partial z} - m ψ_{y z} = 0$ (21)

$\frac{\partial ψ_{y}}{\partial t} - \frac{\partial ψ_{x y}}{\partial x} - \frac{\partial ψ_{0}}{\partial y} + \frac{\partial ψ_{y z}}{\partial z} - m ψ_{y z} = 0$ (22)

$\frac{\partial ψ_{z}}{\partial t} + \frac{\partial ψ_{z x}}{\partial x} - \frac{\partial ψ_{y z}}{\partial y} - \frac{\partial ψ_{0}}{\partial z} - m ψ_{x y} = 0$ (23)

$- \frac{\partial ψ_{x y}}{\partial t} + \frac{\partial ψ_{y}}{\partial x} - \frac{\partial ψ_{x}}{\partial y} + \frac{\partial ψ_{x y z}}{\partial z} - m ψ_{z} = 0$ (24)

$- \frac{\partial ψ_{y z}}{\partial t} + \frac{\partial ψ_{x y z}}{\partial x} + \frac{\partial ψ_{z}}{\partial y} - \frac{\partial ψ_{y}}{\partial z} - m ψ_{x} = 0$ (25)

$- \frac{\partial ψ_{z x}}{\partial t} - \frac{\partial ψ_{z}}{\partial x} + \frac{\partial ψ_{x y z}}{\partial y} + \frac{\partial ψ_{x}}{\partial z} - m ψ_{y} = 0$ (26)

$- \frac{\partial ψ_{x y z}}{\partial t} + \frac{\partial ψ_{y z}}{\partial x} + \frac{\partial ψ_{z x}}{\partial y} + \frac{\partial ψ_{y x}}{\partial z} - m ψ_{0} = 0$ (27)

But is this correct? And what is the mapping of this wavefunction to the wavefunction obtained by the Dirac equation? So, let’s check in the next chapter.

10. Solution to Dirac Equation in Standard Matrix Algebra

In this chapter, we will not use Geometric Algebra. We will just follow paper [6] to get the equations needed to solve the general Dirac Equation.

In matrix algebra the solution to Dirac equation has this form:

$ψ = (\begin{matrix} ψ_{1} \\ ψ_{2} \\ ψ_{3} \\ ψ_{4} \end{matrix})$ (28)

where the $ψ_{k}$ are complex functions. If we consider that they can be divided in the real and the imaginary part of the function, the wavefunction would have the form:

$ψ = (\begin{matrix} ψ_{1} \\ ψ_{2} \\ ψ_{3} \\ ψ_{4} \end{matrix}) = (\begin{matrix} ψ_{1 r} + i ψ_{1 i} \\ ψ_{2 r} + i ψ_{2 i} \\ ψ_{3 r} + i ψ_{3 i} \\ ψ_{4 r} + i ψ_{4 i} \end{matrix})$ (29)

Now, we apply the Dirac equation in matrix algebra according [6]:

$(\begin{matrix} i \frac{\partial}{\partial t} - m & 0 & i \frac{\partial}{\partial z} & i \frac{\partial}{\partial x} + \frac{\partial}{\partial y} \\ 0 & i \frac{\partial}{\partial t} - m & i \frac{\partial}{\partial x} - \frac{\partial}{\partial y} & - i \frac{\partial}{\partial z} \\ - i \frac{\partial}{\partial z} & - i \frac{\partial}{\partial x} - \frac{\partial}{\partial y} & - i \frac{\partial}{\partial t} - m & 0 \\ - i \frac{\partial}{\partial x} + \frac{\partial}{\partial y} & i \frac{\partial}{\partial z} & 0 & - i \frac{\partial}{\partial t} - m \end{matrix}) (\begin{matrix} ψ_{1} \\ ψ_{2} \\ ψ_{3} \\ ψ_{4} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix})$

Applying the division in real and imaginary parts commented, we have:

$(\begin{matrix} i \frac{\partial}{\partial t} - m & 0 & i \frac{\partial}{\partial z} & i \frac{\partial}{\partial x} + \frac{\partial}{\partial y} \\ 0 & i \frac{\partial}{\partial t} - m & i \frac{\partial}{\partial x} - \frac{\partial}{\partial y} & - i \frac{\partial}{\partial z} \\ - i \frac{\partial}{\partial z} & - i \frac{\partial}{\partial x} - \frac{\partial}{\partial y} & - i \frac{\partial}{\partial t} - m & 0 \\ - i \frac{\partial}{\partial x} + \frac{\partial}{\partial y} & i \frac{\partial}{\partial z} & 0 & - i \frac{\partial}{\partial t} - m \end{matrix}) (\begin{matrix} ψ_{1 r} + i ψ_{1 i} \\ ψ_{2 r} + i ψ_{2 i} \\ ψ_{3 r} + i ψ_{3 i} \\ ψ_{4 r} + i ψ_{4 i} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix})$ (30)

And now, performing the matrix multiplication, we have for the first line:

$(i \frac{\partial}{\partial t} - m) (ψ_{1 r} + i ψ_{1 i}) + (i \frac{\partial}{\partial z}) (ψ_{3 r} + i ψ_{3 i}) + (i \frac{\partial}{\partial x} + \frac{\partial}{\partial y}) (ψ_{4 r} + i ψ_{4 i}) = 0$ (31)

$i \frac{\partial ψ_{1 r}}{\partial t} - \frac{\partial ψ_{1 i}}{\partial t} - m ψ_{1 r} - i m ψ_{1 i} + i \frac{\partial ψ_{3 r}}{\partial z} - \frac{\partial ψ_{3 i}}{\partial z} + i \frac{\partial ψ_{4 r}}{\partial x} - \frac{\partial ψ_{4 i}}{\partial x} + \frac{\partial ψ_{4 r}}{\partial y} + i \frac{\partial ψ_{4 i}}{\partial y} = 0$ (32)

Dividing in two equations, one for the real part and another one for the imaginary part, we get:

$- \frac{\partial ψ_{1 i}}{\partial t} - m ψ_{1 r} - \frac{\partial ψ_{3 i}}{\partial z} - \frac{\partial ψ_{4 i}}{\partial x} + \frac{\partial ψ_{4 r}}{\partial y} = 0$ (33)

$\frac{\partial ψ_{1 r}}{\partial t} - m ψ_{1 i} + \frac{\partial ψ_{3 r}}{\partial z} + \frac{\partial ψ_{4 r}}{\partial x} + \frac{\partial ψ_{4 i}}{\partial y} = 0$ (34)

For the second line of the matrix, we get:

$(i \frac{\partial}{\partial t} - m) (ψ_{2 r} + i ψ_{2 i}) + (i \frac{\partial}{\partial x} - \frac{\partial}{\partial y}) (ψ_{3 r} + i ψ_{3 i}) + (- i \frac{\partial}{\partial z}) (ψ_{4 r} + i ψ_{4 i}) = 0$

$i \frac{\partial ψ_{2 r}}{\partial t} - \frac{\partial ψ_{2 i}}{\partial t} - m ψ_{2 r} - i m ψ_{2 i} + i \frac{\partial ψ_{3 r}}{\partial x} - \frac{\partial ψ_{3 i}}{\partial x} - \frac{\partial ψ_{3 r}}{\partial y} - i \frac{\partial ψ_{3 i}}{\partial y} - i \frac{\partial ψ_{4 r}}{\partial z} + \frac{\partial ψ_{4 i}}{\partial z} = 0$

Again, dividing in two Equations (real and imaginary part):

$- \frac{\partial ψ_{2 i}}{\partial t} - m ψ_{2 r} - \frac{\partial ψ_{3 i}}{\partial x} - \frac{\partial ψ_{3 r}}{\partial y} + \frac{\partial ψ_{4 i}}{\partial z} = 0$ (35)

$\frac{\partial ψ_{2 r}}{\partial t} - m ψ_{2 i} + \frac{\partial ψ_{3 r}}{\partial x} - \frac{\partial ψ_{3 i}}{\partial y} - \frac{\partial ψ_{4 r}}{\partial z} = 0$ (36)

For the third line of the equation, you have

$(- i \frac{\partial}{\partial z}) (ψ_{1 r} + i ψ_{1 i}) + (- i \frac{\partial}{\partial x} - \frac{\partial}{\partial y}) (ψ_{2 r} + i ψ_{2 i}) + (- i \frac{\partial}{\partial t} - m) (ψ_{3 r} + i ψ_{3 i}) = 0$

$- i \frac{\partial ψ_{1 r}}{\partial z} + \frac{\partial ψ_{1 i}}{\partial z} - i \frac{\partial ψ_{2 r}}{\partial x} + \frac{\partial ψ_{2 i}}{\partial x} - \frac{\partial ψ_{2 r}}{\partial y} - i \frac{\partial ψ_{2 i}}{\partial y} - i \frac{\partial ψ_{3 r}}{\partial t} + \frac{\partial ψ_{3 i}}{\partial t} - m ψ_{3 r} - i m ψ_{3 i} = 0$

Dividing into real and imaginary parts, we get:

$\frac{\partial ψ_{1 i}}{\partial z} + \frac{\partial ψ_{2 i}}{\partial x} - \frac{\partial ψ_{2 r}}{\partial y} + \frac{\partial ψ_{3 i}}{\partial t} - m ψ_{3 r} = 0$ (37)

$- \frac{\partial ψ_{1 r}}{\partial z} - \frac{\partial ψ_{2 r}}{\partial x} - \frac{\partial ψ_{2 i}}{\partial y} - \frac{\partial ψ_{3 r}}{\partial t} - m ψ_{3 i} = 0$ (38)

And for the fourth line:

$(- i \frac{\partial}{\partial x} + \frac{\partial}{\partial y}) (ψ_{1 r} + i ψ_{1 i}) + (i \frac{\partial}{\partial z}) (ψ_{2 r} + i ψ_{2 i}) + (- i \frac{\partial}{\partial t} - m) (ψ_{4 r} + i ψ_{4 i}) = 0$

$- i \frac{\partial ψ_{1 r}}{\partial x} + \frac{\partial ψ_{1 i}}{\partial x} + \frac{\partial ψ_{1 r}}{\partial y} + i \frac{\partial ψ_{1 i}}{\partial y} + i \frac{\partial ψ_{2 r}}{\partial z} - \frac{\partial ψ_{2 i}}{\partial z} - i \frac{\partial ψ_{4 r}}{\partial t} + \frac{\partial ψ_{4 i}}{\partial t} - m ψ_{4 r} - i m ψ_{4 i} = 0$

Getting these two equations:

$\frac{\partial ψ_{1 i}}{\partial x} + \frac{\partial ψ_{1 r}}{\partial y} - \frac{\partial ψ_{2 i}}{\partial z} + \frac{\partial ψ_{4 i}}{\partial t} - m ψ_{4 r} = 0$ (39)

$- \frac{\partial ψ_{1 r}}{\partial x} + \frac{\partial ψ_{1 i}}{\partial y} + \frac{\partial ψ_{2 r}}{\partial z} - \frac{\partial ψ_{4 r}}{\partial t} - m ψ_{4 i} = 0$ (40)

Putting all the equations together: