# I Dirac equation as one equation for one function (1 Viewer)

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#### akhmeteli

Previously (see, e.g., https://www.physicsforums.com/threads/klein-gordon-equation-and-particles-with-spin.563974/#post-3690162), I mentioned my article in the Journal of Mathematical Physics where I showed that, in a general case, the Dirac equation is equivalent to a fourth-order partial differential equation for just one complex function (which can be made real by a gauge transform).

tom.stoer criticized me and demanded that I "find how to relate [my] equations to the standard Lorentz transformation for spinors." (https://www.physicsforums.com/threads/klein-gordon-equation-and-particles-with-spin.563974/#post-3693864)

I did not feel that was a high-priority task, but recently I found a relativistically covariant form of the fourth-order partial differential equation for just one function that is generally equivalent to the Dirac equation. This form is applicable for an arbitrary component of the Dirac spinor (with some caveats) and any representation of gamma-matrices (with the standard hermiticity properties). The derivation was published in https://arxiv.org/abs/1502.02351 and (peer-reviewed) in "Quantum Foundations, Probability and Information" (A. Khrennikov, B. Toni, eds., Springer, 2018, pp.1-11).

Let me note that the relativistically covariant form is somewhat unusual and was not easy to derive. As the result was discussed at this forum, maybe its covariant form will be of interest for some people.

• waves and change

#### vanhees71

Gold Member
Without having read your papers, just let me ask the question how your equation for one (real?) field component can be equivalent to the fully general Dirac equation, which describes particles and antiparticles of spin 1/2. My simple argument against this claim is that since you have spin 1/2 and thus you need at least a two-component spinor field, either transforming under the representations (1/2,0) or (0,1/2) of the proper orthochronous Lorentz group with two spin states (where spin is usually defined by the representation of the rotation group in the rest frame of the particle, i.e., the zero-momentum eigenstates) each. To be able to also represent spatial reflections you have to orthogonally add these two representations, and you get to the Dirac-spinor representation of the entire (orthochronous) Lorentz group, describing charged spin-1/2 particles and their anti-particles. That's why you have 4 independent zero-momentum eigen modes (i.e., two spin states for particles and two for anti particles). I don't see a way to put all this information into a single real field.

#### akhmeteli

Without having read your papers, just let me ask the question how your equation for one (real?) field component can be equivalent to the fully general Dirac equation.
Thank you for your interest.

I will try to give more details later, but for now let me ask you a question, just to be sure we are not arguing about definitions.

Let us consider a system of four ordinary differential equations of the first order for four functions $$x_1(t), x_2(t), x_3(t), x_4(t)$$ of one real argument:
$$\frac{dx_1}{dt}=t^2+t, \frac{dx_2}{dt}=x_1, \frac{dx_3}{dt}=x_2, \frac{dx_4}{dt}=x_3.$$
In your book, is this system equivalent to just one equation of the fourth order for one function:
$$\frac{d^4 x_4}{dt^4}=t^2+t$$?

#### A. Neumaier

is this system equivalent to just one equation of the fourth order for one function
No, since the specification of the original functions requires the addition of three more differential equations.

• vanhees71

#### martinbn

What is the motivation?

#### akhmeteli

No, since the specification of the original functions requires the addition of three more differential equations.
OK, you can certainly take such a position. It's a matter of definitions. However, a different definition of equivalency seems to be widely used as well (http://www.math.ualberta.ca/~xichen/math5a00s/final1.pdf, http://www.math.psu.edu/tseng/class/Math251/Notes-LinearSystems.pdf , p.3). What is important for me, the system of the first-order equations and the fourth order equation describe the same "physics". I would say, for example, that Hamilton's equations are typically equivalent to Newton's equations. If you disagree, I'm fine with that.

#### akhmeteli

What is the motivation?
Shroedinger noted ((Nature, v.169, p.538 (1952)) ) that if there is a wave function satisfying the Klein-Gordon equation, one can make it real by a gauge transform, although it is believed that charged particles must be described by complex functions. My immediate motivation was to obtain a similar result for the wave function satisfying the Dirac equation: one cannot make four complex function real by a gauge transform, but one can make it for one complex function. Eventually, however, I need this result for interpretation of quantum theory.

#### Demystifier

2018 Award
is this system equivalent to just one equation of the fourth order for one function:
This is like writing the second order Newton equation
$$m\ddot{x}=F$$
as two first order equations
$$m\dot{v}=F$$
$$v=\dot{x}$$
So yes, it's equivalent.

• pBrane

#### akhmeteli

Without having read your papers, just let me ask the question how your equation for one (real?) field component can be equivalent to the fully general Dirac equation, which describes particles and antiparticles of spin 1/2.
The equation of my work (let us call it "fourth-order Dirac equation") is derived from the Dirac equation by boring algebraic elimination.

Let me note that algebraic elimination was used long ago by others to derive a second-order Dirac equation for just two components of the Dirac spinor (O. Laporte, G. E. Uhlenbeck, Phys. Rev., vol. 37, p. 1380 (1931); R. P. Feynman, M. Gell-Mann, Phys. Rev., vol. 109, p. 193 (1958)). For example, Feynman and Gell-Mann wrote that the solutions of the second-order Dirac equation can be put into one-to-one correspondence with the solutions of the standard Dirac equation. Again, if you don't think those two equations are equivalent, this is just a matter of definitions.
vanhees71 said:
you have 4 independent zero-momentum eigen modes (i.e., two spin states for particles and two for anti particles). I don't see a way to put all this information into a single real field.
Your arguments are reasonable, but they are applicable not just to the fourth-order Dirac equation, but also to the second-order Dirac equation of the article by Feynman and Gell-Mann (FG). Actually, FG consider your arguments and explain why they are not waterproof:

"Why must the wave function have four components? It is usually explained by pointing out that to describe the electron spin we must have two, and we must also
represent the negative-energy states or positrons, requiring two more. Yet this argument is unsatisfactory. For a particle of spin zero we use a wave function of
only one component. The sign of the energy is determined by how the wave function varies in space and time. The Klein-Gordon equation is second order and we need both the function and its time derivative to predict the future. So instead of two components for spin zero we use one, but it satisfies a second order equation. Initial states require specification of that one and its time derivative. Thus for the case of spin 1/2, we would expect to be able to use a simple two-component spinor for the wave function, but have it satisfy a second order differential equation."

It should be obvious that FG's reasoning can be adapted to the fourth-order Dirac equation as well.

Thus, you don't need four complex components, you just need one complex component. To describe the same physics using just one real component, you can fix the gauge.

#### akhmeteli

This is like writing the second order Newton equation
$$m\ddot{x}=F$$
as two first order equations
$$m\dot{v}=F$$
$$v=\dot{x}$$
So yes, it's equivalent.
Thank you. By the way, congratulations on the great success of your soccer team (at the expense of my team:-) )

• Demystifier

#### martinbn

This is like writing the second order Newton equation
$$m\ddot{x}=F$$
as two first order equations
$$m\dot{v}=F$$
$$v=\dot{x}$$
So yes, it's equivalent.
So, then what is the point of the paper?

#### A. Neumaier

the relativistically covariant form is somewhat unusual
It is not a single covariant equation but an infinite family of equations labelled by a 4-vector. For a fixed choice of 4-vector (i.e., for a fixed equation) nothing covariant is left.

#### akhmeteli

So, then what is the point of the paper?
Mathematically, for each $n$-th order differential equation there is an equivalent system of 1-st order differential equations, however, the converse statement is not always true, and for a system of 1-st order differential equations an equivalent single differential equation of a higher order does not necessarily exist. Thus, the result of the article (the Dirac equation, which is a system of four 1-st order differential equations, is equivalent to a single 4-th order differential equation) is non-trivial. Thus, the article seems to have some value for mathematics.

Is this equivalent form useful for physics? Well, generally, different forms of fundamental equations proved useful in different situations. Specifically, the 4-th order Dirac equation shows that the Dirac particle can be described by just one real function, although charged particles are believed to require complex functions. So this form is leaner in some respect, as it requires just one real function instead of four complex functions of the standard Dirac equation. My hope is that this result can be used to algebraically eliminate the Dirac field from Dirac-Maxwell electrodynamics (spinor electrodynamics), the same way as the matter field was algebraically eliminated from Klein-Gordon-Maxwell electrodynamics (scalar electrodynamics) in my article in Eur. Phys. Journ. C (http://link.springer.com/content/pdf/10.1140/epjc/s10052-013-2371-4.pdf)

#### akhmeteli

It is not a single covariant equation but an infinite family of equations labelled by a 4-vector. For a fixed choice of 4-vector (i.e., for a fixed equation) nothing covariant is left.
I agree to some extent. I was not able to derive a more traditional covariant form. However, I still believe this is a relativistically covariant form of the equation.

#### Demystifier

2018 Award
By the way, congratulations on the great success of your soccer team (at the expense of my team:-) )
We were lucky. The result could have easily been the opposite one. #### A. Neumaier

I agree to some extent. I was not able to derive a more traditional covariant form. However, I still believe this is a relativistically covariant form of the equation.
This means nothing. One can take any noncovariant equation invariant under rigid motions and embed it into a family of equation parameterized by a timelike 4-vector such that the result looks covariant as a family. This is done simply by applying arbitrary Lorentz transformations to the original equation. Thus a covariant family is trivial to generate, and has nothing to do with covariance of the individual equation.

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• vanhees71

#### akhmeteli

This means nothing. One can take any noncovariant equation invariant under rigid motions and embed it into a family of equation parameterized by a timelike 4-vector such that the result looks covariant as a family. This is done simply by applying arbitrary Lorentz transformations to the original equation. Thus a covariant family is trivial to generate, and has nothing to do with covariance of the individual equation.
Just to make sure I understand: does "rigid motions" mean three-dimensional rotations, translations, and maybe reflections or something different?

#### A. Neumaier

Just to make sure I understand: does "rigid motions" mean three-dimensional rotations, translations, and maybe reflections or something different?
Yes, the first two. A reflection is not a motion but a discrete operation.

• vanhees71

#### akhmeteli

This means nothing. One can take any noncovariant equation invariant under rigid motions and embed it into a family of equation parameterized by a timelike 4-vector such that the result looks covariant as a family. This is done simply by applying arbitrary Lorentz transformations to the original equation. Thus a covariant family is trivial to generate, and has nothing to do with covariance of the individual equation.
I still am not sure I fully understand your remark (please see below).
A. Neumaier said:
It is not a single covariant equation but an infinite family of equations labelled by a 4-vector. For a fixed choice of 4-vector (i.e., for a fixed equation) nothing covariant is left.
Note that all equations of the "infinite family" are equivalent to each other (or, if you do not accept this term, their solutions can be put into one-to-one correspondence, using Feynman and Gell-Mann's wording). The "parametrizing" spinor can be chosen arbitrarily (provided it satisfies a relativistically covariant condition). So if "This means nothing" to you, I am fine with that, but I still believe the form is relativistically covariant. Additionally, it is important that it is valid for an arbitrary component and an arbitrary presentation of the gamma-matrices (both with caveats).

#### akhmeteli

It is not a single covariant equation but an infinite family of equations labelled by a 4-vector. For a fixed choice of 4-vector (i.e., for a fixed equation) nothing covariant is left.
I forgot to explain that all these equations are equivalent to each other because all of them are equivalent to the Dirac equation.

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