# Dirac equation as one equation for one function

• I
Previously (see, e.g., https://www.physicsforums.com/threa...-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/threa...-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
2021 Award
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.

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.

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?

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.

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
Gold Member
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
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.

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.

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)

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
Gold Member
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
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
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).

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.

A. Neumaier
I still believe the form is relativistically covariant.
But only in a trivial sense, since as I pointed out, every equation invariant under rigid motion can be made ''relativistically covariant'' in your sense. Your designation is a misnomer since one can make the nonrelativistic Schroedinger equation for a free particle ''relativistically covariant'' in your sense (by embedding it into a family of equations all equivalent to each other).

But only in a trivial sense, since as I pointed out, every equation invariant under rigid motion can be made ''relativistically covariant'' in your sense. Your designation is a misnomer since one can make the nonrelativistic Schroedinger equation for a free particle ''relativistically covariant'' in your sense (by embedding it into a family of equations all equivalent to each other).
Again, I would like to make sure I understand you correctly (please see below).
A. Neumaier said:
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.
1. How do you Lorentz-transform the wave function in your approach for the nonrelativistic Schroedinger equation? As a scalar?
2. Let us consider the nonrelativistic Schroedinger equation in two inertial reference frames and get two infinite families of equations using your Lorentz-transformation approach. Will all the equations of the two families be equivalent to each other? This does not seem obvious.

A. Neumaier
Will all the equations of the two families be equivalent to each other?
Of course, by a combination of a Lorentz boost and a Galilei boost. In general, whatever you get from a single equation by an invertible transformation will be equivalent to the original, by its derivation.
1. How do you Lorentz-transform the wave function in your approach for the nonrelativistic Schroedinger equation? As a scalar?
Either as a scalar, or as the 0-component of a vector. One gets different families of equations, but each resulting is equivalent to the original one.

Of course, by a combination of a Lorentz boost and a Galilei boost. In general, whatever you get from a single equation by an invertible transformation will be equivalent to the original, by its derivation.
I am afraid I don't understand that. Let us consider, e.g., a one-dimensional free Schroedinger equation (the wave function depends only on $x$ and $t$). Then let us assume that the wave function transforms as a scalar under Lorentz transformations and consider this equation after a Lorentz boost (the wave function will depend only on $x'$ and $t'$). The resulting equation will be different from what we would have if we wrote the Schroedinger equation in terms of $x'$ and $t'$, as the Schroedinger equation is not Lorentz-invariant. Or maybe I don't understand your approach?

Let me also note that the eigenvector of $\gamma^5$ in my work does not parametrize Lorentz transformations, rather it specifies the component (of the spinor wave function) for which we write the equation.

A. Neumaier
maybe I don't understand your approach?
Yes. So I'll be very explicit and give the result of my suggested exercise:

The family of equations $$\Big(p^2+2m\xi\cdot p-(\xi\cdot p)^2\Big)\psi =0,$$ where ##p=i\hbar\nabla## and ##\xi## is a timelike vector with ##\xi^2=1## (metric ##+---##) is covariant according to your criterion.

But It specializes to the Schrödinger equation for a free nonrelativistic particle when ##\xi=(1,0,0,0)##. Boosting the family just changes ##\xi##.

Thus a covariantly parameterized family expresses nothing relativistic, in general.

• Demystifier
Yes. So I'll be very explicit and give the result of my suggested exercise:

The family of equations $$\Big(p^2+2m\xi\cdot p-(\xi\cdot p)^2\Big)\psi =0,$$ where ##p=i\hbar\nabla## and ##\xi## is a timelike vector with ##\xi^2=1## (metric ##+---##) is covariant according to your criterion.

But It specializes to the Schrödinger equation for a free nonrelativistic particle when ##\xi=(1,0,0,0)##. Boosting the family just changes ##\xi##.

Thus a covariantly parameterized family expresses nothing relativistic, in general.
Thank you very much for the clarification.

So let us consider your family of equations. It is pretty obvious that the sets of their solutions generally do not coincide. Let us prove that.

I assume that your ##p## is, up to a constant factor, a 4-dimensional gradient (otherwise your family of equations is not even apparently covariant). Let us also assume that we use such a system of units that ##\hbar=c=m=1##. Let us consider such solutions ##\psi## that are eigenvectors of all components of ##p## with the following eigenvalues: ##(p_0,0,0,p_3)## (their 4-momentum is well-defined). Let us also consider only ##\xi## having the following form: ##\xi=(\xi^0,0,0,\xi^3),## where $$(\xi^0)^2-(\xi^3)^2=1. (1)$$ Then we obtain for the components of 4-momentum:
$$(p_0)^2-(p_3)^2+2(\xi^0 p_0+\xi^3 p_3)-(\xi^0 p_0+\xi^3 p_3)^2=0. (2)$$
If ##\xi=(1,0,0,1)## (components with upper indices), then we obtain
$$(p_3)^2=2p_0.$$
This equation has, among others, the following solution: ##p_0=2,p_3=2.## Let us substitute this solution in equation (2):
$$4-4+2(2\xi^0+2\xi^3)-(2\xi^0+2\xi^3)^2=0$$
or ##\xi^0+\xi^3=(\xi^0+\xi^3)^2.##
In view of equation (1), we obtain ##\xi^0=1,\xi^3=0##. Thus, solution ##p_0=2,p_3=2## does not satisfy equation (2) for an arbitrary combination ##\xi^0,\xi^3## satisfying equation (1). Therefore, equations from your family of equations generally have different solutions depending on ##\xi##. So this family of equations is not consistent. This is what I suspected in my previous post.

So how the equations of my work are different? One can choose some eigenvector of ##\gamma^5## and write down the equation for the relevant component of the Dirac spinor (equation (27) in https://arxiv.org/abs/1502.02351). For each solution of this equation one can restore the Dirac spinor using well-defined rules, and the resulting set of solutions will be the same as for the initial Dirac equation. Therefore, no matter what eigenvector one chooses, one obtains the same set of solutions of the Dirac equation. Therefore, the family of equations is consistent.

A. Neumaier
For each solution of this equation one can restore the Dirac spinor using well-defined rules, and the resulting set of solutions will be the same as for the initial Dirac equation.
The same can be said of my equations. The well-defined rules for restoring the same set of solutions are to apply a Lorentz boost that maps ##\xi## to (1,0,0,0).

True invariance requires the solution set to be unchanged!

The same can be said of my equations. The well-defined rules for restoring the same set of solutions are to apply a Lorentz boost that maps ##\xi## to (1,0,0,0).

True invariance requires the solution set to be unchanged!
So you seem to agree that the set of solutions in your equations does change when you change vector ##\xi##. Thank you for that. Let me note however that the solution set for the Dirac spinor in the equations of my work ("fourth-order Dirac equation") is indeed unchanged when I change eigenvector ##\xi## (which is actually a spinor in my case). Thus, I do have true invariance... Please let me explain.

The original Dirac equation is an equation for the Dirac spinor ##\psi##, which has four complex components. On the other hand, the fourth-order Dirac equation is an equation for just one component ##\bar{\xi}\psi## of the spinor. So, after you solve this equation, you have just one component of the spinor. On the other hand, you need all the components of the spinor to do physics (for example, to determine the current for the solution). Fortunately, if you know just one component of the spinor, you can (subject to some "transversality" condition for the electromagnetic field) restore the entire spinor.

Let us consider just one spinor solution ##\psi## of the original Dirac equation. Note that it satisfies the fourth-order Dirac equation for any eigenvector ##\xi##! Just this fact strongly suggests that the set of spinor solutions does not change when you use a different eigenvector ##\xi##. And indeed, after you solve the fourth-order Dirac equation you get a set of solutions for one component of the spinor, then you restore the set of solutions for the entire spinor, and that set of solutions does not depend on the choice of ##\xi##! You may say that technically the set of solutions of the fourth-order Dirac equation does depend on ##\xi##. Yes, but those are sets of solutions for different components of the spinor, and these sets of solutions produce the same set of solutions for the spinor.

You may dislike the fact that I need to restore the spinor after I find a solution for the fourth-order Dirac equation. However, one can probably avoid this, as one can define current as a function of just one component ##\bar{\xi}\psi## and the eigenvector ##\xi## and forget all other components. So the set of solutions for the current does not depend on the choice of ##\xi##!

A. Neumaier
Thus, I do have true invariance...
No, since for different ##\xi## you get different linear combinations of the full covariant spinor, and the recipe needed to make predictions therefore depends on ##\xi##.

Thank you very much for a prompt reply.
No, since for different ##\xi## you get different linear combinations of the full covariant spinor
I am afraid I don't understand that. I insist that the set of solutions for the full covariant spinor ##\psi## does not depend on the choice of ##\xi##.

Let us consider two different values of the eigenvector, ##\xi_a## and ##\xi_b##. Let us choose one solution ##\bar{\xi_a}\psi## of the fourth-order Dirac equation (FODE) for eigenvector ##\xi_a##. Based on this solution, we can restore the spinor ##\psi##, and its component ##\bar{\xi_b}\psi## will be a solution of FODE for eigenvector ##\xi_b##, and this component restores to the same spinor ##\psi## as component ##\bar{\xi_a}\psi##. Thus, FODE for eigenvectors ##\xi_a## and ##\xi_b## have the same set of solutions for the covariant spinor.

Of course, you can choose such solutions of FODE for eigenvectors ##\xi_a## and ##\xi_b## that restore to different covariant spinors, but how is this different from the fact that you can choose two different solutions for the original Dirac equation?

Let me emphasize again that ##\xi## for FODE does not parametrize Lorentz transforms, it parametrizes components of the covariant spinor.
A. Neumaier said:
and the recipe needed to make predictions therefore depends on ##\xi##.
Yes, the recipe needed to make predictions depends on ##\xi##, but predictions do not depend on ##\xi##.